Is it true that for any integer n,n3-n is divisible by 6?

Asked by Vivek Menon | 28th Jun, 2013, 07:47: PM

Expert Answer:

n^3 - n

Factor out an 'n':
n(n^2 - 1)

Use difference of squares:
n(n+1)(n-1)

Put the factors in order:
(n-1)*n*(n+1)

Proof that it's divisible by 3:
The number n can leave a remainder of 0, 1, or 2 when divided by 3. If n leaves a remainder of 0, it's divisible by 3, and so is the whole expression. If n leaves a remainder of 1, then n-1 is divisible by 3, and the whole expression is, too. If n leaves a remainder of 2, then n+1 leaves a remainder of 3, which rolls back to 0...so, again, the whole expression is divisible by 3. Therefore, no matter what, the whole expression is divisible by 3.

(This is a restatement of the semi-obvious fact that, amongst any 3 consecutive integers, exactly 1 of them is divsible by 3.)

Divisibility by 2:
n is either odd or even. If n is even, we're done. If n is odd, n+1 is even, and, again, it will be divisible by 2.

Divisibility by 6:
A number is divisibile by 6 iff it's divisible by both 3 and 2.

Answered by  | 29th Jun, 2013, 05:27: PM

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