Find all polynomials $P(x), Q(x)$ with real coefficients such that for all real numbers $a$, $P(a)$ is a root of the equation $x^{2023}+Q(a)x^2+(a^{2024}+a)x+a^3+2025a=0.$
Problem
Source: 2024 Vietnam National Olympiad - Problem 2
Tags: algebra, polynomial
05.01.2024 09:19
wassupevery1 wrote: Find all polynomials $P(x), Q(x)$ with real coefficients such that for all real numbers $a$, $P(a)$ is a root of the equation $x^{2023}+Q(a)x^2+(a^{2024}+a)x+a^3+2025a=0.$ So $P(x)^{2023}+Q(x)P(x)^2+x(x^{2023}+1)P(x)+x(x^2+2025)=0$ $\forall x$ So $P(x)|x(x^2+2025)$ Note also that $x \not|P(x)$ (since this would imply $x^2|2025 x$ and two cases : 1) $P(x)=c$ constant $\forall x$ for some nonzero $ c$ This implies $\boxed{P(x)=c,Q(x)=-\left(\frac 1cx^{2024}+\frac 1{c^2}x^3+(\frac {2025}{c^2}+\frac 1c)x+c^{2021}\right)\quad\forall x}$, which indeed fits, whatever is $c\in\mathbb R\setminus\{0\}$ 2) $P(x)=c(x^2+2025)$ $\forall x$ for some nonzero $ c$ This implies $P(x)^{2023}+Q(x)P(x)^2+x(x^{2023}+1)P(x)+\frac 1cxP(x)=0$ And so $P(x)^{2022}+Q(x)P(x)+x(x^{2023}+1)+\frac 1cx=0$ And so $P(x)|x^{2024}+\frac{c+1}cx$, impossible if $c\in\mathbb R$ (just set here $x=i\sqrt{2025}$)
05.01.2024 10:41
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05.01.2024 12:01