For which $n, 120\mid n^5-n$ is true?
I could show that $30\mid n^5-n$ is always true by following way:
$n^5-n = (n-1)n(n+1)(n^2+1)$
Since the product of three consecutive integers is a factor of $n^5-n$, the number is divisible by $2$ and $3.$
For divisibility by $5$: $n \equiv 0,1,2,3,4 \pmod 5 \implies n^2 \equiv 0,1,4,4,1 \pmod 5\implies n^2+1 \equiv 1,2,0,0,2 \pmod 5$.
It follows that $n^5-n$ is divisible by $5$.
Combining everything together, we conclude that $n^5-n $ is divisible by $30.$
For divisibility by $120$, since, the number is divisible by $3\cdot5=15,$ consider divisibility by $8$:
$n \equiv 0,1,2,3,4,5,6,7 \pmod 8 \implies n^2 \equiv 0,1,4,1,0,1,4,1 \pmod 8\implies n^2+1 \equiv 1,2,5,2,1,2,5,2 \pmod 8$
Therefore $n^2+1$ is never divisible by $8.$ So, $n^5-n$ will be divisible by $120,$ when $n \equiv 0,1,7 \pmod 8.$
The answer key says that the statement is true for all odd values of $n$. Help me.