Parity of Integer equals Parity of its Square

Theorem
Let $p \in \Z$ be an integer.

Then $p$ is even $p^2$ is even.

Proof
Let $p$ be an integer.

By the Division Theorem, there are unique integers $k$ and $r$ such that $p = 2k + r$ and $0 \le r < 2$.

That is, $r = 0$ or $r = 1$, where $r = 0$ corresponds to the case of $p$ being even and $r = 1$ corresponds to the case of $p$ being odd.

Odd case
Therefore, if it is not the case that an integer is even, then it is not the case that its square is even.

Conversely, if it is the case that an integer is even (and also a perfect square), then it is the case that its square root is even.