Parity of Integer equals Parity of its Square/Even
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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