Polynomials mapping factorials to factorials

Question

I'm looking for all polynomials $P(x)$ with integer coefficients such that for every $n \in \Bbb N$ there is an $m \in \Bbb N$ such that $P(n!)=m$!. The only solutions seem to be the constant polynomials and $P(x)=x$. Any ideas?

EDIT: When $P$ is linear, i. e. $P(x)=ax+b,a \ne 0$, we easily see that we must have $(n+1) n! > P(n!) > (n-1)!$ for large enough $n$, which yields $P(x)=x$. Thus we can restrivt ourselves to the case deg P>1. A divisibility argument now shows that $P(0)=0$ (by the way, $0 \not \in \Bbb N$). Any suggestions how to proceed further?

Answer 1

Edit: I delete my original answer, because it's unnecessarily complicated.

Denote $P(x)=a_dx^d+\dots +a_0,$ where $d>1, a_d\geq1, a_i\in \mathbb Z.$

For a large enough integer $n$, we have $$n!<\frac{1}2 (n!)^d<P(n!)=m!<(n!)^{d+1}<((d+1)~n)!,$$ hence $n<m<(d+1)~n$. $$d=\lim_{n\to \infty}\dfrac{\ln m!}{\ln n!}=\lim_{n\to \infty}\dfrac{m\ln m}{n\ln n}=\dfrac{m}{n}.$$

$$1=\dfrac{m!}{P(n!)}=\lim_{n\to \infty}\dfrac{(d~n)!}{a_d(n!)^d} \dfrac{a_d(n!)^d}{a_d(n!)^d+\dots +a_0}=\infty,$$ a contradiction.

Answer 2

I find this problem very interesting and gave it a little thought, here is a straightforward and certainly not very beautiful proof. I am sure, you will find some bugs, but I hope, the general idea works.

Let's assume $P(x)$ is as desired and $\deg P>1$. Then, we can write $$P(x)=a_kx^k + \dots + a_lx^l$$ with $a_k,a_l\neq 0$ and $k\geq l$ and $k\geq 2$. We also already know, that $l\geq 1$.

1) First, we exclude the case, where $k=l$. It is easy to see, that this only works for $a_l=1$ and $l=1$, which leads to the constant polynomial.

2) Since we want to have positive numbers, we can safely assume, that $a_k$ is positive. So $P$ is unbounded, there are arbitrarily big pairs $(m,n)\in\mathbb{N}^2$, such that $P(n!)=m!$.

3) In particular for every prime $p\in\mathbb{P}$ there are $m,n\in\mathbb{N}$ with $P(n!)=m!$, such that $m,n\geq p$ and $(2n)^{2lp}\leq n!$. (For given $p$ and $l$, we can choose $n$ big enough, such that this works).

4) Assume $m!\leq ((n+p)!)^l$ for infinitely many $p\in\mathbb{P}$. $$P(n!)\leq ((n+p)!)^l=(n!\cdot(n+1)\dots(n+p))^l\leq (n!\cdot (2n)^p)^l\leq (n!\cdot \sqrt[2l]{n!})^l=(n!)^{l+\frac{1}{2}}$$ In other words. For infinitely many $x$, we have $P(x)\leq x^{l+0.5}$, which only works if $\deg P=l$, but this is 1).

So the opposite is true for infinitely many primes.

4) Now for every such prime $p$, take $m,n\in\mathbb{N}$ as described and let $z\in\mathbb{N}$ denote the biggest integer, such that $p^z\mid (n!)^l$. Since $m! > ((n+p)!)^l\cdot p$, we get $p^{z+1}\mid m!$. From this, we see in $$P(n!)=a_k(n!)^k + \dots + a_l(n!)^l=m!$$ that $p^{z+1}$ also divides each of summands $k,\dots,l+1$, hence also $$p^{z+1}\mid a_l(n!)^l$$ But $p^{z+1}\nmid (n!)^l$, so $p\mid a_l$. This holds for infinitely many primes, so $a_l=0$, a contradiction.