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\ihead{18.781 (Spring 2016): floor and arithmetic functions}
\ohead{\today}
\begin{document}

\title{18.781 (Spring 2016): Floor and arithmetic functions}
\author{Darij Grinberg}
\date{\today}
\maketitle
\tableofcontents

\subsection*{***}

These are the extended notes for the 18.781 (Introduction to Number Theory)
class on 14 April 2016 (the actual class covered about 1/3 of what is in these
notes). I roughly follow \cite[\S 4.1--\S 4.3]{NiZuMo91}, although not always
using the same notations.

I use the notation $\mathbb{N}$ for $\left\{  0,1,2,\ldots\right\}  $, and the
notation $\mathbb{N}_{+}$ for $\left\{  1,2,3,\ldots\right\}  $.

\section{The floor function}

Let me first recall two basic facts about divisibility of integers:

\begin{proposition}
\label{prop.div.sum}Let $a$ be an integer. Let $u$ and $v$ be two integers
that are both divisible by $a$. Then, their sum $u+v$ must also be divisible
by $a$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.div.sum}.]There exists an integer $p$ such
that $u=ap$ (since $u$ is divisible by $a$). There exists an integer $q$ such
that $v=aq$ (since $v$ is divisible by $a$). Consider these $p$ and $q$. Now,
$\underbrace{u}_{=ap}+\underbrace{v}_{=aq}=ap+aq=a\left(  p+q\right)  $.
Hence, $u+v$ is divisible by $a$. This proves Proposition \ref{prop.div.sum}.
\end{proof}

\begin{proposition}
\label{prop.div.=}Let $u$ and $v$ be two nonnegative integers such that $u\mid
v$ and $v\mid u$. Then, $u=v$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.div.=}.]We have $v\mid u$. In other words,
there exists an integer $w$ such that $u=vw$. Consider this $w$. If $v=0$,
then we have $u=\underbrace{v}_{=0}w=0=v$. Hence, if $v=0$, then Proposition
\ref{prop.div.=} holds. Thus, for the rest of this proof, we can WLOG assume
that $v\neq0$. Assume this. Thus, $v>0$ (since $v$ is nonnegative).

We have $u\mid v$. In other words, there exists an integer $z$ such that
$v=uz$. Consider this $z$. Since $uz=v\neq0$, we have $z\neq0$. If $u=0$, then
we have $v=\underbrace{u}_{=0}z=0=u$ and thus $u=v$. Hence, if $u=0$, then
Proposition \ref{prop.div.=} holds. Thus, for the rest of this proof, we can
WLOG assume that $u\neq0$. Assume this. Thus, $u>0$ (since $u$ is nonnegative).

From $uz=v>0$, we obtain $z>0$ (since $u>0$). Hence, $z\geq1$ (since $z$ is an
integer). Hence, $v=u\underbrace{z}_{\geq1}\geq u1$ (since $u>0$) and thus
$v\geq u1=u$. The same argument (with the roles of $u$ and $v$ swapped) yields
$u\geq v$. Combining this with $v\geq u$, we obtain $u=v$. This proves
Proposition \ref{prop.div.=}.
\end{proof}

\subsection{Definition and basic properties}

I shall first discuss the floor function, following \cite[\S 4.1]{NiZuMo91}.

\begin{definition}
\label{def.floor}Let $x$ be a real number. Then, $\left\lfloor x\right\rfloor
$ is defined to be the unique integer $n$ satisfying $n\leq x<n+1$. This
integer $\left\lfloor x\right\rfloor $ is called the \textit{floor} of $x$, or
the \textit{integer part} of $x$.
\end{definition}

\begin{remark}
\label{rmk.floor}\textbf{(a)} Why is $\left\lfloor x\right\rfloor $
well-defined? I mean, why does the unique integer $n$ in Definition
\ref{def.floor} exist, and why is it unique? I will not answer this question
in general (the answer probably depends on how you define real numbers
anyway). However, in the case when $x$ is rational, the proof is simple (see
Corollary \ref{cor.floor.quorem-welldef} below).

\textbf{(b)} What we call $\left\lfloor x\right\rfloor $ is typically called
$\left[  x\right]  $ in older books (such as \cite{NiZuMo91}). I suggest
avoiding the notation $\left[  x\right]  $ wherever possible; it has too many
different meanings (whereas $\left\lfloor x\right\rfloor $ almost always means
the floor of $x$).

\textbf{(c)} The map $\mathbb{R}\rightarrow\mathbb{Z},\ x\mapsto\left\lfloor
x\right\rfloor $ is called the \textit{floor function} or the \textit{greatest
integer function}. There is also a \textit{ceiling function}, which sends each
$x\in\mathbb{R}$ to the unique integer $n$ satisfying $n-1<x\leq n$; this
latter integer is called $\left\lceil x\right\rceil $. The two functions are
connected by the rule $\left\lceil x\right\rceil =-\left\lfloor
-x\right\rfloor $ (for all $x\in\mathbb{R}$).

The floor and the ceiling functions are some of the simplest examples of
discontinuous functions.

\textbf{(d)} Here are some examples of floors:%
\begin{align*}
\left\lfloor n\right\rfloor  &  =n\ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in\mathbb{Z};\\
\left\lfloor 1.32\right\rfloor  &  =1;\ \ \ \ \ \ \ \ \ \ \left\lfloor
\pi\right\rfloor =3;\ \ \ \ \ \ \ \ \ \ \left\lfloor 0.98\right\rfloor =0;\\
\left\lfloor -2.3\right\rfloor  &  =-3;\ \ \ \ \ \ \ \ \ \ \left\lfloor
-0.4\right\rfloor =-1.
\end{align*}


\textbf{(e)} You might have the impression that $\left\lfloor x\right\rfloor $
is \textquotedblleft what remains from $x$ if the digits behind the comma are
removed\textquotedblright. This impression is highly imprecise. For one, it is
completely broken for negative $x$ (for example, $\left\lfloor
-2.3\right\rfloor $ is $-3$, not $-2$). But more importantly, the operation of
\textquotedblleft removing the digits behind the comma\textquotedblright\ from
a number is not well-defined; the periodic decimal representations
$0.999\ldots$ and $1.000\ldots$ belong to the same real number ($1$), but
removing their digits behind the comma leaves us with different integers.

\textbf{(f)} A related map is the map $\mathbb{R}\rightarrow\mathbb{Z}%
,\ x\mapsto\left\lfloor x+\dfrac{1}{2}\right\rfloor $. It sends each real $x$
to the integer that is closest to $x$, choosing the larger one in the case of
a tie. This is one of the many things that are commonly known as
\textquotedblleft rounding\textquotedblright\ a number.
\end{remark}

Floors of rational numbers are directly related to division with remainder:

\begin{proposition}
\label{prop.floor.quorem}Let $a$ and $b$ be integers such that $b>0$. Let $q$
and $r$ be the quotient and the remainder obtained when dividing $a$ by $b$.
Then, $q$ is the unique integer $n$ satisfying $n\leq\dfrac{a}{b}<n+1$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.floor.quorem}.]We know that $q$ and $r$ are
the quotient and the remainder obtained when dividing $a$ by $b$. In other
words, we have $q\in\mathbb{Z}$, $r\in\left\{  0,1,\ldots,b-1\right\}  $ and
$a=qb+r$.

From $r\in\left\{  0,1,\ldots,b-1\right\}  $, we obtain $0\leq r<b$. Now, from
$0\leq r$, we obtain $qb\leq qb+r=a$. Dividing this inequality by $b$, we
obtain $q\leq\dfrac{a}{b}$ (since $b>0$). Also, $a=qb+\underbrace{r}%
_{<b}<qb+b=\left(  q+1\right)  b$. Dividing this inequality by $b$, we obtain
$\dfrac{a}{b}<q+1$ (since $b>0$). Thus, $q\leq\dfrac{a}{b}<q+1$. Hence, $q$ is
an integer $n$ satisfying $n\leq\dfrac{a}{b}<n+1$. It thus remains to show
that $q$ is the \textbf{unique} such integer. In other words, it remains to
show that if $n$ is an integer satisfying $n\leq\dfrac{a}{b}<n+1$, then $n=q$.

So let $n$ be an integer satisfying $n\leq\dfrac{a}{b}<n+1$. We must show that
$n=q$.

We have $n\leq\dfrac{a}{b}<q+1$. Since $n$ and $q$ are integers, this yields
$n\leq\left(  q+1\right)  -1=q$.

We have $q\leq\dfrac{a}{b}<n+1$. Since $q$ and $n$ are integers, this yields
$q\leq\left(  n+1\right)  -1=n$. Combining this with $n\leq q$, we obtain
$n=q$. As we said, this completes our proof.
\end{proof}

\begin{corollary}
\label{cor.floor.quorem-welldef}Let $x$ be a rational number.

\textbf{(a)} The integer $\left\lfloor x\right\rfloor $ is well-defined.

\textbf{(b)} Write $x$ in the form $x=\dfrac{a}{b}$ where $a$ and $b$ are
integers such that $b>0$. Let $q$ and $r$ be the quotient and the remainder
obtained when dividing $a$ by $b$. Then, $\left\lfloor x\right\rfloor =q$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.floor.quorem-welldef}.]Write $x$ in the form
$x=\dfrac{a}{b}$ where $a$ and $b$ are integers such that $b>0$. Let $q$ and
$r$ be the quotient and the remainder obtained when dividing $a$ by $b$. Then,
Proposition \ref{prop.floor.quorem} yields that $q$ is the unique integer $n$
satisfying $n\leq\dfrac{a}{b}<n+1$. In other words, $q$ is the unique integer
$n$ satisfying $n\leq x<n+1$ (since $x=\dfrac{a}{b}$). Thus, the unique
integer $n$ satisfying $n\leq x<n+1$ exists. Thus, $\left\lfloor
x\right\rfloor $ is well-defined. This proves Corollary
\ref{cor.floor.quorem-welldef} \textbf{(a)}.

\textbf{(b)} We have just seen that $q$ is the unique integer $n$ satisfying
$n\leq x<n+1$. But this latter integer has been denoted by $\left\lfloor
x\right\rfloor $. Thus, $\left\lfloor x\right\rfloor =q$. This proves
Corollary \ref{cor.floor.quorem-welldef} \textbf{(b)}.
\end{proof}

\begin{proposition}
\label{prop.floor.adjoint}Let $m$ be an integer, and let $x$ be a real number.
Then, $m\leq x$ holds if and only if $m\leq\left\lfloor x\right\rfloor $ holds.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.floor.adjoint}.]Recall that $\left\lfloor
x\right\rfloor $ is the unique integer $n$ satisfying $n\leq x<n+1$. Thus,
$\left\lfloor x\right\rfloor $ is an integer satisfying $\left\lfloor
x\right\rfloor \leq x<\left\lfloor x\right\rfloor +1$.

If $m\leq x$ holds, then $m\leq\left\lfloor x\right\rfloor $ holds as
well\footnote{\textit{Proof.} Assume that $m\leq x$ holds. We need to prove
that $m\leq\left\lfloor x\right\rfloor $ holds.
\par
Indeed, assume the contrary. Thus, $m>\left\lfloor x\right\rfloor $. Hence,
$m\geq\left\lfloor x\right\rfloor +1$ (since $m$ and $\left\lfloor
x\right\rfloor $ are integers). Thus, $\left\lfloor x\right\rfloor +1\leq
m\leq x$, which contradicts $x<\left\lfloor x\right\rfloor +1$. This
contradiction proves that our assumption was wrong. Hence, $m\leq\left\lfloor
x\right\rfloor $ is proven, qed.}. Conversely, if $m\leq\left\lfloor
x\right\rfloor $ holds, then $m\leq x$ (because $m\leq\left\lfloor
x\right\rfloor \leq x$). Combining these two implications, we conclude that
$m\leq x$ holds if and only if $m\leq\left\lfloor x\right\rfloor $ holds.
Proposition \ref{prop.floor.adjoint} is thus proven.
\end{proof}

\begin{corollary}
\label{cor.floor.largest}Let $x$ be a real number. Then, $\left\lfloor
x\right\rfloor $ is the greatest integer that is smaller or equal to $x$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.floor.largest}.]Clearly, $\left\lfloor
x\right\rfloor $ is the greatest integer that is smaller or equal to
$\left\lfloor x\right\rfloor $. In other words, $\left\lfloor x\right\rfloor $
is the greatest integer $m$ satisfying $m\leq\left\lfloor x\right\rfloor $.
Equivalently, $\left\lfloor x\right\rfloor $ is the greatest integer $m$
satisfying $m\leq x$ (since Proposition \ref{prop.floor.adjoint} shows that
the condition $m\leq\left\lfloor x\right\rfloor $ for an integer $m$ is
equivalent to the condition $m\leq x$). In other words, $\left\lfloor
x\right\rfloor $ is the greatest integer that is smaller or equal to $x$. This
proves Corollary \ref{cor.floor.largest}.
\end{proof}

Corollary \ref{cor.floor.largest} is often used as a definition of
$\left\lfloor x\right\rfloor $. It is also the reason why the map
$\mathbb{R}\rightarrow\mathbb{Z},\ x\mapsto\left\lfloor x\right\rfloor $ is
called the greatest integer function.

Before we come to anything interesting, we shall prove a few more basic
properties of the floor function.

\begin{proposition}
\label{prop.floor.int}Let $m$ be an integer. Then, $\left\lfloor
m\right\rfloor =m$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.floor.int}.]Clearly, $m\leq m<m+1$. But
$\left\lfloor m\right\rfloor $ is the unique integer $n$ satisfying $n\leq
m<n+1$ (because this is how $\left\lfloor m\right\rfloor $ is defined). Hence,
if $n$ is any integer satisfying $n\leq m<n+1$, then $n=\left\lfloor
m\right\rfloor $. Applying this to $n=m$, we obtain $m=\left\lfloor
m\right\rfloor $ (since $m\leq m<m+1$). This proves Proposition
\ref{prop.floor.int}.
\end{proof}

The next fact that we shall prove is \cite[Theorem 4.1 \textbf{(5)}]{NiZuMo91}:

\begin{proposition}
\label{prop.floor.-x}Let $x$ be a real number. Then,%
\[
\left\lfloor x\right\rfloor +\left\lfloor -x\right\rfloor =%
\begin{cases}
0, & \text{if }x\in\mathbb{Z};\\
-1, & \text{if }x\notin\mathbb{Z}%
\end{cases}
.
\]

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.floor.-x}.]We must be in one of the following
two cases:

\textit{Case 1:} We have $x\in\mathbb{Z}$.

\textit{Case 2:} We have $x\notin\mathbb{Z}$.

Let us consider Case 1 first. In this case, we have $x\in\mathbb{Z}$. In other
words, $x$ is an integer. Hence, $-x$ is an integer as well. Thus, Proposition
\ref{prop.floor.int} (applied to $m=-x$) yields that $\left\lfloor
-x\right\rfloor =-x$. But Proposition \ref{prop.floor.int} (applied to $m=x$)
yields $\left\lfloor x\right\rfloor =x$ (since $x$ is an integer). Thus,
$\underbrace{\left\lfloor x\right\rfloor }_{=x}+\underbrace{\left\lfloor
-x\right\rfloor }_{=-x}=x+\left(  -x\right)  =0$. Comparing this with $%
\begin{cases}
0, & \text{if }x\in\mathbb{Z};\\
-1, & \text{if }x\notin\mathbb{Z}%
\end{cases}
=0$ (since $x\in\mathbb{Z}$), we obtain $\left\lfloor x\right\rfloor
+\left\lfloor -x\right\rfloor =%
\begin{cases}
0, & \text{if }x\in\mathbb{Z};\\
-1, & \text{if }x\notin\mathbb{Z}%
\end{cases}
$. Thus, Proposition \ref{prop.floor.-x} is proven in Case 1.

Let us now consider Case 2. In this case, we have $x\notin\mathbb{Z}$. Recall
that $\left\lfloor x\right\rfloor $ is the unique integer $n$ satisfying
$n\leq x<n+1$ (since this is how $\left\lfloor x\right\rfloor $ is defined).
Thus, $\left\lfloor x\right\rfloor $ is an integer and satisfies $\left\lfloor
x\right\rfloor \leq x<\left\lfloor x\right\rfloor +1$. In particular,
$\left\lfloor x\right\rfloor \in\mathbb{Z}$ (since $\left\lfloor
x\right\rfloor $ is an integer).

We cannot have $\left\lfloor x\right\rfloor =x$ (since otherwise, we would
have $\left\lfloor x\right\rfloor =x\notin\mathbb{Z}$, which would contradict
$\left\lfloor x\right\rfloor \in\mathbb{Z}$). Thus, $\left\lfloor
x\right\rfloor \neq x$. Combining this with $\left\lfloor x\right\rfloor \leq
x$, we obtain $\left\lfloor x\right\rfloor <x$.

Now, $x<\left\lfloor x\right\rfloor +1$, so that $-1-\left\lfloor
x\right\rfloor <-x$. Thus, $-1-\left\lfloor x\right\rfloor \leq-x$. On the
other hand, $\left\lfloor x\right\rfloor <x$, so that $-x<-\left\lfloor
x\right\rfloor =\left(  -1-\left\lfloor x\right\rfloor \right)  +1$.

Thus, $-1-\left\lfloor x\right\rfloor \leq-x<\left(  -1-\left\lfloor
x\right\rfloor \right)  +1$. But $\left\lfloor -x\right\rfloor $ is the unique
integer $n$ satisfying $n\leq-x<n+1$ (since this is how $\left\lfloor
-x\right\rfloor $ is defined). Thus, if $n$ is any integer satisfying
$n\leq-x<n+1$, then $n=\left\lfloor -x\right\rfloor $. Applying this to
$n=-1-\left\lfloor x\right\rfloor $, we obtain $-1-\left\lfloor x\right\rfloor
=\left\lfloor -x\right\rfloor $ (since $-1-\left\lfloor x\right\rfloor $ is an
integer satisfying $-1-\left\lfloor x\right\rfloor \leq-x<\left(
-1-\left\lfloor x\right\rfloor \right)  +1$). Hence, $-1=\left\lfloor
x\right\rfloor +\left\lfloor -x\right\rfloor $, so that $\left\lfloor
x\right\rfloor +\left\lfloor -x\right\rfloor =-1$. Comparing this with $%
\begin{cases}
0, & \text{if }x\in\mathbb{Z};\\
-1, & \text{if }x\notin\mathbb{Z}%
\end{cases}
=-1$ (since $x\notin\mathbb{Z}$), we obtain $\left\lfloor x\right\rfloor
+\left\lfloor -x\right\rfloor =%
\begin{cases}
0, & \text{if }x\in\mathbb{Z};\\
-1, & \text{if }x\notin\mathbb{Z}%
\end{cases}
$. Thus, Proposition \ref{prop.floor.-x} is proven in Case 2.

We have now proven Proposition \ref{prop.floor.-x} in each of the two Cases 1
and 2. Thus, Proposition \ref{prop.floor.-x} always holds.
\end{proof}

Now, let us prove \cite[Theorem 4.1 \textbf{(2)}]{NiZuMo91}:

\begin{proposition}
\label{prop.floor.sum1}Let $x$ be a \textbf{nonnegative} real number. Then,
$\left\lfloor x\right\rfloor =\sum\limits_{\substack{m\in\mathbb{N}%
_{+};\\m\leq x}}1$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.floor.sum1}.]First of all, $0\leq x$ (since
$x$ is nonnegative). But Proposition \ref{prop.floor.adjoint} (applied to
$m=0$) shows that $0\leq x$ holds if and only if $0\leq\left\lfloor
x\right\rfloor $ holds. Thus, $0\leq\left\lfloor x\right\rfloor $ holds (since
$0\leq x$ holds). Thus, $\left\lfloor x\right\rfloor \in\mathbb{N}$ (since
$\left\lfloor x\right\rfloor $ is an integer). Hence,
\[
\sum\limits_{\substack{m\in\mathbb{N}_{+};\\m\leq\left\lfloor x\right\rfloor
}}1=\sum\limits_{m=1}^{\left\lfloor x\right\rfloor }1=\left\lfloor
x\right\rfloor .
\]
Hence,%
\[
\left\lfloor x\right\rfloor =\sum\limits_{\substack{m\in\mathbb{N}_{+}%
;\\m\leq\left\lfloor x\right\rfloor }}1=\sum\limits_{\substack{m\in
\mathbb{N}_{+};\\m\leq x}}1
\]
(because Proposition \ref{prop.floor.adjoint} shows that the condition
$m\leq\left\lfloor x\right\rfloor $ for an integer $m$ is equivalent to the
condition $m\leq x$). This proves Proposition \ref{prop.floor.sum1}.
\end{proof}

We shall next prove \cite[Theorem 4.1 \textbf{(3)}]{NiZuMo91}:

\begin{proposition}
\label{prop.floor.x+k}Let $x$ be a real number. Let $k$ be an integer. Then,
$\left\lfloor x+k\right\rfloor =\left\lfloor x\right\rfloor +k$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.floor.x+k}.]Recall that $\left\lfloor
x\right\rfloor $ is the unique integer $n$ satisfying $n\leq x<n+1$. Thus,
$\left\lfloor x\right\rfloor $ is an integer satisfying $\left\lfloor
x\right\rfloor \leq x<\left\lfloor x\right\rfloor +1$.

Also, $\left\lfloor x\right\rfloor +k$ is an integer (since both $\left\lfloor
x\right\rfloor $ and $k$ are integers). Also, $\underbrace{\left\lfloor
x\right\rfloor }_{\leq x}+k\leq x+k$ and $\underbrace{x}_{<\left\lfloor
x\right\rfloor +1}+k<\left\lfloor x\right\rfloor +1+k=\left(  \left\lfloor
x\right\rfloor +k\right)  +1$, so that $\left\lfloor x\right\rfloor +k\leq
x+k<\left(  \left\lfloor x\right\rfloor +k\right)  +1$. Thus, $\left\lfloor
x\right\rfloor +k$ is an integer $n$ satisfying $n\leq x+k<n+1$.

But $\left\lfloor x+k\right\rfloor $ is the unique integer $n$ satisfying
$n\leq x+k<n+1$ (because this is how $\left\lfloor x+k\right\rfloor $ is
defined). Hence, if $n$ is any integer satisfying $n\leq x+k<n+1$, then
$n=\left\lfloor x+k\right\rfloor $. We can apply this to $n=\left\lfloor
x\right\rfloor +k$ (since $\left\lfloor x\right\rfloor +k$ is an integer $n$
satisfying $n\leq x+k<n+1$), and thus obtain $\left\lfloor x\right\rfloor
+k=\left\lfloor x+k\right\rfloor $. Proposition \ref{prop.floor.x+k} is proven.
\end{proof}

\begin{proposition}
\label{prop.floor.n/b}Let $n\in\mathbb{N}$ and $b\in\mathbb{N}_{+}$. Then,
$\sum_{\substack{k\in\left\{  1,2,\ldots,n\right\}  ;\\b\mid k}}1=\left\lfloor
\dfrac{n}{b}\right\rfloor $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.floor.n/b}.]The elements of $\left\{
1,2,\ldots,n\right\}  $ are precisely the elements $k$ of $\mathbb{N}_{+}$
satisfying $k\leq n$. Hence,%
\begin{align*}
\sum_{\substack{k\in\left\{  1,2,\ldots,n\right\}  ;\\b\mid k}}1  &
=\sum_{\substack{k\in\mathbb{N}_{+};\\k\leq n;\\b\mid k}}1=\sum
_{\substack{k\in\mathbb{N}_{+};\\b\mid k;\\k\leq n}}1\\
&  =\sum_{\substack{m\in\mathbb{N}_{+};\\bm\leq n}%
}1\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }bm\text{ for }k\text{ in}\\
\text{the sum (since the }k\in\mathbb{N}_{+}\text{ satisfying}\\
b\mid k\text{ are precisely the integers of}\\
\text{the form }bm\text{ with }m\in\mathbb{N}_{+}\text{)}%
\end{array}
\right) \\
&  =\sum_{\substack{m\in\mathbb{N}_{+};\\m\leq\dfrac{n}{b}}%
}1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }bm\leq n\text{ is equivalent to
}m\leq\dfrac{n}{b}\right) \\
&  =\left\lfloor \dfrac{n}{b}\right\rfloor
\end{align*}
(because Proposition \ref{prop.floor.sum1} (applied to $x=\dfrac{n}{b}$)
yields $\left\lfloor \dfrac{n}{b}\right\rfloor =\sum\limits_{\substack{m\in
\mathbb{N}_{+};\\m\leq\dfrac{n}{b}}}1$). Thus, Proposition
\ref{prop.floor.n/b} is proven.
\end{proof}

The floor function is weakly increasing:

\begin{proposition}
\label{prop.floor.winc}Let $x$ and $y$ be real numbers such that $x\leq y$.
Then, $\left\lfloor x\right\rfloor \leq\left\lfloor y\right\rfloor $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.floor.winc}.]Recall that $\left\lfloor
x\right\rfloor $ is the unique integer $n$ satisfying $n\leq x<n+1$. Thus,
$\left\lfloor x\right\rfloor $ is an integer satisfying $\left\lfloor
x\right\rfloor \leq x<\left\lfloor x\right\rfloor +1$. Hence, $\left\lfloor
x\right\rfloor \leq x\leq y$.

Proposition \ref{prop.floor.adjoint} (applied to $\left\lfloor x\right\rfloor
$ and $y$ instead of $m$ and $x$) shows that $\left\lfloor x\right\rfloor \leq
y$ holds if and only if $\left\lfloor x\right\rfloor \leq\left\lfloor
y\right\rfloor $ holds. Hence, $\left\lfloor x\right\rfloor \leq\left\lfloor
y\right\rfloor $ holds (since $\left\lfloor x\right\rfloor \leq y$ holds).
This proves Proposition \ref{prop.floor.winc}.
\end{proof}

Let us now prove \cite[Theorem 4.1 \textbf{(4)}]{NiZuMo91}:

\begin{proposition}
\label{prop.floor.u+v}Let $u\in\mathbb{R}$ and $v\in\mathbb{R}$. Then,
$\left\lfloor u\right\rfloor +\left\lfloor v\right\rfloor \leq\left\lfloor
u+v\right\rfloor \leq\left\lfloor u\right\rfloor +\left\lfloor v\right\rfloor
+1$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.floor.u+v}.]All of $\left\lfloor
u\right\rfloor $, $\left\lfloor v\right\rfloor $ and $\left\lfloor
u+v\right\rfloor $ are integers (by their definition).

Recall that $\left\lfloor u\right\rfloor $ is the unique integer $n$
satisfying $n\leq u<n+1$. Thus, $\left\lfloor u\right\rfloor $ is an integer
satisfying $\left\lfloor u\right\rfloor \leq u<\left\lfloor u\right\rfloor +1$.

Recall that $\left\lfloor v\right\rfloor $ is the unique integer $n$
satisfying $n\leq v<n+1$. Thus, $\left\lfloor v\right\rfloor $ is an integer
satisfying $\left\lfloor v\right\rfloor \leq v<\left\lfloor v\right\rfloor +1$.

Recall that $\left\lfloor u+v\right\rfloor $ is the unique integer $n$
satisfying $n\leq u+v<n+1$. Thus, $\left\lfloor u+v\right\rfloor $ is an
integer satisfying $\left\lfloor u+v\right\rfloor \leq u+v<\left\lfloor
u+v\right\rfloor +1$.

Proposition \ref{prop.floor.adjoint} (applied to $m=\left\lfloor
u\right\rfloor +\left\lfloor v\right\rfloor $ and $x=u+v$) shows that
$\left\lfloor u\right\rfloor +\left\lfloor v\right\rfloor \leq u+v$ holds if
and only if $\left\lfloor u\right\rfloor +\left\lfloor v\right\rfloor
\leq\left\lfloor u+v\right\rfloor $ holds. Thus, $\left\lfloor u\right\rfloor
+\left\lfloor v\right\rfloor \leq\left\lfloor u+v\right\rfloor $ holds (since
$\underbrace{\left\lfloor u\right\rfloor }_{\leq u}+\underbrace{\left\lfloor
v\right\rfloor }_{\leq v}\leq u+v$ holds).

We know that $\left\lfloor v\right\rfloor +1$ is an integer (since
$\left\lfloor v\right\rfloor $ is an integer). Hence, Proposition
\ref{prop.floor.x+k} (applied to $x=u$ and $k=\left\lfloor v\right\rfloor +1$)
yields $\left\lfloor u+\left\lfloor v\right\rfloor +1\right\rfloor
=\left\lfloor u\right\rfloor +\left\lfloor v\right\rfloor +1$.

But $u+\underbrace{v}_{<\left\lfloor v\right\rfloor +1}<u+\left\lfloor
v\right\rfloor +1$. Hence, Proposition \ref{prop.floor.winc} (applied to
$x=u+v$ and $y=u+\left\lfloor v\right\rfloor +1$) shows that $\left\lfloor
u+v\right\rfloor \leq\left\lfloor u+\left\lfloor v\right\rfloor
+1\right\rfloor =\left\lfloor u\right\rfloor +\left\lfloor v\right\rfloor +1$.
Combining this with $\left\lfloor u\right\rfloor +\left\lfloor v\right\rfloor
\leq\left\lfloor u+v\right\rfloor $, we obtain $\left\lfloor u\right\rfloor
+\left\lfloor v\right\rfloor \leq\left\lfloor u+v\right\rfloor \leq
\left\lfloor u\right\rfloor +\left\lfloor v\right\rfloor +1$.

This proves Proposition \ref{prop.floor.u+v}.
\end{proof}

Finally, let us prove \cite[Theorem 4.1 \textbf{(6)}]{NiZuMo91}:

\begin{proposition}
\label{prop.floor.double}Let $x\in\mathbb{R}$ and $m\in\mathbb{N}_{+}$. Then,
$\left\lfloor \dfrac{\left\lfloor x\right\rfloor }{m}\right\rfloor
=\left\lfloor \dfrac{x}{m}\right\rfloor $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.floor.double}.]Recall that $\left\lfloor
x\right\rfloor $ is the unique integer $n$ satisfying $n\leq x<n+1$. Thus,
$\left\lfloor x\right\rfloor $ is an integer satisfying $\left\lfloor
x\right\rfloor \leq x<\left\lfloor x\right\rfloor +1$.

Recall that $\left\lfloor \dfrac{x}{m}\right\rfloor $ is the unique integer
$n$ satisfying $n\leq\dfrac{x}{m}<n+1$ (by the definition of $\left\lfloor
\dfrac{x}{m}\right\rfloor $. Thus, $\left\lfloor \dfrac{x}{m}\right\rfloor $
is an integer satisfying $\left\lfloor \dfrac{x}{m}\right\rfloor \leq\dfrac
{x}{m}<\left\lfloor \dfrac{x}{m}\right\rfloor +1$.

But $m\in\mathbb{N}_{+}$ and thus $m\geq1>0$. Hence, we can multiply the
inequality $\left\lfloor \dfrac{x}{m}\right\rfloor \leq\dfrac{x}{m}$ by $m$.
We thus obtain $m\left\lfloor \dfrac{x}{m}\right\rfloor \leq x$.

But Proposition \ref{prop.floor.adjoint} (applied to $m\left\lfloor \dfrac
{x}{m}\right\rfloor $ instead of $m$) shows that $m\left\lfloor \dfrac{x}%
{m}\right\rfloor \leq x$ holds if and only if $m\left\lfloor \dfrac{x}%
{m}\right\rfloor \leq\left\lfloor x\right\rfloor $ holds (since $m\left\lfloor
\dfrac{x}{m}\right\rfloor $ is an integer). Thus, $m\left\lfloor \dfrac{x}%
{m}\right\rfloor \leq\left\lfloor x\right\rfloor $ holds (since $m\left\lfloor
\dfrac{x}{m}\right\rfloor \leq x$ holds). Dividing this inequality by $m$, we
obtain $\left\lfloor \dfrac{x}{m}\right\rfloor \leq\dfrac{\left\lfloor
x\right\rfloor }{m}$.

We can divide the inequality $\left\lfloor x\right\rfloor \leq x$ by $m$
(since $m>0$). We thus obtain $\dfrac{\left\lfloor x\right\rfloor }{m}%
\leq\dfrac{x}{m}$. Hence, $\dfrac{\left\lfloor x\right\rfloor }{m}\leq
\dfrac{x}{m}<\left\lfloor \dfrac{x}{m}\right\rfloor +1$.

So we have $\left\lfloor \dfrac{x}{m}\right\rfloor \leq\dfrac{\left\lfloor
x\right\rfloor }{m}<\left\lfloor \dfrac{x}{m}\right\rfloor +1$. In other
words, $\left\lfloor \dfrac{x}{m}\right\rfloor $ is an integer $n$ satisfying
$n\leq\dfrac{\left\lfloor x\right\rfloor }{m}<n+1$.

But $\left\lfloor \dfrac{\left\lfloor x\right\rfloor }{m}\right\rfloor $ is
the unique integer $n$ satisfying $n\leq\dfrac{\left\lfloor x\right\rfloor
}{m}<n+1$ (because this is how $\left\lfloor \dfrac{\left\lfloor
x\right\rfloor }{m}\right\rfloor $ is defined). Hence, if $n$ is any integer
satisfying $n\leq\dfrac{\left\lfloor x\right\rfloor }{m}<n+1$, then
$n=\left\lfloor \dfrac{\left\lfloor x\right\rfloor }{m}\right\rfloor $. We can
apply this to $n=\left\lfloor \dfrac{x}{m}\right\rfloor $ (since $\left\lfloor
\dfrac{x}{m}\right\rfloor $ is an integer $n$ satisfying $n\leq\dfrac
{\left\lfloor x\right\rfloor }{m}<n+1$), and thus obtain $\left\lfloor
\dfrac{x}{m}\right\rfloor =\left\lfloor \dfrac{\left\lfloor x\right\rfloor
}{m}\right\rfloor $. Proposition \ref{prop.floor.double} is proven.
\end{proof}

I refer to \cite[\S 4.1]{NiZuMo91} for further properties of the floor function.

\subsection{Interlude: greatest common divisors}

Before we move on, let me remind you of some basic facts about coprime numbers
and greatest common divisors. First, we recall how greatest common divisors
are defined:

\begin{definition}
\label{def.gcd}Let $b$ and $c$ be two integers. If $\left(  b,c\right)
\neq\left(  0,0\right)  $, then $\gcd\left(  b,c\right)  $ means the greatest
of all common divisors of $b$ and $c$. We also set $\gcd\left(  0,0\right)
=0$. Thus, $\gcd\left(  b,c\right)  $ is defined for any two integers $b$ and
$c$.

If $b$ and $c$ are two integers, then $\gcd\left(  b,c\right)  $ is called the
\textit{greatest common divisor of }$b$ \textit{and }$c$ (even if $\gcd\left(
0,0\right)  $ is not literally the greatest of all common divisors of $0$ and
$0$) or, briefly, the \textit{gcd} of $b$ and $c$. Clearly, $\gcd\left(
b,c\right)  =\gcd\left(  c,b\right)  $ and $\gcd\left(  b,c\right)  \mid b$
and $\gcd\left(  b,c\right)  \mid c$ for any two integers $b$ and $c$. Notice
that $\gcd\left(  b,c\right)  $ is a nonnegative integer (and actually a
positive integer unless $\left(  b,c\right)  =\left(  0,0\right)  $).
\end{definition}

Older books such as \cite{NiZuMo91} tend to denote the gcd of two integers $b$
and $c$ by $\left(  b,c\right)  $ (rather than by $\gcd\left(  b,c\right)  $
as we do); this is a convention that I shall decidedly not follow (since it
risks confusion with the notation $\left(  b,c\right)  $ for the ordered pair
of $b$ and $c$).

The most important property of gcds is \textit{B\'{e}zout's theorem}
(\cite[Theorem 1.3]{NiZuMo91}):

\begin{theorem}
\label{thm.bezout}Let $b$ and $c$ be two integers. Then, there exist integers
$x$ and $y$ such that $\gcd\left(  b,c\right)  =bx+cy$.
\end{theorem}

See \cite[Theorem 1.3]{NiZuMo91} for the proof of Theorem \ref{thm.bezout} in
the case when $\left(  b,c\right)  \neq\left(  0,0\right)  $. In the case when
$\left(  b,c\right)  =\left(  0,0\right)  $, Theorem \ref{thm.bezout}
obviously holds (since we can take $x=0$ and $y=0$).

For another proof of Theorem \ref{thm.bezout}, see the Appendix (Chapter
\ref{chp.bezout}) below.

A basic property of gcds that follows directly from Theorem \ref{thm.bezout}
is the following:

\begin{proposition}
\label{prop.gcd.uniprop}Let $a$, $b$ and $c$ be three integers such that
$a\mid b$ and $a\mid c$. Then, $a\mid\gcd\left(  b,c\right)  $.
\end{proposition}

In words, Proposition \ref{prop.gcd.uniprop} says that any common divisor of
two integers must divide the gcd of these two integers.

\begin{proof}
[Proof of Proposition \ref{prop.gcd.uniprop}.]Theorem \ref{thm.bezout} shows
that there exist integers $x$ and $y$ such that $\gcd\left(  b,c\right)
=bx+cy$. Consider these $x$ and $y$. Now, $a\mid b\mid bx$ and $a\mid c\mid
cy$. Hence, both integers $bx$ and $cy$ are divisible by $a$. Thus, their sum
$bx+cy$ must also be divisible by $a$ (by Proposition \ref{prop.div.sum},
applied to $u=bx$ and $v=cy$). In other words, $a\mid bx+cy$. In other words,
$a\mid\gcd\left(  b,c\right)  $ (since $\gcd\left(  b,c\right)  =bx+cy$). This
proves Proposition \ref{prop.gcd.uniprop}.
\end{proof}

\begin{corollary}
\label{cor.gcd.increase}Let $a$, $b$, $c$ and $d$ be four integers such that
$a\mid c$ and $b\mid d$. Then, $\gcd\left(  a,b\right)  \mid\gcd\left(
c,d\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.gcd.increase}.]We have $\gcd\left(  a,b\right)
\mid a\mid c$ and $\gcd\left(  a,b\right)  \mid b\mid d$. Thus, Proposition
\ref{prop.gcd.uniprop} (applied to $\gcd\left(  a,b\right)  $, $c$ and $d$
instead of $a$, $b$ and $c$) shows that $\gcd\left(  a,b\right)  \mid
\gcd\left(  c,d\right)  $. This proves Corollary \ref{cor.gcd.increase}.
\end{proof}

We shall now use Theorem \ref{thm.bezout} to derive a slight generalization of
\cite[Theorem 1.8]{NiZuMo91}:

\begin{proposition}
\label{prop.gcd(ab,m)}Let $a$, $b$ and $m$ be three integers such that
$\gcd\left(  a,m\right)  =1$. Then, $\gcd\left(  b,m\right)  =\gcd\left(
ab,m\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.gcd(ab,m)}.]Theorem \ref{thm.bezout} (applied
to $a$ and $m$ instead of $b$ and $c$) shows that there exist integers $x$ and
$y$ such that $\gcd\left(  a,m\right)  =ax+my$. Consider these $x$ and $y$. We
have $ax+my=\gcd\left(  a,m\right)  =1$.

