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Find all polynomials $()$ with integer coefficients such that for all natural number $$, the number $()$ is natural and divides the number $!$.

My work:

  1. If $P(x)=x$ and $P(x)=1$ are such then satisfy the condition of the problem.

  2. Let's rewrite the polynomial $()$ in the form $^ ()$, where $()$ is not divisible by $$, i.e. $(0)≠0$. We immediately notice that $0≤≤1$, because if $≥2$, then $2!⋮2^(2)⋮4$, which is impossible.

Thinh Dinh
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Roman83
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  • What does $2!⋮2(2)⋮4$ mean? Isn't it obvious that $d$ can't be $≥2$? $P(2)\in {\pm 1, \pm 2}$ so... – lulu May 06 '25 at 12:03
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    It means $4 \mid 2^d Q(2) \mid 2!$ (the $2d$ was a typo, and the dots are a notation for writing $\mid$ but with arguments the other way round). – Daniel May 06 '25 at 12:17
  • There is a strong (young theorem) that certainly helps you: Let $F\in\mathbb{Z}[x]$ be non--constant and $F(x) \not =\pm x$ then there exists a constant $C_F>0$ such that for all distinct $m,n\in\mathbb{N}$, $$ \gcd\bigl(F(m),F(n)\bigr)\le C_F $$ – Sara May 06 '25 at 17:10
  • @Sara For any constant $k$, $\gcd(m^k,n^k)$ for $m\ne n$ is unbounded (e.g., take $m=2n$). So something is seriously wrong with your theorem. – Emil Jeřábek May 06 '25 at 17:52
  • @EmilJeřábek right. I am sure I read something like this somewhere! I think the condition is $F(x) \not = cx^d$. (I apologize for the oversight. I'm currently traveling on a train!) – Sara May 06 '25 at 19:26
  • @Sara The conclusion has to be something different. As it is written now, it can only hold when $F$ is a constant polynomial, as $\gcd(F(n),F(m))=F(n)$ whenever $m\equiv n\pmod{F(n)}$. – Emil Jeřábek May 06 '25 at 20:57
  • @Sara I wonder whether you were perhaps thinking of something like https://math.stackexchange.com/questions/3142427/proof-of-identity-for-gcd-fd-d-geq-1-for-polynomials-fx-in-mathbb/ or https://math.stackexchange.com/questions/4300742/does-gcd-of-values-of-n-th-degree-primitive-integer-polynomial-always-divid/ . – Emil Jeřábek May 07 '25 at 11:39

1 Answers1

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Fact 1: If $n \in \mathbb{N}$, then $n \mid P(x)$ if and only if $n \mid P(x + n)$.

Fact 2: If $P$ is non-constant, then there are infinitely many primes which divide a term of the sequence $P(0), P(1), P(2), \ldots$.

For the second fact, the proof is the following:

Suppose for sake of contradiction that $p_1, \ldots, p_k$ are the only primes which divide a term of the sequence. Let $N$ be large. Then there are $< O((\log N)^k)$ numbers $< N$ which only have prime factors lying in $p_1, \ldots, p_k$. But there are $\approx O(N^{\frac{1}{\deg P}})$ numbers in the sequence $P(0), P(1), \ldots$ which are $< N$. The latter expression is much bigger than the former, provided $N$ is large enough. This gives the desired contradiction.

Now we will prove that if $P$ is non-constant, then $P(0) = 0$. We do this by showing $P(0)$ is divisible by infinitely many primes. Let $p$ be a prime that divides a term of the sequence. Let $x > 0$ be the smallest number such that $p \mid P(x)$. By Fact 1, $x \le p$, but also $p \mid x!$ since $p \mid P(x)$ and $P(x) \mid x!$. Hence $x = p$. Then by Fact 1, $p \mid P(0)$. Applying this for infinitely many primes gives $P(0) = 0$.

So assuming $P$ is non-constant, let $P(x) = xQ(x)$. Note $Q$ satisfies the same property that $P$ did, so if $Q$ is non-constant then it is divisible by $x$. But by your work so far, this is not possible. Hence $Q$ would have to be constant.

This leaves us with the result that either $P$ is constant or $P(x) = cx$ for some constant $c$.

It is then easy to check that $P(x)$ must be one of $-1$, $1$, $x$ or $-x$.

Daniel
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