The sequence $x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}, x_0>0$ is bounded? Consider the sequence $$x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}, x_0>0.$$
How can you prove that this sequence is bounded or unbounded for those values ​​of $x_0$ for which it is defined.
With a number generator I noticed that all terms of the positive out of range $[-2,2]$
 form finite sets in strictly decreasing order. We found that it can be periodically. For $x_0=\frac{2}{\sqrt{3}}$ its terms  repeat of 2 by 2.
I tried to prove boundedness to the method of reduction to absurdity but we did. 
Thanks so much for any suggestion
 A: In the comments above, one year ago, I was mentioning my answer to a nearly duplicate question, "for a full solution when $x_0=3$ and (for) clear indications about the other cases". The other page is now unavailable to users with not enough reputation hence I reproduce the question and my answer below. (Revised version following a remark by @Puzzled417.)


Is the sequence defined by $x_1=3$ and $x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}$ for every $n\geqslant1$, bounded?

A key remark is that the transformation $x_n\to x_{n+1}$ has two fixed points $\pm2i$. Even though these are not real, one can make use of them, introducing $$z_n=\frac{x_{n}-2i}{x_{n}+2i}.$$ Then, a simple computation shows that, as long as $x_{n+1}$ exists, $z_n$ is on the unit circle and $z_{n+1}=z_n^2$, hence 
$$
z_n=(z_0)^{2^n}.
$$
Note that, if the sequence $(x_n)$ stops to be defined because some $x_n$ is $0$, then the corresponding $z_n$ is $-1$. Furthermore, the sequence $(x_n)$ is bounded if and only if $(z_n)$ stays away from $1$, that is, the sequence $(x_n)$ is unbounded if and only if $1$ is a limit point of $(z_n)$.
To solve the case $x_0=3$, note that $z_0=(5-12i)/13$ hence $z_0=e^{i \theta_0}$ with $$\theta_0=\arccos\left(5/13\right).$$ One knows that the only rational multiples of $\pi$ with a rational cosine have cosine $0$ or $\pm1/2$ or $\pm1$. Thus, $\theta_0/\pi$ is irrational, hence $z_n\ne1$ for every $n$ and every $x_n$ is well defined.
As is well known, for every irrational number $\alpha$, the set of fractional parts of $n\alpha$ is dense in $[0,1]$, however this does not say that the set of fractional parts of $2^n\alpha$ is dense in $[0,1]$, and actually there exists some irrational numbers $\alpha$ such that this set is not dense in $[0,1]$, hence we cannot conclude only using the fact that $\theta/\pi$ is irrational.
However... Consider some neighborhood $N_\epsilon=(-\epsilon,\epsilon)$ of $0$ in $\mathbb R/\mathbb Z$ of Lebesgue measure $2\epsilon$. Its preimage $d^{-1}(N_\epsilon)$ by the doubling map $d:\alpha\mapsto2\alpha$ is made of two intervals of length $\epsilon$, one of them included in $N_\epsilon$, hence the measure of $N_\epsilon\cup d^{-1}(N_\epsilon)$ is at most $2\epsilon+\epsilon$. Likewise, the measure of $N_\epsilon\cup d^{-1}(N_\epsilon)\cup\cdots\cup d^{-k}(N_\epsilon)$ is at most $2\epsilon+\epsilon+\cdots+2^{1-k}\epsilon$ and the measure of the set of points whose orbit passes by $N_\epsilon$ is at most $4\epsilon$.
This shows that the set of points $\alpha$ in $\mathbb R/\mathbb Z$ whose orbit's closure contains $0$ has Lebesgue measure $0$ (or course this set is not empty since it contains all the dyadic numbers, whose orbit contains $0$).
...Which is of course still not enough to rule the case of our $\theta_0/\pi$.
A: Let $x_n=au_n$. Then $au_{n+1}=\dfrac{au_n}{2}-\dfrac{2}{au_n}$, or
$$u_{n+1}=\dfrac{1}{2}\left(u_n-\dfrac{4}{a^2u_n}\right).$$
Choose $a=2$, and the recurrence becomes
$$u_{n+1}=\dfrac{1}{2}\left(u_n-\dfrac{1}{u_n}\right).$$
Consider the identity $\cot2\theta=\dfrac{1}{2}\left(\cot\theta-\dfrac{1}{\cot\theta}\right)$ or, more generally,
$$\cot2^{n+1}\theta=\dfrac{1}{2}\left(\cot2^n\theta-\dfrac{1}{\cot2^n\theta}\right).$$
Starting with $x_0=2u_0=2\cot\theta$, we see that $x_n=2u_n=2\cot2^n\theta$ satisfies the original recurrence relation. Thus $x_n$ is bounded as long as $2^n\theta$ is bounded away from multiples of $\pi$. The analysis of which values of $\theta$ satisfy this condition isn't simple.