Let $g=\gcd\left(  b,m\right)  $ and $h=\gcd\left(  ab,m\right)  $. Thus, $g$
and $h$ are nonnegative integers.

We have $g=\gcd\left(  b,m\right)  \mid b\mid ab$ and $g=\gcd\left(
b,m\right)  \mid m$. Thus, Proposition \ref{prop.gcd.uniprop} (applied to $g$,
$ab$ and $m$ instead of $a$, $b$ and $c$) shows that $g\mid\gcd\left(
ab,m\right)  $. In other words, $g\mid h$ (since $h=\gcd\left(  ab,m\right)  $).

On the other hand, $h=\gcd\left(  ab,m\right)  \mid ab\mid abx$ and
$h=\gcd\left(  ab,m\right)  \mid m\mid mby$. Thus, both integers $abx$ and
$mby$ are divisible by $h$. Therefore, the sum of these two integers must also
be divisible by $h$ (by Proposition \ref{prop.div.sum}, applied to $abx$,
$mby$ and $h$ instead of $u$, $v$ and $a$). In other words, $h\mid abx+mby$.
Since%
\[
abx+mby=b\underbrace{\left(  ax+my\right)  }_{=1}=b,
\]
this rewrites as $h\mid b$. So we have $h\mid b$ and $h\mid m$. Thus,
Proposition \ref{prop.gcd.uniprop} (applied to $h$, $b$ and $m$ instead of
$a$, $b$ and $c$) shows that $h\mid\gcd\left(  b,m\right)  $. In other words,
$h\mid g$ (since $g=\gcd\left(  b,m\right)  $).

Now, we can apply Proposition \ref{prop.div.=} to $u=g$ and $v=h$ (since $g$
and $h$ are nonnegative integers satisfying $g\mid h$ and $h\mid g$), and thus
we obtain $g=h$. Hence, $\gcd\left(  b,m\right)  =g=h=\gcd\left(  ab,m\right)
$. This proves Proposition \ref{prop.gcd(ab,m)}.
\end{proof}

Now, let us recall how coprime integers are defined: We say that an integer
$a$ is \textit{coprime} to an integer $b$ if $\gcd\left(  a,b\right)  =1$. The
relation of being coprime is clearly symmetric (i.e., an integer $a$ is
coprime to an integer $b$ if and only if $b$ is coprime to $a$), because
$\gcd\left(  a,b\right)  =\gcd\left(  b,a\right)  $. Thus, instead of saying
that \textquotedblleft$a$ is coprime to $b$\textquotedblright, we can also say
\textquotedblleft the integers $a$ and $b$ are coprime\textquotedblright.

The following corollary is precisely \cite[Theorem 1.8]{NiZuMo91}:

\begin{corollary}
\label{cor.prod-coprime.ab}Let $a$, $b$ and $m$ be three integers. Assume that
$a$ is coprime to $m$, and assume that $b$ is coprime to $m$. Then, $ab$ is
coprime to $m$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.prod-coprime.ab}.]We know that $a$ is coprime to
$m$. In other words, $\gcd\left(  a,m\right)  =1$. Also, we have assumed that
$b$ is coprime to $m$. In other words, $\gcd\left(  b,m\right)  =1$. Now,
Proposition \ref{prop.gcd(ab,m)} yields $\gcd\left(  b,m\right)  =\gcd\left(
ab,m\right)  $. Thus, $\gcd\left(  ab,m\right)  =\gcd\left(  b,m\right)  =1$.
In other words, $ab$ is coprime to $m$. This proves Corollary
\ref{cor.prod-coprime.ab}.
\end{proof}

The following corollary generalizes Corollary \ref{cor.prod-coprime.ab} to $n$
integers instead of the two integers $a$ and $b$:

\begin{corollary}
\label{cor.prod-coprime.n}Let $c_{1},c_{2},\ldots,c_{n}$ be $n$ integers. Let
$m$ be an integer. Assume that $c_{u}$ is coprime to $m$ for every
$u\in\left\{  1,2,\ldots,n\right\}  $. Then, $c_{1}c_{2}\cdots c_{n}$ is
coprime to $m$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.prod-coprime.n}.]We shall prove that%
\begin{equation}
c_{1}c_{2}\cdots c_{g}\text{ is coprime to }m\ \ \ \ \ \ \ \ \ \ \text{for
every }g\in\left\{  0,1,\ldots,n\right\}  . \label{pf.cor.prod-coprime.n.1}%
\end{equation}


\textit{Proof of (\ref{pf.cor.prod-coprime.n.1}):} We shall prove
(\ref{pf.cor.prod-coprime.n.1}) by induction on $g$:

\textit{Induction base:} We have $c_{1}c_{2}\cdots c_{0}=\left(  \text{empty
product}\right)  =1$ (since empty products are $1$ by definition). But $1$ is
clearly coprime to $m$ (since $\gcd\left(  1,m\right)  \mid1$ and thus
$\gcd\left(  1,m\right)  =1$). In other words, $c_{1}c_{2}\cdots c_{0}$ is
coprime to $m$ (since $c_{1}c_{2}\cdots c_{0}=1$). In other words,
(\ref{pf.cor.prod-coprime.n.1}) holds for $g=0$. This completes the induction base.

\textit{Induction step:} Let $G\in\left\{  0,1,\ldots,n\right\}  $ be
positive. Assume that (\ref{pf.cor.prod-coprime.n.1}) holds for $g=G-1$. We
must prove that (\ref{pf.cor.prod-coprime.n.1}) holds for $g=G$.

We have $G\in\left\{  0,1,\ldots,n\right\}  $, and thus $G\in\left\{
1,2,\ldots,n\right\}  $ (since $G$ is positive).

We know that $c_{1}c_{2}\cdots c_{G-1}$ is coprime to $m$ (since we assumed
that (\ref{pf.cor.prod-coprime.n.1}) holds for $g=G-1$). Also, we assumed that
$c_{u}$ is coprime to $m$ for every $u\in\left\{  1,2,\ldots,n\right\}  $.
Applying this to $u=G$, we see that $c_{G}$ is coprime to $m$. Now, Corollary
\ref{cor.prod-coprime.ab} (applied to $c_{1}c_{2}\cdots c_{G-1}$ and $c_{G}$
instead of $a$ and $b$) shows that $\left(  c_{1}c_{2}\cdots c_{G-1}\right)
c_{G}$ is coprime to $m$. In other words, $c_{1}c_{2}\cdots c_{G}$ is coprime
to $m$ (since $\left(  c_{1}c_{2}\cdots c_{G-1}\right)  c_{G}=c_{1}c_{2}\cdots
c_{G}$). In other words, (\ref{pf.cor.prod-coprime.n.1}) holds for $g=G$. This
completes the induction step, and thus (\ref{pf.cor.prod-coprime.n.1}) is proven.

Now, applying (\ref{pf.cor.prod-coprime.n.1}) to $g=n$, we conclude that
$c_{1}c_{2}\cdots c_{n}$ is coprime to $m$. This proves Corollary
\ref{cor.prod-coprime.n}.
\end{proof}

A further consequence of Proposition \ref{prop.gcd(ab,m)} is the following
fact (\cite[Theorem 1.10]{NiZuMo91}):

\begin{proposition}
\label{prop.prod-coprime.cancel}Let $x$, $y$ and $z$ be three integers such
that $x\mid yz$ and $\gcd\left(  x,y\right)  =1$. Then, $x\mid z$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.prod-coprime.cancel}.]We have $x\mid yz$ and
$x\mid x$. Hence, Proposition \ref{prop.gcd.uniprop} (applied to $x$, $yz$ and
$x$ instead of $a$, $b$ and $c$) shows that $x\mid\gcd\left(  yz,x\right)  $.

We have $\gcd\left(  y,x\right)  =\gcd\left(  x,y\right)  =1$. Hence,
Proposition \ref{prop.gcd(ab,m)} (applied to $a=y$, $b=z$ and $m=x$) shows
that $\gcd\left(  z,x\right)  =\gcd\left(  yz,x\right)  $. Now, $x\mid
\gcd\left(  yz,x\right)  =\gcd\left(  z,x\right)  \mid z$. This proves
Proposition \ref{prop.prod-coprime.cancel}.
\end{proof}

Another important result on gcds is the following fact:

\begin{proposition}
\label{prop.gcd.g}Let $g$ be a positive integer. Let $a$ and $b$ be two
integers. Then, $g\gcd\left(  a,b\right)  =\gcd\left(  ga,gb\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.gcd.g}.]Both $\gcd\left(  a,b\right)  $ and
$g$ are nonnegative integers; hence, $g\gcd\left(  a,b\right)  $ is a
nonnegative integer.

We have $g\underbrace{\gcd\left(  a,b\right)  }_{\mid a}\mid ga$ and
$g\underbrace{\gcd\left(  a,b\right)  }_{\mid b}\mid gb$. Thus, Proposition
\ref{prop.gcd.uniprop} (applied to $g\gcd\left(  a,b\right)  $, $ga$ and $gb$
instead of $a$, $b$ and $c$) shows that $g\gcd\left(  a,b\right)  \mid
\gcd\left(  ga,gb\right)  $.

On the other hand, Theorem \ref{thm.bezout} (applied to $a$ and $b$ instead of
$b$ and $c$) shows that there exist integers $x$ and $y$ such that
$\gcd\left(  a,b\right)  =ax+by$. Consider these $x$ and $y$. We have
$\gcd\left(  ga,gb\right)  \mid ga\mid gax$ and $\gcd\left(  ga,gb\right)
\mid gb\mid gby$. Thus, both integers $gax$ and $gby$ are divisible by
$\gcd\left(  ga,gb\right)  $. Therefore, their sum $gax+gby$ is also divisible
by $\gcd\left(  ga,gb\right)  $ (by Proposition \ref{prop.div.sum}, applied to
$\gcd\left(  ga,gb\right)  $, $gax$ and $gby$ instead of $a$, $u$ and $v$). In
other words, we have $\gcd\left(  ga,gb\right)  \mid gax+gby$. Since
$gax+gby=g\underbrace{\left(  ax+by\right)  }_{=\gcd\left(  a,b\right)
}=g\gcd\left(  a,b\right)  $, this rewrites as $\gcd\left(  ga,gb\right)  \mid
g\gcd\left(  a,b\right)  $.

But we can apply Proposition \ref{prop.div.=} to $u=g\gcd\left(  a,b\right)  $
and $v=\gcd\left(  ga,gb\right)  $ (since $g\gcd\left(  a,b\right)  $ and
$\gcd\left(  ga,gb\right)  $ are nonnegative integers satisfying $g\gcd\left(
a,b\right)  \mid\gcd\left(  ga,gb\right)  $ and $\gcd\left(  ga,gb\right)
\mid g\gcd\left(  a,b\right)  $), and thus we obtain $g\gcd\left(  a,b\right)
=\gcd\left(  ga,gb\right)  $. This proves Proposition \ref{prop.gcd.g}.
\end{proof}

Here is another property of gcds, which we will use later:

\begin{proposition}
\label{prop.gcd.gcd(u,mn)}Let $m$ and $n$ be two coprime positive integers.
Let $u$ be an integer. Then, $\gcd\left(  u,m\right)  \cdot\gcd\left(
u,n\right)  =\gcd\left(  u,mn\right)  $.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.gcd.gcd(u,mn)}.]Set $h=\gcd\left(
u,mn\right)  $, $v=\gcd\left(  u,m\right)  $ and $w=\gcd\left(  u,n\right)  $.
We shall prove that $vw=h$.

We have $v=\gcd\left(  u,m\right)  \in\mathbb{N}_{+}$ (since $m$ is positive)
and $w=\gcd\left(  u,n\right)  \in\mathbb{N}_{+}$ (since $n$ is positive).
Thus, both $v$ and $w$ are positive integers; hence, $vw$ is a positive
integer. Also, $mn$ is positive (since $m$ and $n$ are positive). Now,
$h=\gcd\left(  u,mn\right)  \in\mathbb{N}_{+}$ (since $mn$ is positive).

We have $v=\gcd\left(  u,m\right)  \mid m$ and $w=\gcd\left(  u,n\right)  \mid
n$. Hence, Corollary \ref{cor.gcd.increase} (applied to $v$, $w$, $m$ and $n$
instead of $a$, $b$, $c$ and $d$) yields $\gcd\left(  v,w\right)  \mid
\gcd\left(  m,n\right)  =1$ (since $m$ and $n$ are coprime). Hence,
$\gcd\left(  v,w\right)  =1$.

We have $w=\gcd\left(  u,n\right)  \mid u$ and thus $\dfrac{u}{w}\in
\mathbb{Z}$. Now, $v=\gcd\left(  u,m\right)  \mid u=w\cdot\dfrac{u}{w}$. Thus,
Proposition \ref{prop.prod-coprime.cancel} (applied to $v$, $w$ and $\dfrac
{u}{w}$ instead of $x$, $y$ and $z$) shows that $v\mid\dfrac{u}{w}$ (since
$\gcd\left(  v,w\right)  =1$). In other words, $\dfrac{u}{w}/v\in\mathbb{Z}$.
Now, $\dfrac{u}{vw}=\dfrac{u}{w}/v\in\mathbb{Z}$, so that $vw\mid u$.

But $v=\gcd\left(  u,m\right)  \mid m$ and thus $\dfrac{m}{v}\in\mathbb{Z}$.
Also, $w=\gcd\left(  u,n\right)  \mid n$ and thus $\dfrac{n}{w}\in\mathbb{Z}$.
Now, $\dfrac{mn}{vw}=\dfrac{m}{v}\cdot\dfrac{n}{w}$ is the product of two
integers (since $\dfrac{m}{v}$ and $\dfrac{n}{w}$ are integers), and thus
itself an integer. In other words, $vw\mid mn$.

So we have $vw\mid u$ and $vw\mid mn$. Proposition \ref{prop.gcd.uniprop}
(applied to $vw$, $u$ and $mn$ instead of $a$, $b$ and $c$) thus yields
$vw\mid\gcd\left(  u,mn\right)  =h$.

Proposition \ref{prop.gcd.g} (applied to $n$, $u$ and $m$ instead of $g$, $a$
and $b$) shows that $n\gcd\left(  u,m\right)  =\gcd\left(  nu,nm\right)  $.
Thus, $\gcd\left(  nu,nm\right)  =n\underbrace{\gcd\left(  u,m\right)  }%
_{=v}=nv=vn$.

Now, $h=\gcd\left(  u,mn\right)  \mid mn=nm$ and $h=\gcd\left(  u,mn\right)
\mid u\mid nu$. Hence, Proposition \ref{prop.gcd.uniprop} (applied to $h$,
$nu$ and $nm$ instead of $a$, $b$ and $c$) shows that $h\mid\gcd\left(
nu,nm\right)  $. In other words, $h\mid vn$ (since $\gcd\left(  nu,nm\right)
=vn$).

Proposition \ref{prop.gcd.g} (applied to $v$, $u$ and $n$ instead of $g$, $a$
and $b$) shows that $v\gcd\left(  u,n\right)  =\gcd\left(  vu,vn\right)  $.
Thus, $\gcd\left(  vu,vn\right)  =v\underbrace{\gcd\left(  u,n\right)  }%
_{=w}=vw$.

Now, $h\mid u\mid vu$ and $h\mid vn$. Thus, Proposition \ref{prop.gcd.uniprop}
(applied to $h$, $vu$ and $vn$ instead of $a$, $b$ and $c$) shows that
$h\mid\gcd\left(  vu,vn\right)  $. In other words, $h\mid vw$ (since
$\gcd\left(  vu,vn\right)  =vw$). Combining this with $vw\mid h$, we obtain
$h=vw$ (by Proposition \ref{prop.div.=} (applied to $h$ and $vw$ instead of
$u$ and $v$), since $h$ and $vw$ are positive integers). Now,%
\[
\underbrace{\gcd\left(  u,m\right)  }_{=v}\cdot\underbrace{\gcd\left(
u,n\right)  }_{=w}=vw=h=\gcd\left(  u,mn\right)  .
\]
This proves Proposition \ref{prop.gcd.gcd(u,mn)}.
\end{proof}

Let us derive an easy corollary from Corollary \ref{cor.prod-coprime.n}:

\begin{corollary}
\label{cor.pow-coprimes}Let $a$ and $b$ be two coprime integers. Let
$n\in\mathbb{N}$.

\textbf{(a)} The integer $a^{n}$ is coprime to $b$.

\textbf{(b)} Let $m\in\mathbb{N}$. Then, the integer $a^{n}$ is coprime to
$b^{m}$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.pow-coprimes}.]\textbf{(a)} For each
$i\in\left\{  1,2,\ldots,n\right\}  $, the integer $a$ is coprime to $b$.
Thus, Corollary \ref{cor.prod-coprime.n} (applied to $c_{i}=a$ and $m=b$)
shows that $\underbrace{aa\cdots a}_{n\text{ times}}$ is coprime to $b$. In
other words, $a^{n}$ is coprime to $b$ (since $a^{n}=\underbrace{aa\cdots
a}_{n\text{ times}}$). This proves Corollary \ref{cor.pow-coprimes}
\textbf{(a)}.

\textbf{(b)} Corollary \ref{cor.pow-coprimes} \textbf{(a)} shows that the
integer $a^{n}$ is coprime to $b$. In other words, $b$ and $a^{n}$ are two
coprime integers. Hence, Corollary \ref{cor.pow-coprimes} \textbf{(a)}
(applied to $b$, $a^{n}$ and $m$ instead of $a$, $b$ and $n$) shows that the
integer $b^{m}$ is coprime to $a^{n}$. In other words, the integer $a^{n}$ is
coprime to $b^{m}$. This proves Corollary \ref{cor.pow-coprimes} \textbf{(b)}.
\end{proof}

Let us next use the above theory of greatest common divisors to prove some
properties of primes. We begin with the probably most basic one:

\begin{proposition}
\label{prop.prime.if-ndiv-then-cop}Let $p$ be a prime. Let $a$ be an integer
such that $p\nmid a$. Then, $a$ is coprime to $p$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.prime.if-ndiv-then-cop}.]If we had
$\gcd\left(  a,p\right)  =p$, then we would have $p=\gcd\left(  a,p\right)
\mid a$, which would contradict $p\nmid a$. Hence, we cannot have $\gcd\left(
a,p\right)  =p$.

We have $\left(  a,p\right)  \neq\left(  0,0\right)  $ (since $p\neq0$). Thus,
$\gcd\left(  a,p\right)  $ is a positive integer. Hence, $\gcd\left(
a,p\right)  $ is a positive divisor of $p$ (since $\gcd\left(  a,p\right)
\mid p$). But the only positive divisors of $p$ are $1$ and $p$ (since $p$ is
prime). Hence, $\gcd\left(  a,p\right)  $ must be either $1$ or $p$ (since
$\gcd\left(  a,p\right)  $ is a positive divisor of $p$). Thus, we must have
$\gcd\left(  a,p\right)  =1$ (since we cannot have $\gcd\left(  a,p\right)
=p$). In other words, $a$ is coprime to $p$. This proves Proposition
\ref{prop.prime.if-ndiv-then-cop}.
\end{proof}

\begin{corollary}
\label{cor.prime-pow-coprime}Let $s$ and $t$ be two distinct primes.

\textbf{(a)} The integers $s$ and $t$ are coprime.

\textbf{(b)} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Then, the integers
$s^{n}$ and $t^{m}$ are coprime.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.prime-pow-coprime}.]\textbf{(a)} Assume the
contrary. Thus, the integers $s$ and $t$ are not coprime. In other words, $s$
is not coprime to $t$.

If we had $s\nmid t$, then $s$ would be coprime to $t$ (by Proposition
\ref{prop.prime.if-ndiv-then-cop}, applied to $p=s$ and $a=t$), which would
contradict the fact that $s$ is not coprime to $t$. Thus, we cannot have
$s\nmid t$. Hence, we have $s\mid t$. The same argument (with the roles of $s$
and $t$ switched) yields $t\mid s$. Thus, Proposition \ref{prop.div.=}
(applied to $u=s$ and $v=t$) yields $s=t$. But this contradicts the fact that
$s$ and $t$ are distinct. This contradiction shows that our assumption was
wrong. Hence, Corollary \ref{cor.prime-pow-coprime} \textbf{(a)} is proven.

\textbf{(b)} Corollary \ref{cor.prime-pow-coprime} \textbf{(a)} shows that the
integers $s$ and $t$ are coprime. Hence, Corollary \ref{cor.pow-coprimes}
\textbf{(b)} (applied to $a=s$ and $b=t$) shows that the integer $s^{n}$ is
coprime to $t^{m}$. In other words, the integers $s^{n}$ and $t^{m}$ are
coprime. This proves Corollary \ref{cor.prime-pow-coprime} \textbf{(b)}%
.\footnote{Here is a different way to prove Corollary
\ref{cor.prime-pow-coprime} \textbf{(b)} using the uniqueness of prime
factorization:
\par
Assume the contrary. Thus, the integers $s^{n}$ and $t^{m}$ are not coprime.
In other words, $\gcd\left(  s^{n},t^{m}\right)  >1$. Hence, the integer
$\gcd\left(  s^{n},t^{m}\right)  $ has a prime divisor $q$. Consider this $q$.
\par
But $s$ is a prime. Thus, the prime factorization of $s^{n}$ is $s^{n}%
=\underbrace{s\cdot s\cdot\cdots\cdot s}_{n\text{ times}}$. Hence, the only
prime divisor of $s^{n}$ is $s$.
\par
But $q\mid\gcd\left(  s^{n},t^{m}\right)  \mid s^{n}$. Thus, $q$ is a prime
divisor of $s^{n}$ (since $q$ is a prime and divides $s^{n}$). Since the only
prime divisor of $s^{n}$ is $s$, this shows that $q=s$. The same argument
(applied to $t$ instead of $s$) shows that $q=t$. Hence, $s=q=t$. This
contradicts the fact that $s$ and $t$ are distinct. This contradiction proves
that our assumption was wrong, qed.}
\end{proof}

\begin{proposition}
\label{prop.prime-div-prod}Let $p$ be a prime. Let $a_{1},a_{2},\ldots,a_{n}$
be $n$ integers. Assume that $p\mid a_{1}a_{2}\cdots a_{n}$. Then, there
exists an $i\in\left\{  1,2,\ldots,n\right\}  $ such that $p\mid a_{i}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.prime-div-prod}.]Assume the contrary. Thus,
there exists no $i\in\left\{  1,2,\ldots,n\right\}  $ such that $p\mid a_{i}$.
In other words, for each $u\in\left\{  1,2,\ldots,n\right\}  $, we have
$p\nmid a_{u}$. Hence, for each $u\in\left\{  1,2,\ldots,n\right\}  $, the
number $a_{u}$ is coprime to $p$ (by Proposition
\ref{prop.prime.if-ndiv-then-cop}, applied to $a=a_{u}$). In other words,
$a_{u}$ is coprime to $p$ for each $u\in\left\{  1,2,\ldots,n\right\}  $.
Hence, Corollary \ref{cor.prod-coprime.n} (applied to $m=p$ and $c_{i}=a_{i}$)
shows that $a_{1}a_{2}\cdots a_{n}$ is coprime to $p$. In other words,
$\gcd\left(  p,a_{1}a_{2}\cdots a_{n}\right)  =1$.

But $p\mid p$ and $p\mid a_{1}a_{2}\cdots a_{n}$. Hence, Proposition
\ref{prop.gcd.uniprop} (applied to $a=p$, $b=p$ and $c=a_{1}a_{2}\cdots a_{n}%
$) yields $p\mid\gcd\left(  p,a_{1}a_{2}\cdots a_{n}\right)  =1$. Combined
with $1\mid p$, this yields $p=1$ (by Proposition \ref{prop.div.=}, applied to
$u=p$ and $v=1$). This is absurd, since $p$ is a prime. This contradiction
shows that our assumption was wrong; hence, Proposition
\ref{prop.prime-div-prod} is proven.
\end{proof}

\begin{remark}
\label{rmk.prime-div-prod-word}Proposition \ref{prop.prime-div-prod} can be
restated as follows: If a prime $p$ divides a product of finitely many
integers, then $p$ must divide (at least) one of these integers.
\end{remark}

\subsection{de Polignac's formula}

One of the most useful applications of the floor function is computing the
$p$-adic valuation of factorials. Let us first define our notations:

\begin{definition}
\label{def.vp}Let $p$ be a prime. Let $n$ be a nonzero integer. Then,
$v_{p}\left(  n\right)  $ is defined to be the highest nonnegative integer $k$
such that $p^{k}\mid n$. This nonnegative integer $v_{p}\left(  n\right)  $ is
called the $p$\textit{-adic valuation} of $n$.
\end{definition}

\begin{remark}
Some authors use the notation $e_{p}\left(  n\right)  $ instead of
$v_{p}\left(  n\right)  $.

Another way to characterize $v_{p}\left(  n\right)  $ in Definition
\ref{def.vp} is by the following statement: The number $\dfrac{n}%
{p^{v_{p}\left(  n\right)  }}$ is an integer not divisible by $p$.

Yet another (probably simpler) way to define $v_{p}\left(  n\right)  $ is the
following: $v_{p}\left(  n\right)  $ is the exponent with which $p$ occurs in
the prime factorization of $n$.\ \ \ \ \footnotemark\ (This is clearly
equivalent to the definition of $v_{p}\left(  n\right)  $ above.) While I will
try to avoid using prime factorizations wherever I can, there should be
nothing stopping you from using them; in general, the prime factorization of
$n$ is probably the quickest way to get an intuition for $v_{p}\left(
n\right)  $ (although not the quickest way to compute it!).

Often, the definition of $v_{p}\left(  n\right)  $ is extended to all rational
numbers $n$. Then, one defines $v_{p}\left(  n\right)  $ to be the unique
integer $k$ (not necessarily nonnegative) such that the rational number
$\dfrac{n}{p^{k}}$ can be written as a fraction whose numerator and
denominator are both integers coprime to $p$. This works when $n\neq0$. In the
case of $n=0$, one commonly defines $v_{p}\left(  0\right)  $ to be $-\infty$;
here, $-\infty$ is a symbol which (when it comes to comparing it with
integers) is smaller than every integer.
\end{remark}

\footnotetext{This exponent should be understood as $0$ if $p$ does not occur
in the prime factorization of $n$ at all.}

\begin{theorem}
[de Polignac's formula]\label{thm.de-polignac}Let $p$ be a prime. Let
$n\in\mathbb{N}$. Then,%
\[
v_{p}\left(  n!\right)  =\sum_{i=1}^{\infty}\left\lfloor \dfrac{n}{p^{i}%
}\right\rfloor .
\]
(The sum on the right hand side is infinite, but only finitely many of its
terms are nonzero, and thus it is a well-defined integer.)
\end{theorem}

Before we prove this theorem, here are two simple lemmas:

\begin{lemma}
\label{lem.vp.sum}Let $p$ be a prime. Let $n$ be a nonzero integer. Then,%
\[
v_{p}\left(  n\right)  =\sum_{\substack{i\in\mathbb{N}_{+};\\p^{i}\mid n}}1.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.vp.sum}.]Recall that $v_{p}\left(  n\right)  $ is
defined as the highest nonnegative integer $k$ such that $p^{k}\mid n$. Thus,
$v_{p}\left(  n\right)  $ is a nonnegative integer satisfying $p^{v_{p}\left(
n\right)  }\mid n$.

Every $i\in\mathbb{N}_{+}$ that satisfies $i\leq v_{p}\left(  n\right)  $ must
satisfy $p^{i}\mid n$ (since $i\leq v_{p}\left(  n\right)  $ leads to
$p^{i}\mid p^{v_{p}\left(  n\right)  }\mid n$). Conversely, every
$i\in\mathbb{N}_{+}$ satisfying $p^{i}\mid n$ must satisfy $i\leq v_{p}\left(
n\right)  $ (since $v_{p}\left(  n\right)  $ is the \textbf{highest}
nonnegative integer $k$ such that $p^{k}\mid n$). Thus, the $i\in
\mathbb{N}_{+}$ that satisfy $i\leq v_{p}\left(  n\right)  $ are exactly the
$i\in\mathbb{N}_{+}$ that satisfy $p^{i}\mid n$. Consequently,
\[
\sum_{\substack{i\in\mathbb{N}_{+};\\i\leq v_{p}\left(  n\right)  }%
}1=\sum_{\substack{i\in\mathbb{N}_{+};\\p^{i}\mid n}}1.
\]
Hence, $\sum_{\substack{i\in\mathbb{N}_{+};\\p^{i}\mid n}}1=\sum
_{\substack{i\in\mathbb{N}_{+};\\i\leq v_{p}\left(  n\right)  }}1=\sum
_{i=1}^{v_{p}\left(  n\right)  }1=v_{p}\left(  n\right)  $. This proves Lemma
\ref{lem.vp.sum}.
\end{proof}

\begin{lemma}
\label{lem.vp.prod}Let $p$ be a prime. Let $a_{1},a_{2},\ldots,a_{n}$ be
finitely many nonzero integers. Then,%
\[
v_{p}\left(  a_{1}a_{2}\cdots a_{n}\right)  =v_{p}\left(  a_{1}\right)
+v_{p}\left(  a_{2}\right)  +\cdots+v_{p}\left(  a_{n}\right)  .
\]

\end{lemma}

This lemma is fairly obvious if you follow the \textquotedblleft exponent in
prime factorization\textquotedblright\ interpretation of $v_{p}\left(
n\right)  $. The proof below avoids this interpretation (for the sake of
greater generalizability).

\begin{proof}
[Proof of Lemma \ref{lem.vp.prod}.]Lemma \ref{lem.vp.prod} can be proven
straightforwardly by induction over $n$, provided that the following two
claims are shown:

\begin{statement}
\textit{Claim 1:} We have $v_{p}\left(  1\right)  =0$.
\end{statement}

\begin{statement}
\textit{Claim 2:} We have $v_{p}\left(  ab\right)  =v_{p}\left(  a\right)
+v_{p}\left(  b\right)  $ whenever $a$ and $b$ are two nonzero integers.
\end{statement}

(In fact, Claim 1 settles the induction base, while Claim 2 is used in the
induction step.)

Claim 1 is obvious. It thus remains to prove Claim 2.

\textit{Proof of Claim 2:} Let $a$ and $b$ be two nonzero integers. Recall
that $v_{p}\left(  a\right)  $ is defined as the highest nonnegative integer
$k$ such that $p^{k}\mid a$. Thus, $v_{p}\left(  a\right)  $ is a nonnegative
integer satisfying $p^{v_{p}\left(  a\right)  }\mid a$, but $p^{v_{p}\left(
a\right)  +1}\nmid a$. We have $p^{v_{p}\left(  a\right)  }\mid a$; thus, we
can write $a$ in the form $a=p^{v_{p}\left(  a\right)  }g$ for some
$g\in\mathbb{Z}$. Consider this $g$. If we had $p\mid g$, then we would have
$p^{v_{p}\left(  a\right)  +1}=p^{v_{p}\left(  a\right)  }\underbrace{p}_{\mid
g}\mid p^{v_{p}\left(  a\right)  }g=a$; this would contradict $p^{v_{p}\left(
a\right)  +1}\nmid a$. Hence, we cannot have $p\mid g$. We thus must have
$p\nmid g$. Hence, Proposition \ref{prop.prime.if-ndiv-then-cop} (applied to
$g$ instead of $a$) shows that $g$ is coprime to $p$ (since $p$ is a prime).

Thus, we have found a $g\in\mathbb{Z}$ which is coprime to $p$ and satisfies
$a=p^{v_{p}\left(  a\right)  }g$. The same argument (but made for $b$ instead
of $a$) shows that there exists an $h\in\mathbb{Z}$ which is coprime to $p$
and satisfies $b=p^{v_{p}\left(  b\right)  }h$. Consider this $h$.

Both $g$ and $h$ are coprime to $p$. Hence, $gh$ is also coprime to $p$ (by
Corollary \ref{cor.prod-coprime.ab}, applied to $g$, $h$ and $p$ instead of
$a$, $b$ and $m$). Therefore, $\left(  gh\right)  ^{1}$ is also coprime to
$p^{v_{p}\left(  ab\right)  }$ (by Corollary \ref{cor.pow-coprimes}
\textbf{(b)}, applied to $gh$, $p$, $1$ and $v_{p}\left(  ab\right)  $ instead
of $a$, $b$, $n$ and $m$). In other words, $gh$ is coprime to $p^{v_{p}\left(
ab\right)  }$ (since $\left(  gh\right)  ^{1}=gh$). In other words,
$\gcd\left(  p^{v_{p}\left(  ab\right)  },gh\right)  =1$.

Multiplying the equalities $a=p^{v_{p}\left(  a\right)  }g$ and $b=p^{v_{p}%
\left(  b\right)  }h$, we obtain $ab=p^{v_{p}\left(  a\right)  }%
gp^{v_{p}\left(  b\right)  }h=p^{v_{p}\left(  a\right)  }p^{v_{p}\left(
b\right)  }gh=p^{v_{p}\left(  a\right)  +v_{p}\left(  b\right)  }gh$. Hence,
$p^{v_{p}\left(  a\right)  +v_{p}\left(  b\right)  }\mid ab$.

On the other hand, recall that $v_{p}\left(  ab\right)  $ is defined as the
highest nonnegative integer $k$ such that $p^{k}\mid ab$. Thus, $v_{p}\left(
ab\right)  $ is a nonnegative integer and satisfies $p^{v_{p}\left(
ab\right)  }\mid ab=p^{v_{p}\left(  a\right)  +v_{p}\left(  b\right)
}gh=ghp^{v_{p}\left(  a\right)  +v_{p}\left(  b\right)  }$. Since $\gcd\left(
p^{v_{p}\left(  ab\right)  },gh\right)  =1$, this entails that $p^{v_{p}%
\left(  ab\right)  }\mid p^{v_{p}\left(  a\right)  +v_{p}\left(  b\right)  }$
(by Proposition \ref{prop.prod-coprime.cancel}, applied to $x=p^{v_{p}\left(
ab\right)  }$, $y=gh$ and $z=p^{v_{p}\left(  a\right)  +v_{p}\left(  b\right)
}$). Therefore, $v_{p}\left(  ab\right)  \leq v_{p}\left(  a\right)
+v_{p}\left(  b\right)  $.

