Question

If m is an odd positive integer, then show that $m^2-1$ is divisible by $8$.

Answer 1

Given, m is any odd positive integer.
$\therefore$ m will leave a remainder $1$ or $3$ when divided by $4$,
$\therefore$ m$=4q+1$ or $4q+3$ for some integer q.
(i) When $m=4q+1$
$\Rightarrow m^2=(4q+1{)}^2=16q^2+8q+1$
$\Rightarrow m^2-1=8(2q^2+q)=8k$
(where $k=2q^2+q$)
$\therefore m^2-1$ is divisible by $8$.
(ii) When $m=4q+3$
$\Rightarrow m^2=(4q+3{)}^2=16q^2+24q+9$
$\Rightarrow m^2=16q^2+24q+8+1$
$\Rightarrow m^2-1=16q^2+24q+8$
$=8(2q^2+3q+1)$
$\Rightarrow m^2-1=8k$(where $k=2q^2+3q+1$)
$\therefore m^2-1$ is divisible by $8$.
Hence, $m^2-1$ is divisible by $8$ for any positive old integer m.
Hence proved.