Concerning the nonlinear functional equation $f(x)f(x) =x+1+f(x+1)$ There's a problem I've been working on for awhile that involves some hefty functional equations. For example, I may have something along the lines of 
$$
f(x)f(x) =x+1+f(x+1)
$$
I've tried several different methods of attack (the farthest I ever got was probably with a power series which didn't yield a recurrence relation)  but it never amounts to much. As if that wasn't bad enough, I don't actually know any values of $f(x)$, other than that $\lim_{x\to\infty}f(x)=\infty$. The thing is, I don't actually care about $f(x)$, I only want to know $f(0)$ (analytically) but I can never seem to get two equations for a given point.
I was wondering if anyone had any ideas on how to solve this, or even just some insight into whether or not it can be solved. Thanks!
Edit: Additional facts


*

*It can be required that $1<f(0)<2$

*$f(x)$ is strictly increasing

*$f(x)$ is non-negative

 A: Your constraints are not strong enough to impose a unique value to $f(0)$.  In fact, I show below that there for every value $a$ between $1.8$ and $2$ there is a solution with $f(0)=a$ (in fact  for a given $a$ there are uncoutably many such solutions, as our construction will show).
Given a function $\phi : [0,1[ \to {\mathbb R}$, there is a unique solution $f: [0,+\infty[ \to {\mathbb R}$ to the functional equation that coincides with $\phi$ on $[0,1[$.  Indeed, we will have
$$
\begin{array}{lcl}
f(x)&=&\phi(x-1)^2-x=\phi_1(x) \ \ (\text{for }\ x\in[1,2[) \\
f(x)&=&\phi_1(x-1)^2-x=\phi_2(x) \ \ (\text{for }\ x\in[2,3[) \\
\end{array}\tag{1}
$$
and so on.
Now, let $a \in [1.8,2]$ and let $\phi : [0,1] \to {\mathbb R}$ be a continuous, strictly increasing map satisfying the boundary conditions $\phi(0)=a, \phi(1)=a^2-1$ (so for example you could take $\phi$ affine, or trigonometric, etc).
Then the $f$ uniquely defined by (1) will also be continuous and strictly increasing (in particular, $f$ will be positive). 
So all we need to show is that $f \to +\infty$ at $+\infty$ ; since $f$ is strictly increasing, it suffices to show that $f(n) \to +\infty$ when $n$ is an integer.
We have 
$$
\begin{array}{lcl}
f(1)&=&f(0)^2-1 \geq 1.8^2-1=2.24 \\
f(2)&=&f(1)^2-2 \geq 2.24^2-2=3.0176 \geq 3 \\
f(3)&=&f(2)^2-3 \geq 3^2-2=6 \\
\end{array}\tag{1}
$$
We can then use mathematical induction : let us show that $f(k) \geq k$, for every $k\geq 3$.
The $k=3$ case has just been checked. Suppose that the hypothesis is true for some $k\geq 3$.
  We have 
$$f(k+1)=f(k)^2-(k+1) \geq k^2-(k+1)=k-2+(k-1)^2 \geq k-2+4=k+2 \geq k+1 $$
So the hypothesis holds on the next level. This shows that $\lim_{+\infty}(f)=+\infty$ as wished.
