Odd Square Modulo 8

Theorem
Let $x \in \Z$ be an odd square.

Then $x \equiv 1 \pmod 8$.

Proof
Let $x \in \Z$ be an odd square.

Then $x = n^2$ where $n$ is also odd.

Thus $n$ can be expressed as $2 k + 1$ for some $k \in \Z$.

Hence $x = n^2 = \left({2 k + 1}\right)^2 = 4 k^2 + 4 k + 1 = 4 k \left({k + 1}\right) + 1$.

But $k$ and $k + 1$ are of opposite parity and can therefore be expressed as $2 r, 2 s + 1$ (either way round).

Hence $x = 4 k \left({k + 1}\right) + 1 = 4 \left({2 r}\right) \left({2 s + 1}\right) + 1 = 8 r \left({2 s + 1}\right) + 1$.

Hence the result.