% LaTeX solution template for Math 235
% -----------------------------------------------------------------

% Like most advanced LaTeX files, this one begins with a lot of
% boilerplate. You don't need to understand (or even read) most of it.
% All you need to do is fill in your name, email address,
% and the number of the pset. (Search for "METADATA" to find the place
% for this.) Then, you can go straight to the "EXERCISE 1"
% section and start writing your solutions.
% The "VARIOUS USEFUL COMMANDS" section is probably worth taking a
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\usepackage[T1]{fontenc} % Use 8-bit encoding that has 256 glyphs
\usepackage[english]{babel} % English language/hyphenation
\usepackage{amsmath,amsfonts,amsthm,amssymb} % Math packages
\usepackage{mathrsfs}    % More math packages
\usepackage{sectsty}  % Allows customizing section commands
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                      % and defines \url and \href commands.
\usepackage{graphicx} % For embedding graphics files.
\usepackage{framed}   % For the "leftbar" environment used below.
\usepackage{ifthen}   % Used for the \powset command below.
\usepackage{lastpage} % for counting the number of pages
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\usepackage[usenames,dvipsnames]{xcolor}
\usepackage{tikz}     % This is a powerful tool to draw vector
                      % graphics inside LaTeX. In particular, you can
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\usepackage{verbatim} % For the "verbatim" environment, in which
                      % special symbols can be used freely without
                      % confusing the compiler. (And it's typeset in
                      % a constant-width font.)
                      % Useful, e.g., for quoting code (or ASCII art).

%\numberwithin{table}{section} % Number tables within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)

\setlength\parindent{20pt} % Makes indentation for paragraphs longer.
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%	VARIOUS USEFUL COMMANDS
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% The commands below might be convenient. For example, you probably
% prefer to write $\powset[2]{V}$ for the set of $2$-element subsets
% of $V$, rather than writing $\mathcal{P}_2(V)$.
% Notice that you can easily define your own commands like this.
% Caveat: Some of these commands need to be properly "guarded" when
% they occur in subscripts or superscripts. So you should not write
% $K_\CC$, but rather $K_{\CC}$.
\newcommand{\CC}{\mathbb{C}} % complex numbers
\newcommand{\RR}{\mathbb{R}} % real numbers
\newcommand{\QQ}{\mathbb{Q}} % rational numbers
\newcommand{\NN}{\mathbb{N}} % nonnegative integers
\newcommand{\Z}[1]{\mathbb{Z}/#1\mathbb{Z}} % integers modulo k
                                            % (syntax: "\Z{k}")
\newcommand{\ZZ}{\mathbb{Z}} % integers
\newcommand{\id}{\operatorname{id}} % identity map
\newcommand{\lcm}{\operatorname{lcm}}
% Lowest common multiple. For historical reasons, LaTeX has a \gcd
% command built in, but not an \lcm command. The preceding line
% rectifies that.
\newcommand{\rev}{\operatorname{rev}} % reversal of a walk
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% $\powset[k]{S}$ stands for the set of all $k$-element subsets of
% $S$. The argument $k$ is optional, and if not provided, the result
% is the whole powerset of $S$.
\newcommand{\set}[1]{\left\{ #1 \right\}}
% $\set{...}$ compiles to {...} (set-brackets).
\newcommand{\abs}[1]{\left| #1 \right|}
% $\abs{...}$ compiles to |...| (absolute value, or size of a set).
\newcommand{\tup}[1]{\left( #1 \right)}
% $\tup{...}$ compiles to (...) (parentheses, or tuple-brackets).
\newcommand{\ive}[1]{\left[ #1 \right]}
% $\ive{...}$ compiles to [...] (Iverson bracket, aka truth value).
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
% $\floor{...}$ compiles to |_..._| (floor function).
\newcommand{\verts}[1]{\operatorname{V}\left( #1 \right)}
% $\verts{...}$ compiles to V(...) (vertex set of a graph/digraph).
\newcommand{\edges}[1]{\operatorname{E}\left( #1 \right)}
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% $\lf{...1}{...2}$ compiles to $...1^{\underline{...2}}$.
% This is a notation for the falling factorial.
\newcommand{\underbrack}[2]{\underbrace{#1}_{\substack{#2}}}
% $\underbrack{...1}{...2}$ yields
% $\underbrace{...1}_{\substack{...2}}$. This is useful for doing
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% justifications. For example, try this out:
% $ \underbrack{(a+b)^2}{= a^2 + 2ab + b^2 \\ \text{(by the binomial formula)}} $
\newcommand{\horrule}[1]{\rule{\linewidth}{#1}} % Create horizontal rule command with 1 argument of height
\newcommand{\nnn}{\nonumber\\} % Don't number this line in an "align" environment, and move on to the next line.
\newcommand{\explain}[1]{\tup{\hspace{-0.5pc}\text{\begin{tabular}{c}#1\end{tabular}\hspace{-0.5pc}}}}
% $\explain{...}$ interprets the "..." as text (you can
% nest some math inside by putting it in $...$'s as you
% would in any other text block) and puts it in parentheses.

