If $f(x)^2=x+(x+1)f(x+2)$, what is $f(1)$? Suppose $f$: $\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ and $f(x)^2 = x + (x+1)f(x+2)$, what is $f(1)$? Or more in general, what is $f(x)$?
The motivation behind this problem is that I want to find what the number of this nested radical $\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8...}}}}$. This can be written more generally as $f(x)=\sqrt{x+(x+1)f(x+2)}$ where $x=1$. This is where the problem arises from. If anybody can find an expression for the nested radical or find $f(x)$ I would be very happy!

The functional equation for $f(x)$ actually implies functional
relations also for all its derivatives
$$
\left\{ \matrix{
  f(x)^{\,2}  = x + \left( {x + 1} \right)f(x + 2) \hfill \cr 
  2f(x)f'(x) = 1 + f(x + 2) + \left( {x + 1} \right)f'(x + 2) \hfill \cr 
  2f'(x)^{\,2}  + 2f(x)f''(x) = 2f'(x + 2) + \left( {x + 1} \right)f''(x + 2) \hfill \cr 
  \quad \quad  \vdots  \hfill \cr}  \right.
$$
so that
$$
\left\{ \begin{gathered}
  f(2) = f(0)^{\,2}  \hfill \\
  f'(2) = 2f(0)f'(0) - f(0)^{\,2}  - 1 \hfill \\
  f''(2) = 2f'(0)^{\,2}  + 2f(0)f''(0) - 4f(0)f'(0) + 2f(0)^{\,2}  + 2 \hfill \\
  \quad \quad  \vdots  \hfill \\ 
\end{gathered}  \right.
$$
Therefore, being $f(x)$ continuous, we are not free to fix $f(x)\quad |\;0\le x < 2$
equal to whatever continuous function respecting only $f(2)=f(0)^2$.
Instead it shall be such as to respect the functional relation, at $x$ and $x+2$, for all the derivatives. 
