Suppose we start with a rational number $a_0$, and define $a_{n+1}=2a_n^2-1$ for $n\geq 0$. For what $a_0$ will it be the case that $a_i=a_j$ for some $i\neq j$?
We can start with something like $a_0=1$, then $a_1=1$ so $a_0=a_1$.
If $a_0=0$, we get $0, -1, 1, 1, \ldots$
Likewise if $a_0=-1$, we get $-1,1,1,\ldots$.
But how can we find all $a_0$?
Let $a_{0} = \cos \theta$. Then it is easy to check that $a_{n} = \cos (2^{n}\theta)$. So if $a_{i} = a_{j}$ for some $i \neq j$, then we must have
\begin{align*} \cos(2^{i}\theta) = \cos(2^{j}\theta) &\quad \Longleftrightarrow \quad 2^{i}\theta = 2n\pi \pm 2^{j}\theta, \quad n \in \Bbb{Z} \\ &\quad \Longleftrightarrow \quad \theta = \frac{2n\pi}{2^{i} \pm 2^{j}}, \quad n \in \Bbb{Z} \end{align*}
Thus the problem reduces to find the condition of $(i, j, n, \pm)$ such that
$$ \cos \left( \frac{2n\pi}{2^{i} \pm 2^{j}} \right) \in \Bbb{Q}. $$
Referring to this posting, this is possible if and only if
\begin{align*} \theta \equiv 0, \pm \frac{\pi}{3}, \pm \frac{\pi}{2}, \pm \frac{2\pi}{3} \pmod{2\pi} \end{align*}
This corresponds to $a_{0} \in \{0, \pm \frac{1}{2}, \pm 1 \}$.
In general
$$a_{n+1}:=2a_n^2-1=2a_m^2-1=:a_m\iff a_n=\pm a_m$$
If we choose $\;n\;$ to be the minimal index s.t. $\;a_{n+1}=a_{m+1}\;$ , for some $\;m\neq n\;$ , the above means that
$$a_n=-a_m\iff 2a_{n-1}^2-1=-2a_{m-1}^2+1\iff a_{n-1}^2+a_{m-1}^2=1\ldots$$
Try to take it from here.
