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

Let $1+x=y^2$, where $y>0$ and $1+y=z^2$, where $z>0$.
Hence, $1+z=x^2$, where $x>0$.
Thus, $x-y=(y-z)(y+z)$ and $y-z=(z-x)(z+x)$.


*

*Let $x>y$. Hence, $y>z$, which says that $z>x$. It's contradiction.

*Let $x<y$. Hence, $y<z$, which says that $z<x$. It's contradiction again.
Id est, $x=y=z$ and $x=\frac{1+\sqrt5}{2}$. 
