Find $x\in \Bbb R,$ solving $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$ 
Given: $x\in \mathbb R$, $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$
Find: numeric value of $x$

Problem from a math contest. Sorry if it is a duplicate, but could not find anything similar using the search tool.
My attempt: I squared both sides, and developed the resulting expression, but I'm getting to nowhere.
Hints and/or answers please.
Is there any standard way to approach problems like these?
 A: Clearly any solution of the given equation is a positive real number.
Lemma. The map $x\mapsto \sqrt{1+x}$ is a contraction over $[0,+\infty)$, since for every $x\geq 0$ we have $\frac{d}{dx}\sqrt{x+1}\leq\frac{1}{2}<1$.
Corollary. By the Banach fixed point Theorem, $x=\sqrt{1+x}$ has a unique real solution. Simple algebra gives that such a solution is provided by the golden ratio $\varphi=\frac{1+\sqrt{5}}{2}$.
Lemma. Since $f(x)$ is a contraction over $[0,+\infty)$, $f(f(f(x)))$ is a contraction a fortiori.
Corollary. By the Banach fixed point Theorem, the given equation has a unique real solution. $x=\varphi$ is a solution, hence it is the only one.
A: Essentially the solution is unique(which you can find out yourself). Then there is another function has the same solution to this one: 
$x=\sqrt{1+x} $
Done
A: We know that the Golden ratio $\phi = \frac{1+\sqrt5}{2}$ and $~\bar{\phi}= \frac{1 -\sqrt5}{2}$ satisfy  the equation 
$$x^2 = 1+x  \Longleftrightarrow x= \sqrt {1 + x}~~x>0~ \implies x= f(x)$$ where, $f(x)=  \sqrt {1 + x},~~x>0.$
This means that $ \phi $ is  the only fix points of the function $f(x)=  \sqrt {1 + x},~x>0$
But $$ \sqrt {1 + \sqrt {1 + \sqrt{1+x}}} = f\circ f\circ f(x)= f^3(x) .$$

Therefore your equation reduces to 
  $$x= f^3(x)$$

Which mean that $x$ is fix point of $f^3$. 
On the other hand, $|f'(x)| =\frac{1}{2\sqrt{x+1}} \le \frac12$ then,
$$|(f^3(x))'| =  3|f'(x)\cdot f'(f^2)(x)| \le \frac 34<1$$
then $f^3$ is a contraction and hence,  $f^3$ has  a unique fix point satisfying $x=f^3(x)$ . Whereas, 
 $$ \color{red}{x= f^3(x) \implies  f(x) = f^3(f(x)) \implies x= f(x) }$$
since we observse that $f(x)$ is also a point fix of $f^3$
 by unicity? we get $x=f(x).$ 
But $\phi $ is the only positive number satisfying $x=f(x)$.

Hence, the only solution to $x=f^3(x)$ is $$x=\phi = \frac{1+\sqrt5}{2}~$$

A: Let $f(x) = \sqrt{1 + x}$. The only thing we need is $f(x)$ is a strictly increasing function in $x$. 
When $f(x) \ne x$, we have two posibilities:


*

*$f(x) > x \implies f(f(x)) > f(x) \implies f(f(f(x))) > f(f(x))$.
Combine these, we have
$$f(x) > x \implies f(f(f(x))) > f(f(x)) > f(x) > x \implies f(f(f(x)) > x$$

*$f(x) < x \implies f(f(x)) < f(x) \implies f(f(f(x))) < f(f(x))$.
Combine these, we have
$$f(x) < x \implies f(f(f(x))) < f(f(x)) < f(x) < x \implies f(f(f(x))) < x$$
In both cases, we have $f(f(f(x))) \ne x$. What this means is in order to solve
$$x = \sqrt{1+\sqrt{1+\sqrt{1+x}}} = f(f(f(x)))$$
we need to have 
$$x = f(x) = \sqrt{1+x} \quad\iff\quad x^2 = 1 + x\quad\implies\quad x = \frac{1+\sqrt{5}}{2}$$
Please note that the other root of the quadratic polynomial has been ruled out because it is negative.
