Prove: For every integer n, if n^3-n is not divisble by 4, then n is even. Solution Note that n3 - n = n(n2 -1) = n(n-1)(n+1) = (n-1)n(n+1). Hence it is a product of three consecutive integers. Now since every alternate even integer is a multiple of 4, it follows that neither n-1 nor n+1 must be even since if one of them were, then one of these two should be a multiple of 4 which is not allowed by assumption. In particular, n-1, n+1 are both odd, or equivalently, n is even..