<s>[INST] real values of $x$ which satisfy the equation $\sqrt{1+\sqrt{1+\sqrt{1+x}}}=x$ 
All real values of $x$ which satisfy the equation $\sqrt{1+\sqrt{1+\sqrt{1+x}}}=x$

$\bf{My\; Try::}$ Here $\sqrt{1+\sqrt{1+\sqrt{1+x}}} = x>0$
Now Let $f(x)=\sqrt{1+x}\;,$ Then equation convert into $f(f(f(x)))=x$
Now Here $f(x)=x$ be one function which satisfy above equation.
My question is how can we calculate other function which satisfy above functional equation.
Help required, Thanks
 [/INST] With your remark $x=\sqrt{1+x}$ should lead to a solution.
$$x=\sqrt{1+x} \iff x^2=1+x  \iff x^2-x-1=0 \iff x=\frac{1\pm\sqrt5}{2}$$
But $x=\frac{1+\sqrt5}{2}$ is the only positive solution. We verify that it is indeed solution by using long division :
$$\sqrt{1+\sqrt{1+\sqrt{1+x}}} = x\iff \sqrt{1+\sqrt{1+x}}=x^2-1 \iff \sqrt{1+x}=x^4-2x^2\\\iff x+1=x^8-4x^6+4x^4 \iff x^8-4x^6+4x^4-x-1=0$$
$$x^8-4x^6+4x^4-x-1=(x^2-x-1)(x^6+x^5-2x^4-x^3+x^2+1)$$
Let $g(x)=x^6+x^5-2x^4-x^3+x^2+1$, notice that $\forall x>0, g'(x)>0$, so $g$ is strictly increasing and $g(0)=1>0$, so $g(x)=0$ has no solution on $\mathbb{R}^{*+}$.
Finally $x=\frac{1+\sqrt5}{2}$ is the only solution to the problem.

Notes  : 
The functional equation $\;(f \circ f \circ f)(x) = x,\forall x>0$ only has one continuous solution : $f(x)=x$.

By studying the domain, you have $f : \mathbb{R}^{*+} \rightarrow
 \mathbb{R}^{*+}$. 
$\forall x \in  \mathbb{R}^{*+}$ the image by $f$ of $f(f(x))$ is $x$ so
  $f$ is onto. Also if $f(a)=f(b)$, then $a=f(f(f(a)))=f(f(f(b)))=b$, so
  $f$ is injective. So $f$ is one to one. If $f$ is decreasing $f\circ f$ is increasing and $f\circ f\circ f$ is decrasing, but $id$ is
  increasing, it is absurd, so $f$ is strictly increasing.
Suppose $f(x)>x$, since $f$ is strictly increasing $f(f(x))>f(x)>x$,
  so $x=f(f(f(x))>f(f(x))>f(x)>x$, it is impossible. In the same way
   $f(x)<x$ is impossible, so $\forall x, f(x)=x$. So $f=id$.

</s>