Confused by proof of the irrationality of root 2: if $p^2$ is divisible by $2$, then so is $p$. In typical proofs of the irrationality of $\sqrt{2}$, I have seen the following logic:

If $p^2$ is divisible by $2$, then $p$ is divisible by $2$.

Perhaps I am being over-analytical, but how do we know this to be true? IE. do we require a proof of this implication, or is it simply fact?

$\, n\,$ odd $\Rightarrow n^2$ odd $ $ by $(1\!+\!2i)^2 = 1\!+\!2(2i\!+\!2i^2).\, $ So $\,n^2$ even $\,\Rightarrow n\,$ even by contraposition.
