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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.

Gesbesgue
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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 $2.$

Will Jagy
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  • Which is the other one? Like I said, I am more interested in the motivation. I know it takes this form. – Gesbesgue May 03 '17 at 20:17
  • See e.g. this – Robert Israel May 03 '17 at 20:20
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    $f(x) = x^2 - 2$ is conjugate to the Chebyshev polynomial $T_2(t) = 2 t^2 - 1$. Slightly more generally, $a_{n+1} = 2 c a_n^2 + 4 d a_n + \frac{2d^2-d-1}{c}$ has the solution $a_n = (\cosh(2^n b)-d)/c$. – Robert Israel May 03 '17 at 20:38
  • @RobertIsrael right, although not anything I knew. I think this OP specifically wants the cases when the sequence recurs. I suppose it is a contest problem from times past. – Will Jagy May 03 '17 at 20:41