Find the $P G C D$ of all numbers of the form $n^{13}-n$.

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We will find all primes that divide $n^{13}-n$ for all $n$, i.e., $p \mid (n^{12}-1)$ for all $n$ coprime with $p$. Fermat's Little Theorem states that $p \mid (n^{p-1}-1)$, so if $(p-1) \mid 12$, then $p \mid (n^{12}-1)$ for all $n$ coprime with $p$. We have at least $p=2,3,5,7$ and 13. Now let's show that $p^{2}$ does not divide $n^{13}-n$ for all $n$ with a specific case: $p^{2}$ divides $p^{13}$ but not $p$, so $p^{2}$ does not divide $p^{13}-p$. The last step is to show that there are no other prime factors. The ideal solution would be to prove the converse of Fermat's theorem, but here it suffices to observe that $2^{13}-2=8190=2 \cdot 3^{2} \cdot 5 \cdot 7 \cdot 13$. The greatest common divisor is therefore $2 \cdot 3 \cdot 5 \cdot 7 \cdot 13=2730$.