Problem 1897

We are going to prove that $f(x)=a x^{m}$ for some nonnegative integers $a$ and $m$. If $f(x)$ is the zero polynomial we are done, so assume that $f(x)$ has at least one positive coefficient. In particular $f(1)>0$.

Let $p$ be a prime number. The condition is that $f(n) \equiv 0\ (\bmod\ p)$ implies
$$f\left(n^{\operatorname{rad}(n)}\right) \equiv 0 \quad(\bmod\ p)$$
Since $\operatorname{rad}\left(n^{\operatorname{rad}(n)^{k}}\right)=\operatorname{rad}(n)$ for all $k$, repeated applications of the preceding implication show that if $p$ divides $f(n)$ then
$$f\left(n^{\operatorname{rad}(n)^{k}}\right) \equiv 0 \quad(\bmod\ p) \quad \text{ for all } k$$
The idea is to construct a prime $p$ and a positive integer $n$ such that $p-1$ divides $n$ and $p$ divides $f(n)$. In this case, for $k$ large enough $p-1$ divides $\operatorname{rad}(n)^{k}$. Hence if $(p, n)=1$ then $n^{\operatorname{rad}(n)^{k}} \equiv 1\ (\bmod\ p)$ by Fermat's little theorem, so that
$$f(1) \equiv f\left(n^{\operatorname{rad}(n)^{k}}\right) \equiv 0 \quad(\bmod\ p)$$
Suppose that $f(x)=g(x) x^{m}$ with $g(0) \neq 0$. Let $t$ be a positive integer, $p$ any prime factor of $g(-t)$ and $n=(p-1) t$. So $p-1$ divides $n$ and $f(n)=f((p-1) t) \equiv f(-t) \equiv 0\ (\bmod\ p)$, hence either $(p, n)>1$ or the previous congruence holds. If $(p,(p-1) t)>1$ then $p$ divides $t$ and $g(0) \equiv g(-t) \equiv 0\ (\bmod\ p)$, meaning that $p$ divides $g(0)$.

In conclusion we proved that each prime factor of $g(-t)$ divides $g(0) f(1) \neq 0$, and thus the set of prime factors of $g(-t)$ when $t$ ranges through the positive integers is finite. This is known to imply that $g(x)$ is a constant polynomial, and so $f(x)=a x^{m}$.

Let $f(x)$ be a polynomial with integer coefficients (not necessarily nonnegative) such that $\operatorname{rad}(f(n))$ divides $\operatorname{rad}\left(f\left(n^{\operatorname{rad}(n)}\right)\right)$ for any nonnegative integer $n$. We give a complete description of all polynomials with this property. More precisely, we claim that if $f(x)$ is such a polynomial and $\xi$ is a root of $f(x)$ then so is $\xi^{d}$ for every positive integer $d$.

Therefore each root of $f(x)$ is zero or a root of unity. In particular, if a root of unity $\xi$ is a root of $f(x)$ then $1=\xi^{d}$ is a root too (for some positive integer $d$). In the original problem $f(x)$ has nonnegative coefficients. Then either $f(x)$ is the zero polynomial or $f(1)>0$ and $\xi=0$ is the only possible root. In either case $f(x)=a x^{m}$ with $a$ and $m$ nonnegative integers.

To prove the claim let $\xi$ be a root of $f(x)$, and let $g(x)$ be an irreducible factor of $f(x)$ such that $g(\xi)=0$. If 0 or 1 are roots of $g(x)$ then either $\xi=0$ or $\xi=1$ (because $g(x)$ is irreducible) and we are done. So assume that $g(0), g(1) \neq 0$. By decomposing $d$ as a product of prime numbers, it is enough to consider the case $d=p$ prime. We argue for $p=2$. Since $\operatorname{rad}\left(2^{k}\right)=2$ for every $k$, we have
$$\operatorname{rad}\left(f\left(2^{k}\right)\right) \mid \operatorname{rad}\left(f\left(2^{2 k}\right)\right).$$
Now we prove that $g(x)$ divides $f\left(x^{2}\right)$. Suppose that this is not the case. Then, since $g(x)$ is irreducible, there are integer-coefficient polynomials $a(x), b(x)$ and an integer $N$ such that
$$a(x) g(x)+b(x) f\left(x^{2}\right)=N$$
Each prime factor $p$ of $g\left(2^{k}\right)$ divides $f\left(2^{k}\right)$, so by $\operatorname{rad}\left(f\left(2^{k}\right)\right) \mid \operatorname{rad}\left(f\left(2^{2 k}\right)\right)$ it also divides $f\left(2^{2 k}\right)$. From the equation above with $x=2^{k}$ it follows that $p$ divides $N$.

In summary, each prime divisor of $g\left(2^{k}\right)$ divides $N$, for all $k \geq 0$. Let $p_{1}, \ldots, p_{n}$ be the odd primes dividing $N$, and suppose that
$$g(1)=2^{\alpha} p_{1}^{\alpha_{1}} \cdots p_{n}^{\alpha_{n}}$$
If $k$ is divisible by $\varphi\left(p_{1}^{\alpha_{1}+1} \cdots p_{n}^{\alpha_{n}+1}\right)$ then
$$2^{k} \equiv 1 \quad\left(\bmod p_{1}^{\alpha_{1}+1} \cdots p_{n}^{\alpha_{n}+1}\right)$$
yielding
$$g\left(2^{k}\right) \equiv g(1) \quad\left(\bmod p_{1}^{\alpha_{1}+1} \cdots p_{n}^{\alpha_{n}+1}\right).$$
It follows that for each $i$ the maximal power of $p_{i}$ dividing $g\left(2^{k}\right)$ and $g(1)$ is the same, namely $p_{i}^{\alpha_{i}}$. On the other hand, for large enough $k$, the maximal power of 2 dividing $g\left(2^{k}\right)$ and $g(0) \neq 0$ is the same. From the above, for $k$ divisible by $\varphi\left(p_{1}^{\alpha_{1}+1} \cdots p_{n}^{\alpha_{n}+1}\right)$ and large enough, we obtain that $g\left(2^{k}\right)$ divides $g(0) \cdot g(1)$. This is impossible because $g(0), g(1) \neq 0$ are fixed and $g\left(2^{k}\right)$ is arbitrarily large.

In conclusion, $g(x)$ divides $f\left(x^{2}\right)$. Recall that $\xi$ is a root of $f(x)$ such that $g(\xi)=0$; then $f\left(\xi^{2}\right)=0$, i.e. $\xi^{2}$ is a root of $f(x)$.

Likewise if $\xi$ is a root of $f(x)$ and $p$ an arbitrary prime then $\xi^{p}$ is a root too. The argument is completely analogous, in the proof above just replace 2 by $p$ and "odd prime" by "prime different from $p$."