Prove n^3-n is multiple of 6 for all n
factoring gives form n(n+1)(n-1) from here it should be realised that if n is not a multiple of 3 then n-1 is or n+1 is as multiples of 3 go up in 3s. Hence one of the values either n,n-1 or n+1 must be multiple of 3.
AN
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