2025 International Mathematical Olympiad China National Team Selection Test

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 $(\ast 1),\cdots ,({\ast }_{2025})$. 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: $$\forall n\ge 1,X_{n-1}+(1-{\omega }^{-1}p)X_n+X_{n+1}=0.$$ 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 $(\ast {)}_1$ 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 $(\ast {)}_{2025}$ 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. $\square$