Prove if 2 divides $a^2$, then 2 divides $a$. If 2 divides $a^2$, then 2 divides a.
I know that 2 divides $a^2$ means there is some integer $n$ such that $a^2 = 2n$, 
and similarly, 2 divides $a$ means there is some integer $m$ such that $a = 2m$
I thought I could rewrite $a^2 = 2n$ into this $=  a = 2(n/a)$ but I don't think that helps, because I'm not sure $n/a$ is an integer.  
Thank you for any help!

By division algorithm, $a=2q+r$ where $r=0\ \text{or}\ 1$ and $q\in\mathbb{Z}$. Now $a^2=4q^2+4qr+r^2$. Since $2|a^2$ it follows that $2|r^2$, whence $r=0$. 
OR use the fact that if $p$ is a prime such that $p$ divides $ab$ then $p$ divides $a \ \text{or}\ b$.
