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
pco
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 $x=i\sqrt{2025}$)
RootofUnityfilter
05.01.2024 10:41
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wassupevery1
05.01.2024 12:01
The answer is $(P, Q) = \boxed{ \left( c, -\frac{c(x^{2024}+x)+x^3+2025x+c^{2023}}{c^2} \right) }$ for some non-zero $c$. It is easy to check that these pairs of polynomials satisfy the problem.
The problem condition is equivalent to $$P(x)^{2023}+Q(x)P(x)^2+(x^{2024}+x)P(x)+x^3+2025x=0,$$for all reals $x$.
The above equation implies $P(x)$ divides $x^3+2025x$.
Now consider the cases:
Case 1. $P$ is constant. Let $P(x)=c$ for $c \neq 0$, then it follows that $Q(x) = \frac{-c(x^{2024}+x)-x^3-2025x-c^{2023}}{c^2}$ for some $c \neq 0$.
Case 2. $x$ divides $P$. Write $P(x)=xR(x)$ for some polynomial $R$. Then the conditions imply $$x^{2023}R(x)^{2023} + x^2Q(x)R(x)^2 + x^2(x^{2023}+1)R(x)+x^3+2025x=0.$$But this equation implies $x^2$ divides $2025x$, absurd.
Case 3. $P$ divides $x^2+2025$. Since $x^2+2025$ is irreducible, it follows that $P(x) = C(x^2+2025)$ for some non-zero constant $C$. Plugging back gives $$C^{2023}(x^2+2025)^{2023}+C^2Q(x)(x^2+2025)^2+C(x^{2024}+x)(x^2+2025)+x(x^2+2025)=0, \forall x \in \mathbb{R}$$$$\Leftrightarrow C^{2023}(x^2+2025)^{2022}+C^2Q(x)(x^2+2025)+C(x^{2024}+x)+x=0, \forall x \in \mathbb{R}.$$This gives $x^2+2025 \mid x(C(x^{2023}+1)+1) \Rightarrow x^2+2025 \mid C(x^{2023}+1)+1$.
Write $C(x^{2023}+1)+1 = R(x)(x^2+2025)$ for some polynomial $R$. Plugging in $x=45i$ gives $C = \frac{-1}{(-2025)^{1011} \cdot 45i+1}$ which is non-real, impossible.
We have consider all cases, and the proof is complete.