What is the function $f(x)$ which is differentiable everywhere and $f(x-1)f(x-2)+1=f(x)$? 
What is the function $f(x)$ which is differentiable everywhere and $f(x-1)f(x-2)+1=f(x)$ and $f(1)=f(2)=1$ ?

I've been wondering about this problem for about $1 \frac{1}{2}$ years.
I don't know the tools to solve this problem. So, if you could show me how to find a solution, I would like that.
I found the values of $f(x)$ from $0$ to $10$:
$$0,1,1,2,3,7,22,155,3411,528706,1803416167,\dots,f(n)$$
I realized that at $f(-1)$ can't be found just using $f(x-1)f(x-2)+1=f(x)$ because $n\times f(0)=f(1)$ has infinite solutions. $f(-1)$ can't be zero because then $f(-2)$ wouldn't be defined because $n\times 0=-1$. So maybe if I add $f(x)$ must differentiable everywhere I could get an answer.
 A: Inspired by Milten's comment, here's some dirty asymptotic calculation:

*

*Approximate the original equation
$$f(x) \approx f(x-1)f(x-2)$$
Given the terms you have written, this approximation should be pretty good everywhere except for the first few terms


*Let $g(x) = \log f(x)$. Then
$$g(x) = g(x-1) + g(x-2)$$
is a Fibonacci sequence. The explicit form of this sequence is given by
$$g(x) = \frac{\phi^x-\psi^x}{\sqrt5}$$
where $\phi$ and $\psi$ are golden ratio and its square-conjugate. Since $\psi\approx-0.6$, it will asymptotically tend to zero, so the asymptotic behaviour of the approximated sequence will be
$$f(x) \approx \exp \frac{\phi^x}{\sqrt 5}$$
Probably this approach can be cleaned up by considering very precisely the first few initial values of the sequence $g(x)$ to obtain expressions precise to a desired order. Analytic closed-form expression likely does not exist, because the recursive identity combines both addition and multiplication.
A: I will prove the
Statement
There is no continuous function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that it satisfies following conditions:

*

*$\exists x \in \mathbb{R}$ s.t. $f\left(x\right) = 0$;

*$\forall x \in \mathbb{R}$
$$ f\left(x - 1\right)f\left(x\right) + 1 = f\left(x + 1\right)$$
Proof
Assume that there is such $f$. Let $\mathbb{U}$ be the set of zeros of $f$. $\mathbb{U}$ is nonemtpy.
Let $u \in \mathbb{U}$. Then we have that
$$ f\left(u - 2\right)f\left(u - 1\right) = -1$$
Meaning that both $f\left(u - 2\right)$ and $f\left(u - 1\right)$ are nonzero and have different signs. By continuity there exists
$$\mathbb{U} \ni u' \in \left(u - 2, u - 1\right)$$
Thus we have shown that for any $x \in \mathbb{R}$ there is $u \in \mathbb{U}$ such $u < x$.
If $u \in \mathbb{U}$, then we have the following equalities:
$$\begin{align*}
f\left(u + 0\right) &= 0 \\
f\left(u + 1\right) &= 1 \\
f\left(u + 2\right) &= 1 \\
f\left(u + 3\right) &= 2 \\
f\left(u + n\right) &= a_{n + 1},\quad n \in \mathbb{N} \tag{1} \label{u+n}
\end{align*}$$
, where $a_{n}$ defined here. From this sequence we only need to know that $a_{n}$ is non-decreasing and
$$ \lim\limits_{n \rightarrow \infty} a_{n} = +\infty \tag{2} \label{a_n}$$
By the Weierstrass theorem $f$ reaches its maximum $\mathsf{M} \in \mathbb{R}$ on $\left[0, 1\right]$. By \eqref{a_n} we can find $n \in \mathbb{N}$ such that $a_{n} > \mathsf{M}$. Then we find $u \in \mathbb{U}$ and $m \in \mathbb{N}$ s.t. $-m \leq u \leq -m + 1 \leq -n$. It follows that $u + m \in \left[0, 1\right]$ and $m > n$. But by \eqref{u+n}
$$\begin{align*} \mathsf{M} &\geq f\left(u + m\right) \\
&\geq a_{m + 1} \\ 
&\geq a_{n} \\
&> \mathsf{M} \end{align*}$$
This completes the proof.
It follows that there is no function you are looking for, if it's defined like $f : \mathbb{R} \rightarrow \mathbb{R}$.
P. S. English is not my native language. I apologize for any possible mistakes when writing the answer in English.
A: I am not sure whether this will be of any help for solving this problem, but here it is:
We know $$f(x-1).f(x-2)+1 = f(x)\longrightarrow(1)$$ holds for every $x$ as per the question. Therefore,
$$f((x+1)-1).f((x+1)-2)+1 = f(x+1)\\
=f(x).f(x-1)+1 = f(x+1)\longrightarrow(2)$$
and
$$f((x+2)-1).f((x+2)-2)+1 = f(x+2)\\
=f(x+1).f(x)+1 = f(x+2)\longrightarrow(3)$$
$(2)- (1) =$
$$ f(x + 1) - f(x) = f(x - 1).(f(x) - f(x - 2))\longrightarrow(4)$$
Putting $x = 0$ in $(4)$, we get:
$$f(1) - f(0) = f(-1).(0 - f(-2))\\
1 = -f(-1).f(-2)\\
\implies f(-1) = \frac{-1}{f(-2)} \space\text{or}\space f(-2) = \frac{-1}{f(-1)}
\implies f(-1), f(-2) \space \text{could be non-zero real numbers}
$$
I tried to find their values, but ended up in erroneous calculations (I was half asleep when I did all these yesterday, a few minutes before midnight, so I had no sense of my path).
I feel that this is an additional clue to the values of $f(-1)$ and $f(-2)$, but I have no idea how much this would help.
I'd suggest that you wait till a proper answer is found. And please don't downvote my answer, I had to work even during midnight to find a way.
EDIT $1$ :There was a simpler way to arrive here (see the comments below), but I described the path I took as well as just one inference I had.
EDIT $2$: Refer the answer below for a better answer.
