Prove that if $m^2$ is even then $m$ is even

Question

$m, n \in \mathbb{Z}$ $m^2 = 2n^2 \implies m = 2k$ for some $k \in \mathbb{Z}$

In other words, the first statement implies $m$ is divisible by 2. Why?

My professor used this without proving it.

My idea is:

$m = 2(\frac{n^2}{m})$, but this is not enlightening, since $\frac{n^2}{m} \in \mathbb{Q}$, and it must be an integer for divisibility.

Another idea is:

Since $\mathbb{Z}$ has closure under multiplication, we can say this implies that $m^2$ is divisible by 2, but again, the same issue arises.

Answer 1

Hint: What do you get when you multiply two odd integers together?