But $v_{p}\left(  ab\right)  $ is the \textbf{highest} nonnegative integer $k$
such that $p^{k}\mid ab$. Hence, every nonnegative integer $k$ such that
$p^{k}\mid ab$ must satisfy $k\leq v_{p}\left(  ab\right)  $. Applying this to
$k=v_{p}\left(  a\right)  +v_{p}\left(  b\right)  $, we obtain $v_{p}\left(
a\right)  +v_{p}\left(  b\right)  \leq v_{p}\left(  ab\right)  $ (since
$v_{p}\left(  a\right)  +v_{p}\left(  b\right)  $ satisfies $p^{v_{p}\left(
a\right)  +v_{p}\left(  b\right)  }\mid ab$). Combined with $v_{p}\left(
ab\right)  \leq v_{p}\left(  a\right)  +v_{p}\left(  b\right)  $, this yields
$v_{p}\left(  ab\right)  =v_{p}\left(  a\right)  +v_{p}\left(  b\right)  $.
This proves Claim 2; and as we said, this completes the proof of Lemma
\ref{lem.vp.prod}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.de-polignac}.]We have%
\begin{align*}
v_{p}\left(  \underbrace{n!}_{=1\cdot2\cdot\cdots\cdot n}\right)   &
=v_{p}\left(  1\cdot2\cdot\cdots\cdot n\right)  =v_{p}\left(  1\right)
+v_{p}\left(  2\right)  +\cdots+v_{p}\left(  n\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.vp.prod}, applied to
}a_{i}=i\right)  \\
&  =\underbrace{\sum_{k=1}^{n}}_{=\sum_{\substack{k\in\mathbb{N}_{+};\\k\leq
n}}}\underbrace{v_{p}\left(  k\right)  }_{\substack{=\sum_{\substack{i\in
\mathbb{N}_{+};\\p^{i}\mid k}}1\\\text{(by Lemma \ref{lem.vp.sum}, applied to
}k\\\text{instead of }n\text{)}}}=\underbrace{\sum_{\substack{k\in
\mathbb{N}_{+};\\k\leq n}}\ \ \sum_{\substack{i\in\mathbb{N}_{+};\\p^{i}\mid
k}}}_{\substack{=\sum_{i\in\mathbb{N}_{+}}\ \ \sum_{\substack{k\in
\mathbb{N}_{+};\\k\leq n;\\p^{i}\mid k}}\\\text{(here, we are interchanging}%
\\\text{the order of summation)}}}1\\
&  =\underbrace{\sum_{i\in\mathbb{N}_{+}}}_{=\sum_{i=1}^{\infty}%
}\underbrace{\sum_{\substack{k\in\mathbb{N}_{+};\\k\leq n;\\p^{i}\mid k}%
}1}_{\substack{=\sum_{\substack{k\in\left\{  1,2,\ldots,n\right\}
;\\p^{i}\mid k}}1\\\text{(since the elements }k\text{ of }\mathbb{N}%
_{+}\\\text{satisfying }k\leq n\text{ are precisely}\\\text{the elements of
}\left\{  1,2,\ldots,n\right\}  \text{)}}}=\sum_{i=1}^{\infty}\underbrace{\sum
_{\substack{k\in\left\{  1,2,\ldots,n\right\}  ;\\p^{i}\mid k}}1}%
_{\substack{=\left\lfloor \dfrac{n}{p^{i}}\right\rfloor \\\text{(by
Proposition \ref{prop.floor.n/b},}\\\text{applied to }b=p^{i}\text{)}}}\\
&  =\sum_{i=1}^{\infty}\left\lfloor \dfrac{n}{p^{i}}\right\rfloor .
\end{align*}
This proves Theorem \ref{thm.de-polignac}.
\end{proof}

As an application of Theorem \ref{thm.de-polignac}, we can check that binomial
coefficients are integers (as long as the inputs are nonnegative integers):

\begin{corollary}
\label{cor.de-polignac.bin}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Then,
$\dbinom{n}{m}\in\mathbb{Z}$.
\end{corollary}

Of course, Corollary \ref{cor.de-polignac.bin} can be proven in various simple
ways -- for example, by induction using the recurrence relation of the
binomial coefficients, or combinatorially by interpreting $\dbinom{n}{m}$ as
the number of $m$-element subsets of a given $n$-element set. But let us prove
it using Theorem \ref{thm.de-polignac}, just to show how to use the latter:

\begin{lemma}
\label{lem.vp.int}Let $a$ and $b$ be two nonzero integers. Assume that
$v_{p}\left(  a\right)  \geq v_{p}\left(  b\right)  $ for every prime $p$.
Then, $b\mid a$.
\end{lemma}

\begin{proof}
[First proof of Lemma \ref{lem.vp.int}.]Let $\operatorname*{P}$ be the set of
all primes. Every nonzero integer $n$ satisfies $n=\pm\prod_{p\in
\operatorname*{P}}p^{v_{p}\left(  n\right)  }$\ \ \ \ \footnote{Indeed, for
any prime $p$, we know that $v_{p}\left(  n\right)  $ is the exponent with
which the prime $p$ appears in the prime factorization of $n$. Hence, the
prime factorization of $n$ is $\pm\prod_{p\in\operatorname*{P}}p^{v_{p}\left(
n\right)  }$. (The $\pm$ sign is due to the fact that $n$ can be negative.)}.
Thus,
\begin{equation}
a=\pm\prod_{p\in\operatorname*{P}}p^{v_{p}\left(  a\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ b=\pm\prod_{p\in
\operatorname*{P}}p^{v_{p}\left(  b\right)  }. \label{pf.lem.vp.int.pf1.1}%
\end{equation}
(The two $\pm$ signs may and may not be equal.)

But every $p\in\operatorname*{P}$ satisfies $p^{v_{p}\left(  b\right)  }\mid
p^{v_{p}\left(  a\right)  }$ (since $v_{p}\left(  a\right)  \geq v_{p}\left(
b\right)  $). Hence, $\prod_{p\in\operatorname*{P}}p^{v_{p}\left(  b\right)
}\mid\prod_{p\in\operatorname*{P}}p^{v_{p}\left(  a\right)  }$. In light of
(\ref{pf.lem.vp.int.pf1.1}), this becomes $b\mid a$ (indeed, the $\pm$ signs
clearly have no effect on the divisibility). This proves Lemma
\ref{lem.vp.int}.
\end{proof}

\begin{proof}
[Second proof of Lemma \ref{lem.vp.int}.]Let me, again, give a proof which
avoids the use of prime factorizations. As before, this comes at the cost of
brevity (but again, it leads to more generality).

Let $g=\gcd\left(  a,b\right)  $. Then, $g=\gcd\left(  a,b\right)  \mid a$.
Hence, there exists some $a^{\prime}\in\mathbb{Z}$ such that $a=ga^{\prime}$.
Consider this $a^{\prime}$. Clearly, $a^{\prime}$ is nonzero (since
$ga^{\prime}=a$ is nonzero).

Also, $g=\gcd\left(  a,b\right)  \mid b$. Hence, there exists some $b^{\prime
}\in\mathbb{Z}$ such that $b=gb^{\prime}$. Consider this $b^{\prime}$.
Clearly, $b^{\prime}$ is nonzero (since $gb^{\prime}=b$ is nonzero). Of
course, $g$ is also nonzero (since $g=\gcd\left(  a,b\right)  $ with $a$ and
$b$ being nonzero). Moreover, $g$ is nonnegative (since $g=\gcd\left(
a,b\right)  $) and therefore positive (since $g$ is nonzero).

Proposition \ref{prop.gcd.g} (applied to $a^{\prime}$ and $b^{\prime}$ instead
of $a$ and $b$) shows that $g\gcd\left(  a^{\prime},b^{\prime}\right)
=\gcd\left(  \underbrace{ga^{\prime}}_{=a},\underbrace{gb^{\prime}}%
_{=b}\right)  =\gcd\left(  a,b\right)  =g$. Cancelling $g$ from this equality
(since $g$ is nonzero), we obtain $\gcd\left(  a^{\prime},b^{\prime}\right)
=1$.

Let $p$ be any prime dividing $b^{\prime}$. We shall derive a contradiction
(thus concluding that no such primes exist).

We have $p\mid b^{\prime}$ (since $p$ is a prime dividing $b^{\prime}$). But
recall that $v_{p}\left(  b^{\prime}\right)  $ is defined as the highest
nonnegative integer $k$ such that $p^{k}\mid b^{\prime}$. Thus, every
nonnegative integer $k$ such that $p^{k}\mid b^{\prime}$ must satisfy $k\leq
v_{p}\left(  b^{\prime}\right)  $. Applying this to $k=1$, we obtain $1\leq
v_{p}\left(  b^{\prime}\right)  $ (since $p^{1}=p\mid b^{\prime}$). Hence,
$v_{p}\left(  b^{\prime}\right)  \geq1$.

Now, Lemma \ref{lem.vp.prod} (applied to $n=2$, $a_{1}=g$ and $a_{2}%
=a^{\prime}$) yields $v_{p}\left(  ga^{\prime}\right)  =v_{p}\left(  g\right)
+v_{p}\left(  a^{\prime}\right)  $. Thus, $v_{p}\left(  \underbrace{a}%
_{=ga^{\prime}}\right)  =v_{p}\left(  ga^{\prime}\right)  =v_{p}\left(
g\right)  +v_{p}\left(  a^{\prime}\right)  $. The same argument (used for $b$
and $b^{\prime}$ instead of $a$ and $a^{\prime}$) yields $v_{p}\left(
b\right)  =v_{p}\left(  g\right)  +v_{p}\left(  b^{\prime}\right)  $. But by
assumption, we have $v_{p}\left(  a\right)  \geq v_{p}\left(  b\right)  $.
Thus, $v_{p}\left(  g\right)  +v_{p}\left(  a^{\prime}\right)  =v_{p}\left(
a\right)  \geq v_{p}\left(  b\right)  =v_{p}\left(  g\right)  +v_{p}\left(
b^{\prime}\right)  $. Since $v_{p}\left(  g\right)  $ is an integer, we can
cancel $v_{p}\left(  g\right)  $ from this inequality, and obtain
$v_{p}\left(  a^{\prime}\right)  \geq v_{p}\left(  b^{\prime}\right)  \geq1$.

Recall that $v_{p}\left(  a^{\prime}\right)  $ is defined as the highest
nonnegative integer $k$ such that $p^{k}\mid a^{\prime}$. Thus, $p^{v_{p}%
\left(  a^{\prime}\right)  }\mid a^{\prime}$. But $v_{p}\left(  a^{\prime
}\right)  \geq1$, whence $p\mid p^{v_{p}\left(  a^{\prime}\right)  }\mid
a^{\prime}$.

Now, $p\mid a^{\prime}$ and $p\mid b^{\prime}$. Hence, Proposition
\ref{prop.gcd.uniprop} (applied to $p$, $a^{\prime}$ and $b^{\prime}$ instead
of $a$, $b$ and $c$) shows that $p\mid\gcd\left(  a^{\prime},b^{\prime
}\right)  $, so that $p\mid1$ (since $\gcd\left(  a^{\prime},b^{\prime
}\right)  =1$). This is absurd (since $p$ is a prime).

Now, forget that we fixed $p$. Thus, we have obtained a contradiction for
every prime $p$ dividing $b^{\prime}$. Therefore, there exist no such primes
$p$. Therefore, $b^{\prime}$ is either $1$ or $-1$. In either case,
$b^{\prime}\mid1$. Hence, $b=g\underbrace{b^{\prime}}_{\mid1}\mid g1=g\mid a$.
This proves Lemma \ref{lem.vp.int}.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.de-polignac.bin} (sketched).]If $m>n$, then
$\dbinom{n}{m}=0$; thus, Corollary \ref{cor.de-polignac.bin} is obviously
correct in this case. Hence, we WLOG assume that we don't have $m>n$.
Therefore, $m\leq n$. Consequently, a well-known formula shows that
$\dbinom{n}{m}=\dfrac{n!}{m!\left(  n-m\right)  !}$. Hence, in order to prove
that $\dbinom{n}{m}\in\mathbb{Z}$, it suffices to show that $m!\left(
n-m\right)  !\mid n!$. In light of Lemma \ref{lem.vp.int} (applied to $a=n!$
and $b=m!\left(  n-m\right)  !$), we can achieve this by showing that
\begin{equation}
v_{p}\left(  n!\right)  \geq v_{p}\left(  m!\left(  n-m\right)  !\right)
\ \ \ \ \ \ \ \ \ \ \text{for every prime }p. \label{pf.cor.de-polignac.bin.1}%
\end{equation}


\textit{Proof of (\ref{pf.cor.de-polignac.bin.1}):} Let $p$ be a prime. Lemma
\ref{lem.vp.prod} yields%
\begin{align*}
&  v_{p}\left(  m!\left(  n-m\right)  !\right) \\
&  =\underbrace{v_{p}\left(  m!\right)  }_{\substack{=\sum_{i=1}^{\infty
}\left\lfloor \dfrac{m}{p^{i}}\right\rfloor \\\text{(by Theorem
\ref{thm.de-polignac})}}}+\underbrace{v_{p}\left(  \left(  n-m\right)
!\right)  }_{\substack{=\sum_{i=1}^{\infty}\left\lfloor \dfrac{n-m}{p^{i}%
}\right\rfloor \\\text{(by Theorem \ref{thm.de-polignac})}}}\\
&  =\sum_{i=1}^{\infty}\left\lfloor \dfrac{m}{p^{i}}\right\rfloor +\sum
_{i=1}^{\infty}\left\lfloor \dfrac{n-m}{p^{i}}\right\rfloor =\sum
_{i=1}^{\infty}\underbrace{\left(  \left\lfloor \dfrac{m}{p^{i}}\right\rfloor
+\left\lfloor \dfrac{n-m}{p^{i}}\right\rfloor \right)  }_{\substack{\leq
\left\lfloor \dfrac{m}{p^{i}}+\dfrac{n-m}{p^{i}}\right\rfloor \\\text{(by the
formula }\left\lfloor u\right\rfloor +\left\lfloor v\right\rfloor
\leq\left\lfloor u+v\right\rfloor \\\text{from Proposition
\ref{prop.floor.u+v})}}}\\
&  \leq\sum_{i=1}^{\infty}\left\lfloor \underbrace{\dfrac{m}{p^{i}}%
+\dfrac{n-m}{p^{i}}}_{=\dfrac{n}{p^{i}}}\right\rfloor =\sum_{i=1}^{\infty
}\left\lfloor \dfrac{n}{p^{i}}\right\rfloor =v_{p}\left(  n!\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.de-polignac}}\right)  .
\end{align*}
This proves (\ref{pf.cor.de-polignac.bin.1}).

As we know, this completes the proof of Corollary \ref{cor.de-polignac.bin}.
\end{proof}

Note that Corollary \ref{cor.de-polignac.bin} also holds for all
$n\in\mathbb{Z}$ (not just for all $n\in\mathbb{N}$); but this would require a
different method of proof\footnote{The easiest way to reduce the
$n\in\mathbb{Z}$ case to the $n\in\mathbb{N}$ case is by using the upper
negation formula $\dbinom{n}{m}=\left(  -1\right)  ^{m}\dbinom{m-n-1}{m}$.}.

Our proof of Corollary \ref{cor.de-polignac.bin} using Theorem
\ref{thm.de-polignac} was a slight overkill (as I said, there are easier and
better ways to achieve the same goal); however, the method is useful, as it
also allows proving other results which are harder to obtain in other ways.
Here are two examples of such results (without proof):

\begin{proposition}
\label{prop.de-polignac.binrat}Let $a\in\mathbb{Z}$, $b\in\mathbb{Z}%
\setminus\left\{  0\right\}  $ and $m\in\mathbb{N}$. Then, $\dbinom{a/b}{m}$
is a rational number which can be written as a ratio of two integers whose
denominator is a power of $b$. More precisely, $b^{2m-1}\dbinom{a/b}{m}%
\in\mathbb{Z}$ when $m>0$ (and $\dbinom{a/b}{m}=1\in\mathbb{Z}$ when $m=0$).
\end{proposition}

\begin{proposition}
\label{prop.de-polignac.2m}Let $m\in\mathbb{N}$ and $n\in\mathbb{N}$. Then,
$\dfrac{\left(  2m\right)  !\left(  2n\right)  !}{m!n!\left(  m+n\right)
!}\in\mathbb{Z}$.
\end{proposition}

\section{Arithmetic functions}

\subsection{Arithmetic functions}

Next, I will discuss the notion of arithmetic functions, and some examples
thereof; here I will not really follow \cite[\S 4.2]{NiZuMo91} but rather
build up the same theory from my perspective.

\begin{definition}
\label{def.arifun}An \textit{arithmetic function} shall mean a function from
$\mathbb{N}_{+}$ to $\mathbb{C}$.
\end{definition}

My Definition \ref{def.arifun} appears to be slightly incompatible with the
definition in \cite[\S 4.2]{NiZuMo91}; indeed, the latter defines an
arithmetic function to be a function from $\mathbb{N}_{+}$ to a subset of
$\mathbb{C}$. However, Niven, Zuckerman and Montgomery never specify the
target of the arithmetic functions they introduce in \cite[\S 4.2]{NiZuMo91};
thus, I believe that my Definition \ref{def.arifun} is the definition they
have actually meant. Anyway, most people are cavalier about the target of an
arithmetic function, and prefer to equate any two arithmetic functions which
differ only in the choice of target.

Let us define a bunch of arithmetic functions:\footnote{Recall that a positive
integer $n$ is said to be \textit{squarefree} if no perfect square other than
$1$ divides $n$. Equivalently, a positive integer $n$ is squarefree if and
only if $n$ is a product of \textbf{distinct} primes. Equivalently, a positive
integer $n$ is squarefree if and only if every prime $p$ satisfies
$v_{p}\left(  n\right)  \leq1$.}

\begin{definition}
\label{def.arifun.funcs}We define the following arithmetic functions:

\begin{itemize}
\item The function $\phi:\mathbb{N}_{+}\rightarrow\mathbb{C}$ shall send every
$n\in\mathbb{N}_{+}$ to the number of all $k\in\left\{  1,2,\ldots,n\right\}
$ coprime to $n$. This function $\phi$ is called the \textit{Euler totient
function}, or the \textit{phi function} (and is often denoted by $\varphi$ as well).

\item The function $\operatorname*{d}:\mathbb{N}_{+}\rightarrow\mathbb{C}$
shall send every $n\in\mathbb{N}_{+}$ to the number of positive divisors of
$n$. This function $\operatorname*{d}$ is called the \textit{divisor function}.

\item The function $\underline{1}:\mathbb{N}_{+}\rightarrow\mathbb{C}$ shall
send every $n\in\mathbb{N}_{+}$ to $1$.

\item The function $\underline{0}:\mathbb{N}_{+}\rightarrow\mathbb{C}$ shall
send every $n\in\mathbb{N}_{+}$ to $0$.

\item The function $\iota:\mathbb{N}_{+}\rightarrow\mathbb{C}$ shall send
every $n\in\mathbb{N}_{+}$ to $n$.

\item The function $\sigma:\mathbb{N}_{+}\rightarrow\mathbb{C}$ shall send
every $n\in\mathbb{N}_{+}$ to the sum of all positive divisors of $n$.

\item For each $k\in\mathbb{Z}$, the function $\sigma_{k}:\mathbb{N}%
_{+}\rightarrow\mathbb{C}$ shall send every $n\in\mathbb{N}_{+}$ to the sum of
the $k$-th powers of all positive divisors of $n$. Note that $\sigma
_{0}=\operatorname*{d}$ and $\sigma_{1}=\sigma$.

\item The function $\omega:\mathbb{N}_{+}\rightarrow\mathbb{C}$ shall send
every $n\in\mathbb{N}_{+}$ to the number of all distinct primes dividing $n$.
(For example, $\omega\left(  12\right)  =2$, since the primes dividing $12$
are $2$ and $3$.)

\item The function $\Omega:\mathbb{N}_{+}\rightarrow\mathbb{C}$ shall send
every $n\in\mathbb{N}_{+}$ to the number of all prime factors of $n$ counted
with multiplicity. In other words, if $\Omega\left(  n\right)  $ is the
$k\in\mathbb{N}$ such that $n$ can be written as a product of $k$ primes (not
necessarily distinct primes). (For example, $\Omega\left(  12\right)  =3$,
since $12=2\cdot2\cdot3$.)

\item The function $\mu:\mathbb{N}_{+}\rightarrow\mathbb{C}$ shall send every
$n\in\mathbb{N}_{+}$ to $%
\begin{cases}
\left(  -1\right)  ^{\omega\left(  n\right)  }, & \text{if }n\text{ is
squarefree;}\\
0, & \text{otherwise}%
\end{cases}
$. This function $\mu$ is called the \textit{M\"{o}bius mu function}.

\item The function $\lambda:\mathbb{N}_{+}\rightarrow\mathbb{C}$ shall send
every $n\in\mathbb{N}_{+}$ to $\left(  -1\right)  ^{\Omega\left(  n\right)  }%
$. This function $\lambda$ is called \textit{Liouville's lambda function}.

\item The function $\varepsilon:\mathbb{N}_{+}\rightarrow\mathbb{C}$ shall
send every $n\in\mathbb{N}_{+}$ to $%
\begin{cases}
1, & \text{if }n=1;\\
0, & \text{if }n\neq1
\end{cases}
$.
\end{itemize}
\end{definition}

Of course, you can come up with more examples easily. Most arithmetic
functions that anyone cares about tend to have their images belong to
$\mathbb{Z}$, but the added generality of allowing any complex numbers as
images does not hurt, so I see no point in restricting it.

We introduce one more standard notation:

\begin{definition}
\label{def.sumdiv}Any summation sign of the form \textquotedblleft$\sum_{d\mid
n}$\textquotedblright\ (where $n$ is a given positive integer) will be
understood to mean \textquotedblleft sum over all \textbf{positive} divisors
$d$ of $n$\textquotedblright. This similarly applies when there are further
conditions under the summation sign; for instance, \textquotedblleft%
$\sum\limits_{\substack{d\mid n;\\d\leq3}}$\textquotedblright\ means
\textquotedblleft sum over all positive divisors $d$ of $n$ satisfying
$d\leq3$\textquotedblright.
\end{definition}

\begin{remark}
Some of the functions defined in Definition \ref{def.arifun.funcs} can easily
be reexpressed using the notation from Definition \ref{def.sumdiv}: Namely,
for every $n\in\mathbb{N}_{+}$, we have%
\begin{align*}
\operatorname*{d}\left(  n\right)   &  =\sum_{d\mid n}%
1;\ \ \ \ \ \ \ \ \ \ \sigma\left(  n\right)  =\sum_{d\mid n}d;\\
\sigma_{k}\left(  n\right)   &  =\sum_{d\mid n}d^{k}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{for every }k\in\mathbb{Z}\right)  .
\end{align*}

\end{remark}

Some of the arithmetic functions defined above can be written explicitly in
terms of the prime factorization of $n$. I will first state some of the
explicit representations before I show a method for proving them.

\begin{definition}
\label{def.PF}If $n$ is an integer, then $\operatorname*{PF}n$ will denote the
set of all prime factors of $n$. Note that this set $\operatorname*{PF}n$ is
finite whenever $n$ is nonzero.
\end{definition}

\begin{theorem}
\label{thm.phi.explicit}For every $n\in\mathbb{N}_{+}$, we have%
\[
\phi\left(  n\right)  =\prod_{p\in\operatorname*{PF}n}\left(  p^{v_{p}\left(
n\right)  -1}\left(  p-1\right)  \right)  .
\]

\end{theorem}

\begin{theorem}
\label{thm.d.explicit}For every $n\in\mathbb{N}_{+}$, we have%
\[
\operatorname*{d}\left(  n\right)  =\prod_{p\in\operatorname*{PF}n}\left(
v_{p}\left(  n\right)  +1\right)  .
\]

\end{theorem}

\begin{theorem}
\label{thm.sigmak.explicit}For every $n\in\mathbb{N}_{+}$ and every nonzero
$k\in\mathbb{Z}$, we have%
\[
\sigma_{k}\left(  n\right)  =\prod_{p\in\operatorname*{PF}n}\dfrac{p^{k\left(
v_{p}\left(  n\right)  +1\right)  }-1}{p^{k}-1}.
\]

\end{theorem}

Theorem \ref{thm.phi.explicit} appears in \cite[Theorem 2.19]{NiZuMo91} (in a
slightly restated form). Theorem \ref{thm.d.explicit} is \cite[Theorem
4.3]{NiZuMo91}, and Theorem \ref{thm.sigmak.explicit} is a straightforward
generalization of \cite[Theorem 4.5]{NiZuMo91}. There are simple and
elementary ways to prove each of these theorems; I will give a more abstract
approach to highlight the theory.

\subsection{Multiplicative functions}

\begin{definition}
\label{def.arifun.mult}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be an
arithmetic function.

\textbf{(a)} The function $f$ is said to be \textit{multiplicative} if and
only if it satisfies $f\left(  1\right)  =1$ and%
\[
f\left(  mn\right)  =f\left(  m\right)  f\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for any two coprime }m\in\mathbb{N}_{+}\text{ and
}n\in\mathbb{N}_{+}.
\]


\textbf{(b)} The function $f$ is said to be \textit{totally multiplicative} if
and only if it satisfies $f\left(  1\right)  =1$ and%
\[
f\left(  mn\right)  =f\left(  m\right)  f\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for any two }m\in\mathbb{N}_{+}\text{ and }%
n\in\mathbb{N}_{+}.
\]
(Another word for \textquotedblleft totally multiplicative\textquotedblright%
\ is \textquotedblleft completely multiplicative\textquotedblright.)
\end{definition}

It turns out that totally multiplicative functions are somewhat rare, but
multiplicative functions abound in number theory. Here are some examples:

\begin{proposition}
\label{prop.arifun.mult.exa}Consider the functions defined in Definition
\ref{def.arifun.funcs}.

\textbf{(a)} The function $\phi$ is multiplicative.

\textbf{(b)} The function $\operatorname*{d}$ is multiplicative.

\textbf{(c)} The function $\underline{1}$ is totally multiplicative and multiplicative.

\textbf{(d)} The function $\iota$ is totally multiplicative and multiplicative.

\textbf{(e)} For every $k\in\mathbb{Z}$, the function $\sigma_{k}$ is
multiplicative. In particular, the function $\sigma$ is multiplicative.

\textbf{(f)} The function $\mu$ is multiplicative.

\textbf{(g)} The function $\lambda$ is totally multiplicative and multiplicative.

\textbf{(h)} The function $\varepsilon$ is totally multiplicative and multiplicative.

\textbf{(i)} Every totally multiplicative function is multiplicative.

\textbf{(j)} Let $f\in\mathbb{Z}\left[  x\right]  $ be a polynomial. Let
$N_{P}:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be the function which sends every
$n\in\mathbb{N}_{+}$ to the number of solutions of the congruence $f\left(
x\right)  \equiv0\operatorname{mod}n$. Then, the function $N_{P}$ is multiplicative.

\textbf{(k)} For every integer $u$, the function $\mathbb{N}_{+}%
\rightarrow\mathbb{C},\ n\mapsto\gcd\left(  u,n\right)  $ is multiplicative.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.arifun.mult.exa} (sketched).]\textbf{(i)} This
is obvious (since the requirements for a totally multiplicative function
clearly encompass the requirements for a multiplicative function).

\textbf{(a)} We know that $\phi\left(  1\right)  =1$. We thus only need to
show that $\phi\left(  mn\right)  =\phi\left(  m\right)  \phi\left(  n\right)
$ for any two coprime $m\in\mathbb{N}_{+}$ and $n\in\mathbb{N}_{+}$. But this
is precisely the statement of \cite[first sentence of Theorem 2.19]{NiZuMo91}.
(Here is a brief reminder of the proof: For every $N\in\mathbb{N}_{+}$, let
$\mathcal{R}\left(  N\right)  $ denote the set of all $k\in\left\{
1,2,\ldots,N\right\}  $ coprime to $N$. Now, let $m\in\mathbb{N}_{+}$ and
$n\in\mathbb{N}_{+}$ be coprime. Then, there is a bijection $\mathcal{R}%
\left(  mn\right)  \rightarrow\mathcal{R}\left(  m\right)  \times
\mathcal{R}\left(  n\right)  $ which sends every $k\in\mathcal{R}\left(
mn\right)  $ to $\left(  k^{\prime},k^{\prime\prime}\right)  \in
\mathcal{R}\left(  m\right)  \times\mathcal{R}\left(  n\right)  $, where
$k^{\prime}$ is the unique element of $\mathcal{R}\left(  m\right)  $
congruent to $k$ modulo $m$, and where $k^{\prime\prime}$ is the unique
element of $\mathcal{R}\left(  n\right)  $ congruent to $k$ modulo $n$. The
fact that this map is well-defined and a bijection can be proven using the
Chinese Remainder Theorem. Having this bijection in place, we immediately
conclude that $\left\vert \mathcal{R}\left(  mn\right)  \right\vert
=\left\vert \mathcal{R}\left(  m\right)  \times\mathcal{R}\left(  n\right)
\right\vert =\left\vert \mathcal{R}\left(  m\right)  \right\vert
\cdot\left\vert \mathcal{R}\left(  n\right)  \right\vert $. Since $\left\vert
\mathcal{R}\left(  N\right)  \right\vert =\phi\left(  N\right)  $ for every
$N\in\mathbb{N}_{+}$, this rewrites as $\phi\left(  mn\right)  =\phi\left(
m\right)  \phi\left(  n\right)  $, qed.) Proposition
\ref{prop.arifun.mult.exa} \textbf{(a)} is thus proven.

Proposition \ref{prop.arifun.mult.exa} \textbf{(j)} is essentially
\cite[second sentence of Theorem 2.20]{NiZuMo91}, and is proven in a similar
way as Proposition \ref{prop.arifun.mult.exa} \textbf{(a)}.

Parts \textbf{(c)}, \textbf{(d)} and \textbf{(h)} of Proposition
\ref{prop.arifun.mult.exa} are completely straightforward.

\textbf{(g)} We claim that the following two assertions hold:

\begin{statement}
\textit{Assertion 1:} We have $\lambda\left(  1\right)  =1$.
\end{statement}

\begin{statement}
\textit{Assertion 2:} We have $\lambda\left(  mn\right)  =\lambda\left(
m\right)  \lambda\left(  n\right)  $ for any two $m\in\mathbb{N}_{+}$ and
$n\in\mathbb{N}_{+}$.
\end{statement}

\begin{proof}
[Proof of Assertion 1.]The number $1$ can be written as a product of $0$
primes (because the empty product equals $1$). Hence, $\Omega\left(  1\right)
=0$ (by the definition of $\Omega$).

The definition of $\lambda$ yields $\lambda\left(  1\right)  =\left(
-1\right)  ^{\Omega\left(  1\right)  }=\left(  -1\right)  ^{0}$ (since
$\Omega\left(  1\right)  =0$). Thus, $\lambda\left(  1\right)  =\left(
-1\right)  ^{0}=1$. This proves Assertion 1.
\end{proof}

\begin{proof}
[Proof of Assertion 2.]Let $m\in\mathbb{N}_{+}$ and $n\in\mathbb{N}_{+}$.

Write $m$ as a product of primes; i.e., write $m$ in the form $m=p_{1}%
p_{2}\cdots p_{k}$ for some primes $p_{1},p_{2},\ldots,p_{k}$ (which may and
may not be distinct). Thus, $m$ is a product of $k$ primes; hence,
$\Omega\left(  m\right)  =k$ (by the definition of $\Omega$).

Write $n$ as a product of primes; i.e., write $n$ in the form $n=q_{1}%
q_{2}\cdots q_{\ell}$ for some primes $q_{1},q_{2},\ldots,q_{\ell}$ (which may
and may not be distinct). Thus, $n$ is a product of $\ell$ primes; hence,
$\Omega\left(  n\right)  =\ell$ (by the definition of $\Omega$).

Now, multiplying the equalities $m=p_{1}p_{2}\cdots p_{k}$ and $n=q_{1}%
q_{2}\cdots q_{\ell}$, we obtain $mn=\left(  p_{1}p_{2}\cdots p_{k}\right)
\left(  q_{1}q_{2}\cdots q_{\ell}\right)  =p_{1}p_{2}\cdots p_{k}q_{1}%
q_{2}\cdots q_{\ell}$. Hence, $mn$ is a product of $k+\ell$ primes (since all
of $p_{1},p_{2},\ldots,p_{k},q_{1},q_{2},\ldots,q_{\ell}$ are primes).
Therefore, $\Omega\left(  mn\right)  =k+\ell$ (by the definition of $\Omega$).
Hence,%
\[
\Omega\left(  mn\right)  =\underbrace{k}_{=\Omega\left(  m\right)
}+\underbrace{\ell}_{=\Omega\left(  n\right)  }=\Omega\left(  m\right)
+\Omega\left(  n\right)  .
\]


Now, the definition of $\lambda$ yields $\lambda\left(  m\right)  =\left(
-1\right)  ^{\Omega\left(  m\right)  }$ and $\lambda\left(  n\right)  =\left(
-1\right)  ^{\Omega\left(  n\right)  }$ and $\lambda\left(  mn\right)
=\left(  -1\right)  ^{\Omega\left(  mn\right)  }$. Hence,%
\begin{align*}
\lambda\left(  mn\right)   &  =\left(  -1\right)  ^{\Omega\left(  mn\right)
}=\left(  -1\right)  ^{\Omega\left(  m\right)  +\Omega\left(  n\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Omega\left(  mn\right)
=\Omega\left(  m\right)  +\Omega\left(  n\right)  \right) \\
&  =\underbrace{\left(  -1\right)  ^{\Omega\left(  m\right)  }}_{=\lambda
\left(  m\right)  }\underbrace{\left(  -1\right)  ^{\Omega\left(  n\right)  }%
}_{=\lambda\left(  n\right)  }=\lambda\left(  m\right)  \lambda\left(
n\right)  .
\end{align*}
This proves Assertion 2.
\end{proof}

Now, the function $\lambda$ is totally multiplicative if and only if
Assertions 1 and 2 hold (by the definition of \textquotedblleft totally
multiplicative\textquotedblright). Thus, the function $\lambda$ is totally
multiplicative (since Assertions 1 and 2 hold). Consequently, $\lambda$ is
multiplicative (since every totally multiplicative function is multiplicative
(by Proposition \ref{prop.arifun.mult.exa} \textbf{(i)})). This proves
Proposition \ref{prop.arifun.mult.exa} \textbf{(g)}.

\textbf{(k)} We leave the proof of Proposition \ref{prop.arifun.mult.exa}
\textbf{(k)} to the reader. (It is completely straightforward using
Proposition \ref{prop.gcd.gcd(u,mn)} and the fact that $\gcd\left(
u,1\right)  =1$.)

We defer the proofs of parts \textbf{(b)} and \textbf{(e)} until later.
(Actually, we shall also give a second proof of part \textbf{(a)} later.) It
thus remains to prove Proposition \ref{prop.arifun.mult.exa} \textbf{(f)}.

\textbf{(f)} The definition of $\omega\left(  1\right)  $ shows that
$\omega\left(  1\right)  $ is the number of all distinct primes dividing $1$.
But the latter number is clearly $0$ (since there are no primes dividing $1$).
Hence, $\omega\left(  1\right)  =0$.

The integer $1$ is squarefree; hence, the definition of $\mu$ yields
\begin{align*}
\mu\left(  1\right)   &  =\left(  -1\right)  ^{\omega\left(  1\right)
}=\left(  -1\right)  ^{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\omega\left(  1\right)  =0\right) \\
&  =1.
\end{align*}
Hence, we only need to show that $\mu\left(  mn\right)  =\mu\left(  m\right)
\mu\left(  n\right)  $ for any two coprime $m\in\mathbb{N}_{+}$ and
$n\in\mathbb{N}_{+}$. So let us show this.

Let $m\in\mathbb{N}_{+}$ and $n\in\mathbb{N}_{+}$ be coprime. We must prove
the equality $\mu\left(  mn\right)  =\mu\left(  m\right)  \mu\left(  n\right)
$. If $mn$ is not squarefree, then this equality
holds\footnote{\textit{Proof.} Assume that $mn$ is not squarefree. Thus, there
is some integer $g>1$ such that $g^{2}\mid mn$. Consider this $g$.
\par
There exists a prime $p$ such that $p\mid g$ (since $g>1$). Consider such a
$p$. The prime $p$ cannot divide both $m$ and $n$ (since $m$ and $n$ are
coprime). Hence, either $p\nmid m$ or $p\nmid n$ (or both). We WLOG assume
that $p\nmid m$ (since otherwise, we can simply switch $m$ with $n$). Thus,
$m$ is coprime to $p$ (since $p$ is prime). In other words, $p$ is coprime to
$m$. In other words, $\gcd\left(  p,m\right)  =1$.
\par
We have $p\mid g$, so that $p^{2}\mid g^{2}\mid mn$. Hence, $p\mid p^{2}\mid
mn$. Hence, Proposition \ref{prop.prod-coprime.cancel} (applied to $p$, $m$
and $n$ instead of $x$, $y$ and $z$) shows that $p\mid n$. In other words,
$n=pn^{\prime}$ for some integer $n^{\prime}$. Consider this $n^{\prime}$.
\par
Now, $pp=p^{2}\mid m\underbrace{n}_{=pn^{\prime}}=mpn^{\prime}=pmn^{\prime}$.
We can cancel $p$ from this divisibility (since $p\neq0$) and thus conclude
$p\mid mn^{\prime}$. Hence, Proposition \ref{prop.prod-coprime.cancel}
(applied to $p$, $m$ and $n^{\prime}$ instead of $x$, $y$ and $z$) shows that
$p\mid n^{\prime}$. Hence, $pp\mid pn^{\prime}=n$. In other words, $p^{2}\mid
n$ (since $pp=p^{2}$). Hence, $n$ is not squarefree (since $p^{2}$ is a
perfect square other than $1$). Therefore, $\mu\left(  n\right)  =0$ (by the
definition of $\mu$). The definition of $\mu$ also shows that $\mu\left(
mn\right)  =0$ (since $mn$ is not squarefree). Now, comparing $\mu\left(
mn\right)  =0$ with $\mu\left(  m\right)  \underbrace{\mu\left(  n\right)
}_{=0}=0$, we obtain $\mu\left(  mn\right)  =\mu\left(  m\right)  \mu\left(
n\right)  $, qed.}. Hence, we WLOG assume that $mn$ is squarefree. Therefore,
$m$ is also squarefree (since any perfect square dividing $m$ would also
divide $mn$). Similarly, $n$ is squarefree.

Since $m$ is squarefree, we have $\mu\left(  m\right)  =\left(  -1\right)
^{\omega\left(  m\right)  }$ (by the definition of $\mu$). Since $n$ is
squarefree, we have $\mu\left(  n\right)  =\left(  -1\right)  ^{\omega\left(
n\right)  }$ (by the definition of $\mu$). Since $mn$ is squarefree, we have
$\mu\left(  mn\right)  =\left(  -1\right)  ^{\omega\left(  mn\right)  }$ (by
the definition of $\mu$).

But $\omega\left(  mn\right)  $ is the number of all distinct primes dividing
$mn$ (by the definition of $\omega$). Thus,%
\begin{align}
\omega\left(  mn\right)   &  =\left(  \text{the number of distinct primes
dividing }mn\right) \nonumber\\
&  =\left(  \text{the number of distinct primes dividing }m\text{ or dividing
}n\right)  \label{pf.prop.arifun.mult.exa.f.1}%
\end{align}
(since the primes dividing $mn$ are precisely the primes dividing $m$ or
dividing $n$).

