Prove if $n^2$ is even, then $n$ is even. I am just learning maths, and would like someone to verify my proof.
Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.
Is this valid?

That's right. Here is another proof:
$\forall n\in\mathbb{Z}$ if $n$ is odd, then $n^2$ is odd .
Let $m$ be an integer such that $m^2$ is even. Then $m$ must be even, because if $m$ is odd then $m^2$ must be odd, contrary to the hypothesis.
