<s>[INST] infinitely many negative and infinitely many positive numbers Suppose that 
$$x_1=\frac{1}{4}, \ x_{n+1}=x_{n}^3-3x_n.$$
Show that the sequence has infinitely many negative and infinitely many positive numbers.
My idea: Suppose that it has finitely many negative numbers. then all the numbers after some index, must be larger than $\sqrt{3}$. I want to show that the sequence cannot escape some interval.
 [/INST] The desired claim follows from the following two observations:

Claim. If $x_n \in \{-1/4, 1/4\}$, then $x_n \neq 0$ for all $n \geq 1$.

Proof. Let $p_1 = \operatorname{sign}(x_1)$ and $p_{n+1} = p_n^3 - 3p_n 4^{2 \cdot 3^{n-1}}$. Then we inductively check that $p_n$ is always odd and $ x_n = p_n / 4^{3^{n-1}}$. Since the numerator is always odd integer, it cannot vanish, and the claim follows.

Claim 2. If either $x_n \geq 0$ for any sufficiently large $n$ or $x_n \leq 0$ for any sufficiently $n$, then we actually have $x_n = 0$ for any sufficiently large $n$.

Proof. Write $x_n = 2\cos(2 \pi f_n)$. Then
$$ \cos(2 \pi f_{n+1}) = \frac{x_{n+1}}{2} = \frac{x_n^3 - 3x_n}{2} = 4\cos^3(2 \pi f_n) - 3\cos(2 \pi f_n) = \cos(2 \pi \cdot 3f_n). $$
So it follows that $\cos(2\pi f_{N+n}) = \cos(2\pi \cdot 3^n f_N)$. Now, replacing $(x_n)$ by $(-x_n)$ if necessary, we may assume that $x_n \geq 0$ for all sufficiently large $n$. In other words, there exists $N$ so that $x_{N+n} \geq 0$ for all $n \geq 0$. Then each $3^n f_N$ must avoid the sets $(\frac{1}{4}, \frac{3}{4}) + \mathbb{Z}$. So it follows that
$$ f_N \in \mathbb{R} \setminus \bigcup_{n=0}^{\infty} \bigcup_{k\in\mathbb{Z}} \left( \frac{4k+1}{4 \cdot 3^n}, \frac{4k+3}{4 \cdot 3^n} \right) = \left\{ k \pm \frac{1}{4\cdot 3^n} : n \geq 0 \right\}. $$
This implies that $3^{n_0} f_N = \pm 1$ for some $n_0 \geq 0$, and hence $x_{N+n} = 0$ for all $n \geq n_0$. This proves the desired claim.
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