Croatian Mathematical Competitions

Subtracting the given equations, we obtain $y - x + x^2 - y^2 = x^3 - y^3$, which is equivalent to
$$(x - y)(-1 + x + y) = (x - y)(x^2 + xy + y^2),$$
implying that either $x - y = 0$ or $x + y - 1 = x^2 + xy + y^2$.

In the latter case, $x + y - 1 = x^2 + xy + y^2$ is equivalent to
$$y^2 + (x-1)y + x^2 - x + 1 = 0. \quad (*)$$
This is a quadratic equation in $y$. Its discriminant is
$$D = (x - 1)^2 - 4(x^2 - x + 1) = -3x^2 + 2x - 3,$$
which is a quadratic polynomial in $x$ – its discriminant in turn is $-32 < 0$, so the starting discriminant is $D < 0$ for every real number $x$, which implies that the equation $(*)$ does not have any real solution $y$. Hence there are no solutions in this case.

In the remaining case, $y = x$ so both equations of a given system are reduced to $x + x^2 = x^3$, i.e. $x(x^2 - x - 1) = 0$. This equation has three solutions $x_1 = 0$, $x_{2,3} = \frac{1 \pm \sqrt{5}}{2}$.

The only solutions are $(x, y) \in \{(0, 0), (\frac{1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}), (\frac{1-\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2})\}$.