Golden functions?

Question

If one seeks the functional solutions of $f^2(x)=f(x)+1=f(f(x))$

Are they called 'golden functions' ?

Are they always of the form of rational functions ?

Answer 1

At least on the (unbounded and infinite!) set $\operatorname{im}(f)$, we see that $f(t)=t+1$. Hence the only rational function solution is just $f(x)=x+1$.

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In fact, the following will describe all solutions $f\colon \Bbb R\to\Bbb R$ of the functional equation $$ \forall x\in \Bbb R\colon f(f(x))=f(x)+1$$

Pick a non-empty set $A\subseteq \Bbb R$ that is closed under "$+1$". In other words, we have $A+1\subseteq A$, or $A$ can be written as $A=\bigcup_{a\in A_0}(\Bbb N+a)$. Let $g\colon \Bbb R\setminus A\to A$ be arbitrary and define $$ f(x)=\begin{cases}x+1&x\in A\\g(x)&x\notin A\end{cases}$$ Indeed, with such $f$, we have $f(x)\in A$ for all $x$ and hence $f(f(x))=f(x)+1$.

Here's a non-trivial example of a smooth such function: $$ f(x)=\begin{cases}x+1&x\ge 0\\x^2+x+1+e^{1/x}&x<0\end{cases}$$