Prove: $2730|(n^{13}-n),n\in\mathbb N$

Question

$2730=2\cdot3\cdot5\cdot7\cdot13$

$n^{13}-n=n(n-1)(n+1)(n^2+1)(n^8+n^4+1)$

Divisibility by $2$ and $3$ follows from the product of two and three successive terms.

Divisibility by $5$ follows from Fermat's little theorem: $$n^{5-1}\equiv1(\mod 5)\Rightarrow 5|(n^{13}-n)=(n^4-1)(n^8+n^4+1)$$

How to prove divisibility by $7$ and $13$?

Answer 1

• You have also by Fermat's little theorem :

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

because $13$ is prime.

• Be careful, you have $n(n^4-1)(n^8+n^4+1)=n^{13}-n$, I think you made a typo.

• Dr Sonnhard Graubner treat the case $7$ perfectly !

Answer 2

note that $$n \equiv 0,1,2,3,4,5,6 \mod 7$$ then $$n^{13}\equiv 0,1,2,3,4,5,6 \mod 7$$