I came to the conclusion that f is periodic with period |f(x0)| for x0 non-zero. But I don't see how the periodicity in domain translates to a bound in the codomain. It probably has something to do with the fact that the image intervenes with the periodicity, but I can't put my hand precisely on it and the solution doesn't provide this explanation.
Can someone please explain this part to me? Thank you!
What I have until now : By substituting $y = f(x)$ we obtain $f(0) = 0$ (1) By substituting $y = 0$ we obtain $f(f(x) = f(0) - f(f(x))$ which implies that $f(f(x)) = 0$ pour tout $x \in \mathbb{Z}$ (2) By substituting $x = 0$, we obtain $f(-y) = f(y)$ (3)
By switching the sign of $y$ we obtain : $$f(f(x)+y) = f(-y) - f(f(x))$$ Which we can simplify with (3) and (2) $$f(y+f(x)) = f(y)$$ We take any $f(x)$ that is non-zero. Let $k$ be this value of $x$ and $|f(k)| = a$. We have $f(y+a)=f(y)$ which means f(y+a) has a period of $a = |f(k)| $