$n^2-1$ is divisible by 8 for odd $n$

Factorization. $n^2-1=(n-1)(n+1)$.

Parity argument. Since $n$ is odd, both $n-1$ and $n+1$ are even. They are consecutive even integers, so one of them is divisible by $2$ and the other by $4$. Hence their product is divisible by $8$.

More formally, there exists $k$ such that $\{n-1,n+1\}=\{2k,2k+2\}=2\{k,k+1\}$. Among $k$ and $k+1$, one is even, so $4\mid (n-1)(n+1)$. Together with the factor $2$ from each, we get $8\mid n^2-1$.

Remark. This is a classical parity result.