NT7

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Prove that n^{3}-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\cdot 3))


Alternatively, n^{3}-n=n(n^{2}-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