Parity of Integer equals Parity of its Square

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

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

Even case
Suppose first that $r = 0$, so $p = 2 k$.

Then:
 * $p^2 = \left({2 k}\right)^2 = 4 k^2 = 2 \left({2 k^2}\right)$

and so $p^2$ is even.

Odd case
Suppose instead that $r = 1$, so $p = 2 k + 1$.

Then:


 * $p^2 = \left({2 k + 1}\right)^2 = 4 k^2 + 4 k + 1 = 2 \left({2 k^2 + 2 k}\right) + 1$

and so $p^2$ is odd.

Therefore, if it is not the case that a number is even, then it is not the case that its square is even.

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