If $n$ is an odd integer,then show that $n^{2}-1$ is divisible by $8$.

(N/A) Let $n$ be an odd integer. Any odd integer can be represented in the form $n = 2k + 1$ for some integer $k$.
Substituting this into the expression $n^{2} - 1$:
$n^{2} - 1 = (2k + 1)^{2} - 1$
$= (4k^{2} + 4k + 1) - 1$
$= 4k^{2} + 4k$
$= 4k(k + 1)$
Since $k$ and $k + 1$ are consecutive integers,one of them must be even. Therefore,the product $k(k + 1)$ is divisible by $2$.
Let $k(k + 1) = 2m$ for some integer $m$.
Then,$n^{2} - 1 = 4(2m) = 8m$.
Since $n^{2} - 1$ is a multiple of $8$,it is divisible by $8$ for any odd integer $n$.