Given $f$ an entire function so that $f\circ f$ don't have fixed points.
1) Prove that there is a constant $c\in \mathbb{C}$ with $c\not=0,1$ so that $f(f(z))-z=c(f(z)-z)$, $\forall z\in\mathbb{C}$
2) Prove that $f'\circ f$ is a constant function (using the derivative of the expression before).
3) Show that $f$ is a translation $f(z)=z+b$ with $b\not=0$
I have seen similar problems but not exactly this one. For 1) I have used the little Picard theorem with the function $\frac{f(f(z))-z}{f(z)-z}$. As this function doesn't assume the values 0 and 1, it is constant. For 2) I have $f'(f(z))f'(z)-1=c(f'(z)-1)$, but I don't know how to conclude that $f'\circ f$ is constant. For 3) I don't have any ideas. Hints? Thank you very much