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

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The number n is part of the set N which is the set of all natural numbers 1, 2, 3...

I do not see how n^2 - n is always divisible by 8. For example if we take n = 3, n^2 - n = 9 - 3 = 6 which is not divisible by 8.

Similarly if n = 2, n^2 - n = 4 - 2 = 2 which is not divisible by 3.

**The expression n^2 - n where n belongs to N is not always divisible either by 8 or by 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.**