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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.

Bill Dubuque
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SARTHAK GUPTA
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    You missed the remainders $n\equiv 3\pmod{8}$ and $n\equiv5\pmod{8}$. For these $n^5-n\equiv0\pmod{8}$. After some point you started writing $n^5-5$ instead of $n^5-n$, as it was at the beginning of the question. –  Mar 21 '20 at 14:15
  • My apologies, I edited the question. – SARTHAK GUPTA Mar 21 '20 at 14:18
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    It looks like you checked only the factor $n^2+1$, but $n^5-n=n(n^2+1)(n^2-1)$. For remainders $3$ and $5$ the factors $n$ and $n^2+1$ are not divisible by $8$, but $n^2-1$ is. –  Mar 21 '20 at 14:20
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    You have already shown that $3$ and $5$ are always a divisor. Hence $120|n^5-n$ holds if and only if $8|n^5-n$ holds which can easily be checked by inspection (or even easier if we consider $x^2\equiv 1\mod 8$ for every odd $x$) – Peter Mar 21 '20 at 14:22

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