\documentclass[12pt]{amsart}

\usepackage{amsmath, amssymb, amsthm, enumerate}

\newtheorem{theorem}{Theorem}

\begin{document}

\begin{theorem}
Let $a, b, c \in \mathbb{Z}$. 
If $a|b$ and $a|c$, then $(a^2)|(bc)$.
\end{theorem}

\begin{proof}
Suppose $a|b$, $a|c$, and $a,b,c \in \mathbb{Z}$.
Then $an = b$ and $am = c$ for $m,n \in \mathbb{Z}$.
And $(an)(am) = bc = a^2 nm = a^2 d$ where $d \in \mathbb{Z}$.
Therefore $(a^2)|(bc)$.
\end{proof}

\end{document}