Moreover, there is no overlap between the primes dividing $m$ and the primes
dividing $n$ (since $m$ and $n$ are coprime). Hence,%
\begin{align*}
&  \left(  \text{the number of distinct primes dividing }m\text{ or dividing
}n\right) \\
&  =\underbrace{\left(  \text{the number of distinct primes dividing
}m\right)  }_{\substack{=\omega\left(  m\right)  \\\text{(since this is how
}\omega\left(  m\right)  \text{ is defined)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \text{the number of distinct
primes dividing }n\right)  }_{\substack{=\omega\left(  n\right)
\\\text{(since this is how }\omega\left(  n\right)  \text{ is defined)}}}\\
&  =\omega\left(  m\right)  +\omega\left(  n\right)  .
\end{align*}


Thus, (\ref{pf.prop.arifun.mult.exa.f.1}) becomes%
\begin{align}
\omega\left(  mn\right)   &  =\left(  \text{the number of distinct primes
dividing }m\text{ or dividing }n\right) \nonumber\\
&  =\omega\left(  m\right)  +\omega\left(  n\right)  .
\label{pf.prop.arifun.mult.exa.f.2}%
\end{align}


Now,%
\begin{align*}
\mu\left(  mn\right)   &  =\left(  -1\right)  ^{\omega\left(  mn\right)
}=\left(  -1\right)  ^{\omega\left(  m\right)  +\omega\left(  n\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.arifun.mult.exa.f.2}%
)}\right) \\
&  =\underbrace{\left(  -1\right)  ^{\omega\left(  m\right)  }}_{=\mu\left(
m\right)  }\underbrace{\left(  -1\right)  ^{\omega\left(  n\right)  }}%
_{=\mu\left(  n\right)  }=\mu\left(  m\right)  \mu\left(  n\right)  .
\end{align*}
This completes the proof of $\mu\left(  mn\right)  =\mu\left(  m\right)
\mu\left(  n\right)  $. Thus, Proposition \ref{prop.arifun.mult.exa}
\textbf{(f)} holds.
\end{proof}

Note that the function $\underline{0}:\mathbb{N}_{+}\rightarrow\mathbb{C}$ is
not multiplicative; in fact, it fails to satisfy $\underline{0}\left(
1\right)  =1$.

The pointwise product of multiplicative functions is multiplicative, and the
same holds for totally multiplicative functions:

\begin{proposition}
\label{prop.arifun.mult.pwprod}Let $g:\mathbb{N}_{+}\rightarrow\mathbb{C}$ and
$h:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be two arithmetic functions. Let
$f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be the function defined by%
\begin{equation}
f\left(  n\right)  =g\left(  n\right)  h\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}_{+}.
\label{eq.prop.arifun.mult.pwprod.f(n)=}%
\end{equation}


\textbf{(a)} If the functions $g$ and $h$ are multiplicative, then the
function $f$ is multiplicative.

\textbf{(b)} If the functions $g$ and $h$ are totally multiplicative, then the
function $f$ is totally multiplicative.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.arifun.mult.pwprod}.]\textbf{(a)} Assume that
the functions $g$ and $h$ are multiplicative. We have to prove that the
function $f$ is multiplicative.

The function $g$ is multiplicative. In other words, it satisfies $g\left(
1\right)  =1$, and%
\begin{equation}
g\left(  mn\right)  =g\left(  m\right)  g\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for any two coprime }m\in\mathbb{N}_{+}\text{ and
}n\in\mathbb{N}_{+}. \label{pf.prop.arifun.mult.pwprod.a.1}%
\end{equation}


The function $h$ is multiplicative. In other words, it satisfies $h\left(
1\right)  =1$, and%
\begin{equation}
h\left(  mn\right)  =h\left(  m\right)  h\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for any two coprime }m\in\mathbb{N}_{+}\text{ and
}n\in\mathbb{N}_{+}. \label{pf.prop.arifun.mult.pwprod.a.2}%
\end{equation}


Now, we want to prove that $f$ is multiplicative. In order to do so, we shall
prove the following two assertions:

\begin{statement}
\textit{Assertion 1:} We have $f\left(  1\right)  =1$.
\end{statement}

\begin{statement}
\textit{Assertion 2:} We have $f\left(  mn\right)  =f\left(  m\right)
f\left(  n\right)  $ for any two coprime $m\in\mathbb{N}_{+}$ and
$n\in\mathbb{N}_{+}$.
\end{statement}

\textit{Proof of Assertion 1:} Applying
(\ref{eq.prop.arifun.mult.pwprod.f(n)=}) to $n=1$, we obtain $f\left(
1\right)  =\underbrace{g\left(  1\right)  }_{=1}\underbrace{h\left(  1\right)
}_{=1}=1$. This proves Assertion 1.

\textit{Proof of Assertion 2:} Let $m\in\mathbb{N}_{+}$ and $n\in
\mathbb{N}_{+}$ be coprime. Then, (\ref{eq.prop.arifun.mult.pwprod.f(n)=})
(applied to $m$ instead of $n$) yields $f\left(  m\right)  =g\left(  m\right)
h\left(  m\right)  $. Also, (\ref{eq.prop.arifun.mult.pwprod.f(n)=}) shows
that $f\left(  n\right)  =g\left(  n\right)  h\left(  n\right)  $. But
(\ref{eq.prop.arifun.mult.pwprod.f(n)=}) (applied to $mn$ instead of $n$)
yields%
\begin{align*}
f\left(  mn\right)   &  =\underbrace{g\left(  mn\right)  }%
_{\substack{=g\left(  m\right)  g\left(  n\right)  \\\text{(by
(\ref{pf.prop.arifun.mult.pwprod.a.1}))}}}\underbrace{h\left(  mn\right)
}_{\substack{=h\left(  m\right)  h\left(  n\right)  \\\text{(by
(\ref{pf.prop.arifun.mult.pwprod.a.2}))}}}=\left(  g\left(  m\right)  g\left(
n\right)  \right)  \left(  h\left(  m\right)  h\left(  n\right)  \right) \\
&  =\underbrace{\left(  g\left(  m\right)  h\left(  m\right)  \right)
}_{=f\left(  m\right)  }\underbrace{\left(  g\left(  n\right)  h\left(
n\right)  \right)  }_{=f\left(  n\right)  }=f\left(  m\right)  f\left(
n\right)  .
\end{align*}
This proves Assertion 2.

Now we know that Assertions 1 and 2 hold. In other words, the function $f$ is
multiplicative (by the definition of \textquotedblleft
multiplicative\textquotedblright). This proves Proposition
\ref{prop.arifun.mult.pwprod} \textbf{(a)}.

\textbf{(b)} The proof of Proposition \ref{prop.arifun.mult.pwprod}
\textbf{(b)} is completely analogous to our above proof of Proposition
\ref{prop.arifun.mult.pwprod} \textbf{(a)}. (More precisely, we have to
replace every \textquotedblleft multiplicative\textquotedblright\ by
\textquotedblleft totally multiplicative\textquotedblright\ in our proof of
Proposition \ref{prop.arifun.mult.pwprod} \textbf{(a)}, and remove the word
\textquotedblleft coprime\textquotedblright\ everywhere it appears; this
results in a proof of Proposition \ref{prop.arifun.mult.pwprod} \textbf{(b)}.)
\end{proof}

The reader might have already noticed that there is a relation between the two
arithmetic functions $\omega$ and $\Omega$ and the squarefreeness of a number.
Let us state this explicitly:

\begin{proposition}
\label{prop.arifun.wW}Let $n\in\mathbb{N}_{+}$.

\textbf{(a)} We have $\Omega\left(  n\right)  \geq\omega\left(  n\right)  $.

\textbf{(b)} We have $\Omega\left(  n\right)  =\omega\left(  n\right)  $ if
and only if $n$ is squarefree.

\textbf{(c)} We have $\mu\left(  n\right)  =0^{\Omega\left(  n\right)
-\omega\left(  n\right)  }\cdot\lambda\left(  n\right)  $. (In particular, the
expression \textquotedblleft$0^{\Omega\left(  n\right)  -\omega\left(
n\right)  }$\textquotedblright\ is well-defined, since $\Omega\left(
n\right)  -\omega\left(  n\right)  \geq0$.)
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.arifun.wW}.]Write $n$ as a product of primes;
i.e., write $n$ in the form $n=p_{1}p_{2}\cdots p_{k}$ for some primes
$p_{1},p_{2},\ldots,p_{k}$ (which may and may not be distinct). Then, the
primes dividing $n$ are precisely $p_{1},p_{2},\ldots,p_{k}$%
\ \ \ \ \footnote{This is because if a prime $q$ divides $n$, then $q\mid
n=p_{1}p_{2}\cdots p_{k}$, and therefore $q\mid p_{i}$ for some $i\in\left\{
1,2,\ldots,k\right\}  $ (since $q$ is prime); but this entails $q=p_{i}$ for
this $i$.}. In other words, $\left(  \text{the set of all primes dividing
}n\right)  =\left\{  p_{1},p_{2},\ldots,p_{k}\right\}  $. Now, the definition
of $\omega$ yields%
\begin{align}
\omega\left(  n\right)   &  =\left(  \text{the number of all distinct primes
dividing }n\right) \nonumber\\
&  =\left\vert \underbrace{\left(  \text{the set of all primes dividing
}n\right)  }_{=\left\{  p_{1},p_{2},\ldots,p_{k}\right\}  }\right\vert
\nonumber\\
&  =\left\vert \left\{  p_{1},p_{2},\ldots,p_{k}\right\}  \right\vert
\label{pf.prop.arifun.wW.1}\\
&  \leq k.\nonumber
\end{align}
But $n$ can be written as a product of $k$ primes (because $n=p_{1}p_{2}\cdots
p_{k}$, and the factors $p_{1},p_{2},\ldots,p_{k}$ in this product are
primes). Hence, the definition of $\Omega$ yields $\Omega\left(  n\right)
=k$. Now, $\omega\left(  n\right)  \leq k=\Omega\left(  n\right)  $. This
proves Proposition \ref{prop.arifun.wW} \textbf{(a)}.

\textbf{(b)} We shall prove the following two statements:

\begin{statement}
\textit{Statement 1:} If $\Omega\left(  n\right)  =\omega\left(  n\right)  $,
then $n$ is squarefree.
\end{statement}

\begin{statement}
\textit{Statement 2:} If $n$ is squarefree, then $\Omega\left(  n\right)
=\omega\left(  n\right)  $.
\end{statement}

[\textit{Proof of Statement 1:} Assume that $\Omega\left(  n\right)
=\omega\left(  n\right)  $. We must prove that $n$ is squarefree.

Assume the contrary. Thus, $n$ is not squarefree. Hence, there exists a
perfect square other than $1$ that divides $n$ (by the definition of
\textquotedblleft squarefree\textquotedblright). In other words, there exists
some integer $h>1$ such that $h^{2}\mid n$. Consider this $h$.

The integer $h$ is larger than $1$. Thus, there exists a prime $q$ such that
$q\mid h$. Consider this $q$. From $q\mid h$, we obtain $q^{2}\mid h^{2}\mid
n$.

From (\ref{pf.prop.arifun.wW.1}), we have $\left\vert \left\{  p_{1}%
,p_{2},\ldots,p_{k}\right\}  \right\vert =\omega\left(  n\right)
=\Omega\left(  n\right)  =k$. In other words, the set $\left\{  p_{1}%
,p_{2},\ldots,p_{k}\right\}  $ has exactly $k$ elements. Hence, the $k$
elements $p_{1},p_{2},\ldots,p_{k}$ must be distinct (since otherwise, the set
$\left\{  p_{1},p_{2},\ldots,p_{k}\right\}  $ would have fewer than $k$ elements).

But $q\mid q^{2}\mid n=p_{1}p_{2}\cdots p_{k}$. Since $q$ is a prime, this
entails that $q$ divides at least one of the $k$ integers $p_{1},p_{2}%
,\ldots,p_{k}$ (because of Remark \ref{rmk.prime-div-prod-word}, applied to
$q$ instead of $p$). In other words, at least one $i\in\left\{  1,2,\ldots
,k\right\}  $ satisfies $q\mid p_{i}$. Consider this $i$.

Thus, $q$ is a prime divisor of $p_{i}$ (since $q$ is a prime and satisfies
$q\mid p_{i}$). But the only prime divisor of $p_{i}$ is $p_{i}$ itself (since
$p_{i}$ is a prime). Thus, $q$ must be $p_{i}$ itself. In other words,
$q=p_{i}$.

Now,%
\begin{align*}
qq  &  =q^{2}\mid n=p_{1}p_{2}\cdots p_{k}=\underbrace{p_{i}}_{=q}\cdot\left(
p_{1}p_{2}\cdots p_{i-1}p_{i+1}p_{i+2}\cdots p_{k}\right) \\
&  =q\cdot\left(  p_{1}p_{2}\cdots p_{i-1}p_{i+1}p_{i+2}\cdots p_{k}\right)  .
\end{align*}
We can cancel $q$ from this divisibility (since $q\neq0$), and thus obtain
\[
q\mid p_{1}p_{2}\cdots p_{i-1}p_{i+1}p_{i+2}\cdots p_{k}.
\]
Since $q$ is a prime, this entails that $q$ divides at least one of the $k-1$
integers $p_{1},p_{2},\ldots,p_{i-1},p_{i+1},p_{i+2},\ldots,p_{k}$ (because of
Remark \ref{rmk.prime-div-prod-word}, applied to $q$ instead of $p$). In other
words, at least one $j\in\left\{  1,2,\ldots,k\right\}  \setminus\left\{
i\right\}  $ satisfies $q\mid p_{j}$. Consider this $j$.

We have $j\neq i$ (since $j\in\left\{  1,2,\ldots,k\right\}  \setminus\left\{
i\right\}  $) and thus $p_{j}\neq p_{i}$ (since the $k$ primes $p_{1}%
,p_{2},\ldots,p_{k}$ are distinct).

But $q$ is a prime divisor of $p_{j}$ (since $q$ is a prime and satisfies
$q\mid p_{j}$). But the only prime divisor of $p_{j}$ is $p_{j}$ itself (since
$p_{j}$ is a prime). Thus, $q$ must be $p_{j}$ itself. In other words,
$q=p_{j}$. Hence, $q=p_{j}\neq p_{i}=q$. This is absurd. This contradiction
shows that our assumption was false; hence, $n$ is squarefree. This proves
Statement 1.]

[\textit{Proof of Statement 2:} Assume that $n$ is squarefree. We must prove
that $\Omega\left(  n\right)  =\omega\left(  n\right)  $.

The $k$ primes $p_{1},p_{2},\ldots,p_{k}$ are
distinct\footnote{\textit{Proof.} Assume the contrary. Thus, two of these $k$
primes are equal. In other words, there exist two distinct elements $i$ and
$j$ of $\left\{  1,2,\ldots,k\right\}  $ such that $p_{i}=p_{j}$. Consider
these $i$ and $j$. Clearly, $p_{i}$ and $p_{j}$ are two distinct factors in
the product $p_{1}p_{2}\cdots p_{k}$ (since $i\neq j$). Hence, $p_{i}p_{j}\mid
p_{1}p_{2}\cdots p_{k}=n$.
\par
We have $p_{i}>1$ (since $p_{i}$ is a prime). Thus, $p_{i}^{2}$ is a perfect
square other than $1$. Moreover,%
\[
p_{i}^{2}=p_{i}\underbrace{p_{i}}_{=p_{j}}=p_{i}p_{j}\mid n.
\]
Thus, a perfect square other than $1$ (namely, $p_{i}^{2}$) divides $n$. In
other words, $n$ is not squarefree (by the definition of \textquotedblleft
squarefree\textquotedblright). This contradicts our assumption that $n$ is
squarefree. This contradiction shows that our assumption was false. Qed.}.
Hence, $\left\vert \left\{  p_{1},p_{2},\ldots,p_{k}\right\}  \right\vert =k$.
Now, (\ref{pf.prop.arifun.wW.1}) becomes $\omega\left(  n\right)  =\left\vert
\left\{  p_{1},p_{2},\ldots,p_{k}\right\}  \right\vert =k=\Omega\left(
n\right)  $. In other words, $\Omega\left(  n\right)  =\omega\left(  n\right)
$.\ This proves Statement 2.]

Combining Statement 1 with Statement 2, we conclude that we have
$\Omega\left(  n\right)  =\omega\left(  n\right)  $ if and only if $n$ is
squarefree. This proves Proposition \ref{prop.arifun.wW} \textbf{(b)}.

\textbf{(c)} Proposition \ref{prop.arifun.wW} \textbf{(a)} yields
$\Omega\left(  n\right)  \geq\omega\left(  n\right)  $. Hence, $\Omega\left(
n\right)  -\omega\left(  n\right)  \geq0$. Thus, the expression
\textquotedblleft$0^{\Omega\left(  n\right)  -\omega\left(  n\right)  }%
$\textquotedblright\ is well-defined. It remains to prove that $\mu\left(
n\right)  =0^{\Omega\left(  n\right)  -\omega\left(  n\right)  }\cdot
\lambda\left(  n\right)  $.

We are in one of the following two cases:

\textit{Case 1:} The positive integer $n$ is squarefree.

\textit{Case 2:} The positive integer $n$ is not squarefree.

Let us first consider Case 1. In this case, the positive integer $n$ is
squarefree. Hence, Proposition \ref{prop.arifun.wW} \textbf{(b)} yields that
$\Omega\left(  n\right)  =\omega\left(  n\right)  $. Hence, $\Omega\left(
n\right)  -\omega\left(  n\right)  =0$ and therefore $0^{\Omega\left(
n\right)  -\omega\left(  n\right)  }=0^{0}=1$. But the definition of $\mu$
yields%
\[
\mu\left(  n\right)  =%
\begin{cases}
\left(  -1\right)  ^{\omega\left(  n\right)  }, & \text{if }n\text{ is
squarefree;}\\
0, & \text{otherwise}%
\end{cases}
=\left(  -1\right)  ^{\omega\left(  n\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }n\text{ is squarefree}\right)  .
\]
Comparing this with%
\begin{align*}
\underbrace{0^{\Omega\left(  n\right)  -\omega\left(  n\right)  }}_{=1}%
\cdot\lambda\left(  n\right)   &  =\lambda\left(  n\right)  =\left(
-1\right)  ^{\Omega\left(  n\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\lambda\right) \\
&  =\left(  -1\right)  ^{\omega\left(  n\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\Omega\left(  n\right)  =\omega\left(  n\right)  \right)  ,
\end{align*}
we obtain $\mu\left(  n\right)  =0^{\Omega\left(  n\right)  -\omega\left(
n\right)  }\cdot\lambda\left(  n\right)  $. Thus, $\mu\left(  n\right)
=0^{\Omega\left(  n\right)  -\omega\left(  n\right)  }\cdot\lambda\left(
n\right)  $ is proven in Case 1.

Let us next consider Case 2. In this case, the positive integer $n$ is not squarefree.

If we had $\Omega\left(  n\right)  =\omega\left(  n\right)  $, then $n$ would
be squarefree (by Proposition \ref{prop.arifun.wW} \textbf{(b)}), which would
contradict the fact that $n$ is not squarefree. Hence, we cannot have
$\Omega\left(  n\right)  =\omega\left(  n\right)  $. Thus, we have
$\Omega\left(  n\right)  \neq\omega\left(  n\right)  $.

But Proposition \ref{prop.arifun.wW} \textbf{(a)} yields $\Omega\left(
n\right)  \geq\omega\left(  n\right)  $. Combining this with $\Omega\left(
n\right)  \neq\omega\left(  n\right)  $, we find $\Omega\left(  n\right)
>\omega\left(  n\right)  $. Hence, $\Omega\left(  n\right)  -\omega\left(
n\right)  >0$, so that $0^{\Omega\left(  n\right)  -\omega\left(  n\right)
}=0$.

Now, the definition of $\mu$ yields%
\[
\mu\left(  n\right)  =%
\begin{cases}
\left(  -1\right)  ^{\omega\left(  n\right)  }, & \text{if }n\text{ is
squarefree;}\\
0, & \text{otherwise}%
\end{cases}
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\text{ is not squarefree}\right)
.
\]
Comparing this with%
\[
\underbrace{0^{\Omega\left(  n\right)  -\omega\left(  n\right)  }}_{=0}%
\cdot\lambda\left(  n\right)  =0,
\]
we obtain $\mu\left(  n\right)  =0^{\Omega\left(  n\right)  -\omega\left(
n\right)  }\cdot\lambda\left(  n\right)  $. Thus, $\mu\left(  n\right)
=0^{\Omega\left(  n\right)  -\omega\left(  n\right)  }\cdot\lambda\left(
n\right)  $ is proven in Case 2.

We have now proven $\mu\left(  n\right)  =0^{\Omega\left(  n\right)
-\omega\left(  n\right)  }\cdot\lambda\left(  n\right)  $ in both Cases 1 and
2. Hence, $\mu\left(  n\right)  =0^{\Omega\left(  n\right)  -\omega\left(
n\right)  }\cdot\lambda\left(  n\right)  $ always holds. This completes the
proof of Proposition \ref{prop.arifun.wW} \textbf{(c)}.
\end{proof}

\subsection{The Dirichlet convolution}

Let us now define a way to produce a new arithmetic function from two given
ones: the \textit{Dirichlet convolution}.

\begin{definition}
\label{def.dirichlet}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ and
$g:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be two arithmetic functions. We define
a new arithmetic function $f\star g:\mathbb{N}_{+}\rightarrow\mathbb{C}$ by%
\[
\left(  f\star g\right)  \left(  n\right)  =\sum_{d\mid n}f\left(  d\right)
g\left(  \dfrac{n}{d}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in\mathbb{N}_{+}.
\]
This new function $f\star g$ is called the \textit{Dirichlet convolution} of
$f$ and $g$.
\end{definition}

Here is a more symmetric way to rewrite the definition of $f\star g$:

\begin{remark}
\label{rmk.dirichlet.rewrite}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ and
$g:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be two arithmetic functions. Let
$n\in\mathbb{N}_{+}$. Then,%
\[
\left(  f\star g\right)  \left(  n\right)  =\sum_{\substack{d\in\mathbb{N}%
_{+};\ e\in\mathbb{N}_{+};\\de=n}}f\left(  d\right)  g\left(  e\right)  .
\]

\end{remark}

\begin{proof}
[Proof of Remark \ref{rmk.dirichlet.rewrite}.]Let us first show two simple claims:

\begin{statement}
\textit{Claim 1:} For each $d\in\mathbb{N}_{+}$ satisfying $d\mid n$, we have
\[
\sum_{\substack{e\in\mathbb{N}_{+};\\de=n}}g\left(  e\right)  =g\left(
\dfrac{n}{d}\right)  .
\]

\end{statement}

\begin{statement}
\textit{Claim 2:} For each $d\in\mathbb{N}_{+}$ satisfying $d\nmid n$, we
have
\[
\sum_{\substack{e\in\mathbb{N}_{+};\\de=n}}g\left(  e\right)  =0.
\]

\end{statement}

[\textit{Proof of Claim 1:} Let $d\in\mathbb{N}_{+}$ be such that $d\mid n$.
Then, $\dfrac{n}{d}\in\mathbb{Z}$ (since $d\mid n$). Thus, $\dfrac{n}{d}%
\in\mathbb{N}_{+}$ (since $n\in\mathbb{N}_{+}$ and $d\in\mathbb{N}_{+}$).
Thus, there is exactly one $e\in\mathbb{N}_{+}$ satisfying $de=n$, namely
$e=\dfrac{n}{d}$. Hence, the sum $\sum_{\substack{e\in\mathbb{N}_{+}%
;\\de=n}}g\left(  e\right)  $ contains exactly one addend, namely the addend
for $e=\dfrac{n}{d}$. Thus, $\sum_{\substack{e\in\mathbb{N}_{+};\\de=n}%
}g\left(  e\right)  =g\left(  \dfrac{n}{d}\right)  $. Claim 1 is proven.]

[\textit{Proof of Claim 2:} Let $d\in\mathbb{N}_{+}$ be such that $d\nmid n$.
Then, there exists no $e\in\mathbb{N}_{+}$ satisfying $de=n$ (since $d\nmid
n$). Therefore, the sum $\sum_{\substack{e\in\mathbb{N}_{+};\\de=n}}g\left(
e\right)  $ is empty and thus equals $0$. This proves Claim 2.]

Now,%
\begin{align*}
&  \underbrace{\sum_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+}%
;\\de=n}}}_{=\sum_{d\in\mathbb{N}_{+}}\ \ \sum_{\substack{e\in\mathbb{N}%
_{+};\\de=n}}}f\left(  d\right)  g\left(  e\right)  \\
&  =\sum_{d\in\mathbb{N}_{+}}\ \ \sum_{\substack{e\in\mathbb{N}_{+}%
;\\de=n}}f\left(  d\right)  g\left(  e\right)  =\sum_{d\in\mathbb{N}_{+}%
}f\left(  d\right)  \ \ \sum_{\substack{e\in\mathbb{N}_{+};\\de=n}}g\left(
e\right)  \\
&  =\sum_{\substack{d\in\mathbb{N}_{+};\\d\mid n}}f\left(  d\right)
\underbrace{\sum_{\substack{e\in\mathbb{N}_{+};\\de=n}}g\left(  e\right)
}_{\substack{=g\left(  \dfrac{n}{d}\right)  \\\text{(by Claim 1)}}%
}+\sum_{\substack{d\in\mathbb{N}_{+};\\d\nmid n}}f\left(  d\right)
\underbrace{\sum_{\substack{e\in\mathbb{N}_{+};\\de=n}}g\left(  e\right)
}_{\substack{=0\\\text{(by Claim 2)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }d\in\mathbb{N}_{+}\text{ satisfies either }d\mid n\text{ or
}d\nmid n\\
\text{(but not both at the same time)}%
\end{array}
\right)  \\
&  =\sum_{\substack{d\in\mathbb{N}_{+};\\d\mid n}}f\left(  d\right)  g\left(
\dfrac{n}{d}\right)  +\underbrace{\sum_{\substack{d\in\mathbb{N}_{+};\\d\nmid
n}}f\left(  d\right)  0}_{=0}=\sum_{\substack{d\in\mathbb{N}_{+};\\d\mid
n}}f\left(  d\right)  g\left(  \dfrac{n}{d}\right)  =\sum_{d\mid n}f\left(
d\right)  g\left(  \dfrac{n}{d}\right)  \\
&  =\left(  f\star g\right)  \left(  n\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{because this is how }\left(  f\star g\right)  \left(  n\right)  \text{
was defined}\right)  .
\end{align*}
This proves Remark \ref{rmk.dirichlet.rewrite}.
\end{proof}

\begin{remark}
\label{rmk.dirichlet.polys} Here is a little digression which might make the
Dirichlet convolution $f \star g$ appear less mysterious (but might also
confuse you). I claim that Dirichlet convolution of arithmetic functions is
``like multiplication of power series, but with (some) additions replaced by
multiplications''. Here is what I mean by this:

A power series (say, with complex coefficients) is defined as a sequence
$\left(  a_{0}, a_{1}, a_{2}, \ldots\right)  $ of complex numbers. This
sequence is usually written in the form $\sum_{i \in\mathbb{N}} a_{i} X^{i}$.
(For us here, ``power series'' means ``formal power series in one
indeterminate $X$ with complex coefficients''; we do not care about any
questions of convergence.) The product of two power series $\sum_{i
\in\mathbb{N}} a_{i} X^{i}$ and $\sum_{i \in\mathbb{N}} b_{i} X^{i}$ is
defined by
\[
\left(  \sum_{i \in\mathbb{N}} a_{i} X^{i}\right)  \left(  \sum_{i
\in\mathbb{N}} b_{i} X^{i}\right)  = \sum_{i \in\mathbb{N}} c_{i} X^{i} ,
\]
where
\begin{equation}
c_{n} = \sum_{m = 0}^{n} a_{m} b_{n-m} = \sum_{\substack{d \in\mathbb{N}; \ e
\in\mathbb{N}; \\d+e = n}} a_{d} b_{e} . \label{eq.rmk.dirichlet.polys.1}%
\end{equation}


To every arithmetic function $f : \mathbb{N}_{+} \to\mathbb{C}$, we can assign
a power series $\widehat{f}$ with constant term $0$, defined by $\widehat{f} =
\sum_{i \in\mathbb{N}_{+}} f\left(  i\right)  X^{i}$. This assignment is a
1-to-1 correspondence between the arithmetic functions and the power series
(in one indeterminate) with constant term $0$. In other words, every power
series with constant term $0$ can be written as $\widehat{f}$ for a unique
arithmetic function $f$. Thus, we can define a ``Dirichlet convolution`` on
the set of all power series with constant term $0$, by setting $\widehat{f}
\star\widehat{g} = \widehat{f\star g}$ for every two arithmetic functions $f$
and $g$. Explicitly, this Dirichlet convolution of power series is given by
\[
\left(  \sum_{i \in\mathbb{N}_{+}} a_{i} X^{i}\right)  \star\left(  \sum_{i
\in\mathbb{N}_{+}} b_{i} X^{i}\right)  = \sum_{i \in\mathbb{N}_{+}} d_{i}
X^{i} ,
\]
where
\begin{equation}
d_{n} = \sum_{d \mid n} a_{d} b_{n/d} = \sum_{\substack{d \in\mathbb{N}_{+};
\ e \in\mathbb{N}_{+}; \\de = n}} a_{d} b_{e} .
\label{eq.rmk.dirichlet.polys.2}%
\end{equation}


The similarities between the equations (\ref{eq.rmk.dirichlet.polys.1}) and
(\ref{eq.rmk.dirichlet.polys.2}) should be palpable. Roughly speaking,
(\ref{eq.rmk.dirichlet.polys.2}) is a \textquotedblleft
multiplicative\textquotedblright\ variant of (\ref{eq.rmk.dirichlet.polys.1}):
Whereas the sum $\sum_{\substack{d\in\mathbb{N};\ e\in\mathbb{N}%
;\\d+e=n}}a_{d}b_{e}$ in (\ref{eq.rmk.dirichlet.polys.1}) runs over all
decompositions of $n$ into a \textbf{sum} of two nonnegative integers $d$ and
$e$, the analogous sum $\sum_{\substack{d\in\mathbb{N}_{+};\ e\in
\mathbb{N}_{+};\\de=n}}a_{d}b_{e}$ in (\ref{eq.rmk.dirichlet.polys.2}) runs
over all decompositions of $n$ into a \textbf{product} of two positive
integers $d$ and $e$. (Yes, the multiplicative analogue of nonnegative
integers in this context are positive integers.) So, roughly speaking,
Dirichlet convolution is like multiplication of power series, except that two
monomials $X^{m}$ and $X^{n}$ are taken to $X^{mn}$ and not to $X^{m+n}$.

This analogy has a consequence: It suggests that Dirichlet convolution should
be associative and commutative, and that this should be provable in the same
way as one proves the associativity and the commutativity of the
multiplication of power series. And, indeed, this is the case: see Theorem
\ref{thm.dirichlet.props} below.

See also \cite[\S 8.2]{NiZuMo91} for the notion of \textit{Dirichlet series},
which are \textquotedblleft formal expressions\textquotedblright\ of the form
$\sum_{i\in\mathbb{N}_{+}}\dfrac{a_{i}}{i^{s}}$ for an \textquotedblleft
indeterminate\textquotedblright\ exponent $s$. If we replace the term
$\dfrac{a_{i}}{i^{s}}$ by $a_{i}X^{i}$, then these Dirichlet series turn into
standard formal power series with constant term $0$, but their product turns
into the Dirichlet convolution of power series.
\end{remark}

\begin{theorem}
\label{thm.dirichlet.props}\textbf{(a)} We have $\varepsilon\star
f=f\star\varepsilon=f$ for every arithmetic function $f$.

\textbf{(b)} We have $f\star\left(  g\star h\right)  =\left(  f\star g\right)
\star h$ for every three arithmetic functions $f$, $g$ and $h$.

\textbf{(c)} We have $f\star g=g\star f$ for every two arithmetic functions
$f$ and $g$.
\end{theorem}

\begin{remark}
If you know the notion of a monoid, then you will be able to restate Theorem
\ref{thm.dirichlet.props} as follows: The set of all arithmetic functions is a
commutative monoid under the operation $\star$ with neutral element
$\varepsilon$.

Actually, we can also define an addition operation on arithmetic functions
(namely, pointwise addition: $\left(  f+g\right)  \left(  n\right)  =f\left(
n\right)  +g\left(  n\right)  $). The addition operation $+$ and the Dirichlet
convolution $\star$ turn the set of arithmetic functions into a commutative ring.
\end{remark}

\begin{proof}
[Proof of Theorem \ref{thm.dirichlet.props}.]\textbf{(c)} Let $f$ and $g$ be
two arithmetic functions. Let $n\in\mathbb{N}_{+}$. Remark
\ref{rmk.dirichlet.rewrite} (applied to $g$ and $f$ instead of $f$ and $g$)
yields%
\begin{equation}
\left(  g\star f\right)  \left(  n\right)  =\sum_{\substack{d\in\mathbb{N}%
_{+};\ e\in\mathbb{N}_{+};\\de=n}}g\left(  d\right)  f\left(  e\right)  .
\label{pf.thm.dirichlet.props.c.1}%
\end{equation}
Remark \ref{rmk.dirichlet.rewrite} yields%
\begin{align*}
\left(  f\star g\right)  \left(  n\right)   &  =\sum_{\substack{d\in
\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\de=n}}f\left(  d\right)  g\left(
e\right)  =\underbrace{\sum_{\substack{e\in\mathbb{N}_{+};\ d\in\mathbb{N}%
_{+};\\ed=n}}}_{\substack{=\sum_{\substack{d\in\mathbb{N}_{+};\ e\in
\mathbb{N}_{+};\\ed=n}}\\=\sum_{\substack{d\in\mathbb{N}_{+};\ e\in
\mathbb{N}_{+};\\de=n}}}}\underbrace{f\left(  e\right)  g\left(  d\right)
}_{=g\left(  d\right)  f\left(  e\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have renamed the summation}\\
\text{indices }d\text{ and }e\text{ as }e\text{ and }d
\end{array}
\right) \\
&  =\sum_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\de=n}}g\left(
d\right)  f\left(  e\right)  .
\end{align*}
Comparing this with (\ref{pf.thm.dirichlet.props.c.1}), we obtain $\left(
f\star g\right)  \left(  n\right)  =\left(  g\star f\right)  \left(  n\right)
$.

Now, forget that we fixed $n$. We thus have shown that $\left(  f\star
g\right)  \left(  n\right)  =\left(  g\star f\right)  \left(  n\right)  $ for
each $n\in\mathbb{N}_{+}$. In other words, $f\star g=g\star f$. This proves
Theorem \ref{thm.dirichlet.props} \textbf{(c)}.

\textbf{(a)} Let $f$ be an arithmetic function. Every $n\in\mathbb{N}_{+}$
satisfies%
\begin{align*}
\left(  \varepsilon\star f\right)  \left(  n\right)   &  =\sum_{d\mid
n}\underbrace{\varepsilon\left(  d\right)  }_{\substack{=%
\begin{cases}
1, & \text{if }d=1;\\
0, & \text{if }d\neq1
\end{cases}
\\\text{(by the definition of }\varepsilon\text{)}}}f\left(  \dfrac{n}%
{d}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\varepsilon\star f\right) \\
&  =\sum_{d\mid n}%
\begin{cases}
1, & \text{if }d=1;\\
0, & \text{if }d\neq1
\end{cases}
f\left(  \dfrac{n}{d}\right) \\
&  =\underbrace{%
\begin{cases}
1, & \text{if }1=1;\\
0, & \text{if }1\neq1
\end{cases}
}_{=1}\underbrace{f\left(  \dfrac{n}{1}\right)  }_{=f\left(  n\right)  }%
+\sum_{\substack{d\mid n;\\d\neq1}}\underbrace{%
\begin{cases}
1, & \text{if }d=1;\\
0, & \text{if }d\neq1
\end{cases}
}_{\substack{=0\\\text{(since }d\neq1\text{)}}}f\left(  \dfrac{n}{d}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }d=1\text{ from the sum}\\
\text{(since }1\text{ is a positive divisor of }n\text{)}%
\end{array}
\right) \\
&  =f\left(  n\right)  +\underbrace{\sum_{\substack{d\mid n;\\d\neq
1}}0f\left(  \dfrac{n}{d}\right)  }_{=0}=f\left(  n\right)  .
\end{align*}
In other words, $\varepsilon\star f=f$. But Theorem \ref{thm.dirichlet.props}
\textbf{(c)} (applied to $g=\varepsilon$) yields $f\star\varepsilon
=\varepsilon\star f$. Thus, $f\star\varepsilon=\varepsilon\star f=f$. This
proves Theorem \ref{thm.dirichlet.props} \textbf{(a)}.

\textbf{(b)} Let us first make a general observation: If $F$ and $G$ are two
arithmetic functions, and if $N\in\mathbb{N}_{+}$, then%
\begin{equation}
\left(  F\star G\right)  \left(  N\right)  =\sum_{\substack{D\in\mathbb{N}%
_{+};\ E\in\mathbb{N}_{+};\\DE=N}}F\left(  D\right)  G\left(  E\right)  .
\label{pf.thm.dirichlet.props.b.uppercase}%
\end{equation}
(This is simply Remark \ref{rmk.dirichlet.rewrite}, with the letters
$f,g,n,d,e$ renamed as $F,G,N,D,E$. We are playing this renaming game in order
to avoid collisions between notations.)

