Since $f(n) | (n!)^k$ we see that all prime factors of $f(n)$ is $\leq n $. It is well known that there are infinitely many primes $p$ such that there exist $m$ satisfying $f(m) \equiv 0 \pmod p$ if $f(x)$ is non-constant. For any such $p$, we can assume $1 \leq m \leq p$ but it follows that $m = p$. So for all such $p$ one has that $p | a_0$ where $a_0$ is the constant coefficient. Then $a_0 = 0$ and we can just consider $g(x) = \frac{f(x)}{x}$ which satisfies the same condition as $f(x)$. From this we get that $f(x) = cx^m$ for some integer $c$ and $m$ and it follows that $c = 1$ or $-1$ from checking $n = 1$ and that it sufficies to have $m \leq k$ by checking $n = p$.

First note the trivial solution $f(x)=1$. Now suppose the polynomial is non-constant. Now we prove a lemma: Lemma : For all primes $p$, we have $p \mid f(n) \implies p \mid n$. Proof: We use strong induction on $n$. The base couple cases are trivial by exhausting the few possibilities. Now it suffices to check all $p<n$. Now take some $p<n, p \nmid n$. Then take $n \equiv n_0 \pmod{p}$ where $0<n_0<p$. After that it is clear that $f(n) \equiv f(n_0) \not \equiv 0 \pmod{p}$ so we are done. This means that all nonconstant polynomials satisfying the conditions have a constant term of zero. This means that $n \mid f(n)$. Dividing through by $n$ we receive a new function satisfying the same requirements as $f(n)$, implying that our only solutions are of the form $f(n)= \pm n^j$ for some non-negative $j \le k$. $\Box$ Darn I didn't reload the page in so long that I got massively sniped

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I claim the only solutions are $f(x)=\pm x^a$ for $a=0,1,\ldots, k$. To see that these work, note that $n|n!$ for all positive integer $n$ . If $f$ is constant it must be $\pm 1$, so assume otherwise. I claim only prime factors of $n$ may divide $f(n)$. Note that $a\equiv b\pmod{c}\implies f(a)\equiv f(b)\pmod{c}$ for integer polynomials, so let $d$ be equivalent to $n$ modulo some prime $p$ such that $1\leq d<p$ and thus $f(n)\equiv f(d)\pmod{p}$. But $f(d)|(d!)^{k}$ so it is not divisible by $p$, so $f(n)$ is not divisible by any prime $p$ that does not divide $n$ and the claim is proven. Now, $f(p)=\pm p^l$ for some $0\leq l\leq k$, but as $f$ is nonconstant, we have $l\geq 1$ for infinitely many primes $p$. As $p$ obviously divides all the nonconstant terms of $f(p)$, we then know that the constant term of $f$ is divisible by infinitely many primes, and thus $f$'s constant term is zero. But then we can factor an $n$ out of $f(n)$ and our arguments still hold, so by induction we get that $f$ is $f(x)=\pm x^a$ and we are done.

We infer that $p|f(n) \implies p\leq n$.This means that $(p,f(1))=(p,f(2))=\cdots (p,f(p-1))=1$.Then let $\mathbb{S}_p$ be the set of prime divisors of $f$.If $p \in \mathbb{S}_p$ then $p|f(p) \implies p|f(0)\implies f(0)=0$.Write $f(n)=n^t.P(n)$.Using similar arguments , we will get that $P(0)=0$.Thus , we can reduce the degree and ultimately get $$f(n)=a.n^t$$Since $f(1)|1$ , we have that $a= \pm1$.Also $\pm p^t|p^k \implies t\leq k$

