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?

According to Euclid's lemma if $p$ is a prime and $a,b$ are integers, then $p \mid ab$ implies that $p \mid a$ or $p \mid b$.
Since $2$ is prime and it divides $p^2 = p\cdot p$ then either $2 \mid p$ or $2 \mid p$. In either case, $2 \mid p$.
