Question

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

Answer 1

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

Case-I

When n = 2q,
n²+n=(2q)²+2q
=4q²+2q
= 2q(2q+1)
which is divisible by 2

Case-II

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