Difference between Odd Squares is Divisible by 8

Theorem
Let $a$ and $b$ be odd integers.

Then $a^2 - b^2$ is divisible by $8$.

Proof
Let $a = 2 m + 1$, $b = 2 n + 1$.

Then:

Suppose $m - n$ is even such that $m - n = 2 k$.

Then:
 * $a^2 - b^2 = 4 \paren {2 k} \paren {m + n + 1} = 8 k \paren {m + n + 1}$

and so is divisible by $8$.

Suppose $m - n$ is odd such that $m - n = 2 k + 1$.

Then:
 * $m + n + 1 = m + \paren {2 k + 1 + m} + 1 = 2 m + 2 k = 2 \paren {m + k}$

and so:
 * $a^2 - b^2 = 4 \paren {2 k + 1} 2 \paren {m + k} = 8 \paren {2 k + 1} \paren {m + k}$

and so is again divisible by $8$.

The result follows.