Let $Q(x) = a_{2023}x^{2023}+a_{2022}x^{2022}+\dots+a_{1}x+a_{0} \in \mathbb{Z}[x]$ be a polynomial with integer coefficients. For an odd prime number $p$ we define the polynomial $Q_{p}(x) = a_{2023}^{p-2}x^{2023}+a_{2022}^{p-2}x^{2022}+\dots+a_{1}^{p-2}x+a_{0}^{p-2}.$ Assume that there exist infinitely primes $p$ such that $$\frac{Q_{p}(x)-Q(x)}{p}$$is an integer for all $x \in \mathbb{Z}$. Determine the largest possible value of $Q(2023)$ over all such polynomials $Q$. Proposed by Nikola Velov
Problem
Source: 2023 Macedonian Team Selection Test P5
Tags: algebra, polynomial, abstract algebra
L567
21.05.2023 16:38
Solved with starchan, AdhityaMV, Siddharth03, mueller.25 The maximum value of $Q(2023)$ is $\frac{2023^{2024}-1}{2022}$, achieved when $Q(x) = x^{2023}+x^{2022}+\dots + 1$. We show that each $a_i \in \{-1, 0, 1\}$ after which it is clear that we cannot do better than the value prescribed above. Construct the polynomial $R \in \mathbb{Q}[x]$ with \[R(x) = \sum_{i, a_i \ne 0} \left(a_i-\frac{1}{a_i}\right) x^i\]Using the given hypotheses for large primes $p$, we observe that $R(x)$ for any fixed $x \in \mathbb{Z}$ is divisible by infinitely many large primes, implying that $R$ is all zero polynomial. Hence each $a_i \ne 0$ is one of $-1$ and $1$. We are done. NOTE: The below is copy pasting my solution, ignore it.
starchan
21.05.2023 16:39
NOTE: The above guy is trying to copy my solution. Beware the 567 Solved with L567, AdhityaMV, Siddharth03, mueller.25
The maximum value of $Q(2023)$ is $\frac{2023^{2024}-1}{2022}$, achieved when $Q(x) = x^{2023}+x^{2022}+\dots + 1$. We show that each $a_i \in \{-1, 0, 1\}$ after which it is clear that we cannot do better than the value prescribed above. Construct the polynomial $R \in \mathbb{Q}[x]$ with \[R(x) = \sum_{i, a_i \ne 0} \left(a_i-\frac{1}{a_i}\right) x^i\]Using the given hypotheses for large primes $p$, we observe that $R(x)$ for any fixed $x \in \mathbb{Z}$ is divisible by infinitely many large primes, implying that $R$ is all zero polynomial. Hence each $a_i \ne 0$ is one of $-1$ and $1$. We are done.