Bulgarian Mathematical Competitions

Solution:
If $x \neq -1$ and $y \neq -1$, we easily get that $y = a$. Plugging it in the second equation gives
$$a x^{2} - (a^{2} - 2a - 1)x + a^{3} = 0$$
If $a = 0$ the system has a unique solution $(0 ; 0)$. If $a \neq 0$, we consider the following two cases.

Case 1. $-1$ is a root of $(*)$. Then $(*)$ gives that $a = 1$ or $-1$. In both cases the system has no a solution.

Case 2. The equation $(*)$ has a double root. Then $a = -1$, $a = 1$ or $a = -\frac{1}{3}$. In the first two cases we get that $y = -1$ or $x = -1$, i.e., the system has no a solutions. If $a = -\frac{1}{3}$ the system has a unique solution $(x ; y) = \left(\frac{1}{3} ; -\frac{1}{3}\right)$.

Thus the desired values of $a$ are $0$ and $-\frac{1}{3}$.