Square Has Same Parity Proof

Proof By Contradiction

Assume $a=2k+1$ and $a^2$ is an even integer $2j$ where $k,j\in Z$ Thus, $a^2=(2k+1{)}^2=4k^2+4k+1=2(2k^2+2k)+1$ Since, $a^2$ cannot be represented as an even number, $a^2$ must be odd for an odd $a$. Thus, for $a^2$ to be even, $a$ must be even aswell. $\square$