Proving two infinite nested radicals from a condition Kindly please tell me how to prove the following problem:  
Solve completely : $$ x^2 =y+a ; y^2=z+a ; z^2 =x+a ...(1) $$ and show that (with or without using the preceding question): $$\sqrt{11-2\sqrt{11+2\sqrt{11-...}}} = 1+4\sin 10^\circ$$ and $$\sqrt{23-2\sqrt{23+2\sqrt{23+2\sqrt{23-...}}}} = 1+4\sqrt{3}\sin 20^\circ$$. The signs under the outward-most radical sign have period of three; they are respectively; $-,+,- ; -,+,+$. I think that equation $(1)$ implies that $x$ satisfies a polynomial equation of degree $8$ but I'm not sure. Any help is appreciated.
 A: We need re-scaling:
\begin{align*}
  bx &= \color{red}{-2}\sqrt{11\color{green}{-2}\sqrt{11 \color{blue}{+2}\sqrt{11 \color{red}{-2}\sqrt{11-\ldots}}}} \\
  by &= \color{green}{-2}\sqrt{11 \color{blue}{+2} \sqrt{11 \color{red}{-2} \sqrt{11 \color{green}{-2}\sqrt{11-\ldots}}}} \\
  bz &= \color{blue}{+2}\sqrt{11 \color{red}{-2} \sqrt{11 \color{green}{-2} \sqrt{11 \color{blue}{+2}\sqrt{11-\ldots}}}} \\
  b^2x^2 &= 4(11+by) \\
  b^2 &= 4b \\
  b &= 4 \\
  a &= \frac{11}{4} \\
\end{align*}
Note that $$\fbox{$y<x<0<z$}$$
Next is plug and play:
\begin{align*}
  \xi &= -\frac{1}{2}-2\sin 10^{\circ} \\
  \eta &= \xi^2-\frac{11}{4} \\
  &= -\frac{1}{2}+2\sin 10^{\circ}-2\cos 20^{\circ} \\
  \zeta &= \eta^2-\frac{11}{4} \\
  &= -\frac{1}{2}+2\sin 10^{\circ}+2\cos 40^{\circ} \\
  \zeta^2-\frac{11}{4}
  &= -\frac{1}{2}+16\sin 10^{\circ}-18\cos 20^{\circ}+18\cos 40^{\circ} \\
  &= -\frac{1}{2}+16\sin 10^{\circ}-36\sin 10^{\circ}\sin 30^{\circ} \\
  &= -\frac{1}{2}-2\sin 10^{\circ} \\
  \xi &= \zeta^2-\frac{11}{4}
\end{align*}
Also 
\begin{align*}
  \xi-\eta
  &= 2\cos 20^{\circ}-4\sin 10^{\circ} \\
  &= 2(\cos 20^{\circ}-\cos 80^{\circ})-2\sin 10^{\circ} \\
  &= 4\sin 30^{\circ} \sin 50^{\circ}-2\sin 10^{\circ} \\
  &= 2(\sin 50^{\circ}-\sin 10^{\circ}) \\
  &> 0
\end{align*}
Now $$\fbox{$\eta<\xi<0<\zeta$}$$
Therefore $$(x,y,z)=(\xi, \eta, \zeta)$$
That is
$$\fbox{$
    \sqrt{11\color{green}{-2}\sqrt{11 \color{blue}{+2}
    \sqrt{11 \color{red}{-2}\sqrt{11-\ldots}}}}
    = 1+4\sin 10^{\circ}
  $}$$
Furthermore,
\begin{align*}
  t &=
  \left[
    \left(
      t^2-\frac{11}{4}
    \right)^2-\frac{11}{4} 
  \right]^2-\frac{11}{4} \\
  0 &=
  \left(
    t^2-t-\frac{11}{4}
  \right)
  \left(
    t^3-\frac{1}{2}t^2-\frac{17t}{4}+\frac{25}{8}
  \right)
  \left(
    t^3+\frac{3}{2}t^2-\frac{9t}{4}-\frac{19}{8}
  \right)
\end{align*}


*

*The first factor corresponds to $\pm \sqrt{a \pm \sqrt{a \pm \ldots}}$

*The second factor corresponds to the cyclic permutation of $++-++-\ldots$

*The third factor corresponds to the cyclic permutation of $--+--+\ldots$
The second case is left as an exercise.
