Let $f : C \rightarrow C$ such that $f(z)f(z+1)=f(z^2+z+1)$ for all complex numbers $z$.
Let $A$ the set of the zeros of the function. If the set $A$ is non-empty and it has a finite number of elements, then prove that $A=\{i,-i\}$.
What I have done: we have $f(0)f(1)=f(1)$. If $f(1)$ is zero, then so is $f(3)$ (put $z=1$ in the hypotesis), and so on, we get that A is infinite, contradiction. Thus, f(1) is not zero, so f(0)=1. Plugging in $z=-1$ we get that $f(-1)=f(1)$. What do I do nexr?
Generalize your argument. If $z$ is a root of $f$, then so is $z^2 + z + 1$. Define the sequence $a_0 = z, a_{n+1} = a_n^2 + a_n + 1$. Every element of it is a root. Show that unless $a_0 = \pm i$, that sequence has infinitely many distinct entries.