Question

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

Answer 1

Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p.

Let $n=4p+1$,

$(n^2–1)=(4p+1{)}^2–1=16p^2+8p+1=16p^2+8p=8p(2p+1)$

$\Rightarrow (n^2–1)$ is divisible by 8.

$(n^2–1)=(4p+3{)}^2–1=16p^2+24p+9–1=16p^2+24p+8=8(2p^2+3p+1)$

$\Rightarrow n^2–1$ is divisible by 8.

Therefore, $n^2–1$ is divisible by 8 if n is an odd positive integer.