Solve the following system of equations, in which $a$ is a given number satisfying $|a| > 1$:
$\begin{matrix} x_{1}^2 = ax_2 + 1 \\ x_{2}^2 = ax_3 + 1 \\ \ldots \\ x_{999}^2 = ax_{1000} + 1 \\ x_{1000}^2 = ax_1 + 1 \\ \end{matrix}$
An shortened solution for this one (I spent too much time on it and I'm a bit fed up with it now ) :
Wlog, $a > 1$, by transforming $a \rightarrow -a$ and $x_i \rightarrow -x_i$
It's not difficult to prove (once you have guessed it) that either $0 > x_i > -1 \ \forall i$ either $x_i \geq 1 \ \forall i$.
In the first case, if $x_i \geq x_j$ then $x_{i+1} \leq x_{j+1}$ from which we have then
$x_1 = x_3 = .... = x_{999}$ and $x_2 = x_4 = .... = x_{1000}$
or $x_{2i+1} = (-a \pm \sqrt{4-3a^2})/2$ and $x_{2i+1} = (-a \mp \sqrt{4-3a^2})/2$
In the other case, they are all equal $= (a + \sqrt{a^2+4})/2$