Beautiful. Suppose $P$ is non-constant. It is well-known (e.g. by Schur) that, there are infinitely many primes dividing a member of the set $\{P(n):n\in\mathbb{N}\}$. To that end, let $q$ be a prime, such that $q>\sum_{k=1}^{2019} a_k$; and that, $P(a)\equiv 0\pmod{q}$ for some $a\in\mathbb{N}$. Note that, $P(n)\equiv P(a)\equiv 0\pmod{q}$ if $n\equiv a\pmod{q}$. Now, the idea is to select $n\equiv a\pmod{q}$ and $n\equiv 0\pmod{q-1}$. Since $q$ and $q-1$ are coprime, such an $n$ indeed exists by the Chinese remainder theorem. Now, for this $n$, observe that since $q\nmid a_i$ for every $i$ (recall that $q>\sum_k a_k$), Fermat's theorem yield $a_k^{q-1}\equiv 1\pmod{q}$. In particular, for any $n$ divisible by $q-1$, we have $a_k^n\equiv 1\pmod{q}$, that is, $\sum_{k=1}^{2019}a_k^n\equiv 2019\pmod{q}$. Since $q\mid P(n)$, we must therefore have $q\mid 2019$, which clearly is impossible. Thus, the initial hypothesis was wrong, and that, $P$ is non-constant.

@grupyorum Very nice solution! Do you happen to have a reference for this Schur's theorem that you cite? EDIT: It appears that Lemma 1.5 of this paper proves it. https://www.isibang.ac.in/~sury/polys.pdf

yayups wrote: @grupyorum Very nice solution! Do you happen to have a reference for this Schur's theorem that you cite? EDIT: It appears that Lemma 1.5 of this paper proves it. https://www.isibang.ac.in/~sury/polys.pdf The same idea for Schur works in ISL 2009, which it seems you've solved already

Seriously?

Assume $P$ is nonconstantWe know by Schur's lemma that infinitely many primes $p$ divide $P(n)$ as we vary $n$. So let $p\mid P(n)$ for some $n$ such that $p\nmid a_1,\ldots,a_{2019}$. We have \[ p\mid P(n) \mid a_1^n+\cdots+a_{2019}^n \implies a_1^n+\cdots+a_{2019}^n \equiv 0 \pmod{p}.\]But $p\mid P(n) $, so $p \mid P(n+p)$ as well, hence \[ p\mid P(n+p) \mid a_1^{n+p}+\cdots + a_{2019}^{n+p}.\]However, \[ 0\equiv a_1^{n+p}+\cdots+a_{2019}^{n+p} \equiv a_1^{n+1} + \cdots + a_{2019}^{n+1} \pmod{p}. \]Continuing in this logic, we get $a_1^N + \cdots +a_{2019}^N \equiv 0 \pmod{p}$ for any $N\ge n$. Now plug in $N=100n(p-1)>n$ into the aforementioned assertion, and we get $a_i^N = a_i^{100n(p-1)} \equiv 1 \pmod{p}$ by FLT, since we assumed $a_i \not \equiv 0 \pmod{p}$ for any $i$. Hence $0\equiv 2019 \pmod{p}$, which means there are finitely many possibilities for $p$, contradicting Schur's lemma. Therefore, $P$ is constant.

Assume $P$ was nonconstant. First, I claim there are infinitely many primes that divide some $P(x)$ for an integer $x$, this is because if there were finitely many primes $p_{1}, p_{2}, \ldots p_{k}$, first we must have $P(0)\neq 0$ (or else $x | P(x)$ and then just take $x = p$ for prime $P$), then take $M = P(0)p_{1}\ldots p_{k}$ and $P(M)$ can not be divided by $p_{1}, \ldots p_{k}$, contradiction. Now, consider some $p$ that divides $P(n)$, $n\equiv 0\mod p-1$, and $p > 2019$. This is possible, because if we have some $x$ such that $p | P(x)$, then by CRT we want $n\equiv x\mod p, n\equiv 0\mod p-1$, which is possible. Then, \[a_{1}^{n} + \ldots a_{2019}^{n} \leq 2019 \neq p \mod p\]but this is a contradiction, since $p | P(n) | a_{1}^{n} + \ldots + a_{2019}^{n}$ Thus $P$ must be constant.

a sufficiently large prime number must be chosen according to the shur theorem.

This problem goofy. Assume $P$ is non-constant. We claim that $P(n)$ is divisible by infinitely many primes as $n$ ranges. Indeed, if we assume that $p_1,p_2, \ldots, p_k$ are all the primes dividing $P(n)$ then setting $n=p_1 \cdot p_2 \cdots p_k \cdot P(0)$ gives $P(n)$ to be not divisible by any $p_i$. Contradiction. Thus, let $q>a_1,a_2,...,a_{2019},2019$ divide $P(m)$ and not any $a_i$. This implies that $q$ divides $a_1^m+a_2^m+ \cdots +a_{2019}^m$. Notice that $q|P(m) \Rightarrow q|P(m+q)$ which implies \[ a_1^{m+q}+a_2^{m+q}+ \cdots +a_{2019}^{m+q} \equiv a_1^{m+1}+a_2^{m+1}+ \cdots +a_{2019}^{m+1} \equiv 0 \text{ }(\text{mod } q) \]So, keep adding $1$ to each exponent until we get \[ 0 \equiv a_1^{q-1}+ \cdots +a_{2019}^{q-1} \equiv 2019 \text{ }(\text{mod } q)\]So $q$ divides $2019$ which is a contradiction of $q>2019$. $\blacksquare$

By Schur's theorem, take a prime divisor of $P$ greater than $a_1,a_2 \cdots a_{2019},2019$ Let $p \mid P(n)$ then $p \mid P(n+pk)$ now just choose $k$ s.t. $p-1 \mid n+pk$ then we get $0 \equiv a_1^n + a_2^n \cdots a_{2019}^n \equiv 2019 (\mod p)$ which is a contradiction. Hence $P$ is constant polynomial.