%----------------------------------------------------------------------------------------
%	MAKING SUMMATION SIGNS ALWAYS PUT THEIR BOUNDS ABOVE AND BELOW
%	THE SIGN
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% The following are hacks to ensure that sums (such as
% $\sum_{k=1}^n k$) always put their bounds (i.e., the $k=1$ and the
% $n$) underneath and above the sign, as opposed to on its right.
% Same for products (\prod), set unions (\bigcup) and set
% intersections (\bigcap). Remove the 8 lines below if you do not want
% this behavior.
\let\sumnonlimits\sum
\let\prodnonlimits\prod
\let\cupnonlimits\bigcup
\let\capnonlimits\bigcap
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
\renewcommand{\bigcup}{\cupnonlimits\limits}
\renewcommand{\bigcap}{\capnonlimits\limits}

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%	ENVIRONMENTS
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% The incantations below define how theorem environments
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\newtheoremstyle{plainsl}% <name>
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  {}% <Theorem head spec (can be left empty, meaning `normal')>

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\theoremstyle{plainsl}
  \newtheorem{theorem}{Theorem}[section]
  \newtheorem{proposition}[theorem]{Proposition}
  \newtheorem{lemma}[theorem]{Lemma}
  \newtheorem{corollary}[theorem]{Corollary}
  \newtheorem{conjecture}[theorem]{Conjecture}
% Environments that don't:
\theoremstyle{definition}
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  \newtheorem{example}[theorem]{Example}
  \newtheorem{exercise}[theorem]{Exercise}
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\newenvironment{statement}{\begin{quote}}{\end{quote}}
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%----------------------------------------------------------------------------------------
%	METADATA
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\newcommand{\myname}{Darij Grinberg} % ENTER YOUR NAME HERE
\newcommand{\mymail}{darij.grinberg@drexel.edu} % ENTER YOUR EMAIL HERE
\newcommand{\psetnumber}{5} % ENTER THE NUMBER OF THIS PSET HERE

%----------------------------------------------------------------------------------------
%	HEADER AND FOOTER
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\ihead{Solutions to homework set \#\psetnumber} % Page header left
\ohead{page \thepage\ of \pageref{LastPage}} % Page header right
\ifoot{\myname} % left footer
\ofoot{\mymail} % right footer

%----------------------------------------------------------------------------------------
%	TITLE SECTION
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\title{	
\normalfont \normalsize 
\textsc{Drexel University, Department of Mathematics} \\ [25pt] % Your university, school and/or department name(s)
\horrule{0.5pt} \\[0.4cm] % Thin top horizontal rule
\huge Math 235: Mathematical Problem Solving, \\
Fall 2023:
Homework \psetnumber\\% The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
}
\author{\myname}

\begin{document}

\maketitle % This command causes the title part to be printed.

%----------------------------------------------------------------------------------------
%	EXERCISE 1
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 1}

\subsection{Problem}

Let $a, b, c \in \ZZ$. Prove that:

\begin{enumerate}

\item[\textbf{(a)}] We have
\[
\gcd \tup{ b, c } \cdot \gcd \tup{ c,a }
\cdot \gcd \tup{ a, b }
= \gcd \tup{ a, b, c } \cdot \gcd \tup{ bc, ca, ab }  .
\]


\item[\textbf{(b)}] We have
\[
\lcm \tup{ b, c } \cdot \lcm \tup{ c,a }
\cdot \lcm \tup{ a, b }
= \lcm \tup{ a, b, c } \cdot \lcm \tup{ bc, ca, ab }  .
\]


\item[\textbf{(c)}] Assume that $a, b, c$ are nonzero.
Then,
\[
\dfrac{\gcd \tup{ bc, ca, ab }}{\gcd \tup{ a, b, c }}
= \dfrac{\lcm \tup{ bc, ca, ab }}{\lcm \tup{ a, b, c }}
\in \ZZ.
\]

\end{enumerate}

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 2
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 2}

\subsection{Problem}

Prove that $\gcd \tup{a, b} = \gcd \tup{a+b, \lcm \tup{a,b}}$
for any two integers $a$ and $b$.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 3
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 3}

\subsection{Problem}

Fix a prime $p$. Let $a$ and $b$ be two integers satisfying
$\gcd \tup{ a, b } = p$. What values can $\gcd \tup{ a^5,
b^{13} }$ take?

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 4
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 4}

\subsection{Problem}

Let $n, a, b, c \in \NN$ such that $n$ is odd. Prove
that
\[
\dfrac{ \tup{ na } ! \tup{ nb } ! \tup{ nc } !}
{a! b! c! \tup{ \tup{b+c} ! \tup{c+a} ! \tup{a+b} !
}^{\tup{n-1}/2}}
\]
is an integer.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 5
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 5}

\subsection{Problem}

Let $n \in \NN$. Prove that
\[
\tup{n+1} \lcm\tup{ \dbinom{n}{0}, \dbinom{n}{1}, \ldots, \dbinom{n}{n}}
= \lcm \tup{1, 2, \ldots, n+1} .
\]

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 6
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 6}

\subsection{Problem}

Let $n$ be any positive integer. Let $a, b, c$ be three integers.
Assume that $bc \mid a^n$ and $ca \mid b^n$ and $ab \mid c^n$.
Show that $abc \mid \tup{a+b+c}^{n+1}$.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 7
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 7}

\subsection{Problem}

Let $k$ and $n$ be any two positive integers. Prove that an
expression of the form
\[
\pm \dfrac{1}{k} \pm \dfrac{1}{k+1} \pm \dfrac{1}{k+2}
\pm \cdots \pm \dfrac{1}{k+n}
\]
(where each $\pm$ sign is either a $+$ or a $-$ sign) will
never be an integer, no matter what the $\pm$ signs are.

