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\newtheorem*{definition}{Definition}
\newtheorem*{example}{Example}
\newtheorem*{exercise}{Exercise}
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\newtheorem*{lemma}{Lemma}
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\begin{document}
\title{A Short Lemma}
\begin{lemma} 
If $a^2$ is an even nonnegative integer, then so is $a$.
\end{lemma}
\begin{proof}
The proof is by a contrapositive argument.  We will show that if $a$ \textit{is not} an even nonnegative integer, then $a^2$ is not an even nonnegative integer.  
\par\vspace{0.2 cm}
Suppose $a$ is a nonnegative integer that is \textit{not} even.  Then by definition, $a$ can be written in the form
\[
a=2k+1
\vspace{0.2 cm}
\]
for some nonnegative integer $k$, and therefore
\[
a^2=(2k+1)^2=4k^2+4k+1\mbox{   for some nonnegative integer } k
\vspace{0.2 cm}
\]
Let $j=2k^2+2k$. Then $j$ is also a nonnegative integer and by substitution
\[
a^2=4k^2+4k+1=2(2k^2+2)+1=2j+1 
\vspace{0.2 cm}
\]
and so by definition $a^2$ is an odd nonnegative integer, and therefore \textit{not} and even nonnegative integer.
\par\vspace{0.2 cm}\noindent
This completes the proof.
\end{proof} 
\end{document}