Now, let $f$, $g$ and $h$ be three arithmetic functions. Let $n\in
\mathbb{N}_{+}$. We have%
\begin{align}
&  \left(  \left(  f\star g\right)  \star h\right)  \left(  n\right)
\nonumber\\
&  =\sum_{\substack{D\in\mathbb{N}_{+};\ E\in\mathbb{N}_{+};\\DE=n}}\left(
f\star g\right)  \left(  D\right)  h\left(  E\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.dirichlet.props.b.uppercase}), applied to }F=f\star g\text{,
}G=h\text{ and }N=n\right) \nonumber\\
&  =\sum_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\de=n}%
}\underbrace{\left(  f\star g\right)  \left(  d\right)  }_{\substack{=\sum
_{\substack{D\in\mathbb{N}_{+};\ E\in\mathbb{N}_{+};\\DE=d}}f\left(  D\right)
g\left(  E\right)  \\\text{(by (\ref{pf.thm.dirichlet.props.b.uppercase}),
applied to }F=f\text{, }G=g\text{ and }N=d\text{)}}}h\left(  e\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation
indices }D\text{ and }E\text{ as }d\text{ and }e\right) \nonumber\\
&  =\sum_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\de=n}}\left(
\sum_{\substack{D\in\mathbb{N}_{+};\ E\in\mathbb{N}_{+};\\DE=d}}f\left(
D\right)  g\left(  E\right)  \right)  h\left(  e\right) \nonumber\\
&  =\underbrace{\sum_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}%
_{+};\\de=n}}\ \ \sum_{\substack{D\in\mathbb{N}_{+};\ E\in\mathbb{N}%
_{+};\\DE=d}}}_{\substack{=\sum_{d\in\mathbb{N}_{+}}\ \ \sum_{\substack{D\in
\mathbb{N}_{+};\ E\in\mathbb{N}_{+};\\DE=d}}\ \ \sum_{\substack{e\in
\mathbb{N}_{+};\\de=n}}\\\text{(here, we are interchanging the order of
summation)}}}f\left(  D\right)  g\left(  E\right)  h\left(  e\right)
\nonumber\\
&  =\sum_{d\in\mathbb{N}_{+}}\ \ \sum_{\substack{D\in\mathbb{N}_{+}%
;\ E\in\mathbb{N}_{+};\\DE=d}}\underbrace{\sum_{\substack{e\in\mathbb{N}%
_{+};\\de=n}}}_{\substack{=\sum_{\substack{e\in\mathbb{N}_{+};\\DEe=n}%
}\\\text{(since }d=DE\text{)}}}f\left(  D\right)  g\left(  E\right)  h\left(
e\right) \nonumber\\
&  =\underbrace{\sum_{d\in\mathbb{N}_{+}}\ \ \sum_{\substack{D\in
\mathbb{N}_{+};\ E\in\mathbb{N}_{+};\\DE=d}}}_{=\sum_{D\in\mathbb{N}%
_{+};\ E\in\mathbb{N}_{+}}}\ \ \sum_{\substack{e\in\mathbb{N}_{+}%
;\\DEe=n}}f\left(  D\right)  g\left(  E\right)  h\left(  e\right)
=\underbrace{\sum_{D\in\mathbb{N}_{+};\ E\in\mathbb{N}_{+}}\ \ \sum
_{\substack{e\in\mathbb{N}_{+};\\DEe=n}}}_{=\sum_{\substack{D\in\mathbb{N}%
_{+};\ E\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\DEe=n}}}f\left(  D\right)
g\left(  E\right)  h\left(  e\right) \nonumber\\
&  =\sum_{\substack{D\in\mathbb{N}_{+};\ E\in\mathbb{N}_{+};\ e\in
\mathbb{N}_{+};\\DEe=n}}f\left(  D\right)  g\left(  E\right)  h\left(
e\right) \nonumber\\
&  =\sum_{\substack{c\in\mathbb{N}_{+};\ d\in\mathbb{N}_{+};\ e\in
\mathbb{N}_{+};\\cde=n}}f\left(  c\right)  g\left(  d\right)  h\left(
e\right) \label{pf.thm.dirichlet.props.b.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation
indices }D\text{ and }E\text{ as }c\text{ and }d\right)  .\nonumber
\end{align}
On the other hand,%
\begin{align*}
&  \left(  f\star\left(  g\star h\right)  \right)  \left(  n\right) \\
&  =\sum_{\substack{D\in\mathbb{N}_{+};\ E\in\mathbb{N}_{+};\\DE=n}}f\left(
D\right)  \left(  g\star h\right)  \left(  E\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.dirichlet.props.b.uppercase}), applied to }F=f\text{, }G=g\star
h\text{ and }N=n\right) \\
&  =\sum_{\substack{c\in\mathbb{N}_{+};\ d\in\mathbb{N}_{+};\\cd=n}}f\left(
c\right)  \underbrace{\left(  g\star h\right)  \left(  d\right)
}_{\substack{=\sum_{\substack{D\in\mathbb{N}_{+};\ E\in\mathbb{N}_{+}%
;\\DE=d}}g\left(  D\right)  h\left(  E\right)  \\\text{(by
(\ref{pf.thm.dirichlet.props.b.uppercase}), applied to }F=g\text{, }G=h\text{
and }N=d\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation
indices }D\text{ and }E\text{ as }c\text{ and }d\right) \\
&  =\sum_{\substack{c\in\mathbb{N}_{+};\ d\in\mathbb{N}_{+};\\cd=n}}f\left(
c\right)  \left(  \sum_{\substack{D\in\mathbb{N}_{+};\ E\in\mathbb{N}%
_{+};\\DE=d}}g\left(  D\right)  h\left(  E\right)  \right) \\
&  =\underbrace{\sum_{\substack{c\in\mathbb{N}_{+};\ d\in\mathbb{N}%
_{+};\\cd=n}}\ \ \sum_{\substack{D\in\mathbb{N}_{+};\ E\in\mathbb{N}%
_{+};\\DE=d}}}_{\substack{=\sum_{d\in\mathbb{N}_{+}}\ \ \sum_{\substack{D\in
\mathbb{N}_{+};\ E\in\mathbb{N}_{+};\\DE=d}}\ \ \sum_{\substack{c\in
\mathbb{N}_{+};\\cd=n}}\\\text{(here, we are interchanging the order of
summation)}}}f\left(  c\right)  g\left(  D\right)  h\left(  E\right) \\
&  =\sum_{d\in\mathbb{N}_{+}}\ \ \sum_{\substack{D\in\mathbb{N}_{+}%
;\ E\in\mathbb{N}_{+};\\DE=d}}\underbrace{\sum_{\substack{c\in\mathbb{N}%
_{+};\\cd=n}}}_{\substack{=\sum_{\substack{c\in\mathbb{N}_{+};\\cDE=n}%
}\\\text{(since }d=DE\text{)}}}f\left(  c\right)  g\left(  D\right)  h\left(
E\right) \\
&  =\underbrace{\sum_{d\in\mathbb{N}_{+}}\ \ \sum_{\substack{D\in
\mathbb{N}_{+};\ E\in\mathbb{N}_{+};\\DE=d}}}_{=\sum_{D\in\mathbb{N}%
_{+};\ E\in\mathbb{N}_{+}}}\ \ \sum_{\substack{c\in\mathbb{N}_{+}%
;\\cDE=n}}f\left(  c\right)  g\left(  D\right)  h\left(  E\right)
=\underbrace{\sum_{D\in\mathbb{N}_{+};\ E\in\mathbb{N}_{+}}\ \ \sum
_{\substack{c\in\mathbb{N}_{+};\\cDE=n}}}_{=\sum_{\substack{c\in\mathbb{N}%
_{+};\ D\in\mathbb{N}_{+};\ E\in\mathbb{N}_{+};\\cDE=n}}}f\left(  c\right)
g\left(  D\right)  h\left(  E\right) \\
&  =\sum_{\substack{c\in\mathbb{N}_{+};\ D\in\mathbb{N}_{+};\ E\in
\mathbb{N}_{+};\\cDE=n}}f\left(  c\right)  g\left(  D\right)  h\left(
E\right)  =\sum_{\substack{c\in\mathbb{N}_{+};\ d\in\mathbb{N}_{+}%
;\ e\in\mathbb{N}_{+};\\cde=n}}f\left(  c\right)  g\left(  d\right)  h\left(
e\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation
indices }D\text{ and }E\text{ as }d\text{ and }e\right)  .
\end{align*}
Comparing this with (\ref{pf.thm.dirichlet.props.b.1}), we obtain $\left(
f\star\left(  g\star h\right)  \right)  \left(  n\right)  =\left(  \left(
f\star g\right)  \star h\right)  \left(  n\right)  $.

Now, forget that we fixed $n$. We thus have proven that $\left(  f\star\left(
g\star h\right)  \right)  \left(  n\right)  =\left(  \left(  f\star g\right)
\star h\right)  \left(  n\right)  $ for every $n\in\mathbb{N}_{+}$. In other
words, $f\star\left(  g\star h\right)  =\left(  f\star g\right)  \star h$.
This proves Theorem \ref{thm.dirichlet.props} \textbf{(b)}.
\end{proof}

\subsection{Examples of Dirichlet convolutions}

Let us see what Dirichlet convolution does to the arithmetic functions we
know. We start with some simple observations:

\begin{proposition}
\label{prop.dirichlet.1*1}We have $\underline{1}\star\underline{1}%
=\operatorname*{d}$. (See Definition \ref{def.arifun.funcs} for the
definitions of $\underline{1}$ and $\operatorname*{d}$.)
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.dirichlet.1*1}.]For every $n\in\mathbb{N}_{+}%
$, we have%
\begin{align*}
\left(  \underline{1}\star\underline{1}\right)  \left(  n\right)   &
=\sum_{d\mid n}\underbrace{\underline{1}\left(  d\right)  }%
_{\substack{=1\\\text{(by the definition of }\underline{1}\text{)}%
}}\underbrace{\underline{1}\left(  \dfrac{n}{d}\right)  }%
_{\substack{=1\\\text{(by the definition of }\underline{1}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\underline{1}%
\star\underline{1}\right) \\
&  =\sum_{d\mid n}1=\underbrace{\left(  \text{the number of positive divisors
of }n\right)  }_{\substack{=\operatorname*{d}\left(  n\right)  \\\text{(since
this is how }\operatorname*{d}\left(  n\right)  \text{ was defined)}}}\cdot1\\
&  =\operatorname*{d}\left(  n\right)  \cdot1=\operatorname*{d}\left(
n\right)  .
\end{align*}
In other words, $\underline{1}\star\underline{1}=\operatorname*{d}$. This
proves Proposition \ref{prop.dirichlet.1*1}.
\end{proof}

\begin{proposition}
\label{prop.dirichlet.sig}\textbf{(a)} We have $\iota\star\underline{1}%
=\sigma$.

\textbf{(b)} Let $k\in\mathbb{Z}$. Let $\iota_{k}:\mathbb{N}_{+}%
\rightarrow\mathbb{C}$ be the function sending each $n\in\mathbb{N}_{+}$ to
$n^{k}$. Then, $\iota_{k}\star\underline{1}=\sigma_{k}$.
\end{proposition}

We leave the proof of Proposition \ref{prop.dirichlet.sig} to the reader.

Before we go on, let us show an auxiliary fact:

\begin{lemma}
\label{lem.divs.n/d}Let $n\in\mathbb{N}_{+}$. Let $\mathcal{D}\left(
n\right)  $ be the set of all positive divisors of $n$. Then, the map%
\[
\mathcal{D}\left(  n\right)  \rightarrow\mathcal{D}\left(  n\right)
,\ \ \ \ \ \ \ \ \ \ d\mapsto n/d
\]
is well-defined and bijective.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.divs.n/d}.]For every $d\in\mathcal{D}\left(
n\right)  $, we have $n/d\in\mathcal{D}\left(  n\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $d\in\mathcal{D}\left(  n\right)  $.
Thus, $d$ is a positive divisor of $n$ (since $\mathcal{D}\left(  n\right)  $
is the set of all positive divisors of $n$). Hence, $d$ is a positive integer
and satisfies $d\mid n$. Now, $n/d$ is an integer (since $d\mid n$) and is
positive (since $n$ and $d$ are positive). Hence, $n/d$ is a positive integer.
Thus, $n/d$ is a positive divisor of $n$ (since $n/d\mid n$). In other words,
$n/d\in\mathcal{D}\left(  n\right)  $ (since $\mathcal{D}\left(  n\right)  $
is the set of all positive divisors of $n$). Qed.}. Thus, we can define a map
$\rho:\mathcal{D}\left(  n\right)  \rightarrow\mathcal{D}\left(  n\right)  $
by%
\[
\rho\left(  d\right)  =n/d\ \ \ \ \ \ \ \ \ \ \text{for every }d\in
\mathcal{D}\left(  n\right)  .
\]
Consider this map $\rho$. This map $\rho$ is the map%
\begin{equation}
\mathcal{D}\left(  n\right)  \rightarrow\mathcal{D}\left(  n\right)
,\ \ \ \ \ \ \ \ \ \ d\mapsto n/d \label{pf.lem.divs.n/d.1}%
\end{equation}
(because $\rho\left(  d\right)  =n/d$ for every $d\in\mathcal{D}\left(
n\right)  $). Thus, the map (\ref{pf.lem.divs.n/d.1}) is well-defined.

We have
\begin{align*}
\left(  \rho\circ\rho\right)  \left(  d\right)   &  =\rho\left(
\underbrace{\rho\left(  d\right)  }_{=n/d}\right)  =\rho\left(  n/d\right)
=n/\left(  n/d\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\rho\right) \\
&  =d=\operatorname*{id}\left(  d\right)
\end{align*}
for every $d\in\mathcal{D}\left(  n\right)  $. In other words, $\rho\circ
\rho=\operatorname*{id}$. Hence, the maps $\rho$ and $\rho$ are mutually
inverse. In particular, the map $\rho$ is invertible, i.e., bijective. In
other words, the map (\ref{pf.lem.divs.n/d.1}) is bijective (since the map
$\rho$ is the map (\ref{pf.lem.divs.n/d.1})). Thus, we have shown that the map
(\ref{pf.lem.divs.n/d.1}) is well-defined and bijective. Lemma
\ref{lem.divs.n/d} is proven.
\end{proof}

Here is a more interesting result:

\begin{proposition}
\label{prop.dirichlet.phi}We have $\phi\star\underline{1}=\iota$.
\end{proposition}

Before we prove this, let us restate it in a more elementary fashion:

\begin{proposition}
\label{prop.dirichlet.phi.explicit}We have
\[
\sum_{d\mid n}\phi\left(  d\right)  =n\ \ \ \ \ \ \ \ \ \ \text{for every
}n\in\mathbb{N}_{+}.
\]

\end{proposition}

Proposition \ref{prop.dirichlet.phi.explicit} is \cite[Theorem 4.6]{NiZuMo91}.
The proof we are going to give for it here is actually the second proof given
for it in \cite{NiZuMo91}:

\begin{proof}
[Proof of Proposition \ref{prop.dirichlet.phi.explicit}.]Fix $n\in
\mathbb{N}_{+}$. Let me first show that%
\begin{equation}
\sum_{\substack{k\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(  k,n\right)
=d}}1=\phi\left(  \dfrac{n}{d}\right)
\label{pf.prop.dirichlet.phi.explicit.1}%
\end{equation}
for every positive divisor $d$ of $n$.

\textit{Proof of (\ref{pf.prop.dirichlet.phi.explicit.1}):} Let $d$ be a
positive divisor of $n$. Define a set $K$ by%
\[
K=\left\{  k\in\left\{  1,2,\ldots,n\right\}  \ \mid\ \gcd\left(  k,n\right)
=d\right\}  .
\]
Thus, $\sum_{\substack{k\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(
k,n\right)  =d}}1=\sum_{k\in K}1=\left\vert K\right\vert \cdot1=\left\vert
K\right\vert $.

On the other hand, define a set $F$ by%
\[
F=\left\{  k\in\left\{  1,2,\ldots,\dfrac{n}{d}\right\}  \ \mid\ k\text{ is
coprime to }\dfrac{n}{d}\right\}  .
\]
Then, $\left\vert F\right\vert $ is the number of all $k\in\left\{
1,2,\ldots,\dfrac{n}{d}\right\}  $ coprime to $\dfrac{n}{d}$; this number is
$\phi\left(  \dfrac{n}{d}\right)  $ (since this is how $\phi\left(  \dfrac
{n}{d}\right)  $ is defined). In other words, $\left\vert F\right\vert
=\phi\left(  \dfrac{n}{d}\right)  $.

But every $u\in K$ satisfies $\dfrac{u}{d}\in F$%
\ \ \ \ \footnote{\textit{Proof.} Let $u\in K$. Thus, $u$ is an element of
$\left\{  1,2,\ldots,n\right\}  $ and satisfies $\gcd\left(  u,n\right)  =d$
(by the definition of $K$). Now, $d=\gcd\left(  u,n\right)  \mid u$, so that
$\dfrac{u}{d}$ is an integer. This integer $\dfrac{u}{d}$ must belong to
$\left\{  1,2,\ldots,\dfrac{n}{d}\right\}  $ (since $u$ belongs to $\left\{
1,2,\ldots,n\right\}  $). But Proposition \ref{prop.gcd.g} (applied to $d$,
$\dfrac{u}{d}$ and $\dfrac{n}{d}$ instead of $g$, $a$ and $b$) yields
$d\gcd\left(  \dfrac{u}{d},\dfrac{n}{d}\right)  =\gcd\left(
\underbrace{d\cdot\dfrac{u}{d}}_{=u},\underbrace{d\cdot\dfrac{n}{d}}%
_{=n}\right)  =\gcd\left(  u,n\right)  =d$. Cancelling $d$ from this equality,
we obtain $\gcd\left(  \dfrac{u}{d},\dfrac{n}{d}\right)  =1$ (since $d\neq0$).
In other words, $\dfrac{u}{d}$ is coprime to $\dfrac{n}{d}$. Thus, we have
shown that $\dfrac{u}{d}$ is an element of $\left\{  1,2,\ldots,\dfrac{n}%
{d}\right\}  $ and is coprime to $\dfrac{n}{d}$. In other words, $\dfrac{u}%
{d}\in F$ (by the definition of $F$), qed.}. Thus, we can define a map%
\[
\alpha:K\rightarrow F,\ \ \ \ \ \ \ \ \ \ u\mapsto\dfrac{u}{d}.
\]


On the other hand, every $v\in F$ satisfies $dv\in K$%
\ \ \ \ \footnote{\textit{Proof.} Let $v\in F$. Thus, $v$ is an element of
$\left\{  1,2,\ldots,\dfrac{n}{d}\right\}  $ and is coprime to $\dfrac{n}{d}$
(by the definition of $F$). Now, $\gcd\left(  v,\dfrac{n}{d}\right)  =1$
(since $v$ is coprime to $\dfrac{n}{d}$). But Proposition \ref{prop.gcd.g}
(applied to $d$, $v$ and $\dfrac{n}{d}$ instead of $g$, $a$ and $b$) shows
that $d\gcd\left(  v,\dfrac{n}{d}\right)  =\gcd\left(  dv,\underbrace{d\cdot
\dfrac{n}{d}}_{=n}\right)  =\gcd\left(  dv,n\right)  $, so that $\gcd\left(
dv,n\right)  =d\underbrace{\gcd\left(  v,\dfrac{n}{d}\right)  }_{=1}=d$. Also,
from $v\in\left\{  1,2,\ldots,\dfrac{n}{d}\right\}  $, we obtain
$dv\in\left\{  1,2,\ldots,n\right\}  $. Hence, we have shown that $dv$ is an
element of $\left\{  1,2,\ldots,n\right\}  $ and satisfies $\gcd\left(
dv,n\right)  =d$. In other words, $dv\in K$ (by the definition of $K$), qed.}.
Thus, we can define a map%
\[
\beta:F\rightarrow K,\ \ \ \ \ \ \ \ \ \ v\mapsto dv.
\]
The two maps $\alpha$ and $\beta$ that we have now defined are mutually
inverse (since one of them divides its input by $d$, whereas the other
multiplies it by $d$). Hence, $\alpha$ is a bijection. Thus, there is a
bijection $K\rightarrow F$ (namely, $\alpha$). Hence, $\left\vert F\right\vert
=\left\vert K\right\vert $. Now, $\phi\left(  \dfrac{n}{d}\right)  =\left\vert
F\right\vert =\left\vert K\right\vert =\sum_{\substack{k\in\left\{
1,2,\ldots,n\right\}  ;\\\gcd\left(  k,n\right)  =d}}1$, and therefore
(\ref{pf.prop.dirichlet.phi.explicit.1}) is proven.

Now, let $\mathcal{D}\left(  n\right)  $ be the set of all positive divisors
of $n$. Then, the summation sign $\sum_{d\in\mathcal{D}\left(  n\right)  }$
means the same thing as $\sum_{d\mid n}$ (namely, a summation over all
positive divisors $d$ of $n$).

But Lemma \ref{lem.divs.n/d} shows that the map%
\[
\mathcal{D}\left(  n\right)  \rightarrow\mathcal{D}\left(  n\right)
,\ \ \ \ \ \ \ \ \ \ d\mapsto n/d
\]
is well-defined and bijective. Thus, this map is a bijection. Hence, we can
substitute $\dfrac{n}{d}$ for $d$ in the sum $\sum_{d\in\mathcal{D}\left(
n\right)  }\phi\left(  d\right)  $. We thus obtain
\begin{align*}
\sum_{d\in\mathcal{D}\left(  n\right)  }\phi\left(  d\right)   &
=\underbrace{\sum_{d\in\mathcal{D}\left(  n\right)  }}_{=\sum_{d\mid n}%
}\underbrace{\phi\left(  \dfrac{n}{d}\right)  }_{\substack{=\sum
_{\substack{k\in\left\{  1,2,\ldots,n\right\}  ;\\\gcd\left(  k,n\right)
=d}}1\\\text{(by (\ref{pf.prop.dirichlet.phi.explicit.1}))}}}=\underbrace{\sum
_{d\mid n}\ \ \sum_{\substack{k\in\left\{  1,2,\ldots,n\right\}
;\\\gcd\left(  k,n\right)  =d}}}_{\substack{=\sum_{k\in\left\{  1,2,\ldots
,n\right\}  }\\\text{(because for every }k\in\left\{  1,2,\ldots,n\right\}
\text{,}\\\text{the number }\gcd\left(  k,n\right)  \text{ is a positive
divisor of }n\text{)}}}1\\
&  =\sum_{k\in\left\{  1,2,\ldots,n\right\}  }1=n.
\end{align*}
Therefore, $n=\underbrace{\sum_{d\in\mathcal{D}\left(  n\right)  }}%
_{=\sum_{d\mid n}}\phi\left(  d\right)  =\sum_{d\mid n}\phi\left(  d\right)
$. This proves Proposition \ref{prop.dirichlet.phi.explicit}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.dirichlet.phi}.]For every $n\in\mathbb{N}_{+}%
$, we have%
\begin{align*}
\left(  \phi\star\underline{1}\right)  \left(  n\right)   &  =\sum_{d\mid
n}\phi\left(  d\right)  \underbrace{\underline{1}\left(  \dfrac{n}{d}\right)
}_{\substack{=1\\\text{(by the definition of }\underline{1}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\phi\star
\underline{1}\right) \\
&  =\sum_{d\mid n}\phi\left(  d\right)  =n\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition \ref{prop.dirichlet.phi.explicit}}\right) \\
&  =\iota\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\iota\left(  n\right)  \text{ is defined to be }n\right)  .
\end{align*}
In other words, $\phi\star\underline{1}=\iota$. This proves Proposition
\ref{prop.dirichlet.phi}.
\end{proof}

Here is another important fact about Dirichlet convolution:

\begin{proposition}
\label{prop.dirichlet.mu}We have $\mu\star\underline{1}=\varepsilon$.
\end{proposition}

Again, we shall first restate it in concrete language before proving it:

\begin{proposition}
\label{prop.dirichlet.mu.explicit}We have
\[
\sum_{d\mid n}\mu\left(  d\right)  =\varepsilon\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}_{+}.
\]

\end{proposition}

Proposition \ref{prop.dirichlet.mu.explicit} is \cite[Theorem 4.7]{NiZuMo91},
and the book gives two proofs for it. Let us sketch a third:\footnote{See
\cite[proof of Proposition 2.6]{Grinbe16} for a detailed version of this
proof.}

\begin{proof}
[Proof of Proposition \ref{prop.dirichlet.mu.explicit} (sketched).]Fix
$n\in\mathbb{N}_{+}$. We must prove the identity
\begin{equation}
\sum_{d\mid n}\mu\left(  d\right)  =\varepsilon\left(  n\right)  .
\label{pf.prop.dirichlet.mu.explicit.goal}%
\end{equation}


First of all, we recall that $\mu\left(  1\right)  =1$. (This has already been
proven in our proof of Proposition \ref{prop.arifun.mult.exa} \textbf{(f)}.)
Hence, $\sum_{d\mid1}\mu\left(  d\right)  =\mu\left(  1\right)  =1=\varepsilon
\left(  1\right)  $ (because $\varepsilon\left(  1\right)  $ is defined to be
$1$). In other words, (\ref{pf.prop.dirichlet.mu.explicit.goal}) is proven for
the case when $n=1$. Thus, we WLOG assume that $n\neq1$ from now on. Hence,
$\varepsilon\left(  n\right)  =0$ (by the definition of $\varepsilon$). Also,
$n$ has at least one prime divisor (since $n\neq1$). Pick any prime divisor
$q$ of $n$.

Let $D$ be the set of all squarefree positive divisors $d$ of $n$ satisfying
$q\nmid d$.

Let $E$ be the set of all squarefree positive divisors $d$ of $n$ satisfying
$q\mid d$.

The map%
\[
D\rightarrow E,\ \ \ \ \ \ \ \ \ \ d\mapsto qd
\]
is well-defined and a bijection\footnote{Check this! (Or see \cite[proof of
Proposition 2.6]{Grinbe16} for the proof.)}. Moreover, every $d\in D$
satisfies%
\begin{equation}
\mu\left(  qd\right)  =-\mu\left(  d\right)
\label{pf.prop.dirichlet.mu.explicit.mu(qd)}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.dirichlet.mu.explicit.mu(qd)}):} Let
$d\in D$. Thus, $d$ is a squarefree positive divisor $d$ of $n$ satisfying
$q\nmid d$ (by the definition of $D$). From $q\nmid d$, we conclude that $q$
is coprime to $d$ (since $q$ is prime). Hence, $\mu\left(  qd\right)
=\mu\left(  q\right)  \mu\left(  d\right)  $ (since the function $\mu$ is
multiplicative).
\par
But $q$ is a prime; thus, $q$ is squarefree. Hence, the definition of $\mu$
yields $\mu\left(  q\right)  =\left(  -1\right)  ^{\omega\left(  q\right)
}=\left(  -1\right)  ^{1}$ (since $\omega\left(  q\right)  =1$ (again since
$q$ is a prime)). Thus, $\mu\left(  qd\right)  =\underbrace{\mu\left(
q\right)  }_{=\left(  -1\right)  ^{1}=-1}\mu\left(  d\right)  =-\mu\left(
d\right)  $, qed.}.

Now,%
\begin{align*}
\sum_{d\mid n}\mu\left(  d\right)   &  =\sum_{\substack{d\mid n;\\d\text{ is
squarefree}}}\mu\left(  d\right)  +\sum_{\substack{d\mid n;\\d\text{ is not
squarefree}}}\underbrace{\mu\left(  d\right)  }_{\substack{=0\\\text{(by the
definition of }\mu\text{,}\\\text{since }d\text{ is not squarefree)}}}\\
&  =\sum_{\substack{d\mid n;\\d\text{ is squarefree}}}\mu\left(  d\right)
+\underbrace{\sum_{\substack{d\mid n;\\d\text{ is not squarefree}}}0}%
_{=0}=\sum_{\substack{d\mid n;\\d\text{ is squarefree}}}\mu\left(  d\right) \\
&  =\underbrace{\sum_{\substack{d\mid n;\\d\text{ is squarefree;}\\q\mid d}%
}}_{\substack{=\sum_{d\in E}\\\text{(by the definition of }E\text{)}}%
}\mu\left(  d\right)  +\underbrace{\sum_{\substack{d\mid n;\\d\text{ is
squarefree;}\\q\nmid d}}}_{\substack{=\sum_{d\in D}\\\text{(by the definition
of }D\text{)}}}\mu\left(  d\right) \\
&  =\sum_{d\in E}\mu\left(  d\right)  +\sum_{d\in D}\mu\left(  d\right)
=\sum_{d\in D}\underbrace{\mu\left(  qd\right)  }_{\substack{=-\mu\left(
d\right)  \\\text{(by (\ref{pf.prop.dirichlet.mu.explicit.mu(qd)}))}}%
}+\sum_{d\in D}\mu\left(  d\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }qd\text{ for }d\text{ in the first sum,
since}\\
\text{the map }D\rightarrow E,\ d\mapsto qd\text{ is a bijection}%
\end{array}
\right) \\
&  =\sum_{d\in D}\left(  -\mu\left(  d\right)  \right)  +\sum_{d\in D}%
\mu\left(  d\right)  =-\sum_{d\in D}\mu\left(  d\right)  +\sum_{d\in D}%
\mu\left(  d\right)  =0\\
&  =\varepsilon\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\varepsilon\left(  n\right)  =0\right)  .
\end{align*}
This proves (\ref{pf.prop.dirichlet.mu.explicit.goal}). Thus, Proposition
\ref{prop.dirichlet.mu.explicit} is proven.
\end{proof}

Our proof of Proposition \ref{prop.dirichlet.mu.explicit} used a standard
technique: In order to prove that a sum is $0$, we split the sum into two
smaller sums, which cancelled each other out term by term (i.e., every term of
one cancelled a term of the other). This kind of proof is widespread in
combinatorics and other disciplines.

\begin{proof}
[Proof of Proposition \ref{prop.dirichlet.mu}.]For every $n\in\mathbb{N}_{+}$,
we have%
\begin{align*}
\left(  \mu\star\underline{1}\right)  \left(  n\right)   &  =\sum_{d\mid n}%
\mu\left(  d\right)  \underbrace{\underline{1}\left(  \dfrac{n}{d}\right)
}_{\substack{=1\\\text{(by the definition of }\underline{1}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mu\star
\underline{1}\right) \\
&  =\sum_{d\mid n}\mu\left(  d\right)  =\varepsilon\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.dirichlet.mu.explicit}}\right)  .
\end{align*}
In other words, $\mu\star\underline{1}=\varepsilon$. This proves Proposition
\ref{prop.dirichlet.mu}.
\end{proof}

The Dirichlet convolutions we have so far computed allow us to compute other
Dirichlet convolutions without actually working with sums, but simply by
applying Theorem \ref{thm.dirichlet.props}. Here is one result we can obtain
in this way:

\begin{proposition}
\label{prop.dirichlet.phi=}We have $\mu\star\iota=\phi$.
\end{proposition}

In concrete language, this says the following:

\begin{proposition}
\label{prop.dirichlet.phi=.explicit}We have
\[
\sum_{d\mid n}\mu\left(  d\right)  \dfrac{n}{d}=\phi\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}_{+}.
\]

\end{proposition}

Instead of proving the concrete version combinatorially and then deriving the
Dirichlet convolution from it, we will go the opposite way this time:

\begin{proof}
[Proof of Proposition \ref{prop.dirichlet.phi=}.]Proposition
\ref{prop.dirichlet.phi} yields $\iota=\phi\star\underline{1}=\underline{1}%
\star\phi$ (by Theorem \ref{thm.dirichlet.props} \textbf{(c)}, applied to
$f=\phi$ and $g=\underline{1}$). Now,%
\begin{align*}
\mu\star\underbrace{\iota}_{=\underline{1}\star\phi}  &  =\mu\star\left(
\underline{1}\star\phi\right)  =\underbrace{\left(  \mu\star\underline{1}%
\right)  }_{\substack{=\varepsilon\\\text{(by Proposition
\ref{prop.dirichlet.mu})}}}\star\phi\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.dirichlet.props}
\textbf{(b)}, applied to }f=\mu\text{, }g=\underline{1}\text{ and }%
h=\phi\right) \\
&  =\varepsilon\star\phi=\phi\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem
\ref{thm.dirichlet.props} \textbf{(a)}, applied to }f=\phi\right)  .
\end{align*}
Thus, Proposition \ref{prop.dirichlet.phi=} is proven.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.dirichlet.phi=.explicit}.]Proposition
\ref{prop.dirichlet.phi=} yields $\phi=\mu\star\iota$. Thus, every
$n\in\mathbb{N}_{+}$ satisfies%
\begin{align*}
\underbrace{\phi}_{=\mu\star\iota}\left(  n\right)   &  =\left(  \mu\star
\iota\right)  \left(  n\right)  =\sum_{d\mid n}\mu\left(  d\right)
\underbrace{\iota\left(  \dfrac{n}{d}\right)  }_{\substack{=\dfrac{n}%
{d}\\\text{(by the definition of }\iota\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\mu\star\iota\right) \\
&  =\sum_{d\mid n}\mu\left(  d\right)  \dfrac{n}{d}.
\end{align*}
This proves Proposition \ref{prop.dirichlet.phi=.explicit}.
\end{proof}

Proposition \ref{prop.dirichlet.phi=.explicit} is \cite[(4.1)]{NiZuMo91}.

\subsection{M\"{o}bius inversion}

A particularly useful consequence of the \textquotedblleft calculus of
Dirichlet convolution\textquotedblright\ we have established is the so-called
\textit{M\"{o}bius inversion formula}:

\begin{theorem}
[M\"{o}bius inversion formula]\label{thm.moebius-inv}Let $f:\mathbb{N}%
_{+}\rightarrow\mathbb{C}$ and $F:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be two
arithmetic functions. Then, we have the following logical equivalence:%
\begin{align*}
&  \ \left(  F\left(  n\right)  =\sum_{d\mid n}f\left(  d\right)  \text{ for
all }n\in\mathbb{N}_{+}\right) \\
&  \Longleftrightarrow\ \left(  f\left(  n\right)  =\sum_{d\mid n}\mu\left(
d\right)  F\left(  \dfrac{n}{d}\right)  \text{ for all }n\in\mathbb{N}%
_{+}\right)  .
\end{align*}

\end{theorem}

Theorem \ref{thm.moebius-inv} is \cite[Theorems 4.8 and 4.9]{NiZuMo91}. It is
merely the most well-known of the many \textquotedblleft M\"{o}bius inversion
formulas\textquotedblright\ that appear in various parts of mathematics; see
\cite{BenGol} or \cite{Rota-F1} or \cite[\S 3.7]{Stanley-EC1} for
introductions into the more general theory of M\"{o}bius functions (of
partially ordered sets).

We shall prove Theorem \ref{thm.moebius-inv} by rewriting it in the following
equivalent form:

\begin{proposition}
\label{prop.moebius-inv.*}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ and
$F:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be two arithmetic functions. Then, we
have the following logical equivalence:%
\[
\left(  F=f\star\underline{1}\right)  \ \Longleftrightarrow\ \left(
f=\mu\star F\right)  .
\]

\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.moebius-inv.*}.]We have $f\star\underline{1}%
=\underline{1}\star f$ (by Theorem \ref{thm.dirichlet.props} \textbf{(c)},
applied to $g=\underline{1}$). Also, $\mu\star\underline{1}=\underline{1}%
\star\mu$ (by Theorem \ref{thm.dirichlet.props} \textbf{(c)}, applied to $\mu$
and $\underline{1}$ instead of $f$ and $g$). But $\mu\star\underline{1}%
=\varepsilon$ (by Proposition \ref{prop.dirichlet.mu}). Hence, $\underline{1}%
\star\mu=\mu\star\underline{1}=\varepsilon$.

We must prove the equivalence $\left(  F=f\star\underline{1}\right)
\ \Longleftrightarrow\ \left(  f=\mu\star F\right)  $. In other words, we must
prove the two implications $\left(  F=f\star\underline{1}\right)
\ \Longrightarrow\ \left(  f=\mu\star F\right)  $ and $\left(  F=f\star
\underline{1}\right)  \ \Longleftarrow\ \left(  f=\mu\star F\right)  $.

\textit{Proof of the implication }$\left(  F=f\star\underline{1}\right)
\ \Longrightarrow\ \left(  f=\mu\star F\right)  $\textit{:} Assume that
$F=f\star\underline{1}$. Then, $F=f\star\underline{1}=\underline{1}\star f$.
Hence,%
\begin{align*}
\mu\star\underbrace{F}_{=\underline{1}\star f}  &  =\mu\star\left(
\underline{1}\star f\right)  =\underbrace{\left(  \mu\star\underline{1}%
\right)  }_{=\varepsilon}\star f\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Theorem \ref{thm.dirichlet.props} \textbf{(b)}, applied to }%
\mu\text{, }\underline{1}\text{ and }f\\
\text{instead of }f\text{, }g\text{ and }h
\end{array}
\right) \\
&  =\varepsilon\star f=f\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem
\ref{thm.dirichlet.props} \textbf{(a)}}\right)  .
\end{align*}
Thus, $f=\mu\star F$. This proves the implication $\left(  F=f\star
\underline{1}\right)  \ \Longrightarrow\ \left(  f=\mu\star F\right)  $.

\textit{Proof of the implication }$\left(  F=f\star\underline{1}\right)
\ \Longleftarrow\ \left(  f=\mu\star F\right)  $\textit{:} Assume that
$f=\mu\star F$. Hence,%
\begin{align*}
f\star\underline{1}  &  =\underline{1}\star\underbrace{f}_{=\mu\star
F}=\underline{1}\star\left(  \mu\star F\right)  =\underbrace{\left(
\underline{1}\star\mu\right)  }_{=\varepsilon}\star F\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Theorem \ref{thm.dirichlet.props} \textbf{(b)}, applied to
}\underline{1}\text{, }\mu\text{ and }F\\
\text{instead of }f\text{, }g\text{ and }h
\end{array}
\right) \\
&  =\varepsilon\star F=F\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem
\ref{thm.dirichlet.props} \textbf{(a)}, applied to }F\text{ instead of
}f\right)  .
\end{align*}
Thus, $F=f\star\underline{1}$. This proves the implication $\left(
F=f\star\underline{1}\right)  \ \Longleftarrow\ \left(  f=\mu\star F\right)  $.

