Conjecture on a functional equality and nested radical of Ramanujan :$3=\sqrt{1+f(2)\sqrt{1+f(3)\sqrt{1+\cdots}}}$ Hi i was wondering something about the great Ramanujan :
I think moreover I'm not the only one who propose this kind of problem (so if you have a link related to this subject).
We have :
$$3=\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots}}}$$
Now the problem :


Let $f(x)$ be a positive,continuous  and differentiable on function$(0,\infty)$ and non-constant  then the functional equation :
    $$3=\sqrt{1+f(2)\sqrt{1+f(3)\sqrt{1+\cdots}}}$$
    Have as unique solution $f(x)=x$


I have tried to build a counter-example without success .First it seems obvious for strictly increasing\decreasing functions such that $f(x)>x$ or $f(x)<x$ . I would like to create an counter-example of the form :
$$f(x)=x+g(x)$$ 
Where $g(x)$ is a periodic function . 
I have tried more general representation without success .
If it's true we can see how Ramanujan was great .
Any helps are highly appreciated .
Thanks a lot .
 A: Define $f$ and $g$ as
$$f\colon (0,\infty)\to\mathbb{R},\;x\mapsto x,$$
and
$$g\colon (0,\infty)\to\mathbb{R},\;g(x)=0\;for\;x\in(0,2]\;and\;$$
$$g(x)=cos(2\pi x)-1\;for \;x\in(2,\infty).$$
Define $h$ as
$$h\colon (0,\infty)\to\mathbb{R},\;x\mapsto g(x)+f(x).$$
$h$ is non-constant. For all $x\in(0,2]$, it holds, that $h(x)=f(x)\gt0$.
For all $x\in(2,\infty)$, it holds, that $f(x)\gt2$ and $g(x)\ge-2$. Hence $h(x)\gt0$ for all $x\in(0,\infty)$. For all $x\in\mathbb N$ follows $g(x)=0$. $f$ and $g$ are differentiable over $(0,2)$ and $(2,\infty)$. Look at
$$\lim_{k\to 0-0} \frac{g(2+k)-g(2)}{k}=\lim_{k\to 0-0} \frac{0-0}{k}=0$$
and 
$$\lim_{k\to 0+0} \frac{g(2+k)-g(2)}{k}=\lim_{k\to 0+0} \frac{cos(2\pi (2+k))-1-cos(2\pi\cdot2) +1}{k}=-2\pi sin(2\pi\cdot2)=0$$
and notice, that $g$ is differentiable at $x=2$. It follows, that $h=f+g$ is also differentiable as a sum of two differentiable functions. Hence there exists a positive, non-constant, differentiable function $h\neq f$, such that $$3=\sqrt{1+h(2)\sqrt{1+h(3)\sqrt{1+\cdots}}}.$$ Therefore this conjecture does not hold.
