\documentclass{article}
\usepackage{amsmath}
\usepackage{amsthm}

\newtheorem{thm}{Theorem}

\title{Divisibility}
\author{Doug Baldwin}
\date{Spring 2021}


\begin{document}

\maketitle

\begin{thm}
If $a$, $b$ and $c$ are integers such that $a|b$ and $a|c$, then $a|(b+c)$.
\end{thm}

\begin{proof}
We assume that $a$, $b$ and $c$ are integers such that $a|b$ and $a|c$, and will show that $a|(b+c)$. Since $a|b$, the definition of divisibility means that $b = xa$ for some integer $x$. Similarly $c = ya$ for some integer $y$. Now
\begin{align*}
b + c &= xa + ya \\
  &= (x+y)a
\end{align*}
So we have shown that $b+c$ is an integer multiple of $a$, since $x + y$ is an integer by closure under addition. Thus we have shown that if $a$, $b$ and $c$ are integers such that $a|b$ and $a|c$, then $a|(b+c)$.
\end{proof}

\end{document}