# Prove if n2n^2 is even, then nn 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?

It could use a little more explanation, but yes, it works. I’d expand it to point out explicitly why $n(n+1)$ is even and that $n=(n^2+n)-n^2$ is then the difference of two even numbers and as such is even (assuming that you already have this fact available to use).
An alternative approach is to show that if $n$ is odd, then $n^2$ is odd; the desired result is the contrapositive and therefore follows at once.