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

We take it as read that all integers are either even or odd.

• Let $p$ be even.

Then by definition it can be expressed as $p = 2 k$ for some $k \in \Z$.

Thus:

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

and so $p^2$ is even.

• Now, suppose $p$ is not even. That is, $p$.

Then by definition it can be expressed as $p = 2 k + 1$ is odd for some $k \in \Z$.

Thus:

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

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.

$\blacksquare$

Sources

• George F. Simmons: Calculus Gems (1992), Chapter $\text {B}.2$