$P(x) = n!$ where $P(x) \in \mathbb Z[x]$ has finitely many $(x,n) \in \mathbb Z^{2}$ assuming $abc$ conjecture.
Consider the following variant: Given $c,d,r,s,k \in \mathbb Z$ and $P(x) = n!$ where $P(x) \in \mathbb Z[x]$ with coefficients of $P(x)$ ~ $O(k^{c})$ and degree($P(x)$) ~ $O(\log^{d} {k})$, is there $(x,n) \in \mathbb Z^{2} :x$ ~ $O(\log^{r} {k})$ and $n$ ~ $O(k^{s})$?
What fraction $\epsilon$ of $n \in \mathbb Z$ would have nice relations like that? Is it something harder than what the $abc$ conjecture (whose proof is not known) could give?
It seems if $\epsilon > 0$ is fixed, then one might be able compute the factorial for most $n'$ quickly as it guarantees an $n$ close enough to $n'$ whose factorial can be found using the diophantine relation (provided one is able to find an appropriate $P(x)$ and its solution quickly) and another constant number of multiplies should suffice to get $n'!$ since $\epsilon > 0$ is fixed.
