How can I show that n^2 - n where n belongs to N always divisible by 8 or 3?

We'll factorize n^2 - n  =
n*(n-1)

We'll give natural values to n, from 1 to
n;

For n = 1 => n*(n-1) = 1*0 =
0

For n = 2 => n*(n-1) = 2*1 = 2, that is not
divisible by 3, nor 8.

Therefore, the
difference n^2 - n is not divisible by 3 or 8, for any natural value of
n.