TITLE: A weird functional equation: $f\big(f(x)\big)=\big(f(x)+1\big)x$
QUESTION [6 upvotes]: Let $f:\mathbb{R}\rightarrow \mathbb{R}$
be a function such that
$$f\big(f(x)\big)=\big(f(x)+1\big)x$$
for all real $x$, $f(-1)=0$ and $f(0)=-1$.
Find all such functions $f$.

REPLY [2 votes]: If that helps:
Let $x_n$ denote the $n^{th}$ iterate of the function $f$. We have the recurrence
$$x_0=x,\\x_{n+2}=(x_{n+1}+1)x_n.$$
For large $x_{n+1}$ we can neglect the term $+1$ and, taking the logarithm, we get a Fibonacci-like recurrence:
$$\log x_{n+2}=\log x_{n+1}+\log x_n.$$
Hence the iterates are asymptotic to $a^{F_n}$ for some $a$ and 
$$f(x)\to x^\phi.$$