Now, both implications are proven; hence, the proof of Proposition
\ref{prop.moebius-inv.*} is complete.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.moebius-inv}.]We have the following chain of
equivalences:%
\begin{align*}
&  \ \left(  F=f\star\underline{1}\right) \\
&  \Longleftrightarrow\ \left(  F\left(  n\right)  =\left(  f\star
\underline{1}\right)  \left(  n\right)  \text{ for all }n\in\mathbb{N}%
_{+}\right) \\
&  \Longleftrightarrow\ \left(  F\left(  n\right)  =\sum_{d\mid n}f\left(
d\right)  \text{ for all }n\in\mathbb{N}_{+}\right)
\end{align*}
(since every $n\in\mathbb{N}_{+}$ satisfies
\begin{align*}
\left(  f\star\underline{1}\right)  \left(  n\right)   &  =\sum_{d\mid
n}f\left(  d\right)  \underbrace{\underline{1}\left(  \dfrac{n}{d}\right)
}_{\substack{=1\\\text{(by the definition of }\underline{1}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }f\star
\underline{1}\right) \\
&  =\sum_{d\mid n}f\left(  d\right)
\end{align*}
). Hence, we have the following chain of equivalences:%
\begin{align*}
&  \ \left(  F\left(  n\right)  =\sum_{d\mid n}f\left(  d\right)  \text{ for
all }n\in\mathbb{N}_{+}\right) \\
&  \Longleftrightarrow\ \left(  F=f\star\underline{1}\right) \\
&  \Longleftrightarrow\ \left(  f=\mu\star F\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.moebius-inv.*}%
}\right) \\
&  \Longleftrightarrow\ \left(  f\left(  n\right)  =\left(  \mu\star F\right)
\left(  n\right)  \text{ for all }n\in\mathbb{N}_{+}\right) \\
&  \Longleftrightarrow\ \left(  f\left(  n\right)  =\sum_{d\mid n}\mu\left(
d\right)  F\left(  \dfrac{n}{d}\right)  \text{ for all }n\in\mathbb{N}%
_{+}\right)
\end{align*}
(since every $n\in\mathbb{N}_{+}$ satisfies%
\[
\left(  \mu\star F\right)  \left(  n\right)  =\sum_{d\mid n}\mu\left(
d\right)  F\left(  \dfrac{n}{d}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\mu\star F\right)
\]
). This proves Theorem \ref{thm.moebius-inv}.
\end{proof}

\subsection{Dirichlet convolution and multiplicativity}

We will now connect the concept of multiplicative functions with the Dirichlet convolution:

\begin{theorem}
\label{thm.dirichlet.mult}Let $f$ and $g$ be two multiplicative arithmetic
functions. Then, the arithmetic function $f\star g$ is also multiplicative.
\end{theorem}

Theorem \ref{thm.dirichlet.mult} is a generalization of \cite[Theorem
4.4]{NiZuMo91}, and the following proof follows the same ideas as the proof of
\cite[Theorem 4.4]{NiZuMo91}.\footnote{Note that the claim of Theorem
\ref{thm.dirichlet.mult} is also the first part of \cite[\S 8.2, problem
1]{NiZuMo91}.}

\begin{proof}
[Proof of Theorem \ref{thm.dirichlet.mult}.]The function $f$ is
multiplicative. In other words, it satisfies $f\left(  1\right)  =1$, and%
\begin{equation}
f\left(  mn\right)  =f\left(  m\right)  f\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for any two coprime }m\in\mathbb{N}_{+}\text{ and
}n\in\mathbb{N}_{+}. \label{pf.thm.dirichlet.mult.multf}%
\end{equation}


The function $g$ is multiplicative. In other words, it satisfies $g\left(
1\right)  =1$, and%
\begin{equation}
g\left(  mn\right)  =g\left(  m\right)  g\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for any two coprime }m\in\mathbb{N}_{+}\text{ and
}n\in\mathbb{N}_{+}. \label{pf.thm.dirichlet.mult.multg}%
\end{equation}


The definition of $f\star g$ yields
\[
\left(  f\star g\right)  \left(  1\right)  =\sum_{d\mid1}f\left(  d\right)
g\left(  \dfrac{1}{d}\right)  =\underbrace{f\left(  1\right)  }_{=1}%
\underbrace{g\left(  \dfrac{1}{1}\right)  }_{=g\left(  1\right)  =1}=1.
\]


Now, we want to prove that $f\star g$ is multiplicative. In order to do so, we
need to verify that $\left(  f\star g\right)  \left(  1\right)  =1$ and that
\begin{equation}
\left(  f\star g\right)  \left(  mn\right)  =\left(  f\star g\right)  \left(
m\right)  \cdot\left(  f\star g\right)  \left(  n\right)
\label{pf.thm.dirichlet.mult.goal}%
\end{equation}
for any two coprime $m\in\mathbb{N}_{+}$ and $n\in\mathbb{N}_{+}$. Since
$\left(  f\star g\right)  \left(  1\right)  =1$ is already proven, it thus
only remains to prove (\ref{pf.thm.dirichlet.mult.goal}).

So let $m\in\mathbb{N}_{+}$ and $n\in\mathbb{N}_{+}$ be coprime. We need to
prove (\ref{pf.thm.dirichlet.mult.goal}).

For any $N\in\mathbb{N}_{+}$, let $\mathcal{D}\left(  N\right)  $ be the set
of all positive divisors of $N$.

Consider the map%
\[
\mathbf{f}:\mathcal{D}\left(  m\right)  \times\mathcal{D}\left(  n\right)
\rightarrow\mathcal{D}\left(  mn\right)  ,\ \ \ \ \ \ \ \ \ \ \left(
d,e\right)  \mapsto de.
\]
This map $\mathbf{f}$ is well-defined (because if $d$ and $e$ are positive
divisors of $m$ and $n$, respectively, then $de$ is a positive divisor of $mn$).

Consider the map%
\[
\mathbf{g}:\mathcal{D}\left(  mn\right)  \rightarrow\mathcal{D}\left(
m\right)  \times\mathcal{D}\left(  n\right)  ,\ \ \ \ \ \ \ \ \ \ u\mapsto
\left(  \gcd\left(  u,m\right)  ,\gcd\left(  u,n\right)  \right)  .
\]
This map $\mathbf{g}$ is well-defined (because if $u$ is a positive divisor of
$mn$, then $\gcd\left(  u,m\right)  $ and $\gcd\left(  u,n\right)  $ are
positive divisors of $m$ and $n$, respectively).

We have $\mathbf{f}\circ\mathbf{g}=\operatorname*{id}$%
\ \ \ \ \footnote{\textit{Proof.} Let $u\in\mathcal{D}\left(  mn\right)  $.
Thus, $u$ is a positive divisor of $mn$. Therefore, $\gcd\left(  u,mn\right)
=u$.
\par
The definition of $\mathbf{g}$ shows that $\mathbf{g}\left(  u\right)
=\left(  \gcd\left(  u,m\right)  ,\gcd\left(  u,n\right)  \right)  $. Now,%
\begin{align*}
\left(  \mathbf{f}\circ\mathbf{g}\right)  \left(  u\right)   &  =\mathbf{f}%
\left(  \underbrace{\mathbf{g}\left(  u\right)  }_{=\left(  \gcd\left(
u,m\right)  ,\gcd\left(  u,n\right)  \right)  }\right)  =\mathbf{f}\left(
\gcd\left(  u,m\right)  ,\gcd\left(  u,n\right)  \right) \\
&  =\gcd\left(  u,m\right)  \cdot\gcd\left(  u,n\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mathbf{f}\right) \\
&  =\gcd\left(  u,mn\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.gcd.gcd(u,mn)}}\right) \\
&  =u.
\end{align*}
\par
Now, forget that we fixed $u$. We thus have proven that $\left(
\mathbf{f}\circ\mathbf{g}\right)  \left(  u\right)  =u$ for every
$u\in\mathcal{D}\left(  mn\right)  $. In other words, $\mathbf{f}%
\circ\mathbf{g}=\operatorname*{id}$.} and $\mathbf{g}\circ\mathbf{f}%
=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  d,e\right)
\in\mathcal{D}\left(  m\right)  \times\mathcal{D}\left(  n\right)  $. Then,
$\mathbf{f}\left(  d,e\right)  =de$ (by the definition of $\mathbf{f}$), and%
\[
\left(  \mathbf{g}\circ\mathbf{f}\right)  \left(  d,e\right)  =\mathbf{g}%
\left(  \underbrace{\mathbf{f}\left(  d,e\right)  }_{=de}\right)
=\mathbf{g}\left(  de\right)  =\left(  \gcd\left(  de,m\right)  ,\gcd\left(
de,n\right)  \right)
\]
(by the definition of $\mathbf{g}$).
\par
We have $\left(  d,e\right)  \in\mathcal{D}\left(  m\right)  \times
\mathcal{D}\left(  n\right)  $. In other words, $d\in\mathcal{D}\left(
m\right)  $ and $e\in\mathcal{D}\left(  n\right)  $. In other words, $d$ is a
positive divisor of $m$, and $e$ is a positive divisor of $n$. Thus, $d\mid m$
and $e\mid n$.
\par
Let $d^{\prime}=\gcd\left(  de,m\right)  $. Then, $d^{\prime}=\gcd\left(
de,m\right)  \mid m$ and $e\mid n$. Hence, Corollary \ref{cor.gcd.increase}
(applied to $d^{\prime}$, $e$, $m$ and $n$ instead of $a$, $b$, $c$ and $d$)
yields $\gcd\left(  d^{\prime},e\right)  \mid\gcd\left(  m,n\right)  =1$
(since $m$ and $n$ are coprime). Hence, $\gcd\left(  d^{\prime},e\right)  =1$.
\par
Note also that $d^{\prime}=\gcd\left(  de,m\right)  \mid de=ed$. Thus,
Proposition \ref{prop.prod-coprime.cancel} (applied to $x=d^{\prime}$, $y=e$
and $z=d$) yields $d^{\prime}\mid d$ (since $\gcd\left(  d^{\prime},e\right)
=1$).
\par
On the other hand, $d\mid de$ and $d\mid m$. Thus, Proposition
\ref{prop.gcd.uniprop} (applied to $d$, $de$ and $m$ instead of $a$, $b$ and
$c$) yields $d\mid\gcd\left(  de,m\right)  $. In other words, $d\mid
d^{\prime}$ (since $d^{\prime}=\gcd\left(  de,m\right)  $).
\par
Now, the integers $d$ and $d^{\prime}$ are positive and thus nonnegative.
Hence, Proposition \ref{prop.div.=} (applied to $u=d$ and $v=d^{\prime}$)
yields $d=d^{\prime}$ (since $d\mid d^{\prime}$ and $d^{\prime}\mid d$). Thus,
$d=d^{\prime}=\gcd\left(  de,m\right)  $. In other words, $\gcd\left(
de,m\right)  =d$. The same argument (with the roles of $m$ and $n$
interchanged, and correspondingly also the roles of $d$ and $e$ interchanged)
shows that $\gcd\left(  ed,n\right)  =e$. Now,%
\[
\left(  \mathbf{g}\circ\mathbf{f}\right)  \left(  d,e\right)  =\left(
\underbrace{\gcd\left(  de,m\right)  }_{=d},\underbrace{\gcd\left(
de,n\right)  }_{=\gcd\left(  ed,n\right)  =e}\right)  =\left(  d,e\right)  .
\]
\par
Now, forget that we fixed $\left(  d,e\right)  $. We thus have shown that
$\left(  \mathbf{g}\circ\mathbf{f}\right)  \left(  d,e\right)  =\left(
d,e\right)  $ for each $\left(  d,e\right)  \in\mathcal{D}\left(  m\right)
\times\mathcal{D}\left(  n\right)  $. In other words, $\mathbf{g}%
\circ\mathbf{f}=\operatorname*{id}$.}. Hence, the maps $\mathbf{f}$ and
$\mathbf{g}$ are mutually inverse. In particular, this shows that the map
$\mathbf{f}$ is a bijection. In other words, the map $\mathcal{D}\left(
m\right)  \times\mathcal{D}\left(  n\right)  \rightarrow\mathcal{D}\left(
mn\right)  ,\ \left(  d,e\right)  \mapsto de$ is a bijection (since this map
is precisely $\mathbf{f}$).

We make two more simple observations:

\begin{enumerate}
\item We have
\begin{equation}
f\left(  de\right)  =f\left(  d\right)  f\left(  e\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }d\in\mathcal{D}\left(  m\right)  \text{ and
}e\in\mathcal{D}\left(  n\right)  . \label{pf.thm.dirichlet.mult.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.dirichlet.mult.4}):} Let
$d\in\mathcal{D}\left(  m\right)  $ and $e\in\mathcal{D}\left(  n\right)  $.
In other words, $d$ is a positive divisor of $m$, and $e$ is a positive
divisor of $n$. Hence, $d\mid m$ and $e\mid n$. Therefore, Corollary
\ref{cor.gcd.increase} (applied to $d$, $e$, $m$ and $n$ instead of $a$, $b$,
$c$ and $d$) yields $\gcd\left(  d,e\right)  \mid\gcd\left(  m,n\right)  =1$
(since $m$ and $n$ are coprime). Hence, $\gcd\left(  d,e\right)  =1$. In other
words, $d$ and $e$ are coprime. Hence, (\ref{pf.thm.dirichlet.mult.multf})
(applied to $d$ and $e$ instead of $m$ and $n$) yields $f\left(  de\right)
=f\left(  d\right)  f\left(  e\right)  $, qed.}

\item We have%
\begin{equation}
g\left(  \dfrac{mn}{de}\right)  =g\left(  \dfrac{m}{d}\right)  g\left(
\dfrac{n}{e}\right)  \ \ \ \ \ \ \ \ \ \ \text{for any }d\in\mathcal{D}\left(
m\right)  \text{ and }e\in\mathcal{D}\left(  n\right)  .
\label{pf.thm.dirichlet.mult.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.dirichlet.mult.5}):} Let
$d\in\mathcal{D}\left(  m\right)  $ and $e\in\mathcal{D}\left(  n\right)  $.
In other words, $d$ is a positive divisor of $m$, and $e$ is a positive
divisor of $n$. Hence, $d\mid m$ and $e\mid n$. This shows that $\dfrac{m}{d}$
and $\dfrac{n}{e}$ are integers. Furthermore, the numbers $\dfrac{m}{d}$ and
$\dfrac{n}{e}$ are positive (since $m$, $d$, $n$ and $e$ are positive). Hence,
$\dfrac{m}{d}$ and $\dfrac{n}{e}$ are positive integers.
\par
Moreover, $\dfrac{m}{d}\mid m$ and $\dfrac{n}{e}\mid n$. Hence, Corollary
\ref{cor.gcd.increase} (applied to $\dfrac{m}{d}$, $\dfrac{n}{e}$, $m$ and $n$
instead of $a$, $b$, $c$ and $d$) yields $\gcd\left(  \dfrac{m}{d},\dfrac
{n}{e}\right)  \mid\gcd\left(  m,n\right)  =1$ (since $m$ and $n$ are
coprime). Hence, $\gcd\left(  \dfrac{m}{d},\dfrac{n}{e}\right)  =1$. In other
words, $\dfrac{m}{d}$ and $\dfrac{n}{e}$ are coprime. Hence,
(\ref{pf.thm.dirichlet.mult.multg}) (applied to $\dfrac{m}{d}$ and $\dfrac
{n}{e}$ instead of $m$ and $n$) yields $g\left(  \dfrac{m}{d}\cdot\dfrac{n}%
{e}\right)  =g\left(  \dfrac{m}{d}\right)  g\left(  \dfrac{n}{e}\right)  $.
Hence, $g\left(  \underbrace{\dfrac{mn}{de}}_{=\dfrac{m}{d}\cdot\dfrac{n}{e}%
}\right)  =g\left(  \dfrac{m}{d}\cdot\dfrac{n}{e}\right)  =g\left(  \dfrac
{m}{d}\right)  g\left(  \dfrac{n}{e}\right)  $, qed.}
\end{enumerate}

Now, the definition of $f\star g$ yields%
\begin{align*}
&  \left(  f\star g\right)  \left(  mn\right)  \\
&  =\sum_{d\mid mn}f\left(  d\right)  g\left(  \dfrac{mn}{d}\right)
=\underbrace{\sum_{u\mid mn}}_{=\sum_{u\in\mathcal{D}\left(  mn\right)  }%
}f\left(  u\right)  g\left(  \dfrac{mn}{u}\right)  \ \ \ \ \ \ \ \ \ \ \left(
%
\begin{array}
[c]{c}%
\text{here, we have renamed the}\\
\text{summation index }d\text{ as }u
\end{array}
\right)  \\
&  =\sum_{u\in\mathcal{D}\left(  mn\right)  }f\left(  u\right)  g\left(
\dfrac{mn}{u}\right)  =\underbrace{\sum_{\left(  d,e\right)  \in
\mathcal{D}\left(  m\right)  \times\mathcal{D}\left(  n\right)  }}%
_{=\sum_{d\in\mathcal{D}\left(  m\right)  }\ \ \sum_{e\in\mathcal{D}\left(
n\right)  }}f\left(  de\right)  g\left(  \dfrac{mn}{de}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }de\text{ for }u\text{ in the sum, since
the}\\
\text{map }\mathcal{D}\left(  m\right)  \times\mathcal{D}\left(  n\right)
\rightarrow\mathcal{D}\left(  mn\right)  ,\ \left(  d,e\right)  \mapsto
de\text{ is a bijection}%
\end{array}
\right)  \\
&  =\underbrace{\sum_{d\in\mathcal{D}\left(  m\right)  }}_{=\sum_{d\mid m}%
}\ \ \underbrace{\sum_{e\in\mathcal{D}\left(  n\right)  }}_{=\sum_{e\mid n}%
}\underbrace{f\left(  de\right)  }_{\substack{=f\left(  d\right)  f\left(
e\right)  \\\text{(by (\ref{pf.thm.dirichlet.mult.4}))}}}\underbrace{g\left(
\dfrac{mn}{de}\right)  }_{\substack{=g\left(  \dfrac{m}{d}\right)  g\left(
\dfrac{n}{e}\right)  \\\text{(by (\ref{pf.thm.dirichlet.mult.5}))}}}\\
&  =\sum_{d\mid m}\ \ \sum_{e\mid n}f\left(  d\right)  f\left(  e\right)
g\left(  \dfrac{m}{d}\right)  g\left(  \dfrac{n}{e}\right)  =\left(
\sum_{d\mid m}f\left(  d\right)  g\left(  \dfrac{m}{d}\right)  \right)
\left(  \sum_{e\mid n}f\left(  e\right)  g\left(  \dfrac{n}{e}\right)
\right)  \\
&  =\left(  \sum_{d\mid m}f\left(  d\right)  g\left(  \dfrac{m}{d}\right)
\right)  \left(  \sum_{d\mid n}f\left(  d\right)  g\left(  \dfrac{n}%
{d}\right)  \right)
\end{align*}
(here, we renamed the summation index $e$ as $d$ in the second sum). Comparing
this with%
\begin{align*}
&  \underbrace{\left(  f\star g\right)  \left(  m\right)  }_{\substack{=\sum
_{d\mid m}f\left(  d\right)  g\left(  \dfrac{m}{d}\right)  \\\text{(by the
definition of }f\star g\text{)}}}\cdot\underbrace{\left(  f\star g\right)
\left(  n\right)  }_{\substack{=\sum_{d\mid n}f\left(  d\right)  g\left(
\dfrac{n}{d}\right)  \\\text{(by the definition of }f\star g\text{)}}}\\
&  =\left(  \sum_{d\mid m}f\left(  d\right)  g\left(  \dfrac{m}{d}\right)
\right)  \left(  \sum_{d\mid n}f\left(  d\right)  g\left(  \dfrac{n}%
{d}\right)  \right)  ,
\end{align*}
we obtain $\left(  f\star g\right)  \left(  mn\right)  =\left(  f\star
g\right)  \left(  m\right)  \cdot\left(  f\star g\right)  \left(  n\right)  $.
Thus, (\ref{pf.thm.dirichlet.mult.goal}) is proven. As we have said, this
completes the proof of Theorem \ref{thm.dirichlet.mult}.
\end{proof}

Notice that Theorem \ref{thm.dirichlet.mult} has no analogue for totally
multiplicative functions: The Dirichlet convolution $f\star g$ of two totally
multiplicative functions might not be totally multiplicative.

We can use Theorem \ref{thm.dirichlet.mult} to prove (and sometimes reprove)
parts of Proposition \ref{prop.arifun.mult.exa}:

\begin{proof}
[Proof of Proposition \ref{prop.arifun.mult.exa} \textbf{(b)}.]The arithmetic
function $\underline{1}$ is clearly multiplicative (and totally
multiplicative). Thus, Theorem \ref{thm.dirichlet.mult} (applied to
$f=\underline{1}$ and $g=\underline{1}$) shows that $\underline{1}%
\star\underline{1}$ is multiplicative. But since $\underline{1}\star
\underline{1}=\operatorname*{d}$ (by Proposition \ref{prop.dirichlet.1*1}),
this shows that $\operatorname*{d}$ is multiplicative. This proves Proposition
\ref{prop.arifun.mult.exa} \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.arifun.mult.exa} \textbf{(e)}.]Let
$k\in\mathbb{Z}$. Define the arithmetic function $\iota_{k}$ as in Proposition
\ref{prop.dirichlet.sig} \textbf{(b)}. This $\iota_{k}$ is clearly
multiplicative (and totally multiplicative). We also know that the arithmetic
function $\underline{1}$ is clearly multiplicative. Thus, Theorem
\ref{thm.dirichlet.mult} (applied to $f=\iota_{k}$ and $g=\underline{1}$)
shows that $\iota_{k}\star\underline{1}$ is multiplicative. But since
$\iota_{k}\star\underline{1}=\sigma_{k}$ (by Proposition
\ref{prop.dirichlet.sig} \textbf{(b)}), this shows that $\sigma_{k}$ is
multiplicative. Applying this to $k=1$, we conclude that $\sigma$ is
multiplicative (since $\sigma_{1}=\sigma$). This completes the proof of
Proposition \ref{prop.arifun.mult.exa} \textbf{(e)}.
\end{proof}

\begin{proof}
[Second proof of Proposition \ref{prop.arifun.mult.exa} \textbf{(a)}.]The
arithmetic function $\mu$ is multiplicative (by Proposition
\ref{prop.arifun.mult.exa} \textbf{(f)}). The arithmetic function $\iota$ is
clearly multiplicative (and totally multiplicative). Hence, Theorem
\ref{thm.dirichlet.mult} (applied to $f=\mu$ and $g=\iota$) shows that
$\mu\star\iota$ is multiplicative. But since $\mu\star\iota=\phi$ (by
Proposition \ref{prop.dirichlet.phi=}), this shows that $\phi$ is
multiplicative. This proves Proposition \ref{prop.arifun.mult.exa}
\textbf{(a)} again.
\end{proof}

As an easy consequence of Theorem \ref{thm.dirichlet.mult}, we can obtain
\cite[Theorem 4.4]{NiZuMo91}:

\begin{corollary}
\label{cor.dirichlet.mult.1}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be a
multiplicative arithmetic function. Define an arithmetic function
$F:\mathbb{N}_{+}\rightarrow\mathbb{C}$ by
\begin{equation}
F\left(  n\right)  =\sum_{d\mid n}f\left(  d\right)
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }n.
\label{eq.cor.dirichlet.mult.1.defF}%
\end{equation}
Then, the function $F$ is multiplicative.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.dirichlet.mult.1}.]The arithmetic function
$\underline{1}$ is clearly multiplicative (and totally multiplicative). Thus,
Theorem \ref{thm.dirichlet.mult} (applied to $g=\underline{1}$) shows that
$f\star\underline{1}$ is multiplicative. But every positive integer $n$
satisfies%
\begin{align*}
\left(  f\star\underline{1}\right)  \left(  n\right)   &  =\sum_{d\mid
n}f\left(  d\right)  \underbrace{\underline{1}\left(  \dfrac{n}{d}\right)
}_{\substack{=1\\\text{(by the definition of }\underline{1}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }f\star
\underline{1}\right) \\
&  =\sum_{d\mid n}f\left(  d\right)  =F\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.cor.dirichlet.mult.1.defF}%
)}\right)  .
\end{align*}
Hence, $f\star\underline{1}=F$. But recall that $f\star\underline{1}$ is
multiplicative. In other words, $F$ is multiplicative (since $f\star
\underline{1}=F$). This proves Corollary \ref{cor.dirichlet.mult.1}.
\end{proof}

\subsection{Explicit formulas from multiplicativity}

One of the nice things about multiplicative arithmetic functions is that, in
order to compute their values, it suffices to compute their values on prime powers:

\begin{proposition}
\label{prop.arifun.mult.ppows}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be a
multiplicative function. Let $n\in\mathbb{N}_{+}$. Then,%
\[
f\left(  n\right)  =\prod_{p\in\operatorname*{PF}n}f\left(  p^{v_{p}\left(
n\right)  }\right)  .
\]
(See Definition \ref{def.PF} for the definition of $\operatorname*{PF}n$.)
\end{proposition}

Applying this proposition to $f=\phi$, $f=\operatorname*{d}$ and $f=\sigma
_{k}$, we easily obtain Theorem \ref{thm.phi.explicit}, Theorem
\ref{thm.d.explicit} and Theorem \ref{thm.sigmak.explicit}, respectively (once
we compute the values $f\left(  p^{v_{p}\left(  n\right)  }\right)  $, but
this is easy in all three cases).

Proposition \ref{prop.arifun.mult.ppows} follows from the following fact:

\begin{proposition}
\label{prop.arifun.mult.prod}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be a
multiplicative function. Let $a_{1},a_{2},\ldots,a_{k}$ be finitely many
pairwise coprime\footnotemark\ positive integers. Then,%
\[
f\left(  a_{1}a_{2}\cdots a_{k}\right)  =f\left(  a_{1}\right)  f\left(
a_{2}\right)  \cdots f\left(  a_{k}\right)  .
\]

\end{proposition}

\footnotetext{We say that $k$ integers $a_{1},a_{2},\ldots,a_{k}$ are
\textit{pairwise coprime} if they have the property that $a_{u}$ is coprime to
$a_{v}$ for any integers $u$ and $v$ satisfying $1\leq u<v\leq k$.}\ Both the
proof of Proposition \ref{prop.arifun.mult.prod} (by induction over $k$) and
the proof of Proposition \ref{prop.arifun.mult.ppows} (using Proposition
\ref{prop.arifun.mult.prod}) are rather straightforward:

\begin{proof}
[Proof of Proposition \ref{prop.arifun.mult.prod}.]The function $f$ is
multiplicative. In other words, it satisfies $f\left(  1\right)  =1$ and%
\begin{equation}
f\left(  mn\right)  =f\left(  m\right)  f\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for any two coprime }m\in\mathbb{N}_{+}\text{ and
}n\in\mathbb{N}_{+} \label{pf.prop.arifun.mult.prod.mult}%
\end{equation}
(by the definition of \textquotedblleft multiplicative\textquotedblright).

The integers $a_{1},a_{2},\ldots,a_{k}$ are pairwise coprime. In other words,
\begin{equation}
a_{u}\text{ is coprime to }a_{v} \label{pf.prop.arifun.mult.prod.copri}%
\end{equation}
for any integers $u$ and $v$ satisfying $1\leq u<v\leq k$.

We shall show that%
\begin{equation}
f\left(  a_{1}a_{2}\cdots a_{i}\right)  =f\left(  a_{1}\right)  f\left(
a_{2}\right)  \cdots f\left(  a_{i}\right)
\label{pf.prop.arifun.mult.prod.goal}%
\end{equation}
for every $i\in\left\{  0,1,\ldots,k\right\}  $.

\textit{Proof of (\ref{pf.prop.arifun.mult.prod.goal}):} We shall prove
(\ref{pf.prop.arifun.mult.prod.goal}) by induction over $i$:

\textit{Induction base:} We have $a_{1}a_{2}\cdots a_{0}=\left(  \text{empty
product}\right)  =1$. Applying the map $f$ to both sides of this equation, we
obtain%
\[
f\left(  a_{1}a_{2}\cdots a_{0}\right)  =f\left(  1\right)  =1.
\]
Comparing this with $f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots
f\left(  a_{0}\right)  =\left(  \text{empty product}\right)  =1$, we obtain
$f\left(  a_{1}a_{2}\cdots a_{0}\right)  =f\left(  a_{1}\right)  f\left(
a_{2}\right)  \cdots f\left(  a_{0}\right)  $. In other words,
(\ref{pf.prop.arifun.mult.prod.goal}) holds for $i=0$. This completes the
induction base.

\textit{Induction step:} Let $j\in\left\{  0,1,\ldots,k\right\}  $ be
positive. Assume that (\ref{pf.prop.arifun.mult.prod.goal}) holds for $i=j-1$.
We must prove that (\ref{pf.prop.arifun.mult.prod.goal}) holds for $i=j$.

We have assumed that (\ref{pf.prop.arifun.mult.prod.goal}) holds for $i=j-1$.
In other words, we have%
\[
f\left(  a_{1}a_{2}\cdots a_{j-1}\right)  =f\left(  a_{1}\right)  f\left(
a_{2}\right)  \cdots f\left(  a_{j-1}\right)  .
\]


But $a_{u}$ is coprime to $a_{j}$ for every $u\in\left\{  1,2,\ldots
,j-1\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $u\in\left\{
1,2,\ldots,j-1\right\}  $. Thus, $u$ is an integer satisfying $1\leq u\leq
j-1$. Now, $1\leq u\leq j-1<j\leq k$ (since $j\in\left\{  0,1,\ldots
,k\right\}  $). Therefore, (\ref{pf.prop.arifun.mult.prod.copri}) (applied to
$v=j$) shows that $a_{u}$ is coprime to $a_{j}$. Qed.}. Hence, Corollary
\ref{cor.prod-coprime.n} (applied to $n=j-1$, $c_{u}=a_{u}$ and $m=a_{j}$)
shows that $a_{1}a_{2}\cdots a_{j-1}$ is coprime to $a_{j}$. Therefore,
(\ref{pf.prop.arifun.mult.prod.mult}) (applied to $m=a_{1}a_{2}\cdots a_{j-1}$
and $n=a_{j}$) yields%
\begin{align*}
f\left(  \left(  a_{1}a_{2}\cdots a_{j-1}\right)  a_{j}\right)   &
=\underbrace{f\left(  a_{1}a_{2}\cdots a_{j-1}\right)  }_{=f\left(
a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(  a_{j-1}\right)  }f\left(
a_{j}\right) \\
&  =\left(  f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{j-1}\right)  \right)  f\left(  a_{j}\right)  =f\left(  a_{1}\right)
f\left(  a_{2}\right)  \cdots f\left(  a_{j}\right)  .
\end{align*}
Comparing this with $f\left(  \left(  a_{1}a_{2}\cdots a_{j-1}\right)
a_{j}\right)  =f\left(  a_{1}a_{2}\cdots a_{j}\right)  $, we obtain $f\left(
a_{1}a_{2}\cdots a_{j}\right)  =f\left(  a_{1}\right)  f\left(  a_{2}\right)
\cdots f\left(  a_{j}\right)  $. In other words,
(\ref{pf.prop.arifun.mult.prod.goal}) holds for $i=j$. Thus, the induction
step is complete, and so (\ref{pf.prop.arifun.mult.prod.goal}) is proven.

Now, we can apply (\ref{pf.prop.arifun.mult.prod.goal}) to $i=k$. As a result,
we obtain $f\left(  a_{1}a_{2}\cdots a_{k}\right)  =f\left(  a_{1}\right)
f\left(  a_{2}\right)  \cdots f\left(  a_{k}\right)  $. This proves
Proposition \ref{prop.arifun.mult.prod}.
\end{proof}

Let us restate Proposition \ref{prop.arifun.mult.prod} in a more convenient
form before we come to the proof of Proposition \ref{prop.arifun.mult.ppows}:

\begin{corollary}
\label{cor.arifun.mult.prodS}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be a
multiplicative function. Let $S$ be a finite set. Let $m_{s}$ be a positive
integer for each $s\in S$. Assume that the integers $m_{s}$ and $m_{t}$ are
coprime whenever $s$ and $t$ are two distinct elements of $S$. Then,%
\[
f\left(  \prod_{s\in S}m_{s}\right)  =\prod_{s\in S}f\left(  m_{s}\right)  .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.arifun.mult.prodS}.]Let $\left(  s_{1}%
,s_{2},\ldots,s_{k}\right)  $ be a list of all elements of $S$ (with each
element appearing exactly once in the list). Then, $\prod_{s\in S}%
m_{s}=m_{s_{1}}m_{s_{2}}\cdots m_{s_{k}}$ and $\prod_{s\in S}f\left(
m_{s}\right)  =f\left(  m_{s_{1}}\right)  f\left(  m_{s_{2}}\right)  \cdots
f\left(  m_{s_{k}}\right)  $.

Also, if $i$ and $j$ are two distinct elements of $\left\{  1,2,\ldots
,k\right\}  $, then the integers $m_{s_{i}}$ and $m_{s_{j}}$ are
coprime\footnote{\textit{Proof.} Let $i$ and $j$ be two distinct elements of
$\left\{  1,2,\ldots,k\right\}  $.
\par
The list $\left(  s_{1},s_{2},\ldots,s_{k}\right)  $ contains no element more
than once (because of its definition). In other words, the elements
$s_{1},s_{2},\ldots,s_{k}$ are pairwise distinct. Hence, $s_{i}\neq s_{j}$
(since $i\neq j$). In other words, the elements $s_{i}$ and $s_{j}$ are
distinct.
\par
But the integers $m_{s}$ and $m_{t}$ are coprime whenever $s$ and $t$ are two
distinct elements of $S$. Applying this to $s=s_{i}$ and $t=s_{j}$, we
conclude that the integers $m_{s_{i}}$ and $m_{s_{j}}$ are coprime (since
$s_{i}$ and $s_{j}$ are distinct). Qed.}. In other words, $m_{s_{1}},m_{s_{2}%
},\ldots,m_{s_{k}}$ are pairwise coprime integers. Hence, Proposition
\ref{prop.arifun.mult.prod} (applied to $a_{i}=m_{s_{i}}$) yields%
\[
f\left(  m_{s_{1}}m_{s_{2}}\cdots m_{s_{k}}\right)  =f\left(  m_{s_{1}%
}\right)  f\left(  m_{s_{2}}\right)  \cdots f\left(  m_{s_{k}}\right)  .
\]
Thus,%
\[
f\left(  \underbrace{\prod_{s\in S}m_{s}}_{=m_{s_{1}}m_{s_{2}}\cdots m_{s_{k}%
}}\right)  =f\left(  m_{s_{1}}m_{s_{2}}\cdots m_{s_{k}}\right)  =f\left(
m_{s_{1}}\right)  f\left(  m_{s_{2}}\right)  \cdots f\left(  m_{s_{k}}\right)
=\prod_{s\in S}f\left(  m_{s}\right)  .
\]
This proves Corollary \ref{cor.arifun.mult.prodS}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.arifun.mult.ppows}.]The prime factorization of
$n$ is $n=\prod_{p\in\operatorname*{PF}n}p^{v_{p}\left(  n\right)  }%
=\prod_{s\in\operatorname*{PF}n}s^{v_{s}\left(  n\right)  }$ (here, we renamed
the index $p$ as $s$ in the product). Clearly, $s^{v_{s}\left(  n\right)  }$
is a positive integer for each $s\in\operatorname*{PF}n$. Furthermore, the
integers $s^{v_{s}\left(  n\right)  }$ and $t^{v_{t}\left(  n\right)  }$ are
coprime whenever $s$ and $t$ are two distinct elements of $\operatorname*{PF}%
n$\ \ \ \ \footnote{\textit{Proof.} Let $s$ and $t$ be two distinct elements
of $\operatorname*{PF}n$. We must prove that the integers $s^{v_{s}\left(
n\right)  }$ and $t^{v_{t}\left(  n\right)  }$ are coprime.
\par
All elements of $\operatorname*{PF}n$ are primes. Hence, $s$ is a prime (since
$s$ is an element of $\operatorname*{PF}n$). Similarly, $t$ is a prime. Hence,
$s$ and $t$ are two distinct primes (since $s$ and $t$ are distinct). Thus,
Corollary \ref{cor.prime-pow-coprime} \textbf{(b)} (applied to $v_{s}\left(
n\right)  $ and $v_{t}\left(  n\right)  $ instead of $n$ and $m$) yields that
the integers $s^{v_{s}\left(  n\right)  }$ and $t^{v_{t}\left(  n\right)  }$
are coprime. Qed.}. Hence, Corollary \ref{cor.arifun.mult.prodS} (applied to
$S=\operatorname*{PF}n$ and $m_{s}=s^{v_{s}\left(  n\right)  }$) yields%
\[
f\left(  \prod_{s\in\operatorname*{PF}n}s^{v_{s}\left(  n\right)  }\right)
=\prod_{s\in\operatorname*{PF}n}f\left(  s^{v_{s}\left(  n\right)  }\right)
=\prod_{p\in\operatorname*{PF}n}f\left(  p^{v_{p}\left(  n\right)  }\right)
\]
(here, we renamed the index $s$ as $p$ in the product). Thus,%
\[
f\left(  \underbrace{n}_{=\prod_{s\in\operatorname*{PF}n}s^{v_{s}\left(
n\right)  }}\right)  =f\left(  \prod_{s\in\operatorname*{PF}n}s^{v_{s}\left(
n\right)  }\right)  =\prod_{p\in\operatorname*{PF}n}f\left(  p^{v_{p}\left(
n\right)  }\right)  .
\]
This proves Proposition \ref{prop.arifun.mult.ppows}.
\end{proof}

\subsection{Pointwise products}

Let us briefly discuss a much simpler way to \textquotedblleft
multiply\textquotedblright\ two arithmetic functions than Dirichlet
convolution: the \textit{pointwise product}.

\begin{definition}
\label{def.pwprod}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ and
$g:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be two arithmetic functions. We define
a new arithmetic function $f\cdot g:\mathbb{N}_{+}\rightarrow\mathbb{C}$ by%
\[
\left(  f\cdot g\right)  \left(  n\right)  =f\left(  n\right)  g\left(
n\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}_{+}.
\]
This new function $f\cdot g$ is called the \textit{pointwise product }of $f$
and $g$.
\end{definition}

We have the following (much simpler) analogue of Theorem
\ref{thm.dirichlet.props}:

\begin{theorem}
\label{thm.pwprod.props}\textbf{(a)} We have $\underline{1}\cdot
f=f\cdot\underline{1}=f$ for every arithmetic function $f$.

