Since this kind or problem occurs so often, I think we should directly solve the following: Determine the greatest common divisior $d_k$ of all the elements of $S_k = \{n^k-n |n \in \mathbb{Z} \}$. The result is $d_k=\prod_{\substack{p \in \mathbb P \\ (p-1)|(k-1)}} p$ (being the denominator of the $(k-1)$-th Bernoulli number if $2|(k-1)$) and follows directly if using primitive roots.

* note to self: work out ZeTaX's solution with primitive roots and move to that chapter if appropriorate *

For the initial one: 2^13-2 factors as 2*3^2*5*7*13 so the only possible factors of the gcd are 2,3(twice),5,7,13 But using Fermat's little theorem we can prove that 2,3,5,7,13|(n^13-n) We now need to consider whether 9|n^13-n for every n, but this does not happen for n=3 so the gcd is 2*3*5*7*13=2730 PostScript In Fermat's little theorem we can only assume that n is not congruent to zero modp(where p prime). But obviously n^13 is congruent to n for every n congruent to zero modp

there will not exist a prime with its exponential more than 1 as when we let $n=p$ then it is clear that $V_{p}p(p^{12}-1)=1$ $ord_p(n)| p-1$ and also $ord_p(n)|12$ so the possible values for $ord_p(n)= 1,2,3,4,6,12$ We must note that p cant be bigger than 12 as there must exist a primitive root so any prime bigger than 12 will not be a common divisor so the only possible primes are $3,5,7,13$ greatest common divisor: $3\times5\times7\times13$