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 $$P_{2024}(x)=t(\ast )$$ there are exactly $2^{2024}$ distinct solutions in the interval $(-a,a)$. Indeed, let $x=a\cos \phi$ ($0<\phi <\pi$). Then $$P_1(a\cos \phi )=P(a\cos \phi )=a(2{\cos }^2\phi -1)=a\cos 2\phi ,$$ $$P_2(a\cos \phi )=P(a\cos 2\phi )=a\cos 2^2\phi ,$$ $$P_{2024}(a\cos \phi )=a\cos 2^{2024}\phi .$$ The equation () becomes $a\cos 2^{2024}\phi =t$. This equation is equivalent to $$2^{2024}\phi =\pm \arccos \frac{t}{a}+2k\pi ,$$ so $$\phi =\frac{\pm \arccos \frac{t}{a}+2k\pi }{2^{2024}}(k\in \mathbb{Z}).$$ Since $0<\phi <\pi$, the equation () has exactly $2^{2024}$ distinct real roots as $$x=a\cos \frac{\arccos \frac{t}{a}+2k\pi }{2^{2024}},k\in \{0,1,\dots ,2^{2023}-1\}$$ and $$x=a\cos \frac{-\arccos \frac{t}{a}+2k\pi }{2^{2024}},k\in \{1,2,\dots ,2^{2023}\}.$$ 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 $$P(x)\in \{x_1,x_2,\dots ,x_{2^k}\}.(\ast )$$ 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 $x_i\ne x_j$ for all $i\ne j$ 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. $\square$