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.