Question

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

Answer 1

Any odd positive integer is in the form of 4p+1 & 4p+3 for n=4p+1

$(n^2-1)=(1p+1{)}^2-1=16p^2+8p+1=16p^2+8p=8p(2p+1)$

$=(n^2-1)$ is divisible by 8.

$(n^2-1)=(4p+3{)}^2-1=8(2p^2+3p+1)$

$\Rightarrow n^2-1$ is divisible by 8

$n^2-1$ is divisible by 8