Let $P\in \Bbb{Z}[X]$ be a polynomial with degree $d>1$.


It is conjectured that for all such $P$, their range for integer inputs $R\_P:=P(\Bbb{Z})$ has finite intersection with the set of factorials $\{n!:n\ge 0\}$.


We say that $P$ is “good” if there does not exist some $Q\in \Bbb{Z}[X]\setminus \{X\}$ such that $P \mid P\circ Q$. Examples: if $P=X^2$ then $Q=2X$ shows $P$ isn’t good; if $P=X^2-1$, then $Q=X^2$ shows $P$ isn’t good.


I was curious if there are any counter-examples to the following stronger claim:


For all such good $P$, $R\_P$ does not contain an infinite sequence $a\_1<a\_2<\dots$ where $a\_i \mid a\_{i+1}$ for $i\ge 1$. Or even stronger, there exists a constant $C=C\_P$ so that $R\_P$ does not contain divisibility chains longer than $C$.


Also, is there a nice characterization for when $P$ is good?



Divisibility chains and polynomials
Every $P$ is a counterexample. Indeed, given a polynomial $P$ consider the recursive sequence $b\_{n+1}=f(b\_n)$ where I take $f(x)=x+P(x)$, say. Then $P(b\_{n+1}) = P(b\_n + P(b\_n)) \equiv P(b\_n) \equiv 0 \bmod P(b\_n)$ since $x-y \mid P(x)-P(y)$ in general. Setting $a\_n=P(b\_n) \in R\_P$ this says that $a\_{n+1}$ is divisible by $a\_n$.


One can replace $f$ by more general polynomials. An important property of $f$ is that it permutes the roots of $P$.


