Balti Tee 1996 · Ülesanne 13

Solution: (a) For $f$ even, we have $f(x-1) = f\left((x-1)^{2} + (x-1) + 1\right) = f\left(x^{2} - x + 1\right) = f\left((-x)^{2} - x + 1\right) = f(-x) = f(x)$ for any $x \in \mathbb{Z}$. Hence $f$ has a constant value; any constant will do.

(b) For $f$ odd, a similar computation yields $f(x-1) = -f(x)$. Since $f(0) = 0$, we see that $f(x) = 0$ for all $x \in \mathbb{Z}$.