Vietnamese MO

From the hypothesis, we have $$P(x{)}^{2023}+Q(x)P(x{)}^2+(x^{2024}+x)P(x)+x(x^2+2025)=0,(1)$$ for all $x\in \mathbb{R}$. From that, we deduce that the polynomial $x(x^2+2025)$ is divisible by the polynomial $P(x)$. Thus, the polynomial $P(x)$ must have one of the following forms $k$, $kx$, $k(x^2+2025)$ or $kx(x^2+2025)$, where $k$ is some real constant. However, if polynomial $P(x)$ is divisible by polynomial $x$, then from equation (1), we deduce that the polynomial $2025x$ is divisible by $x^2$, contradiction. Therefore, the polynomial $P(x)$ must have one of two forms $k$ or $k(x^2+2025)$ where $k$ is some non-zero constant.

If $P(x)=k$ then from equation (1), we have $$k^{2023}+k^{2}Q(x)+k(x^{2024}+x)+x^3+2025x=0,$$ or $$Q(x)=\frac{-k(x^{2024}+x)-x^3-2025x-k^{2023}}{k^2}$$ for all real numbers $x$.

If $P(x)=k(x^2+2025)$ then from equation (1), we have $$k^{2023}(x^2+2025{)}^{2023}+k^2(x^2+2025{)}^{2}Q(x)+k(x^2+2025)x(x^{2023}+1)+x(x^2+2025)=0,\text{ or}$$ $$k^{2023}(x^2+2025{)}^{2022}+k(x^2+2025)Q(x)+kx(x^{2023}+1)+x=0,$$ for all $x\in \mathbb{R}$. We obtain that $(x^2+2025)(kx(x^{2023}+1)+x)$. On the other hand, the two polynomials $x$ and $x^2+2025$ are coprime, so we deduce that $(x^2+2025)|k(x^{2023}+1)+1=k{x}^{2023}+k+1$. Putting $c=-2025$, we have $x^2\equiv c(\mathrm{mod}{x}^2-c)$, therefore $$\begin{gathered} k{x}^{2023}+k+1=k(x^2{)}^{1011}x+k+1 \\ \equiv k{c}^{1011}x+k+1(\mathrm{mod}{x}^2-c). \end{gathered}$$ Since $(x^2-c)|(k{x}^{2023}+k+1)$, we deduce that $(x^2-c)|(k{c}^{1011}x+k+1)$. However, as $\mathrm{deg}(k{c}^{1011}x+k+1)<\mathrm{deg}(x^2-c)$, it follows that $k{c}^{1011}=0$ and $k+1=0$. But this system has no real solutions. Therefore, in this case there do not exist polynomials $P(x),Q(x)$ that satisfy the requirement.

In short, the polynomials satisfy the problem are of the form $P(x)=k$, and $$Q(x)=\frac{-k(x^{2024}+x)-x^3-2025x-k^{2023}}{k^2},$$ where $k$ is some non-zero number. $\square$