Integral polynomials dividing N! Consider a polynomial $P(X)\in\mathbb Z[X]$. Is it true that $P(N)$ divides $N!$ for infinitely many integer $N$?
This question is motivated by the special case where $P(X) = X^2 + 1$ that appeared in a math olympiad.
I was wondering if anyone can point me to references of this question.
 A: In general this is an open problem.  A closely related problem (essentially equivalent) is to ask for the values $P(n)$ for $n$ of size $X$ to be $X$ smooth (i.e. composed only of primes below $X$).  This is known for quadratic polynomials, but already open for general cubic irreducible polynomials.  For some special polynomials (e.g. those of the form $ax^d+b$) such results are known, thanks to Schinzel, Balog and Wooley etc. For a description of related results, and more references, see this paper of Dartyge, Martin and Tenenbaum. 
A: Wow! I have been thinking about this problem, too. I can prove that for every odd natural number $d$ there exists infinitely many $n \in \mathbb{N}$ such that $n^{d}+1 \mid n!$. Here you have my gorgeous proof of the case $d=3$ of the result (cf. https://oeis.org/A270441):
For every $k \in \mathbb{N}$ we have that 
\begin{eqnarray*}(k^{6}+2k^{4}+k^{2}+1)^{3}+1 &=& (k^{2}-k+1)(k^{2}+2)(k^{2}+k+1) &\cdot& (k^{4}-k^{3}+2k^{2}-2k+1)(k^{4}+1)(k^{4}+k^{3}+2k^{2}+2k+1);\end{eqnarray*} 
therefore, it follows that, for every $k \in \mathbb{N} \setminus \{1\}$, the natural number $N_{k}:=k^{6}+2k^{4}+k^{2}+1$ is such that $N_{k}^{3}+1$ divides $N_{k}!$. QED.
I have to finish something else now, but if you are interested in what I have mentioned, I may add more details about the result that I obtained later on.
