Putnam 2018 B4 wrote:

Given a real number $a$ , we define a sequence by $x_0 = 1$ , $x_1 = x_2 = a$ , and $x_{n+1} = 2x_nx_{n-1} - x_{n-2}$ for $n \ge 2$ . Prove that if $x_n = 0$ for some $n$ , then the sequence is periodic.

Let $T_i(x)$ denote the $i^{\text{th}}$ Chebyshev polynomial. Let $F_i$ be the $i^{\text{th}}$ Fibonacci number, where $F_0 = 0, F_1 = 1$ and $F_n = F_{n - 1} + F_{n - 2}$ for all $n \ge 2$ .
Main Claim. For any $i \ge 0, x_i = T_{F_i}(a)$ .
Proof. We will prove this by induction on $i$ .

This is true for $i = 0, i = 1, i = 2$ , since $x_0 = 1 = T_0(a)$ and $x_1 = x_2 = a = T_1(a)$ .
Suppose that this is true for any $i \le j$ for a fixed constant $j \ge 2$ . Therefore,
$\begin{align*} x_{n + 1} &= 2x_n x_{n - 1} - x_{n - 2} \\ &= 2 \cdot T_{F_n}(a) \cdot T_{F_{n - 1}}(a) - T_{F_{n - 2}}(a) \\ &= T_{F_{n + 1}}(a) \end{align*}$ and therefore, we are done by induction.

We will consider two possible cases of $a$ :

If $|a| \le 1$ , then there exists $\theta \in [0, \pi)$ such that $\cos(\theta) = a$ . Therefore, by assumption, there exists $i$ such that
$T_{F_i}(a) = 0 \implies T_{F_i}(\cos(\theta)) \implies \cos(F_i \theta) = 0$ Therefore, $F_i \theta = k \pi$ for some $k \in \mathbb{Z}$ . However, note that Fibonacci numbers are periodic modulo $F_i$ , i.e. there exists $N$ such that $F_{a + N} \equiv F_a \pmod{F_i}$ for all $a \in \mathbb{Z}_{\ge 0}$ . This is because there are finitely possible residue for $(F_{x}, F_{x + 1})$ . Now, we claim that $\{ \cos(F_x \theta) \}_{x \ge 0}$ is also periodic. Indeed, note that
$\cos(F_{a + N} \theta) = \cos(F_a \theta) \ \text{for all } a \in \mathbb{Z}_{\ge 0}.$
If $|a| > 1$ , then there exists $x \in \mathbb{R} \setminus \{ 1, 0 \}$ such that $a = \frac{x + x^{-1}}{2}$ . Therefore, by assumption, there exists $i$ such that
$T_{F_i}(a) = 0 \implies T_{F_i} \left( \frac{x + x^{-1}}{2} \right) = 0 \implies \frac{x^{F_i} + x^{-F_i}}{2} = 0 \implies (x^{F_i})^2 + 1 = 0$ However, since $x \in \mathbb{R}$ , this has no solution.