Vietnamese MO

We will present two solutions for this problem.

First solution:
We will prove that the polynomial $P(x) = \frac{2}{a}x^2 - a$ satisfies the requirements of the problem. Specifically, we will prove the equation
$$P_{2024}(x) = t \quad (*)$$
there are exactly $2^{2024}$ distinct solutions in the interval $(-a, a)$. Indeed, let $x = a \cos \varphi$ ($0 < \varphi < \pi$). Then
$$P_1(a \cos \varphi) = P(a \cos \varphi) = a(2 \cos^2 \varphi - 1) = a \cos 2\varphi,$$
$$P_2(a \cos \varphi) = P(a \cos 2\varphi) = a \cos 2^2\varphi,$$
$$P_{2024}(a \cos \varphi) = a \cos 2^{2024}\varphi.$$
The equation (*) becomes $a \cos 2^{2024}\varphi = t$. This equation is equivalent to
$$2^{2024}\varphi = \pm \arccos \frac{t}{a} + 2k\pi,$$
so
$$\varphi = \frac{\pm \arccos \frac{t}{a} + 2k\pi}{2^{2024}} \quad (k \in \mathbb{Z}).$$
Since $0 < \varphi < \pi$, the equation (*) has exactly $2^{2024}$ distinct real roots as
$$x = a \cos \frac{\arccos \frac{t}{a} + 2k\pi}{2^{2024}}, \quad k \in \{0, 1, \dots, 2^{2023} - 1\}$$
and
$$x = a \cos \frac{-\arccos \frac{t}{a} + 2k\pi}{2^{2024}}, \quad k \in \{1, 2, \dots, 2^{2023}\}.$$
So, $P(x) = \frac{2}{a}x^2 - a$ is a polynomial that satisfies the problem requirements. ■

Second solution:
Consider the polynomial $P(x) = x^2 - c$ with $c$ being some constant in the range $(a, a^2 - a)$. We will prove that, for all positive integers $n$ and for every real number $t \in (-a, a)$, the equation $P_n(x) = t$ has exactly $2^{2024}$ distinct real roots in the interval $(-a, a)$.

With $n = 1$, the equation $P_1(x) = t$ has two distinct real solutions: $-\sqrt{c+t}$ and $\sqrt{c+t}$. Both of these solutions belong to the range $(-a, a)$. Thus, the assertion is true for $n = 1$.

Assuming the assertion is true up to $n = k$, that means the equation $P_k(x) = t$ has exactly $2^k$ distinct solutions in the segment $(-a, a)$ namely $x_1, x_2, \dots, x_{2^k}$. Consider the equation $P_{k+1}(x) = t$. This equation is equivalent to
$$P_k(P(x)) = t,$$
or
$$P(x) \in \{x_1, x_2, \dots, x_{2^k}\}. \quad (*)$$
According to the results of the case $n = 1$, each equation $P(x) = x_i$ has exactly two distinct real roots in the interval $(-a, a)$.

In addition, because $x_i \neq x_j$ for all $i \neq j$ so the solutions of the equation $P(x) = x_i$ are distinct from the solutions of the equation $P(x) = x_j$. Combined with (*), we deduce that the equation $P_{k+1}(x) = t$ has exactly $2^{k+1}$ distinct real roots in the range $(-a, a)$. Thus the conclusion is also true for $n = k + 1$. According to the principle of induction, we have the assertion true for all positive integers $n$. In particular, the equation $P_{2024}(x) = t$ has exactly $2^{2024}$ distinct real roots. $\square$