\textbf{(b)} We have $f\cdot\left(  g\cdot h\right)  =\left(  f\cdot g\right)
\cdot h$ for every three arithmetic functions $f$, $g$ and $h$.

\textbf{(c)} We have $f\cdot g=g\cdot f$ for every two arithmetic functions
$f$ and $g$.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.pwprod.props}.]\textbf{(c)} Let $f$ and $g$ be two
arithmetic functions. Let $n\in\mathbb{N}_{+}$. The definition of $f\cdot g$
yields $\left(  f\cdot g\right)  \left(  n\right)  =f\left(  n\right)
g\left(  n\right)  =g\left(  n\right)  f\left(  n\right)  $. But the
definition of $g\cdot f$ yields $\left(  g\cdot f\right)  \left(  n\right)
=g\left(  n\right)  f\left(  n\right)  $. Comparing these two equalities, we
obtain $\left(  f\cdot g\right)  \left(  n\right)  =\left(  g\cdot f\right)
\left(  n\right)  $.

Now, forget that we fixed $n$. We thus have shown that $\left(  f\cdot
g\right)  \left(  n\right)  =\left(  g\cdot f\right)  \left(  n\right)  $ for
each $n\in\mathbb{N}_{+}$. In other words, $f\cdot g=g\cdot f$. This proves
Theorem \ref{thm.pwprod.props} \textbf{(c)}.

\textbf{(a)} Let $f$ be an arithmetic function. Every $n\in\mathbb{N}_{+}$
satisfies%
\begin{align*}
\left(  \underline{1}\cdot f\right)  \left(  n\right)   &
=\underbrace{\underline{1}\left(  n\right)  }_{\substack{=1\\\text{(by the
definition}\\\text{of }\underline{1}\text{)}}}\cdot f\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\underline{1}\cdot
f\right) \\
&  =f\left(  n\right)
\end{align*}
In other words, $\underline{1}\cdot f=f$. But Theorem \ref{thm.pwprod.props}
\textbf{(c)} (applied to $g=\underline{1}$) yields $f\cdot\underline{1}%
=\underline{1}\cdot f$. Thus, $f\cdot\underline{1}=\underline{1}\cdot f=f$.
This proves Theorem \ref{thm.pwprod.props} \textbf{(a)}.

\textbf{(b)} Let $f$, $g$ and $h$ be three arithmetic functions. Let
$n\in\mathbb{N}_{+}$. The definition of $\left(  f\cdot g\right)  \cdot h$
yields%
\[
\left(  \left(  f\cdot g\right)  \cdot h\right)  \left(  n\right)
=\underbrace{\left(  f\cdot g\right)  \left(  n\right)  }_{\substack{=f\left(
n\right)  g\left(  n\right)  \\\text{(by the definition}\\\text{of }f\cdot
g\text{)}}}h\left(  n\right)  =f\left(  n\right)  g\left(  n\right)  h\left(
n\right)  .
\]
But the definition of $f\cdot\left(  g\cdot h\right)  $ yields%
\[
\left(  f\cdot\left(  g\cdot h\right)  \right)  \left(  n\right)  =f\left(
n\right)  \cdot\underbrace{\left(  g\cdot h\right)  \left(  n\right)
}_{\substack{=g\left(  n\right)  h\left(  n\right)  \\\text{(by the
definition}\\\text{of }g\cdot h\text{)}}}=f\left(  n\right)  g\left(
n\right)  h\left(  n\right)  .
\]
Comparing these two equalities, we obtain $\left(  f\cdot\left(  g\cdot
h\right)  \right)  \left(  n\right)  =\left(  \left(  f\cdot g\right)  \cdot
h\right)  \left(  n\right)  $.

Now, forget that we fixed $n$. We thus have proven that $\left(  f\cdot\left(
g\cdot h\right)  \right)  \left(  n\right)  =\left(  \left(  f\cdot g\right)
\cdot h\right)  \left(  n\right)  $ for every $n\in\mathbb{N}_{+}$. In other
words, $f\cdot\left(  g\cdot h\right)  =\left(  f\cdot g\right)  \cdot h$.
This proves Theorem \ref{thm.pwprod.props} \textbf{(b)}.
\end{proof}

We also get the following essentially obvious analogue of Theorem
\ref{thm.dirichlet.mult}:

\begin{theorem}
\label{thm.pwprod.mult}Let $f$ and $g$ be two multiplicative arithmetic
functions. Then, the arithmetic function $f\cdot g$ is also multiplicative.
\end{theorem}

Theorem \ref{thm.pwprod.mult} is just Proposition
\ref{prop.arifun.mult.pwprod} \textbf{(a)} with different notations, but let
us prove it again:

\begin{proof}
[Proof of Theorem \ref{thm.pwprod.mult}.]The function $f$ is multiplicative.
In other words, it satisfies $f\left(  1\right)  =1$, and%
\begin{equation}
f\left(  mn\right)  =f\left(  m\right)  f\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for any two coprime }m\in\mathbb{N}_{+}\text{ and
}n\in\mathbb{N}_{+}. \label{pf.thm.pwprod.mult.multf}%
\end{equation}


The function $g$ is multiplicative. In other words, it satisfies $g\left(
1\right)  =1$, and%
\begin{equation}
g\left(  mn\right)  =g\left(  m\right)  g\left(  n\right)
\ \ \ \ \ \ \ \ \ \ \text{for any two coprime }m\in\mathbb{N}_{+}\text{ and
}n\in\mathbb{N}_{+}. \label{pf.thm.pwprod.mult.multg}%
\end{equation}


The definition of $f\cdot g$ yields
\[
\left(  f\cdot g\right)  \left(  1\right)  =\underbrace{f\left(  1\right)
}_{=1}\underbrace{g\left(  1\right)  }_{=1}=1.
\]


Now, we want to prove that $f\cdot g$ is multiplicative. In order to do so, we
need to verify that $\left(  f\cdot g\right)  \left(  1\right)  =1$ and that
\begin{equation}
\left(  f\cdot g\right)  \left(  mn\right)  =\left(  f\cdot g\right)  \left(
m\right)  \cdot\left(  f\cdot g\right)  \left(  n\right)
\label{pf.thm.pwprod.mult.goal}%
\end{equation}
for any two coprime $m\in\mathbb{N}_{+}$ and $n\in\mathbb{N}_{+}$. Since
$\left(  f\cdot g\right)  \left(  1\right)  =1$ is already proven, it thus
only remains to prove (\ref{pf.thm.pwprod.mult.goal}).

So let $m\in\mathbb{N}_{+}$ and $n\in\mathbb{N}_{+}$ be coprime. We need to
prove (\ref{pf.thm.pwprod.mult.goal}).

The definition of $f\cdot g$ yields
\[
\left(  f\cdot g\right)  \left(  mn\right)  =\underbrace{f\left(  mn\right)
}_{\substack{=f\left(  m\right)  f\left(  n\right)  \\\text{(by
(\ref{pf.thm.pwprod.mult.multf}))}}}\underbrace{g\left(  mn\right)
}_{\substack{=g\left(  m\right)  g\left(  n\right)  \\\text{(by
(\ref{pf.thm.pwprod.mult.multg}))}}}=f\left(  m\right)  f\left(  n\right)
g\left(  m\right)  g\left(  n\right)  .
\]
Comparing this with%
\begin{align*}
\underbrace{\left(  f\cdot g\right)  \left(  m\right)  }_{\substack{=f\left(
m\right)  g\left(  m\right)  \\\text{(by the definition of }f\cdot g\text{)}%
}}\cdot\underbrace{\left(  f\cdot g\right)  \left(  n\right)  }%
_{\substack{=f\left(  n\right)  g\left(  n\right)  \\\text{(by the definition
of }f\cdot g\text{)}}}  &  =f\left(  m\right)  g\left(  m\right)  f\left(
n\right)  g\left(  n\right) \\
&  =f\left(  m\right)  f\left(  n\right)  g\left(  m\right)  g\left(
n\right)  ,
\end{align*}
we obtain $\left(  f\cdot g\right)  \left(  mn\right)  =\left(  f\cdot
g\right)  \left(  m\right)  \cdot\left(  f\cdot g\right)  \left(  n\right)  $.
Thus, (\ref{pf.thm.pwprod.mult.goal}) is proven. As we have said, this
completes the proof of Theorem \ref{thm.pwprod.mult}.
\end{proof}

\subsection{Lowest common multiples}

Let us next define yet another way of multiplying two arithmetic functions,
the so-called \textit{lcm convolution}. We begin by introducing lowest common
multiples of two positive integers:

\begin{definition}
Let $a$ be an integer. A \textit{multiple} of $a$ means an integer that is
divisible by $a$.
\end{definition}

For instance, $12$ is a multiple of $3$, since $3\mid12$.

\begin{definition}
Let $b$ and $c$ be two integers. A \textit{common multiple} of $b$ and $c$
means an integer that is both a multiple of $b$ and a multiple of $c$.
\end{definition}

For example, $12$ is a common multiple of $4$ and $6$, since $4\mid12$ and
$6\mid12$.

\begin{proposition}
\label{prop.lcm.exists}Let $b$ and $c$ be two positive integers. Then, there
exists a smallest positive common multiple of $b$ and $c$. (In other words,
the set of all positive common multiples of $b$ and $c$ has a smallest element.)
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.lcm.exists}.]We know that $b$ and $c$ are
positive integers. Hence, their product $bc$ is a positive integer as well.
Moreover, this positive integer $bc$ is clearly a multiple of $b$ (since
$b\mid bc$) and a multiple of $c$ (since $c\mid bc$); thus, it is a common
multiple of $b$ and $c$. Hence, $bc$ is a positive common multiple of $b$ and
$c$. Therefore, the set of all positive common multiples of $b$ and $c$ has at
least one element (namely, the element $bc$). Thus, this set is nonempty.
Therefore, this set is a nonempty set of positive integers.

But it is well-known that any nonempty set of positive integers has a minimum
element. Hence, the set of all positive common multiples of $b$ and $c$ has a
minimum element (because this set is a nonempty set of positive integers). In
other words, there exists a smallest positive common multiple of $b$ and $c$.
\end{proof}

\begin{definition}
\label{def.lcm.lcm}Let $b$ and $c$ be two positive integers. Then,
$\operatorname{lcm}\left(  b,c\right)  $ is defined to be the smallest
positive common multiple of $b$ and $c$. (This is well-defined, because
Proposition \ref{prop.lcm.exists} shows that there exists a smallest positive
common multiple of $b$ and $c$.)

This number $\operatorname{lcm}\left(  b,c\right)  $ is called the
\textit{lowest common multiple of }$b$ \textit{and }$c$ or the \textit{least
common multiple of }$b$ \textit{and }$c$ or, briefly, the \textit{lcm} of $b$
and $c$. Clearly, it satisfies $\operatorname{lcm}\left(  b,c\right)
=\operatorname{lcm}\left(  c,b\right)  $ and $b\mid\operatorname{lcm}\left(
b,c\right)  $ and $c\mid\operatorname{lcm}\left(  b,c\right)  $. Notice that
$\operatorname{lcm}\left(  b,c\right)  $ is a positive integer (by its definition).
\end{definition}

We have been slightly lazy here and only defined the lowest common multiple of
two positive integers. We could extend this definition to two arbitrary
integers, or even to several integers. But we will not need this generality in
what follows.

Older books such as \cite{NiZuMo91} often denote the lcm of two integers $b$
and $c$ by $\left[  b,c\right]  $ (rather than by $\operatorname{lcm}\left(
b,c\right)  $ as we do).

The most important property of lcms is the following fact, which is in a sense
a mirror image of Proposition \ref{prop.gcd.uniprop}:

\begin{proposition}
\label{prop.lcm.uniprop}Let $b$ and $c$ be two positive integers. Let $a$ be
an integer such that $b\mid a$ and $c\mid a$. Then, $\operatorname{lcm}\left(
b,c\right)  \mid a$.
\end{proposition}

In words, Proposition \ref{prop.lcm.uniprop} says that any common multiple of
two positive integers must be divisible by the lcm of these two integers.

\begin{proof}
[Proof of Proposition \ref{prop.lcm.uniprop}.]Let $\ell=\operatorname{lcm}%
\left(  b,c\right)  $. Then, $\ell$ is the smallest positive common multiple
of $b$ and $c$ (by the definition of the lcm). Hence, $\ell$ is a positive integer.

Let $q$ and $r$ be the quotient and the remainder obtained when dividing $a$
by $\ell$. Thus, we have $q\in\mathbb{Z}$, $r\in\left\{  0,1,\ldots
,\ell-1\right\}  $ and $a=q\ell+r$ (by the definition of division with
remainder). From $a=q\ell+r$, we obtain $a-q\ell=r$. From $r\in\left\{
0,1,\ldots,\ell-1\right\}  $, we obtain $r\geq0$ and $r\leq\ell-1<\ell$.

We have $b\mid a$ and $b\mid\operatorname{lcm}\left(  b,c\right)  =\ell
\mid-q\ell$. In other words, the two integers $a$ and $-q\ell$ are both
divisible by $b$. Hence, the sum $a+\left(  -q\ell\right)  $ of these two
integers must also be divisible by $b$ (by Proposition \ref{prop.div.sum},
applied to $b$, $a$ and $-q\ell$ instead of $a$, $u$ and $v$). In other words,
$b\mid a+\left(  -q\ell\right)  $. In view of $a+\left(  -q\ell\right)
=a-q\ell=r$, this rewrites as $b\mid r$. The same argument (applied to $c$
instead of $b$) yields $c\mid r$ (since $c\mid\operatorname{lcm}\left(
b,c\right)  =\ell$).

Now, $r$ is an integer that is both a multiple of $b$ (since $b\mid r$) and a
multiple of $c$ (since $c\mid r$). In other words, $r$ is a common multiple of
$b$ and $c$.

Recall that $\ell$ is the smallest positive common multiple of $b$ and $c$.
Hence, each positive common multiple of $b$ and $c$ must be $\geq\ell$. Thus,
if $r$ was positive, then $r$ would be $\geq\ell$ (because $r$ is a common
multiple of $b$ and $c$, and therefore would be a positive common multiple of
$b$ and $c$); but this would contradict $r<\ell$. Hence, $r$ cannot be
positive. In other words, we must have $r\leq0$. Combining this with $r\geq0$,
we obtain $r=0$. Thus, $a=q\ell+\underbrace{r}_{=0}=q\ell$. Now,
$\operatorname{lcm}\left(  b,c\right)  =\ell\mid q\ell=a$. This proves
Proposition \ref{prop.lcm.uniprop}.
\end{proof}

\begin{corollary}
\label{cor.lcm.uniprop-both-ways}Let $b$ and $c$ be two positive integers. Let
$a$ be an integer. Then, we have the logical equivalence%
\[
\left(  b\mid a\text{ and }c\mid a\right)  \ \Longleftrightarrow\ \left(
\operatorname{lcm}\left(  b,c\right)  \mid a\right)  .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.lcm.uniprop-both-ways}.]We have the logical
implication \newline$\left(  \operatorname{lcm}\left(  b,c\right)  \mid
a\right)  \ \Longrightarrow\ \left(  b\mid a\text{ and }c\mid a\right)  $
(because if $\operatorname{lcm}\left(  b,c\right)  \mid a$ holds, then we have
$b\mid\operatorname{lcm}\left(  b,c\right)  \mid a$ and $c\mid
\operatorname{lcm}\left(  b,c\right)  \mid a$, and therefore $\left(  b\mid
a\text{ and }c\mid a\right)  $). But we also have the logical implication
$\left(  b\mid a\text{ and }c\mid a\right)  \ \Longrightarrow\ \left(
\operatorname{lcm}\left(  b,c\right)  \mid a\right)  $ (by Proposition
\ref{prop.lcm.uniprop}). Combining these two implications, we obtain the
equivalence $\left(  b\mid a\text{ and }c\mid a\right)  \ \Longleftrightarrow
\ \left(  \operatorname{lcm}\left(  b,c\right)  \mid a\right)  $. This proves
Corollary \ref{cor.lcm.uniprop-both-ways}.
\end{proof}

The gcd and the lcm of two positive integers $b$ and $c$ are connected by the
equality $\gcd\left(  b,c\right)  \cdot\operatorname{lcm}\left(  b,c\right)
=bc$ (see, for example, \cite[Theorem 1.13]{NiZuMo91} or \cite[Theorem
7]{Conrad19}). We will not have any use for this connection, however. Most
texts on number theory study the lcm as an afterthought of the gcd; however,
as Keith Conrad shows in \cite[Theorem 7]{Conrad19}, it is actually easier to
build up the theory of gcds and lcms (of two positive integers) by starting
with lcms first.

For future use, let us state a trivial property of lcms:

\begin{lemma}
\label{lem.lcm.fin}Let $n\in\mathbb{N}_{+}$. Then, the set $\left\{  \left(
d,e\right)  \in\mathbb{N}_{+}\times\mathbb{N}_{+}\ \mid\ \operatorname{lcm}%
\left(  d,e\right)  =n\right\}  $ is finite.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.lcm.fin}.]If $\left(  d,e\right)  \in\mathbb{N}%
_{+}\times\mathbb{N}_{+}$ satisfies $\operatorname{lcm}\left(  d,e\right)
=n$, then $\left(  d,e\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  d,e\right)  \in\mathbb{N}%
_{+}\times\mathbb{N}_{+}$ be such that $\operatorname{lcm}\left(  d,e\right)
=n$. We must prove that $\left(  d,e\right)  \in\left\{  1,2,\ldots,n\right\}
^{2}$.
\par
We have $\left(  d,e\right)  \in\mathbb{N}_{+}\times\mathbb{N}_{+}$; thus,
$d\in\mathbb{N}_{+}$ and $e\in\mathbb{N}_{+}$. In other words, $d$ and $e$ are
positive integers. Also, $n$ is a positive integer (since $n\in\mathbb{N}_{+}%
$); thus, $n>0$. Now, $d\mid\operatorname{lcm}\left(  d,e\right)  =n$. In
other words, there exists an integer $z$ such that $n=dz$. Consider this $z$.
If we had $z\leq0$, then we would have $n=d\underbrace{z}_{\leq0}\leq d0$
(since $d$ is positive), which would contradict $n>0=d0$. Thus, we cannot have
$z\leq0$. Hence, we have $z>0$. Thus, $z\geq1$ (since $z$ is an integer). Now,
$n=d\underbrace{z}_{\geq1}\geq d1$ (since $d$ is positive), so that $n\geq
d1=d$. In other words, $d\leq n$. Hence, $d\in\left\{  1,2,\ldots,n\right\}  $
(since $d$ is a positive integer). Similarly, $e\in\left\{  1,2,\ldots
,n\right\}  $ (since $e\mid\operatorname{lcm}\left(  d,e\right)  =n$).
Combining $d\in\left\{  1,2,\ldots,n\right\}  $ with $e\in\left\{
1,2,\ldots,n\right\}  $, we find $\left(  d,e\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$. Qed.}. In other words, the set $\left\{  \left(
d,e\right)  \in\mathbb{N}_{+}\times\mathbb{N}_{+}\ \mid\ \operatorname{lcm}%
\left(  d,e\right)  =n\right\}  $ is a subset of $\left\{  1,2,\ldots
,n\right\}  ^{2}$. Therefore, this set $\left\{  \left(  d,e\right)
\in\mathbb{N}_{+}\times\mathbb{N}_{+}\ \mid\ \operatorname{lcm}\left(
d,e\right)  =n\right\}  $ is finite (since the set $\left\{  1,2,\ldots
,n\right\}  ^{2}$ is finite). This proves Lemma \ref{lem.lcm.fin}.
\end{proof}

\subsection{Lcm-convolution}

We can now define another way of \textquotedblleft
multiplying\textquotedblright\ two arithmetic functions:

\begin{definition}
\label{def.lcmconv}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ and
$g:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be two arithmetic functions. We define
a new arithmetic function $f \tilstar g:\mathbb{N}_{+}\rightarrow\mathbb{C}$
by%
\[
\left(  f \tilstar g\right)  \left(  n\right)  =\sum_{\substack{d\in
\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\\operatorname{lcm}\left(  d,e\right)
=n}}f\left(  d\right)  g\left(  e\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}n\in\mathbb{N}_{+}.
\]
(This is well-defined, because the sum on the right hand side of this equality
is finite\footnotemark.)

This new function $f \tilstar g$ is called the \textit{lcm-convolution} of $f$
and $g$.
\end{definition}

\footnotetext{\textit{Proof.} Let $n\in\mathbb{N}_{+}$. We must prove that the
sum $\sum_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+}%
;\\\operatorname{lcm}\left(  d,e\right)  =n}}f\left(  d\right)  g\left(
e\right)  $ is finite. In other words, we must prove that there are only
finitely many pairs $\left(  d,e\right)  \in\mathbb{N}_{+}\times\mathbb{N}%
_{+}$ such that $\operatorname{lcm}\left(  d,e\right)  =n$. In other words, we
must prove that the set $\left\{  \left(  d,e\right)  \in\mathbb{N}_{+}%
\times\mathbb{N}_{+}\ \mid\ \operatorname{lcm}\left(  d,e\right)  =n\right\}
$ is finite. But this follows immediately from Lemma \ref{lem.lcm.fin}. Qed.}

\begin{example}
Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ and $g:\mathbb{N}_{+}%
\rightarrow\mathbb{C}$ be two arithmetic functions. Then,%
\begin{align*}
\left(  f \tilstar g\right)  \left(  1\right)   &  =\sum_{\substack{d\in
\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\\operatorname{lcm}\left(  d,e\right)
=1}}f\left(  d\right)  g\left(  e\right)  =f\left(  1\right)  g\left(
1\right)  ;\\
\left(  f \tilstar g\right)  \left(  p\right)   &  =\sum_{\substack{d\in
\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\\operatorname{lcm}\left(  d,e\right)
=p}}f\left(  d\right)  g\left(  e\right) \\
&  =f\left(  1\right)  g\left(  p\right)  +f\left(  p\right)  g\left(
1\right)  +f\left(  p\right)  g\left(  p\right)  \ \ \ \ \ \ \ \ \ \ \text{for
any prime }p;\\
\left(  f \tilstar g\right)  \left(  6\right)   &  =\sum_{\substack{d\in
\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\\operatorname{lcm}\left(  d,e\right)
=6}}f\left(  d\right)  g\left(  e\right) \\
&  =f\left(  1\right)  g\left(  6\right)  +f\left(  2\right)  g\left(
3\right)  +f\left(  2\right)  g\left(  6\right)  +f\left(  3\right)  g\left(
2\right)  +f\left(  3\right)  g\left(  6\right) \\
&  \ \ \ \ \ \ \ \ \ \ +f\left(  6\right)  g\left(  1\right)  +f\left(
6\right)  g\left(  2\right)  +f\left(  6\right)  g\left(  3\right)  +f\left(
6\right)  g\left(  6\right)  .
\end{align*}

\end{example}

The operation $\tilstar $ on the set of arithmetic functions appears in
\cite[Theorem 1]{Lehmer31} (where $f \tilstar g$ is denoted by $h$) and in
\cite[\S 4.4]{Toth14} (where this operation $\tilstar $ is denoted by $\oplus
$, and generalized to functions of $r$ arguments). It is also known as the
\textit{Lehmer convolution} (or \textit{Lehmer product}) or \textit{von
Sterneck convolution}. We shall prove the following of its properties
(analogues of Theorems \ref{thm.dirichlet.props} and \ref{thm.dirichlet.mult}, respectively):

\begin{theorem}
\label{thm.lcmconv.props}\textbf{(a)} We have $\varepsilon\tilstar
f=f \tilstar \varepsilon=f$ for every arithmetic function $f$.

\textbf{(b)} We have $f \tilstar \left(  g \tilstar h\right)  =\left(  f
\tilstar g\right)  \tilstar h$ for every three arithmetic functions $f$, $g$
and $h$.

\textbf{(c)} We have $f \tilstar g=g \tilstar f$ for every two arithmetic
functions $f$ and $g$.
\end{theorem}

\begin{theorem}
\label{thm.lcmconv.mult}Let $f$ and $g$ be two multiplicative arithmetic
functions. Then, the arithmetic function $f \tilstar g$ is also multiplicative.
\end{theorem}

The proofs will rely on the following result of von Sterneck and Lehmer (see,
e.g., \cite[Theorem 1]{Lehmer31}), which connects the lcm-convolution with the
pointwise product and the Dirichlet convolution:

\begin{theorem}
\label{thm.lcmconv.lehmer}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ and
$g:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be two arithmetic functions. Then,%
\[
\underline{1}\star\left(  f \tilstar g\right)  =\left(  \underline{1}\star
f\right)  \cdot\left(  \underline{1}\star g\right)  .
\]

\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.lcmconv.lehmer}.]Define three arithmetic functions
$F$, $G$ and $H$ by
\[
F=\underline{1}\star f,\ \ \ \ \ \ \ \ \ \ G=\underline{1}\star
g\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ H=\underline{1}\star\left(
f \tilstar g\right)  .
\]


Let $n\in\mathbb{N}_{+}$. Then, $H=\underline{1}\star\left(  f\tilstar
g\right)  =\left(  f\tilstar g\right)  \star\underline{1}$ (by Theorem
\ref{thm.dirichlet.props} \textbf{(c)}, applied to $\underline{1}$ and
$f\tilstar g$ instead of $f$ and $g$). Applying both sides of this equality to
$n$, we obtain%
\begin{align}
H\left(  n\right)   &  =\left(  \left(  f\tilstar g\right)  \star
\underline{1}\right)  \left(  n\right)  \nonumber\\
&  =\sum_{d\mid n}\left(  f\tilstar g\right)  \left(  d\right)
\underbrace{\underline{1}\left(  \dfrac{n}{d}\right)  }%
_{\substack{=1\\\text{(by the definition of }\underline{1}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\left(  f\tilstar
g\right)  \star\underline{1}\right)  \nonumber\\
&  =\sum_{d\mid n}\left(  f\tilstar g\right)  \left(  d\right)
=\underbrace{\sum_{s\mid n}}_{\substack{=\sum_{\substack{s\text{ is a
positive}\\\text{divisor of }n}}\\\text{(by Definition \ref{def.sumdiv})}%
}}\underbrace{\left(  f\tilstar g\right)  \left(  s\right)  }_{\substack{=\sum
_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\\operatorname{lcm}%
\left(  d,e\right)  =s}}f\left(  d\right)  g\left(  e\right)  \\\text{(by the
definition of }f\tilstar g\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}d\text{ as }s\right)  \nonumber\\
&  =\underbrace{\sum_{\substack{s\text{ is a positive}\\\text{divisor of }%
n}}\ \ \sum_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+}%
;\\\operatorname{lcm}\left(  d,e\right)  =s}}}_{=\sum_{\substack{d\in
\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\\operatorname{lcm}\left(  d,e\right)
\text{ is a positive}\\\text{divisor of }n}}}f\left(  d\right)  g\left(
e\right)  \nonumber\\
&  =\sum_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+}%
;\\\operatorname{lcm}\left(  d,e\right)  \text{ is a positive}\\\text{divisor
of }n}}f\left(  d\right)  g\left(  e\right)
.\label{pf.thm.lcmconv.lehmer.Hn=1}%
\end{align}


Now, let $d\in\mathbb{N}_{+}$ and $e\in\mathbb{N}_{+}$ be arbitrary. Then, $d$
and $e$ are two positive integers. Hence, Corollary
\ref{cor.lcm.uniprop-both-ways} (applied to $d$, $e$ and $n$ instead of $b$,
$c$ and $a$) shows that we have the logical equivalence%
\begin{equation}
\left(  d\mid n\text{ and }e\mid n\right)  \ \Longleftrightarrow\ \left(
\operatorname{lcm}\left(  d,e\right)  \mid n\right)  .
\label{pf.thm.lcmconv.lehmer.equiv1}%
\end{equation}


Also, $\operatorname{lcm}\left(  d,e\right)  $ is positive (by the definition
of $\operatorname{lcm}\left(  d,e\right)  $). Thus, $\operatorname{lcm}\left(
d,e\right)  $ is a positive divisor of $n$ if and only if $\operatorname{lcm}%
\left(  d,e\right)  $ is a divisor of $n$. Hence, we have the following chain
of logical equivalences:%
\begin{align*}
&  \ \left(  \operatorname{lcm}\left(  d,e\right)  \text{ is a positive
divisor of }n\right) \\
&  \Longleftrightarrow\ \left(  \operatorname{lcm}\left(  d,e\right)  \text{
is a divisor of }n\right)  \ \Longleftrightarrow\ \left(  \operatorname{lcm}%
\left(  d,e\right)  \mid n\right) \\
&  \Longleftrightarrow\ \left(  d\mid n\text{ and }e\mid n\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.lcmconv.lehmer.equiv1}%
)}\right)  .
\end{align*}


Now, forget that we fixed $d$ and $e$. We thus have proven the logical
equivalence
\[
\left(  \operatorname{lcm}\left(  d,e\right)  \text{ is a positive divisor of
}n\right)  \ \Longleftrightarrow\ \left(  d\mid n\text{ and }e\mid n\right)
\]
for every $d\in\mathbb{N}_{+}$ and $e\in\mathbb{N}_{+}$. Hence, we have the
following equality of summation signs:%
\[
\sum_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\\operatorname{lcm}%
\left(  d,e\right)  \text{ is a positive}\\\text{divisor of }n}}=\sum
_{\substack{d\in\mathbb{N}_{+};\ e\in\mathbb{N}_{+};\\d\mid n\text{ and }e\mid
n}}=\underbrace{\sum_{\substack{d\in\mathbb{N}_{+}\\d\mid n}}}%
_{\substack{=\sum_{\substack{d\text{ is a positive}\\\text{divisor of }%
n}}\\=\sum_{d\mid n}\\\text{(by Definition \ref{def.sumdiv})}}%
}\underbrace{\sum_{\substack{e\in\mathbb{N}_{+}\\e\mid n}}}_{\substack{=\sum
_{\substack{e\text{ is a positive}\\\text{divisor of }n}}\\=\sum_{e\mid
n}\\\text{(by Definition \ref{def.sumdiv})}}}=\sum_{d\mid n}\ \ \sum_{e\mid
n}.
\]
Thus, (\ref{pf.thm.lcmconv.lehmer.Hn=1}) becomes%
\begin{equation}
H\left(  n\right)  =\underbrace{\sum_{\substack{d\in\mathbb{N}_{+}%
;\ e\in\mathbb{N}_{+};\\\operatorname{lcm}\left(  d,e\right)  \text{ is a
positive}\\\text{divisor of }n}}}_{=\sum_{d\mid n}\ \ \sum_{e\mid n}}f\left(
d\right)  g\left(  e\right)  =\sum_{d\mid n}\ \ \sum_{e\mid n}f\left(
d\right)  g\left(  e\right)  .\label{pf.thm.lcmconv.lehmer.Hn=2}%
\end{equation}


On the other hand, $F=\underline{1}\star f=f\star\underline{1}$ (by Theorem
\ref{thm.dirichlet.props} \textbf{(c)}, applied to $\underline{1}$ and $f$
instead of $f$ and $g$). Applying both sides of this equality to $n$, we
obtain
\begin{align}
F\left(  n\right)   &  =\left(  f\star\underline{1}\right)  \left(  n\right)
=\sum_{d\mid n}f\left(  d\right)  \underbrace{\underline{1}\left(  \dfrac
{n}{d}\right)  }_{\substack{=1\\\text{(by the definition of }\underline{1}%
\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
f\star\underline{1}\right)  \nonumber\\
&  =\sum_{d\mid n}f\left(  d\right)  .\label{pf.thm.lcmconv.lehmer.Fn=1}%
\end{align}
The same argument (applied to $G$ and $g$ instead of $F$ and $f$) yields
$G\left(  n\right)  =\sum_{d\mid n}g\left(  d\right)  $. Hence,%
\begin{equation}
G\left(  n\right)  =\sum_{d\mid n}g\left(  d\right)  =\sum_{e\mid n}g\left(
e\right)  \label{pf.thm.lcmconv.lehmer.Gn=1}%
\end{equation}
(here, we have renamed the summation index $d$ as $e$). Now, the definition of
$F\cdot G$ yields%
\begin{align*}
\left(  F\cdot G\right)  \left(  n\right)   &  =\underbrace{F\left(  n\right)
}_{\substack{=\sum_{d\mid n}f\left(  d\right)  \\\text{(by
(\ref{pf.thm.lcmconv.lehmer.Fn=1}))}}}\ \ \underbrace{G\left(  n\right)
}_{\substack{=\sum_{e\mid n}g\left(  e\right)  \\\text{(by
(\ref{pf.thm.lcmconv.lehmer.Gn=1}))}}}=\left(  \sum_{d\mid n}f\left(
d\right)  \right)  \left(  \sum_{e\mid n}g\left(  e\right)  \right)  \\
&  =\sum_{d\mid n}\ \ \sum_{e\mid n}f\left(  d\right)  g\left(  e\right)
=H\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.lcmconv.lehmer.Hn=2})}\right)  .
\end{align*}


Now, forget that we fixed $n$. We thus have proven that $\left(  F\cdot
G\right)  \left(  n\right)  =H\left(  n\right)  $ for each $n\in\mathbb{N}%
_{+}$. In other words, $F\cdot G=H$.

Comparing this with $\underbrace{F}_{=\underline{1}\star f}\cdot
\underbrace{G}_{=\underline{1}\star g}=\left(  \underline{1}\star f\right)
\cdot\left(  \underline{1}\star g\right)  $, we obtain%
\[
\left(  \underline{1}\star f\right)  \cdot\left(  \underline{1}\star g\right)
=H=\underline{1}\star\left(  f \tilstar g\right)  .
\]
This proves Theorem \ref{thm.lcmconv.lehmer}.
\end{proof}

An easy consequence of Theorem \ref{thm.lcmconv.lehmer} and Proposition
\ref{prop.moebius-inv.*}, we now obtain the following:

\begin{corollary}
\label{cor.lcmconv.lehmer-mu}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ and
$g:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be two arithmetic functions. Then,%
\[
f \tilstar g=\mu\star\left(  \left(  \underline{1}\star f\right)  \cdot\left(
\underline{1}\star g\right)  \right)  .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.lcmconv.lehmer-mu}.]Define an arithmetic function
$H$ by $H=\underline{1}\star\left(  f \tilstar g\right)  $. Then,%
\[
H=\underline{1}\star\left(  f \tilstar g\right)  =\left(  \underline{1}\star
f\right)  \cdot\left(  \underline{1}\star g\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.lcmconv.lehmer}}\right)
.
\]
Also, $H=\underline{1}\star\left(  f \tilstar g\right)  =\left(  f
\tilstar g\right)  \star\underline{1}$ (by Theorem \ref{thm.dirichlet.props}
\textbf{(c)}, applied to $\underline{1}$ and $f \tilstar g$ instead of $f$ and
$g$).

But Proposition \ref{prop.moebius-inv.*} (applied to $f \tilstar g$ and $H$
instead of $f$ and $F$) yields that we have the following logical equivalence:%
\[
\left(  H=\left(  f \tilstar g\right)  \star\underline{1}\right)
\ \Longleftrightarrow\ \left(  f \tilstar g=\mu\star H\right)  .
\]
Hence, we have $f \tilstar g=\mu\star H$ (since we have $H=\left(  f
\tilstar g\right)  \star\underline{1}$). Thus,%
\[
f \tilstar g=\mu\star\underbrace{H}_{=\left(  \underline{1}\star f\right)
\cdot\left(  \underline{1}\star g\right)  }=\mu\star\left(  \left(
\underline{1}\star f\right)  \cdot\left(  \underline{1}\star g\right)
\right)  .
\]
This proves Corollary \ref{cor.lcmconv.lehmer-mu}.
\end{proof}

We also have the following simple corollary of Proposition
\ref{prop.moebius-inv.*}:

\begin{corollary}
\label{cor.lcmconv.moebius-inv}Let $f:\mathbb{N}_{+}\rightarrow\mathbb{C}$ and
$g:\mathbb{N}_{+}\rightarrow\mathbb{C}$ be two arithmetic functions such that
$\underline{1}\star f=\underline{1}\star g$. Then, $f=g$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.lcmconv.moebius-inv}.]Define an arithmetic
function $F$ by $F=\underline{1}\star f$. Thus, $F=\underline{1}\star
f=f\star\underline{1}$ (by Theorem \ref{thm.dirichlet.props} \textbf{(c)},
applied to $\underline{1}$ and $f$ instead of $f$ and $g$). But Proposition
\ref{prop.moebius-inv.*} yields that we have the following logical
equivalence:%
\[
\left(  F=f\star\underline{1}\right)  \ \Longleftrightarrow\ \left(
f=\mu\star F\right)  .
\]
Hence, we have $f=\mu\star F$ (since $F=f\star\underline{1}$).

On the other hand, $F=\underline{1}\star f=\underline{1}\star g=g\star
\underline{1}$ (by Theorem \ref{thm.dirichlet.props} \textbf{(c)}, applied to
$\underline{1}$ and $g$ instead of $f$ and $g$). But Proposition
\ref{prop.moebius-inv.*} (applied to $g$ instead of $f$) yields that we have
the following logical equivalence:%
\[
\left(  F=g\star\underline{1}\right)  \ \Longleftrightarrow\ \left(
g=\mu\star F\right)  .
\]
Hence, we have $g=\mu\star F$ (since $F=g\star\underline{1}$).

