$n^2-1$ is divisible by $8$, if $n$ is

The correct option is A: an odd positive integer
Any positive odd integer is of the form of $(4m+1)$ or $(4m+3)$

Let $n=4m+1$

so, $n^2-1=(4m+1{)}^2-1=16m^2+1-(2\times 4m\times 1)-1$

$=16m^2-8m$

$=8m\times (2m-1)$

So, $n^2-1$ is divisible by $8$.

When $n=4m+3$

$n^2-1=(4m+3{)}^2-1=16m^2+3^2+(2\times 4m\times 3)-1$

$=16m^2+24m+8$

$=8(2m^2+3m+1)$

So, $n^2-1$ is divisible by $8$

$\therefore$ $n^2-1$ is divisible by $8$ when n is odd positive integer.