If $p^2$ is even, then $p$ is even.

Proof:

Suppose $p^2$ is even and $p$ is not. Since $p$ is odd, let $p = 2k+1$ where $k$ is an integer. Now, square $p$ so that we have,

$$ \begin{align*} p^2 &= (2k+1)^2 \\ &= 4k^2 + 4k + 1 \\ &= 2(2k^2+2k) + 1. \end{align*} $$

$2k^2 + 2k$ is an integer and so $p^2$ has the form of an odd number so it’s odd! But this is a contradiction since we established earlier that $p^2$ is even. Therefore, $p$ can’t be odd and has to be even. $\blacksquare$