Comparing $f=\mu\star F$ with $g=\mu\star F$, we obtain $f=g$. This proves
Corollary \ref{cor.lcmconv.moebius-inv}.
\end{proof}

Now, proving Theorem \ref{thm.lcmconv.props} and Theorem
\ref{thm.lcmconv.mult} is a child's play:

\begin{proof}
[Proof of Theorem \ref{thm.lcmconv.props}.]\textbf{(c)} Let $f$ and $g$ be two
arithmetic functions. Then, Theorem \ref{thm.pwprod.props} \textbf{(c)}
(applied to $\underline{1}\star f$ and $\underline{1}\star g$ instead of $f$
and $g$) yields $\left(  \underline{1}\star f\right)  \cdot\left(
\underline{1}\star g\right)  =\left(  \underline{1}\star g\right)
\cdot\left(  \underline{1}\star f\right)  $. But Corollary
\ref{cor.lcmconv.lehmer-mu} (applied to $g$ and $f$ instead of $f$ and $g$)
yields%
\begin{equation}
g \tilstar f=\mu\star\left(  \left(  \underline{1}\star g\right)  \cdot\left(
\underline{1}\star f\right)  \right)  . \label{pf.thm.lcmconv.props.c.1}%
\end{equation}
On the other hand, Corollary \ref{cor.lcmconv.lehmer-mu} yields%
\[
f \tilstar g=\mu\star\underbrace{\left(  \left(  \underline{1}\star f\right)
\cdot\left(  \underline{1}\star g\right)  \right)  }_{=\left(  \underline{1}%
\star g\right)  \cdot\left(  \underline{1}\star f\right)  }=\mu\star\left(
\left(  \underline{1}\star g\right)  \cdot\left(  \underline{1}\star f\right)
\right)  =g \tilstar f\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.lcmconv.props.c.1})}\right)  .
\]
This proves Theorem \ref{thm.lcmconv.props} \textbf{(c)}.

\textbf{(a)} Let $f$ be an arithmetic function. Theorem
\ref{thm.dirichlet.props} \textbf{(a)} (applied to $\underline{1}$ instead of
$f$) yields $\varepsilon\star\underline{1}=\underline{1}\star\varepsilon
=\underline{1}$. Theorem \ref{thm.pwprod.props} \textbf{(a)} (applied to
$\underline{1}\star f$ instead of $f$) yields $\underline{1}\cdot\left(
\underline{1}\star f\right)  =\left(  \underline{1}\star f\right)
\cdot\underline{1}=\underline{1}\star f$.

Now, Theorem \ref{thm.lcmconv.lehmer} (applied to $g=\varepsilon$) yields%
\[
\underline{1}\star\left(  f \tilstar \varepsilon\right)  =\left(
\underline{1}\star f\right)  \cdot\underbrace{\left(  \underline{1}%
\star\varepsilon\right)  }_{=\underline{1}}=\left(  \underline{1}\star
f\right)  \cdot\underline{1}=\underline{1}\star f.
\]
Hence, Corollary \ref{cor.lcmconv.moebius-inv} (applied to $f \tilstar
\varepsilon$ and $f$ instead of $f$ and $g$) yields $f \tilstar
\varepsilon=f$.

But Theorem \ref{thm.lcmconv.props} \textbf{(c)} (applied to $g=\varepsilon$)
yields $f \tilstar \varepsilon=\varepsilon\tilstar f$. Hence, $\varepsilon
\tilstar f=f \tilstar \varepsilon=f$. This proves Theorem
\ref{thm.lcmconv.props} \textbf{(a)}.

\textbf{(b)} Let $f$, $g$ and $h$ be three arithmetic functions. Theorem
\ref{thm.lcmconv.lehmer} (applied to $g \tilstar h$ instead of $g$) yields%
\begin{align*}
\underline{1}\star\left(  f \tilstar \left(  g \tilstar h\right)  \right)   &
=\left(  \underline{1}\star f\right)  \cdot\underbrace{\left(  \underline{1}%
\star\left(  g \tilstar h\right)  \right)  }_{\substack{=\left(
\underline{1}\star g\right)  \cdot\left(  \underline{1}\star h\right)
\\\text{(by Theorem \ref{thm.lcmconv.lehmer},}\\\text{applied to }g\text{ and
}h\text{ instead of }f\text{ and }g\text{)}}}=\left(  \underline{1}\star
f\right)  \cdot\left(  \left(  \underline{1}\star g\right)  \cdot\left(
\underline{1}\star h\right)  \right) \\
&  =\left(  \left(  \underline{1}\star f\right)  \cdot\left(  \underline{1}%
\star g\right)  \right)  \cdot\left(  \underline{1}\star h\right)
\end{align*}
(by Theorem \ref{thm.pwprod.props} \textbf{(b)} (applied to $\underline{1}%
\star f$, $\underline{1}\star g$ and $\underline{1}\star h$ instead of $f$,
$g$ and $h$)). On the other hand, Theorem \ref{thm.lcmconv.lehmer} (applied to
$f \tilstar g$ and $h$ instead of $f$ and $g$) yields%
\[
\underline{1}\star\left(  \left(  f \tilstar g\right)  \tilstar
h\right)  =\underbrace{\left(  \underline{1}\star\left(  f \tilstar
g\right)  \right)  }_{\substack{=\left(  \underline{1}\star f\right)
\cdot\left(  \underline{1}\star g\right)  \\\text{(by Theorem
\ref{thm.lcmconv.lehmer})}}}\cdot\left(  \underline{1}\star h\right)  =\left(
\left(  \underline{1}\star f\right)  \cdot\left(  \underline{1}\star g\right)
\right)  \cdot\left(  \underline{1}\star h\right)  .
\]
Comparing these two equalities, we obtain $\underline{1}\star\left(  f
\tilstar \left(  g \tilstar h\right)  \right)  =\underline{1}\star\left(
\left(  f \tilstar g\right)  \tilstar h\right)  $. Hence, Corollary
\ref{cor.lcmconv.moebius-inv} (applied to $f \tilstar
\left(  g \tilstar h\right)  $ and $\left(  f \tilstar g \right)  \tilstar h$
instead of $f$ and $g$) yields $f \tilstar \left(  g \tilstar h\right)
=\left(  f \tilstar g\right)  \tilstar h$. This proves Theorem
\ref{thm.lcmconv.props} \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.lcmconv.mult}.]The function $\underline{1}$ is
multiplicative (by Theorem \ref{prop.arifun.mult.exa} \textbf{(c)}). Also, the
function $\mu$ is multiplicative (by Theorem \ref{prop.arifun.mult.exa}
\textbf{(f)}). Now, the functions $\underline{1}$ and $f$ are multiplicative.
Hence, Theorem \ref{thm.dirichlet.mult} (applied to $\underline{1}$ and $f$
instead of $f$ and $g$) yields that the arithmetic function $\underline{1}%
\star f$ is also multiplicative. The same argument (applied to $g$ instead of
$f$) yields that the arithmetic function $\underline{1}\star g$ is also
multiplicative. Hence, Theorem \ref{thm.pwprod.mult} (applied to
$\underline{1}\star f$ and $\underline{1}\star g$ instead of $f$ and $g$)
yields that the arithmetic function $\left(  \underline{1}\star f\right)
\cdot\left(  \underline{1}\star g\right)  $ is also multiplicative.

Now, the arithmetic functions $\mu$ and $\left(  \underline{1}\star f\right)
\cdot\left(  \underline{1}\star g\right)  $ are multiplicative. Hence, Theorem
\ref{thm.dirichlet.mult} (applied to $\mu$ and $\left(  \underline{1}\star
f\right)  \cdot\left(  \underline{1}\star g\right)  $ instead of $f$ and $g$)
yields that the arithmetic function $\mu\star\left(  \left(  \underline{1}%
\star f\right)  \cdot\left(  \underline{1}\star g\right)  \right)  $ is also multiplicative.

But Corollary \ref{cor.lcmconv.lehmer-mu} yields that $f \tilstar g=\mu
\star\left(  \left(  \underline{1}\star f\right)  \cdot\left(  \underline{1}%
\star g\right)  \right)  $. Hence, the arithmetic function $f \tilstar g$ is
multiplicative (since we know that the arithmetic function $\mu\star\left(
\left(  \underline{1}\star f\right)  \cdot\left(  \underline{1}\star g\right)
\right)  $ is multiplicative). This proves Theorem \ref{thm.lcmconv.mult}.
\end{proof}

\section{\label{chp.bezout}Appendix: a proof of B\'{e}zout's identity}

Let me finally give a proof of Theorem \ref{thm.bezout}, which was used
several times above. Proofs of this theorem abound in the literature; yet I
have never seen the following proof written up. I believe that this proof has
the advantage of being constructive (unlike the proof in \cite[proof of
Theorem 1.3]{NiZuMo91}, which starts out by choosing the least positive
integer in a potentially infinite set) and yet not too messy (unlike some
proofs using the extended Euclidean algorithm). Of course, all the standard
proofs of Theorem \ref{thm.bezout} are \textquotedblleft essentially the
same\textquotedblright, in the sense that they offer different points of view
on one and the same idea (viz., that of the Euclidean algorithm).

We first prepare for our proof by showing some simple lemmas:

\begin{lemma}
\label{lem.bezout.gcd(b,c)=gcd(c,b)}Let $b$ and $c$ be two integers. Then,
$\gcd\left(  b,c\right)  =\gcd\left(  c,b\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.bezout.gcd(b,c)=gcd(c,b)}.]If $\left(  b,c\right)
=\left(  0,0\right)  $, then Lemma \ref{lem.bezout.gcd(b,c)=gcd(c,b)} is
obvious. Hence, for the rest of this proof, we WLOG assume that $\left(
b,c\right)  \neq\left(  0,0\right)  $. Thus, $\left(  c,b\right)  \neq\left(
0,0\right)  $. Hence, $\gcd\left(  c,b\right)  $ is the greatest of all common
divisors of $c$ and $b$ (by the definition of $\gcd\left(  c,b\right)  $). In
other words, $\gcd\left(  c,b\right)  $ is the greatest of all common divisors
of $b$ and $c$ (since the common divisors of $c$ and $b$ are the same as the
common divisors of $b$ and $c$). On the other hand, $\gcd\left(  b,c\right)  $
is the greatest of all common divisors of $b$ and $c$ (by the definition of
$\gcd\left(  b,c\right)  $). Hence, the two numbers $\gcd\left(  c,b\right)  $
and $\gcd\left(  b,c\right)  $ have been characterized in precisely the same
way (namely, as the greatest of all common divisors of $b$ and $c$).
Therefore, these two numbers are equal. In other words, $\gcd\left(
b,c\right)  =\gcd\left(  c,b\right)  $. This proves Lemma
\ref{lem.bezout.gcd(b,c)=gcd(c,b)}.
\end{proof}

\begin{lemma}
\label{lem.bezout.gcd(b,c)=gcd(-b,c)}Let $b$ and $c$ be two integers. Then:

\textbf{(a)} We have $\gcd\left(  b,c\right)  =\gcd\left(  -b,c\right)  $.

\textbf{(b)} We have $\gcd\left(  b,c\right)  =\gcd\left(  b,-c\right)  $.

\textbf{(c)} We have $\gcd\left(  b,c\right)  =\gcd\left(  \left\vert
b\right\vert ,\left\vert c\right\vert \right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)}.]If $\left(  b,c\right)
=\left(  0,0\right)  $, then Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)} is
obvious. Hence, for the rest of this proof, we WLOG assume that $\left(
b,c\right)  \neq\left(  0,0\right)  $.

\textbf{(a)} We make the following two observations:

\begin{statement}
\textit{Observation 1:} Every common divisor of $b$ and $c$ is a common
divisor of $-b$ and $c$.
\end{statement}

\begin{proof}
[Proof of Observation 1.]Let $d$ be a common divisor of $b$ and $c$. We must
prove that $d$ is a common divisor of $-b$ and $c$.

We know that $d$ is a common divisor of $b$ and $c$; hence, $d\mid b$ and
$d\mid c$. Now, $d\mid b\mid\left(  -1\right)  b=-b$. So we know that $d$
divides the two integers $-b$ and $c$ (since $d\mid-b$ and $d\mid c$). Hence,
$d$ is a common divisor of $-b$ and $c$. This completes the proof of
Observation 1.
\end{proof}

\begin{statement}
\textit{Observation 2:} Every common divisor of $-b$ and $c$ is a common
divisor of $b$ and $c$.
\end{statement}

\begin{proof}
[Proof of Observation 2.]Observation 1 (applied to $-b$ instead of $b$) shows
that every common divisor of $-b$ and $c$ is a common divisor of $-\left(
-b\right)  $ and $c$. In other words, every common divisor of $-b$ and $c$ is
a common divisor of $b$ and $c$ (since $-\left(  -b\right)  =b$). This proves
Observation 2.
\end{proof}

Combining Observation 1 with Observation 2, we conclude that the common
divisors of $b$ and $c$ are the same as the common divisors of $-b$ and $c$.

Now, $\left(  -b,c\right)  \neq\left(  0,0\right)  $ (since $\left(
b,c\right)  \neq\left(  0,0\right)  $). Hence, $\gcd\left(  -b,c\right)  $ is
the greatest of all common divisors of $-b$ and $c$ (by the definition of
$\gcd\left(  -b,c\right)  $). In other words, $\gcd\left(  -b,c\right)  $ is
the greatest of all common divisors of $b$ and $c$ (since the common divisors
of $b$ and $c$ are the same as the common divisors of $-b$ and $c$). On the
other hand, $\gcd\left(  b,c\right)  $ is the greatest of all common divisors
of $b$ and $c$ (by the definition of $\gcd\left(  b,c\right)  $). Hence, the
two numbers $\gcd\left(  -b,c\right)  $ and $\gcd\left(  b,c\right)  $ have
been characterized in precisely the same way (namely, as the greatest of all
common divisors of $b$ and $c$). Therefore, these two numbers are equal. In
other words, $\gcd\left(  -b,c\right)  =\gcd\left(  b,c\right)  $. This proves
Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)} \textbf{(a)}.

\textbf{(b)} Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)} \textbf{(b)} can be
proven in the same way as Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)}
\textbf{(a)} (but now we must use the fact that the divisors of $c$ are the
same as the divisors of $-c$).

[An alternative proof of Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)}
\textbf{(b)} proceeds as follows: We have%
\begin{align*}
\gcd\left(  b,c\right)   &  =\gcd\left(  c,b\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.bezout.gcd(b,c)=gcd(c,b)}%
}\right) \\
&  =\gcd\left(  -c,b\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)} \textbf{(a)}, applied to}\\
c\text{ and }b\text{ instead of }b\text{ and }c
\end{array}
\right) \\
&  =\gcd\left(  b,-c\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Lemma \ref{lem.bezout.gcd(b,c)=gcd(c,b)}, applied to}\\
-c\text{ and }b\text{ instead of }b\text{ and }c
\end{array}
\right)  ;
\end{align*}
thus, Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)} \textbf{(b)} is proven.]

\textbf{(c)} Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)} \textbf{(c)} can be
proven in the same way as Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)}
\textbf{(a)} (but now we must use the fact that the divisors of $b$ are the
same as the divisors of $\left\vert b\right\vert $, and that the divisors of
$c$ are the same as the divisors of $\left\vert c\right\vert $).

[Here is an alternative proof of Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)}
\textbf{(c)}: Using Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)} \textbf{(a)}, we
can find that $\gcd\left(  \left\vert b\right\vert ,c\right)  =\gcd\left(
b,c\right)  $\ \ \ \ \footnote{\textit{Proof.} We must prove that $\gcd\left(
\left\vert b\right\vert ,c\right)  =\gcd\left(  b,c\right)  $. If $\left\vert
b\right\vert =b$, then this is obvious. Hence, for the rest of this proof, we
WLOG assume that $\left\vert b\right\vert \neq b$.
\par
Clearly, $\left\vert b\right\vert $ is either $b$ or $-b$. Thus, $\left\vert
b\right\vert =-b$ (since $\left\vert b\right\vert \neq b$). Hence,
$\gcd\left(  \underbrace{\left\vert b\right\vert }_{=-b},c\right)
=\gcd\left(  -b,c\right)  =\gcd\left(  b,c\right)  $ (by Lemma
\ref{lem.bezout.gcd(b,c)=gcd(-b,c)} \textbf{(a)}), qed.}. Using Lemma
\ref{lem.bezout.gcd(b,c)=gcd(-b,c)} \textbf{(b)}, we can find that
$\gcd\left(  \left\vert b\right\vert ,\left\vert c\right\vert \right)
=\gcd\left(  \left\vert b\right\vert ,c\right)  $%
\ \ \ \ \footnote{\textit{Proof.} We must prove that $\gcd\left(  \left\vert
b\right\vert ,\left\vert c\right\vert \right)  =\gcd\left(  \left\vert
b\right\vert ,c\right)  $. If $\left\vert c\right\vert =c$, then this is
obvious. Hence, for the rest of this proof, we WLOG assume that $\left\vert
c\right\vert \neq c$.
\par
Clearly, $\left\vert c\right\vert $ is either $c$ or $-c$. Thus, $\left\vert
c\right\vert =-c$ (since $\left\vert c\right\vert \neq c$).
\par
But Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)} \textbf{(b)} (applied to
$\left\vert b\right\vert $ instead of $b$) yields $\gcd\left(  \left\vert
b\right\vert ,c\right)  =\gcd\left(  \left\vert b\right\vert ,-c\right)  $.
Compared with $\gcd\left(  \left\vert b\right\vert ,\underbrace{\left\vert
c\right\vert }_{=-c}\right)  =\gcd\left(  \left\vert b\right\vert ,-c\right)
$, this yields $\gcd\left(  \left\vert b\right\vert ,\left\vert c\right\vert
\right)  =\gcd\left(  \left\vert b\right\vert ,c\right)  $, qed.}. Hence,
$\gcd\left(  \left\vert b\right\vert ,\left\vert c\right\vert \right)
=\gcd\left(  \left\vert b\right\vert ,c\right)  =\gcd\left(  b,c\right)  $.
This proves Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)} \textbf{(c)}.]
\end{proof}

\begin{lemma}
\label{lem.bezout.gcd(b+uc,c)}Let $b$, $c$ and $u$ be three integers. Then:

\textbf{(a)} We have $\gcd\left(  b,c\right)  =\gcd\left(  b+uc,c\right)  $.

\textbf{(b)} We have $\gcd\left(  b,c\right)  =\gcd\left(  b,ub+c\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.bezout.gcd(b+uc,c)}.]If $\left(  b,c\right)  =\left(
0,0\right)  $, then Lemma \ref{lem.bezout.gcd(b+uc,c)} is obvious (because if
$\left(  b,c\right)  =\left(  0,0\right)  $, then all four integers $b$, $c$,
$b+uc$ and $ub+c$ are $0$). Hence, for the rest of this proof, we WLOG assume
that $\left(  b,c\right)  \neq\left(  0,0\right)  $.

\textbf{(a)} We make the following two observations:

\begin{statement}
\textit{Observation 1:} Every common divisor of $b+uc$ and $c$ is a common
divisor of $b$ and $c$.
\end{statement}

\begin{proof}
[Proof of Observation 1.]Let $d$ be a common divisor of $b+uc$ and $c$. We
must prove that $d$ is a common divisor of $b$ and $c$.

We know that $d$ is a common divisor of $b+uc$ and $c$; hence, $d\mid b+uc$
and $d\mid c$. Now, $d\mid c\mid-uc$. So we know that $d$ divides the two
integers $b+uc$ and $-uc$ (since $d\mid b+uc$ and $d\mid-uc$). Hence, $d$ must
also divide the sum of these two integers. In other words, we have
$d\mid\left(  b+uc\right)  +\left(  -uc\right)  =b$. Now, $d$ divides both $b$
and $c$ (since $d\mid b$ and $d\mid c$). Hence, $d$ is a common divisor of $b$
and $c$. This completes the proof of Observation 1.
\end{proof}

\begin{statement}
\textit{Observation 2:} Every common divisor of $b$ and $c$ is a common
divisor of $b+uc$ and $c$.
\end{statement}

\begin{proof}
[Proof of Observation 2.]Let $d$ be a common divisor of $b$ and $c$. We must
prove that $d$ is a common divisor of $b+uc$ and $c$.

We know that $d$ is a common divisor of $b$ and $c$; hence, $d\mid b$ and
$d\mid c$. Now, $d\mid c\mid uc$. So we know that $d$ divides the two integers
$b$ and $uc$ (since $d\mid b$ and $d\mid uc$). Hence, $d$ must also divide the
sum of these two integers. In other words, we have $d\mid b+uc$. Now, $d$
divides both $b+uc$ and $c$ (since $d\mid b+uc$ and $d\mid c$). Hence, $d$ is
a common divisor of $b+uc$ and $c$. This completes the proof of Observation 2.
\end{proof}

Combining Observation 1 with Observation 2, we conclude that the common
divisors of $b+uc$ and $c$ are the same as the common divisors of $b$ and $c$.

But $\left(  b+uc,c\right)  \neq\left(  0,0\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary (for the sake of
contradiction). Thus, $\left(  b+uc,c\right)  =\left(  0,0\right)  $. Hence,
$b+uc=0$ and $c=0$. Now, $0=b+u\underbrace{c}_{=0}=b$, so that $b=0$. Combined
with $c=0$, this yields $\left(  b,c\right)  =\left(  0,0\right)  $, which
contradicts $\left(  b,c\right)  \neq\left(  0,0\right)  $. This contradiction
shows that our assumption was false, qed.}. Hence, $\gcd\left(  b+uc,c\right)
$ is the greatest of all common divisors of $b+uc$ and $c$ (by the definition
of $\gcd\left(  b+uc,c\right)  $). In other words, $\gcd\left(  b+uc,c\right)
$ is the greatest of all common divisors of $b$ and $c$ (since the common
divisors of $b+uc$ and $c$ are the same as the common divisors of $b$ and
$c$). On the other hand, $\gcd\left(  b,c\right)  $ is the greatest of all
common divisors of $b$ and $c$ (by the definition of $\gcd\left(  b,c\right)
$). Hence, the two numbers $\gcd\left(  b+uc,c\right)  $ and $\gcd\left(
b,c\right)  $ have been characterized in precisely the same way (namely, as
the greatest of all common divisors of $b$ and $c$). Therefore, these two
numbers are equal. In other words, $\gcd\left(  b+uc,c\right)  =\gcd\left(
b,c\right)  $. This proves Lemma \ref{lem.bezout.gcd(b+uc,c)} \textbf{(a)}.

\textbf{(b)} One way to prove Lemma \ref{lem.bezout.gcd(b+uc,c)} \textbf{(b)}
is by arguing similarly to how we argued in our proof of Lemma
\ref{lem.bezout.gcd(b+uc,c)} \textbf{(a)}. Let us, however, proceed
differently: We have%
\begin{align*}
\gcd\left(  b,c\right)   &  =\gcd\left(  c,b\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.bezout.gcd(b,c)=gcd(c,b)}%
}\right) \\
&  =\gcd\left(  \underbrace{c+ub}_{=ub+c},b\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Lemma \ref{lem.bezout.gcd(b+uc,c)} \textbf{(a)}, applied to }c\text{
and }b\\
\text{instead of }b\text{ and }c
\end{array}
\right) \\
&  =\gcd\left(  ub+c,b\right) \\
&  =\gcd\left(  b,ub+c\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Lemma \ref{lem.bezout.gcd(b,c)=gcd(c,b)}, applied to}\\
ub+c\text{ and }b\text{ instead of }b\text{ and }c
\end{array}
\right)  .
\end{align*}
Thus, Lemma \ref{lem.bezout.gcd(b+uc,c)} \textbf{(b)} is proven.
\end{proof}

\begin{lemma}
\label{lem.bezout.gcd(a,0)}Let $a\in\mathbb{N}$.

\textbf{(a)} We have $\gcd\left(  a,0\right)  =a$.

\textbf{(b)} We have $\gcd\left(  0,a\right)  =a$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.bezout.gcd(a,0)}.]\textbf{(a)} We have $\gcd\left(
0,0\right)  =0$. In other words, Lemma \ref{lem.bezout.gcd(a,0)} \textbf{(a)}
holds for $a=0$. Thus, for the rest of the proof of Lemma
\ref{lem.bezout.gcd(a,0)} \textbf{(a)}, we can WLOG assume that $a\neq0$.
Assume this. Thus, $a$ is a positive integer (since $a\in\mathbb{N}$ and
$a\neq0$). Hence, every divisor of $a$ is $\leq a$. Thus, the greatest of all
divisors of $a$ is $a$ itself (since $a$ itself is a divisor of $a$).

We make the following two observations:

\begin{statement}
\textit{Observation 1:} Every divisor of $a$ is a common divisor of $a$ and
$0$.
\end{statement}

\begin{proof}
[Proof of Observation 1.]Let $d$ be a divisor of $a$. We must prove that $d$
is a common divisor of $a$ and $0$.

We have $d\mid a$ (since $d$ is a divisor of $a$) and $d\mid0$ (obviously).
Thus, $d$ divides both $a$ and $0$. Hence, $d$ is a common divisor of $a$ and
$0$. This completes the proof of Observation 1.
\end{proof}

\begin{statement}
\textit{Observation 2:} Every common divisor of $a$ and $0$ is a divisor of
$a$.
\end{statement}

\begin{proof}
[Proof of Observation 2.]Observation 2 is obvious.
\end{proof}

Combining Observation 1 with Observation 2, we see that the common divisors of
$a$ and $0$ are the same as the divisors of $a$.

We have $\left(  a,0\right)  \neq\left(  0,0\right)  $ (since $a\neq0$). Thus,
$\gcd\left(  a,0\right)  $ is the greatest of all common divisors of $a$ and
$0$ (by the definition of $\gcd\left(  a,0\right)  $). In other words,
$\gcd\left(  a,0\right)  $ is the greatest of all divisors of $a$ (since the
common divisors of $a$ and $0$ are the same as the divisors of $a$). In other
words, $\gcd\left(  a,0\right)  $ is $a$ (since the greatest of all divisors
of $a$ is $a$). This proves Lemma \ref{lem.bezout.gcd(a,0)} \textbf{(a)}.

\textbf{(b)} We could prove Lemma \ref{lem.bezout.gcd(a,0)} \textbf{(b)}
similarly how we proved Lemma \ref{lem.bezout.gcd(a,0)} \textbf{(a)}. But we
can just as easily derive Lemma \ref{lem.bezout.gcd(a,0)} \textbf{(b)} from
Lemma \ref{lem.bezout.gcd(a,0)} \textbf{(a)}: We have%
\begin{align*}
\gcd\left(  0,a\right)   &  =\gcd\left(  a,0\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Lemma \ref{lem.bezout.gcd(b,c)=gcd(c,b)}, applied to }0\text{ and
}a\\
\text{instead of }b\text{ and }c
\end{array}
\right) \\
&  =a\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.bezout.gcd(a,0)}
\textbf{(a)}}\right)  .
\end{align*}
This proves Lemma \ref{lem.bezout.gcd(a,0)} \textbf{(b)}.
\end{proof}

Now, we prove the (trivial) particular case of Theorem \ref{thm.bezout} when
$b$ and $c$ are nonnegative integers one of which is $0$:

\begin{lemma}
\label{lem.bezout.0}Let $b\in\mathbb{N}$ and $c\in\mathbb{N}$ be such that
either $b=0$ or $c=0$ (or both). Then, there exist integers $x$ and $y$ such
that $\gcd\left(  b,c\right)  =bx+cy$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.bezout.0}.]We have either $b=0$ or $c=0$. Thus, we
are in one of the following two cases:

\textit{Case 1:} We have $b=0$.

\textit{Case 2:} We have $c=0$.

Let us first consider Case 1. In this case, we have $b=0$. Thus, $\gcd\left(
\underbrace{b}_{=0},c\right)  =\gcd\left(  0,c\right)  =c$ (by Lemma
\ref{lem.bezout.gcd(a,0)} \textbf{(b)}, applied to $a=c$). Compared with
$b0+c1=c1=c$, this yields $\gcd\left(  b,c\right)  =b0+c1$. Hence, there exist
integers $x$ and $y$ such that $\gcd\left(  b,c\right)  =bx+cy$ (namely, $x=0$
and $y=1$). Thus, Lemma \ref{lem.bezout.0} is proven in Case 1.

Let us now consider Case 2. In this case, we have $c=0$. Thus, $\gcd\left(
b,\underbrace{c}_{=0}\right)  =\gcd\left(  b,0\right)  =b$ (by Lemma
\ref{lem.bezout.gcd(a,0)} \textbf{(a)}, applied to $a=b$). Compared with
$b1+c0=b1=b$, this yields $\gcd\left(  b,c\right)  =b1+c0$. Hence, there exist
integers $x$ and $y$ such that $\gcd\left(  b,c\right)  =bx+cy$ (namely, $x=1$
and $y=0$). Thus, Lemma \ref{lem.bezout.0} is proven in Case 2.

Hence, Lemma \ref{lem.bezout.0} is proven in each of the two Cases 1 and 2.
Thus, Lemma \ref{lem.bezout.0} always holds.
\end{proof}

Next, we prove the particular case of Theorem \ref{thm.bezout} when $b$ and
$c$ are nonnegative:

\begin{lemma}
\label{lem.bezout.NN}Let $b\in\mathbb{N}$ and $c\in\mathbb{N}$. Then, there
exist integers $x$ and $y$ such that $\gcd\left(  b,c\right)  =bx+cy$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.bezout.NN}.]We shall prove Lemma \ref{lem.bezout.NN}
by strong induction on $b+c$:

Let $N\in\mathbb{N}$. Assume that Lemma \ref{lem.bezout.NN} holds in the case
when $b+c<N$. We must prove that Lemma \ref{lem.bezout.NN} holds in the case
when $b+c=N$.

We have assumed that Lemma \ref{lem.bezout.NN} holds in the case when $b+c<N$.
In other words, the following holds:

\begin{statement}
\textit{Observation 1:} If $b\in\mathbb{N}$ and $c\in\mathbb{N}$ satisfy
$b+c<N$, then there exist integers $x$ and $y$ such that $\gcd\left(
b,c\right)  =bx+cy$.
\end{statement}

Let now $b\in\mathbb{N}$ and $c\in\mathbb{N}$ be such that $b+c=N$. We are
going to show that
\begin{equation}
\text{there exist integers }x\text{ and }y\text{ such that }\gcd\left(
b,c\right)  =bx+cy. \label{pf.lem.bezout.NN.claim}%
\end{equation}


If we have either $b=0$ or $c=0$ (or both), then (\ref{pf.lem.bezout.NN.claim}%
) is true (by Lemma \ref{lem.bezout.0}). Thus, for the rest of this proof of
(\ref{pf.lem.bezout.NN.claim}), we can WLOG assume that we have neither $b=0$
nor $c=0$. Assume this.

We have neither $b=0$ nor $c=0$. In other words, we have $b\neq0$ and $c\neq
0$. Thus, $b>0$ (since $b\in\mathbb{N}$ and $b\neq0$) and $c>0$ (since
$c\in\mathbb{N}$ and $c\neq0$). Now, we are in one of the following two cases:

\textit{Case 1:} We have $b<c$.

\textit{Case 2:} We have $b\geq c$.

Let us first consider Case 1. In this case, we have $b<c$. Thus, $b\leq c$, so
that $c-b\in\mathbb{N}$. Moreover, $c=\left(  c-b\right)  +\underbrace{b}%
_{>0}>c-b$, so that $c-b<c$ and thus $b+\underbrace{\left(  c-b\right)  }%
_{<c}<b+c=N$. Therefore, we can apply Observation 1 to $c-b$ instead of $c$.
As a result, we conclude that there exist integers $x$ and $y$ such that
$\gcd\left(  b,c-b\right)  =bx+\left(  c-b\right)  y$. Denote these $x$ and
$y$ by $x_{0}$ and $y_{0}$. Hence, $x_{0}$ and $y_{0}$ are integers satisfying
$\gcd\left(  b,c-b\right)  =bx_{0}+\left(  c-b\right)  y_{0}$.

Lemma \ref{lem.bezout.gcd(b+uc,c)} \textbf{(b)} (applied to $u=-1$) yields
\begin{align*}
\gcd\left(  b,c\right)   &  =\gcd\left(  b,\underbrace{\left(  -1\right)
b+c}_{=c-b}\right)  =\gcd\left(  b,c-b\right)  =bx_{0}+\left(  c-b\right)
y_{0}\\
&  =bx_{0}+cy_{0}-by_{0}=b\left(  x_{0}-y_{0}\right)  +cy_{0}.
\end{align*}
Thus, there exist integers $x$ and $y$ such that $\gcd\left(  b,c\right)
=bx+cy$ (namely, $x=x_{0}-y_{0}$ and $y=y_{0}$). Therefore,
(\ref{pf.lem.bezout.NN.claim}) is proven in Case 1.

Let us now consider Case 2. In this case, we have $b\geq c$. Thus,
$b-c\in\mathbb{N}$. Moreover, $b=\left(  b-c\right)  +\underbrace{c}_{>0}%
>b-c$, so that $b-c<b$ and thus $\underbrace{\left(  b-c\right)  }%
_{<b}+c<b+c=N$. Therefore, we can apply Observation 1 to $b-c$ instead of $b$.
As a result, we conclude that there exist integers $x$ and $y$ such that
$\gcd\left(  b-c,c\right)  =\left(  b-c\right)  x+cy$. Denote these $x$ and
$y$ by $x_{0}$ and $y_{0}$. Hence, $x_{0}$ and $y_{0}$ are integers satisfying
$\gcd\left(  b-c,c\right)  =\left(  b-c\right)  x_{0}+cy_{0}$.

Lemma \ref{lem.bezout.gcd(b+uc,c)} \textbf{(a)} (applied to $u=-1$) yields
\begin{align*}
\gcd\left(  b,c\right)   &  =\gcd\left(  \underbrace{b+\left(  -1\right)
c}_{=b-c},c\right)  =\gcd\left(  b-c,c\right)  =\left(  b-c\right)
x_{0}+cy_{0}\\
&  =bx_{0}-cx_{0}+cy_{0}=bx_{0}+c\left(  y_{0}-x_{0}\right)  .
\end{align*}
Thus, there exist integers $x$ and $y$ such that $\gcd\left(  b,c\right)
=bx+cy$ (namely, $x=x_{0}$ and $y=y_{0}-x_{0}$). Therefore,
(\ref{pf.lem.bezout.NN.claim}) is proven in Case 2.

We have now proven (\ref{pf.lem.bezout.NN.claim}) in each of the two Cases 1
and 2. Thus, (\ref{pf.lem.bezout.NN.claim}) always holds (since Cases 1 and 2
cover all possibilities). So we have proven that there exist integers $x$ and
$y$ such that $\gcd\left(  b,c\right)  =bx+cy$.

Now, forget that we fixed $b$ and $c$. We thus have shown that if
$b\in\mathbb{N}$ and $c\in\mathbb{N}$ are such that $b+c=N$, then there exist
integers $x$ and $y$ such that $\gcd\left(  b,c\right)  =bx+cy$. In other
words, Lemma \ref{lem.bezout.NN} holds in the case when $b+c=N$. This
completes the induction step; thus, Lemma \ref{lem.bezout.NN} is proven by
strong induction.
\end{proof}

Now, we can finally deliver the coup-de-gr\^{a}ce to Theorem \ref{thm.bezout}:

\begin{proof}
[Proof of Theorem \ref{thm.bezout}.]We have $\left\vert b\right\vert
\in\mathbb{N}$ and $\left\vert c\right\vert \in\mathbb{N}$. Hence, Lemma
\ref{lem.bezout.NN} (applied to $\left\vert b\right\vert $ and $\left\vert
c\right\vert $ instead of $b$ and $c$) yields that there exist integers $x$
and $y$ such that $\gcd\left(  \left\vert b\right\vert ,\left\vert
c\right\vert \right)  =\left\vert b\right\vert x+\left\vert c\right\vert y$.
Denote these $x$ and $y$ by $x_{0}$ and $y_{0}$. Hence, $x_{0}$ and $y_{0}$
are integers satisfying $\gcd\left(  \left\vert b\right\vert ,\left\vert
c\right\vert \right)  =\left\vert b\right\vert x_{0}+\left\vert c\right\vert
y_{0}$.

But $\left\vert b\right\vert $ is either $b$ or $-b$. In either case,
$\left\vert b\right\vert $ is divisible by $b$ (since both $b$ and $-b$ are
divisible by $b$). Hence, there exists a $\beta\in\mathbb{Z}$ such that
$\left\vert b\right\vert =\beta b$. Similarly, there exists a $\gamma
\in\mathbb{Z}$ such that $\left\vert c\right\vert =\gamma c$. Consider these
$\beta$ and $\gamma$. Now, Lemma \ref{lem.bezout.gcd(b,c)=gcd(-b,c)}
\textbf{(c)} yields%
\[
\gcd\left(  b,c\right)  =\gcd\left(  \left\vert b\right\vert ,\left\vert
c\right\vert \right)  =\underbrace{\left\vert b\right\vert }_{=\beta b}%
x_{0}+\underbrace{\left\vert c\right\vert }_{=\gamma c}y_{0}=\underbrace{\beta
bx_{0}}_{=b\left(  \beta x_{0}\right)  }+\underbrace{\gamma cy_{0}}_{=c\left(
\gamma y_{0}\right)  }=b\left(  \beta x_{0}\right)  +c\left(  \gamma
y_{0}\right)  .
\]
Hence, there exist integers $x$ and $y$ such that $\gcd\left(  b,c\right)
=bx+cy$ (namely, $x=\beta x_{0}$ and $y=\gamma y_{0}$). This proves Theorem
\ref{thm.bezout}.
\end{proof}

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\end{document}