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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 $a_0$?

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  • $\begingroup$ Well, for example for $\;a_0= 1\implies a_1=a_0\;$ ... $\endgroup$ – DonAntonio Nov 12 '13 at 19:37
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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 $a_{0} \in \{0, \pm \frac{1}{2}, \pm 1 \}$.

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  • $\begingroup$ Why did you decide that $\;|a_0|\le 1\;$ , @sos440 ? This follows from my answer, of course, but I can't see any reason why in yours... $\endgroup$ – DonAntonio Nov 13 '13 at 4:26
  • $\begingroup$ @DonAntonio, This solution works if $\theta$ is allowed to have complex values. But a simpler argument is as follow: If $|a_{0}| > 1$ and $a_{0} = \cosh \varphi$, then $\a_{n} = \cosh (2^{n}\varphi)$. $\endgroup$ – Sangchul Lee Nov 13 '13 at 14:10
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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.

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