\subsection{Hint}

Show that one of the numbers $k,k+1,\ldots,k+n$
has a higher $2$-valuation than all of the others.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 8
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 8}

\subsection{Problem}

For any positive integer $n$, we let $\tau \tup{n}$ denote the
number of all positive divisors of $n$, and we let
$\zeta \tup{n}$ denote the product of all positive divisors of
$n$. (For example, $\tau \tup{12} = 6$ and
$\zeta \tup{12} = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 12
= 1728$.)

\begin{enumerate}

\item[\textbf{(a)}] Prove that
$\zeta \tup{n} = n^{\tau\tup{n}/2}$ for each integer $n\geq2$.

\item[\textbf{(b)}] Let $n$ and $m$ be two positive integers
satisfying $\zeta \tup{n} = \zeta \tup{m}$. Prove that $n = m$.

\end{enumerate}

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 9
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 9}

\subsection{Problem}

Let $n$ be a positive integer. Let $d_1, d_2, \ldots, d_k$
be all positive divisors of $n$ (including $1$ and $n$ itself,
as usual). For each $i \in \set{1, 2, \ldots, k}$, let $m_i$
be the number of all positive divisors of $d_i$. Prove that
\[
m_1^3 + m_2^3 + \cdots + m_k^3 = \tup{m_1 + m_2 + \cdots + m_k}^2.
\]

\subsection{Hint}

Feel free to use the classical identity
\[
1^3 + 2^3 + \cdots + a^3 = \tup{1 + 2 + \cdots + a}^2,
\]
which holds for every $a \in \NN$.

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 10
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 10}

\subsection{Problem}

Let $p$ be a prime. Let $n \in \NN$.
Set $n? := \prod_{k=1}^{n} k^k = 1^1 \cdot 2^2 \cdot \cdots \cdot n^n$.
Prove that
\[
v_p \tup{n?}
= \sum_{i \geq 1} p^i \cdot
\dfrac{\floor{n/p^i} \cdot \tup{\floor{n/p^i} + 1}}{2}.
\]

\subsection{Solution}

...

%----------------------------------------------------------------------------------------
%	EXERCISE 11
%----------------------------------------------------------------------------------------
\horrule{0.3pt} \\[0.4cm]

\section{Exercise 11}

\subsection{Problem}

Let $\tup{u_1, u_2, u_3, \ldots}$ be a sequence of nonzero integers
such that
\[
\text{every } a, b \in \set{1, 2, 3, \ldots}
\text{ satisfy } \gcd \tup{u_a, u_b} = \abs{u_{\gcd\tup{a,b}}} .
\]
(We have seen such a sequence in Exercise 2.9.6; another example
is the Fibonacci sequence because of \cite[Exercise 3.7.2]{20f}.)

Prove that there exists a sequence $\abs{v_1, v_2, v_3, \ldots}$
of nonzero integers such that
\[
\text{each } n \in \set{1, 2, 3, \ldots} \text{ satisfies }
u_n = \prod_{d \mid n} v_d.
\]
Here, the symbol ``$\prod_{d \mid n}$'' means a
product over all positive divisors $d$ of $n$. Thus, the
equality $u_n = \prod_{d \mid n} v_d.$ means
\begin{align*}
u_1 &= v_1\ \ \ \ \ \ \ \ \ \ \text{for }n=1;\\
u_2 &= v_1v_2\ \ \ \ \ \ \ \ \ \ \text{for }n=2;\\
u_3 &= v_1v_3\ \ \ \ \ \ \ \ \ \ \text{for }n=3;\\
u_4 &= v_1v_2v_4\ \ \ \ \ \ \ \ \ \ \text{for }n=4;\\
u_5 &= v_1v_5\ \ \ \ \ \ \ \ \ \ \text{for }n=5;\\
u_6 &= v_1v_2v_3v_6\ \ \ \ \ \ \ \ \ \ \text{for }n=6;
\end{align*}
and so on.

\subsection{Solution}

...



\horrule{0.3pt} \\[0.4cm]

\begin{thebibliography}{99999999}                                                                                         %

\bibitem[Grinbe20]{20f}Darij Grinberg, \textit{Math 235: Mathematical Problem
Solving}, 10 August 2021. \newline
\url{https://www.cip.ifi.lmu.de/~grinberg/t/20f/mps.pdf}

\end{thebibliography}

\end{document}

