Balti Tee 2019 · Ülesanne 3

Putting $y = -1$ and $x = 0$, we obtain $f(-1) = 0$. Suppose that $f$ has another zero, i.e., there is an $a \neq -1$ with $f(a) = 0$. Putting $x = 0$ and $y = -a$ gives $f(-a^2) = 0$. Now choosing $y = a$, we obtain $f(x - a) = 0$. Hence, $f(x) = 0$ for all $x$ which clearly is a solution.

Let us assume now that $x = -1$ is the only zero of $f$. Then $x = y - 1$ gives $f((y - 1)f(y) - y^2) = 0$. Consequently, we have $(y - 1)f(y) = y^2 - 1$ for all $y$ which implies $f(x) = x + 1$ for all $x \neq 1$. Setting $y = 2, x = 3$ gives $f(1) = 2$. Hence, $f(x) = x + 1$ for all $x$ which is another solution.

Altogether, the only solutions are $f(x) = 0$ and $f(x) = x + 1$.