Awkward Functional Equation $f(f(x))=(x+1)f(x)$

Question

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(f(x))=(x+1)f(x)$ for all real $x$ and $f$ attains the value $(-1)$ at some point . Find all such functions $f$.

Answer 1

First we prove $ f $ is injective when $ f(x) \neq 0 $.
Assume: $ f(a)=f(b) \implies (a+1)f(a) = f(f(a)) = f(f(b)) = (b+1)f(b) = (b+1)f(a) \implies a=b \lor f(a)=f(b)=0 \implies $

$ f(x) $ is injective everywhere where $ f(x) \neq 0 $.

Next we prove $ f(0)=0 $:
$ (0+1)f(0) = f(f(0)) \implies f(f(0)) = f(0) \implies $ (semi-)injectivity above $ \implies f(0)=0 $.

Next we prove $ f(-1)=0 $:
$ f(0)=0=(-1+1)f(-1)=f(f(-1)) \implies $ (semi-)injectivity above $ \implies f(-1) =0 $.

Now we show that when $ f() $ must attain the value $-1$ somewhere, ( say at $ x=a $ ) this will lead to a contradiction:
Let $ f(a)=-1 : \implies $
$ -(a+1) =(a+1)f(a)=f(f(a))= f(-1) = 0 \implies -(a+1)=0 \implies a = -1 \implies f(-1)=-1 $

This contradicts $ f(-1) =0 $ so there are no solutions.