Parity of Integer equals Parity of its Square

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Theorem

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


Then $p$ is even if and only if $p^2$ 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

Let $r = 0$, 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.

$\Box$


Odd case

Let $r = 1$, 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.

$\Box$


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.

$\blacksquare$


Sources