If $n = 2m - 1$ with $m \in \mathbb{N}$, then $n^2 - 1$ is always divisible by—

For odd $n=2m-1$, $n^2-1=(n-1)(n+1)$ is the product of two consecutive even numbers, one of which is divisible by $4$ and the other by $2$. Their product is therefore always divisible by $8$.