finding the real values of $x$ such that : $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ How to find the real values of $x$ such that : $$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
 A: Elaborating some on what @Fredrik Meyer suggested, one can get:
$$\begin{align*}x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}\hspace{5pt}&\Rightarrow\hspace{5pt} x^2=2+\sqrt{2-\sqrt{2+x}}\hspace{5pt}\Rightarrow\hspace{5pt} x^4-4x^2+4=2-\sqrt{2+x}\\
&\Rightarrow\hspace{5pt}x^8+16x^4+4-8x^6+4x^4-16x^2=2+x\\
&\Rightarrow\hspace{5pt} x^8-8x^6+20x^4-16x^2-x+2=0
\end{align*}$$
The last polynomial can be factored by noticing that it vanishes at $x=2$ and at $x=-1$ into $(x-2) (x+1) (x^3-3 x+1) (x^3+x^2-2 x-1)=0$ (using WA in the end).
Now you can find many roots - but beware: not all of them solve the initial problem: each transition above gives additional assumptions on $x$: first, $x\geq 0$. Then $x^2-2\geq0$ and after that $x^4-4x^2+2\leq0$. Do you see why?
All of those together imply that $\sqrt2\leq x\leq\sqrt{2+\sqrt2}$.
Using some real analysis (or WA :)), one can show that there exists a unique root of that polynomial on $(\sqrt2,\sqrt{2+\sqrt2})$.
NB: This is not the cleanest approach and not very elegant, but it works. I'm almost sure that I have seen much nicer way to solve it, but I can't think of one now.
A: Here is a basic square free analysis approach. Clearly not the elegant approach Dennis Gulko has seen before. Consider the function 
$$f(x)=x-\sqrt{2+\sqrt{2-\sqrt{2+x}}}.$$ 
Its domain is $[-2,2]$, where it is continuous and increasing. Since
$$
f(-2)=-2-\sqrt{2+\sqrt{2}}<0<f(2)=2\sqrt{2}
$$
there exists (intermediate value theorem) a unique (since $f$ is increasing) zero $x_0$ for $f$. Here is a graph to confirm this claim. Using a calculator, one can check that
$$
f(3/2)<0<f(8/5)\quad\Rightarrow\quad \frac{3}{2}=1.5<x_0<1.6=\frac{8}{5}.
$$
If you want to go a little further further in the decimal expansion of $x_0$, you can use dichotomy. Or ask Wolfram.
A: Here's an approach leading to a "closed formula". 
The two innermost square roots are defined only, if $x\in[-2,2]$, so we can write
$x=2\cos\theta$ for some $\theta\in[0,\pi]$. My solution relies on the trig identities
$$
\sqrt{\frac{1+\cos\alpha}2}=\cos\frac\alpha2
$$
and
$$
\sqrt{\frac{1-\cos\alpha}2}=\sin\frac\alpha2
$$
that hold for all $\alpha\in[0,\pi]$ (outside this range we may get differing signs).
Thus
$$
\sqrt{2+x}=\sqrt{2+2\cos\theta}=2\cos\frac\theta2,
$$
and then
$$
\sqrt{2-\sqrt{2+x}}=\sqrt{2-2\cos\frac\theta2}=2\sin\frac\theta4=2\cos(\frac\pi2-\frac\theta4).
$$
Going on we get
$$
2\cos\theta=x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}=\sqrt{2+2\cos(\frac\pi2-\frac\theta4)}=2\cos(\frac\pi4-\frac\theta8).
$$
In the interval $[0,\pi]$ cosine is injective, so we can conclude that
$$
\theta=\frac\pi4-\frac\theta8\Leftrightarrow\theta=\frac{2\pi}9.
$$
The only real solution is thus
$$
x=2\cos\frac{2\pi}9\approx1.53209.
$$
A: I would use the square-and-order method to turn it into a polynomial. The next step would be to find its real roots.