Spamming poly-NT is sooo cool. Solution: The set of solutions is $f(n)=c \cdot n^l$ where $c \in \{-1,1 \}$ and $l \in \{0,1,2,...,k \}$ Proof: If $f$ was constant then it would be $1$ or $-1$ by plugging $n=1$. So now assume that its not constant, hence by Schur we have that we can set a larger enough prime $p$ such that for some $n$ we have $p \mid f(n)$ and now we can set that $n \le p$ and $n \in \mathbb Z^+$ to get $p \mid (n!)^k$ which would mean that $p=n$ hence $p \mid f(p)$ but aince we said we can let $p$ as larger as we want we have that $p \mid a_0$ hence $a_0=0$ which means that the replace $n \cdot g(n)=f(n)$ makes sense, now from here its very easy that we can repeat the same process until we get $n^k \cdot h(n)=f(n)$ in which here $h$ has to be $1$ or $-1$ hence we are done Edit: First solution i post from my new Laptop

Short Answer **Trivial by Schur's Lol ** Long Answer We claim the only solutions are $\boxed{f(n) \equiv \pm n^d}$ (where $0 \leq d \leq k$). Obviously they work. Now we will prove the converse. Assume $f$ is nonconstant (otherwise only solution is $f \equiv 1$ obviously). Define \[f(X):=a_dX^d + \dots + a_0\]Now see that by Schur's Theorem there exists infinitely many primes $p$ such that $p \mid f(n)$ for some $n$. Consider the minimal such $n$. Then we get $p \geq n$ or else $p \mid f(n-p)$ contradicting minimality. But also $p \mid f(n) \mid {\left(n! \right)}^k \implies p \leq n$, and hence $p=n$ and we get \[p \mid f(p) \iff p \mid f(0) \text{ for infinitely many primes $p$ } \iff f(0)=0 \iff a_0=0\]and hence now we can define \[g(X):=\frac{f(X)}{X}\]and repeat the process untill we get $f(X)=a_dX^d$. And then by the relations $f(1) \mid 1$ and $f(2) \mid 2^k$, we get the above solutions, as desired.

v_Enhance wrote: For a fixed integer $k$, determine all polynomials $f(x)$ with integer coefficients such that $f(n)$ divides $(n!)^k$ for every positive integer $n$. $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ If $k=0 \Rightarrow f(n)|1,\forall n\in\mathbb{Z}^+ \Rightarrow f(x)=\pm 1$ Let $p$ be a prime factor of $f(n)$, with $n\in\mathbb{Z}^+$ fixed $\Rightarrow p|f(n)|(n!)^k \Rightarrow p\leq n$ Let $a\in \mathbb{Z}^+_0/ n-ap> 0$($a$ is the maximum that satisfies) $\Rightarrow p|ap=n-(n-ap)|f(n)-f(n-ap)\Rightarrow p|f(n-ap)|(n-ap)!^k \Rightarrow p\leq n-ap \Rightarrow p=n-ap\text{(by definition of a)} \Rightarrow p|n$ Then $p|f(n)\Rightarrow p|n!,\forall p \text{ prime}$ $\Rightarrow f(p)=\pm 1 \text{ or } p|f(p)|p!^k,\forall p\text{ prime}\Rightarrow f(p)\in\{\pm 1, \pm p, \pm p^2, \ldots, \pm p^k\},\forall p\text{ prime}$ Some occur infinite times $\Rightarrow f(x)=\pm x^i, i\in\{0,1,\ldots, k\} \text{ is a solution}$ Combining both cases: $$f(x)=\pm x^i, i\in\{0,1,\ldots, k\} \text{ is a solution}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

Trivial by schur's theorem but cute

We claim $f(x)=x^i,-x^i$ for $0\le i\le k$. The claim is that the prime factors of $f(n)$ are a subset of the prime factors of $n$ (without multiplicity). We prove this by strong induction. It is clearly true for $n=1,2$. For $n>2$ then all prime factors of $f(n)$ are $\le n$, and for each $p<n$ we have $p=n-(n-p)\mid f(n)-f(n-p)$, so $p\mid f(n)\implies p\mid f(n-p)\implies p\mid n-p\implies p\mid n$ as desired. In particular this implies that $f(p)$ is a positive or negative power of $p$ with nonnegative exponent $\le k$ for all $p$. Thus there exist some $0\le i\le k,c\in\{-1,1\}$ for which $f(p)=cp^i$ for infinitely many $p$. Both sides are polynomial in $p$, so since they are equal infinitely many times they are the same polynomial, as desired.