Different Approaches to this recurrence formula $a_0=x,\;a_n=a_{n-1}^2-2$. For integer $m$, find $x$ s.t. $a_m=x$. In one of my first maths lectures, I had an assignment that went like this. Let $a_n$ be a sequence defined by
$$ a_0=x,\;a_n=a_{n-1}^2-2
$$
For $m=2000$ (or any other integer, this doesn't really matter), find all $x$ s.t. $$a_{2000}=x.$$
I am curious to see what other approaches to this problem there are. I'll shortly present one of mine: Write $x$ as $x=c+c^{-1}$ and see that $|c|=1$ has to be fulfilled. Than $x=2*\cos(..)$ and the problem becomes very simple. I chose this approach because I wanted to eliminate the $-2$ in the recurrence formula. I reckon most other approaches will in the end be similar to this one, but the motivations might differ, and that's what I am interested in.

This is one of just two quadratic recurrences that has a closed form. There are constants $A,B$ such that  $AB = 1$ and 
$$ a_n = A^{2^n} + B^{2^n}    $$
So
$$ a_0 = A + B $$
If $x$ is real and $x > 2,$ then one of $A,B$ is larger than $1$ and the sequence grows without bound. If $x=2$ the sequence is constant. 
Analogous if $x < -2$ or $x = -2.$
If $-2 < x < 2,$ I guess $A,B$ are complex. Need to ponder.
There are a countable set of real $x$ for which $A$ is an integer power of a root of unity. We need
$$ \arctan \frac{\sqrt {4-x^2}}{x}  $$
to be a rational multiple of $\pi.$ For example, with $x=1,$ we have $\arctan \sqrt 3 = \frac{\pi}{3}$ and $A$ is a sixth root of unity. 
Next, need to consider the fact that the exponents will be powers of $2.$
