Find the GCD of $S = \{ n^{13} - n \mid n \in \mathbb{Z} \}.$

Question

Determine the greatest common divisor of the elements of the set $S = \{ n^{13} - n \mid n \in \mathbb{Z} \}.$

I put in a few values of $n$ and I know that $10$ is a common factor, but I'm not sure if that's the greatest common factor.

Answer 1

By Fermat's Little Theorem, we know that $13\mid n^{13}-n, 7\mid n^7-n, 5\mid n^5-n,3\mid n^3-n$ and $2\mid n^2-n$. I claim the product $2\cdot3\cdot5\cdot7\cdot13=2730$ is the largest common factor.

Note that $2^{13}-1=8190=3\cdot2730$ and $(3^{13}-3)-(2^{13}-1)=2730\cdot581$ where $\gcd(3,581)=1$ so $2730$ is the answer.

Answer 2

By Fermat, $n^{13} \equiv n \mod p$ if $p-1$ divides $12$. Since the divisors of $12$ are $1,2,3,4, 6, 12$, the corresponding primes are $2, 3, 5, 7, 13$. And indeed the gcd of $2^{13}-2$ and $3^{13}-3$ is $2 \cdot 3 \cdot 5 \cdot 7 \cdot 13$.