Question

Prove that $n^2-n$ is divisible by 2 for every positive integer.

Answer 1

Any positive integer is of the form $2q$ or $2q+1,$ where q is some integer

When$,$ $n=2q,$

$n²+n=\left(2q\right)²+2q$

$=4q²+2q$

$=2q(2q+1)$

which is divisible by 2$

when $n=2q+1$

$n²-n=(2q+1)²+(2q+1)$

$=4q²+4q+1+2q+1$

$=4q²+6q+2$

$=2(2q²+3q+1)$

which is divisible by$2$

hence $n²+n$ is divisible by $2$ for every positive integer n$.$