Parity of Integer equals Parity of its Square/Odd

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Theorem

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

Let $p$ be odd.


Then $p^2$ is also odd.


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
$r = 1$ corresponds to the case of $p$ being odd.


Let $r = 1$, so:

$p = 2 k + 1$

Then:

$p^2 = \paren {2 k + 1}^2 = 4 k^2 + 4 k + 1 = 2 \paren {2 k^2 + 2 k} + 1$

and so $p^2$ is odd.

$\blacksquare$


Sources

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