Question

Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive integer.

Answer 1

Let $n=2k+1$ ($k\in$ Integers)

$n^2-1=(2k+1{)}^2-1=4k^2+4k$

$=4k(k+1)$

If $k$ is even, we are done, if $k$ is odd , then $k+1$ is even and we one done, as $4$ is already there as a factor and another even number will make $8$ a factor.