Question

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

Answer 1

We know that any odd positive integer is of the form $4q+1$ or, $4q+3$ for some integer $q$.

So, we have the following cases:
Case $I$ When $n=4q+1$
In this case, we have

$n^2-1=(4q+1{)}^2-1=16q^2+8q+1-1={16}^2+8q=8q(2q+1)$
$\Rightarrow n^2-1$ is divisible by $8$ [$\because 8q(2q+1)$ is divisible by $8$]
Case $II$ When $n=4q+3$
In this case, we have

$n^2-1=(4q+3{)}^2-1=16q^2+24q+9-1=16q^2+24q+8$
$\Rightarrow n^2-1=8(2q^2+3q+1)=8(2q+1)(q+1)$

$\Rightarrow n^2-1$ is divisible by $8$ [$\because 8(2q+1)(q+1)$ is divisible by $8$]

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