Parity of Integer equals Parity of its Square/Even

Theorem

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

Let $p$ be even.

Then $p^2$ is also even.

Proof

Let $p$ be an integer.

By the Division Theorem, there exist 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.

Let $r = 0$, so:

$p = 2 k$

Then:

$p^2 = \paren {2 k}^2 = 4 k^2 = 2 \paren {2 k^2}$

and so $p^2$ is even.

$\blacksquare$

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes