Exercise 334Number theory › Divisibility and gcdHigh schoolAdded 31 Oct 2026

6 divides n³ − n

#divisibility#factorisation

Statement

Prove that for every integer $n$ , the number $n^3-n$ is divisible by $6$ .

Hint

Factor $n^3-n$ .

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We have $n^3-n=(n-1)\,n\,(n+1)$ , a product of three consecutive integers.

So the product is divisible by $2$ and by $3$ . Since $2$ and $3$ are coprime, it is divisible by $6$ .