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Prove that $390\mid n^{13}-n$ for all $n\in\mathbb Z$

Attempt:

$$390=2\times3\times5\times13$$

Fermat theorem: $a^p\equiv a\pmod p$, here:

$$n^{13}\equiv n \pmod{13}$$

So $\color{red}{13}\mid n^{13}-n$ but what can I say about $\color{red}{390}$?

MaximusFastidiousIrreverence
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Error 404
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  • Hint: $2$ divides $n^{13}-n$ for every $n$. Indeed if $n$ is even is clear and if $n$ is odd is clear as well... – Giovanni De Gaetano Jul 05 '16 at 07:51
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    Hint for $5\mid n^{13}-n$: by Fermat's little thm, $n^5\equiv n (mod ;5)$, so $n^{13}= (n^5)^2\cdot n^3\equiv n^2\cdot n^3=n^5\equiv n$. – cjackal Jul 05 '16 at 07:51
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    Factorization $n(n-1)(n+1)(n^2+n+1)(n^2-n+1)(n^2+1)(n^4-n^2+1)$ – Takahiro Waki Jul 05 '16 at 08:11
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    In addition to the duplicate target this thread also has good answers. I guess it is a matter of taste which set of answers you find more useful. When I closed the latter thread I was applying the simple but crude principle that the older thread has the right of way. This has since been re-evaluated, and you are more than welcome to weigh in in this occasion. Observe that $390\cdot7=2730$. – Jyrki Lahtonen Jul 05 '16 at 08:26
  • Thank you @JyrkiLahtonen $~~~~$ – Error 404 Jul 05 '16 at 08:29

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By Fermat's theorem, $n^{p-1} \equiv 1 \mod p$ when $p$ is a prime and $n$ is prime to $p$. Assume that none of $2,3,5,13$ divide $n$. Then $n \equiv 1\mod 2$ and hence $n^{12} \equiv 1 \mod 2 $, $n^2 \equiv 1 \mod 3$ and hence $n^{12} \equiv 1 \mod 3$, $n^4 \equiv 1 \mod 5$ and consequently, $n^{12} \equiv 1 \mod 5$. Finally, $n^{12} \equiv 1 \mod 13$. Thus in this case, $n^{12} \equiv 1 \mod 390$ and hence $n^{13} \equiv n \mod 390$. If any of the primes $2,3,5,13$ divide $n$, then $n^{13} - n = n(n^{12}-1)$ is also divisible by that prime. This completes the proof.