Vietnamese MO

From the hypothesis, we have
$$P(x)^{2023} + Q(x)P(x)^2 + (x^{2024} + x)P(x) + x(x^2 + 2025) = 0, \quad (1)$$
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 $(x^2 + 2025)|k(x^{2023} + 1) + 1 = kx^{2023} + k + 1$. Putting $c = -2025$, we have $x^2 \equiv c \pmod{x^2 - c}$, therefore
$$kx^{2023} + k + 1 = k(x^2)^{1011}x + k + 1 \\ \equiv kc^{1011}x + k + 1 \pmod{x^2 - c}.$$
Since $(x^2-c)|(kx^{2023}+k+1)$, we deduce that $(x^2-c)|(kc^{1011}x+k+1)$. However, as $\deg(kc^{1011}x+k+1) < \deg(x^2-c)$, it follows that $kc^{1011} = 0$ 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. $\square$