TITLE: Divisors of $5^{n!}-3^{n!}$
QUESTION [6 upvotes]: Find the number of integers $k$ in $\{1, \dots , n \}$ such that $k \mid 5^{n!}-3^{n!}$.

I've been trying to see just by testing the natural numbers n one at a time, in hopes to see a pattern to this, without much success. I am wondering what I should be attempting to figure this out.

REPLY [2 votes]: Even a bit more explicite:
We have by Euler's theorem that $$ a^{\varphi(m)}-b^{\varphi(m)} \equiv 0 \pmod m$$ where $\varphi(m)$ is the Euler's totient function. A simple consequence from this is, that also 
$$ a^{\varphi(m)\cdot x}-b^{\varphi(m) \cdot x}  \equiv 0 \pmod m$$
for any $x \gt 1$ .
We have also  $\varphi(m) \lt m$ .
Now look at the exponents in your formula: because they are factorials, they contain for some chosen $m \le n$ all numbers $k \lt m$. So one of those $k$ must be the $\varphi()$ of m.     
Now proceed in thinking yourself...