Here's a question I recently found from browsin AoPS, but couldn't figure out the answer:
For a prime $p$, let $\nu_p(x)$ denote the highest power of $p$ dividing $x$, for example $\nu_3(8) = 0$, $\nu_2(24)= 3$, $\nu_5(10) = 1$. Is there a positive integer $m$ such that for all primes $p$, we have $$ \nu_p(2^{p-1} - 1) \leq m$$ ? What if $2$ is replaced by some other positive number ?