Exercise 706Number theory › Divisibility and gcdOlympiadAdded 31 Oct 2026

30 divides n⁵ − n

#divisibility#factorisation

Statement

Prove that $n^5-n$ is divisible by $30$ for every integer $n$ .

Hint

Factor $n^5-n=n(n-1)(n+1)(n^2+1)$ and check divisibility by $2$ , $3$ and $5$ separately.

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We have $n^5-n=(n-1)\,n\,(n+1)\,(n^2+1)$ .

By 2 and 3. The product $(n-1)n(n+1)$ of three consecutive integers is divisible by $2$ and by $3$ .
By 5. Look at $n$ modulo $5$ . If $n\equiv0,1$ or $4$ , one of $n$ , $n-1$ , $n+1$ is divisible by $5$ . If $n\equiv2$ or $3$ , then $n^2\equiv4$ , so $n^2+1\equiv0\pmod5$ .

The numbers $2,3,5$ are pairwise coprime, so $n^5-n$ is divisible by $2\cdot3\cdot5=30$ .