NT7

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Prove that n3 − n is divisible by 6.

Evaluate n^3-n (\mbox{mod}\ 2) for n = 0,1

0^3-0\equiv 0
1^3-1\equiv 0

Evaluate n^3-n (\mbox{mod}\ 3) for n = 0,1,2

0^3-0\equiv 0
1^3-1\equiv 0
2^3-2\equiv 0

Thus n^3-n \equiv 0 (\mbox{mod}\ 2) and
n^3-n \equiv 0 (\mbox{mod}\ 3)
so n^3-n \equiv 0 (\mbox{mod}\ (2\cdot3))


Alternatively, n3 − n = n(n2 − 1) = n(n − 1)(n + 1) = (n − 1)n(n + 1)

(n-1)n(n+1) \equiv 0 (\mbox{mod}\ 2) since it is a product of more than two consecutive integers
(n-1)n(n+1) \equiv 0 (\mbox{mod}\ 3) since it is a product of three consecutive integers
Hence as above

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