Find the solution to a system of equations Initially I'd like to solve the following problem:

Find $x$: $$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}$$

I'm already aware of one approach to solve it, but now I wonder what if I set $y=\sqrt{2+\sqrt{2+x}}$
?
That is, to solve the following system:
$$ \begin{cases} x = \sqrt{ 2+ \sqrt{ 2+y }} \\ y= \sqrt{2+\sqrt{2+x}}  \end{cases}$$
 A: Sounds like you're basically hoping to see some "systems of equations"-specific method that solves the system you gave. Unfortunately, methods for solving the system are mostly going to come down to a) solve one of the equations for one of the variables, b) plug that into the other equation and solve from there. In this case, that just gives
$$x = \sqrt{2+\sqrt{2+y}} = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}$$
...which is your original equation again. That's probably the easiest way to solve this system of equations, but that takes the fun out of it since the whole point was to avoid the standard method of solving the original equation.
If you really want to work with the system and not with the original equation, you could try numerical methods like Newton's method using a computer. These methods would basically let you choose a starting guess $(x_0, y_0)$ and then iterate to get closer to the solution (or sometimes they don't converge depending on the quality of your starting guess). It wouldn't give you an exact formula for the answer though, just a list of decimal numbers that get very close to the answer.
A: I think this makes sense since now you have two unknowns and two equations. For your system, all you need to do is just find the relationship between $x$ and $y$ and plug in, then finally you have to solve an equation which is $$\big[[(x^2-2)^2-2]^2-2\big]^2-2=x$$.
However, this doesn't seem so convenient. You can also set $$y:= \sqrt{2+\sqrt{2+\sqrt{2+x}}}$$
This also won't be a convenient idea, since it will finally bring you to solve the equation above.
Thus seemingly your method can't easily solve the problem here.
I guess you already know that you can change the original problem into a recursive sequence. If your method isn't this one. I introduce it. And this method is pretty widely used in dealing with this kind of problem.
$$ x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}$$
This is $$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}$$
We set
$$\begin{aligned} & a_1= \sqrt{2} \\ & a_2={\sqrt{2+\sqrt{2}}} \\  & a_3=\sqrt{2+\sqrt{2+\sqrt{2}}}\\  & \vdots \end{aligned} $$
This is $$\begin{aligned} & a_1= \sqrt{2} \\ & a_2={\sqrt{2+a_1}} \\ & a_3=\sqrt{2+a_2} \\ &\vdots \end{aligned} $$
The sequence $\{a_n\}$ is increasing (clear) and bounded above (bounded by 2 you can verify by induction). Thus it has limit, suppose it's limit is $a$, we have $$ \lim_{n\to \infty}a_n= \lim_{n\to \infty}\sqrt{2+a_{n-1}}\iff a = \sqrt{2+a} \implies a=2,-1$$
It's clear that $a>0$, so $a=2$ which is $x$
A: I've just got an idea.
Let's suppose $x>y$, then $x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt{2+\sqrt{2+y}}<\sqrt{2+\sqrt{2+x}}=y$, which is impossible. When $x<y$ it's the same. Hence $x=y$.
As a result, $x=\sqrt{2+\sqrt2+x}\Rightarrow x^2=2+\sqrt{2+x}\Rightarrow x^4-4x^2+4=x+2\Rightarrow x^4-4x^2-x+2=0\Rightarrow$$\left(x^2+x-1\right)\left(x+1\right)\left(x-2\right)=0\Rightarrow x=2$.
If there's any mistake please tell me.
A: Render $x=2\cos\theta$. Prove that if $f(\theta)=2\cos(\theta)$ then
$f(\theta/2)=\sqrt{2+f(\theta)}.$
Thereby prove that
$\cos(\theta)=\cos(\theta/16)$
and solve this equation by exploiting the periodicity and symmetry of the cosine function.
You should get sixteen distinct values of $x$. Only four of these are actually equal to $y$ in your system of equations.
