Croatian Mathematical Competitions

By summing the given equations we get
$$x + y + z = 0.$$
It follows that $z = -x - y$, and from the first equation we get
$$\begin{aligned} x^2 - y &= (-x - y)^2 \\ x^2 - y &= x^2 + 2xy + y^2 \\ 2xy + y^2 + y &= 0 \\ y(2x + y + 1) &= 0, \end{aligned}$$
so $y = 0$ or $2x + y + 1 = 0$.

If $y = 0$, it follows that $z = -x$ so from the second equation we get $x(x - 1) = 0$, hence $x = 0$ or $x = 1$. The corresponding solutions are $(x, y, z) = (0, 0, 0)$ and $(x, y, z) = (1, 0, -1)$.

If $2x + y + 1 = 0$, we have $y = -2x - 1$ so $z = -x - y = x + 1$. Now from the third equation we get $x(x + 1) = 0$, hence $x = 0$ or $x = -1$. The corresponding solutions are $(x, y, z) = (0, -1, 1)$ and $(x, y, z) = (-1, 1, 0)$.

Thus, the solutions of the given system of equations are
$$(x, y, z) \in \{(0, 0, 0), (1, 0, -1), (0, -1, 1), (-1, 1, 0)\}.$$