International Mathematical Olympiad China National Team Selection Test, 2025

Proof: We first prove a lemma.
Lemma: Let $\beta$ be a complex number. If a sequence $\{a_n\}$ satisfies $a_0 = 0$, $a_1 \ne 0$, and for all $n \ge 1$,
$$a_{n-1} + \beta a_n + a_{n+1} = 0,$$
then $\{a_n\}$ does not converge to $0$.
Proof of Lemma: Let $\alpha_1, \alpha_2$ be roots of $x^2 + \beta x + 1 = 0$, so $\alpha_1\alpha_2 = 1$.
If $\alpha_1 = \alpha_2$, then $\alpha_1 = \alpha_2 = \pm 1$ and $a_n = pn + q$ or $a_n = (-1)^n(pn + q)$. Clearly $a_n$ doesn't converge to $0$.
If $\alpha_1 \ne \alpha_2$, then $a_n = p\alpha_1^n + q\alpha_2^n$ with $p, q \in \mathbb{C}$ and $pq \ne 0$ (since $a_0 = 0$, $a_1 \ne 0$).
Case 1: $|\alpha_1| \ne |\alpha_2|$. Then $a_n$ cannot converge to $0$ since $|\alpha_1||\alpha_2| = 1$.
Case 2: $|\alpha_1| = |\alpha_2| = 1$. Let $\alpha_1 = e^{2\pi i\theta}$, $\alpha_2 = e^{-2\pi i\theta}$.
If $\theta \in \mathbb{Q}$, then $\{a_n\}$ is periodic and non-zero.
If $\theta \notin \mathbb{Q}$, then $\{n\theta\}$ is dense modulo $1$, making $\{a_n\}$ have values dense on some circle.
In all cases, $a_n$ doesn't converge to $0$. $\square$
Now the main proof. Denote the given equations as $(*1), \cdots, (*_{2025})$.
Case 1: $\prod_{i=1}^{2025} p_i \ne 0$.
Define transformed sequences:
$$b_n^{(i)} = \sqrt[2025]{p_{i+1} p_{i+2} \cdots p_{i+2024}} \cdot a_n^{(i)}$$
where indices are cyclic modulo $2025$. These satisfy:
$$b_{n-1}^{(i)} + b_n^{(i)} + b_{n+1}^{(i)} = \sqrt[2025]{p_1 \cdots p_{2025}} \cdot b_n^{(i+1)}.$$
---

Thus we may assume $p_1 = \cdots = p_{2025} = p$. Let $\omega$ be a $2025$th root of unity and define:
$$X_n := \sum_{k=0}^{2024} \omega^k a_n^{(k+1)}.$$
Then:
$$\forall n \ge 1, \quad X_{n-1} + (1 - \omega^{-1}p)X_n + X_{n+1} = 0.$$
Since not all $a_1^{(i)} = 0$, some $X_1 \ne 0$. By the lemma, $\{X_n\}$ doesn't converge to $0$.
Case 2: If there exists some $p_i = 0$. Without loss of generality, assume $p_1 = 0$. Then from $(*)_1$ we know that for any $n \ge 1$ we have $a_{n-1}^{(1)} + a_n^{(1)} + a_{n+1}^{(1)} = 0$.
If $a_1^{(1)} \ne 0$, applying the lemma to the sequence $\{a_n^{(1)}\}$ shows that $\{a_n^{(1)}\}$ does not converge to $0$.
If $a_1^{(1)} = 0$, then $a_n^{(1)} \equiv 0$. Consequently, from $(*)_{2025}$ we know that for any $n \ge 1$ we have $a_{n-1}^{(2025)} + a_n^{(2025)} + a_{n+1}^{(2025)} = 0$.
In this case, if $a_1^{(2025)} \ne 0$, we can apply the lemma to the sequence $\{a_n^{(2025)}\}_{n \ge 0}$. If $a_1^{(2025)} = 0$ then $a_n^{(2025)} \equiv 0$, and similarly we obtain $a_{n-1}^{(2024)} + a_n^{(2024)} + a_{n+1}^{(2024)} = 0$. Continuing this process, by induction we can prove that either all sequences in the problem are identically zero, or there exists at least one sequence that does not converge to $0$. The former case contradicts the problem's assumptions, thus completing the proof. $\square$