Question

Question 6
If n is an odd integer, then show that $n^2-1$ is divisible by 8.

Answer 1

Let $a=n^2-1$.
We know, any odd integer n can be represented by
n = 2q + 1, where q is an integer.
square of n, $n^2=(2q+1{)}^2=4q^2+4q+1=4(q^2+q)+1$
So, $a=n^2-1$ = $4(q^2+q)+1-1$ = $4\left(q\right)(q+1)$
case 1 : q is odd
$\Rightarrow$ (q+1) is even. Then we can write
q + 1 = 2k , for some integer k
then,
$a=4q(q+1)=4q\left(2k\right)=8qk$
$\Rightarrow$ 8 divides a = $n^2-1$

case 2 : q is even
$\Rightarrow$ q = 2k , for some integer k.
then, $a=4q(q+1)=4\left(2k\right)(q+1)=8k(q+1)$
$\Rightarrow$ 8 divides a = $n^2-1$

Therefore for any odd integer n, $n^2-1$ is divisible by 8.