BxMO Team Selection Test

The system of equations is symmetric: if you swap $x$ and $y$, for example, then the third equation stays the same and the first two equations are swapped. Hence, we can assume without loss of generality that $x \ge y \ge z$. Then the system of equations becomes:
$$\begin{aligned} x^2 - yz &= y - z + 1, \\ y^2 - zx &= x - z + 1, \\ z^2 - xy &= x - y + 1. \end{aligned}$$
Subtracting the second equation from the first, we obtain $x^2 - y^2 + z(x - y) = y - x$, or $(x - y)(x + y + z + 1) = 0$. This yields $x = y$ or $x + y + z = -1$. Subtracting the third equation from the second, we obtain $y^2 - z^2 + x(y - z) = y - z$, or $(y - z)(y + z + x - 1) = 0$. This yields $y = z$ or $x + y + z = 1$.

We now distinguish two cases: $x = y$ and $x \neq y$. In the first case, we have $y \neq z$, as otherwise we would have $x = y = z$ for which the first equation becomes $0 = 1$, a contradiction. Now it follows that $x + y + z = 1$, or $2x + z = 1$. Substituting $y = x$ and $z = 1 - 2x$ in the first equation yields $x^2 - x(1 - 2x) = x - (1 - 2x) + 1$, which can be simplified to $3x^2 - x = 3x$, or $3x^2 = 4x$. We get $x = 0$ or $x = \frac{4}{3}$. With $x = 0$, we find $y = 0$, $z = 1$, but does not satisfy our assumption $x \ge y \ge z$. Thus, the only remaining possibility is $x = \frac{4}{3}$, which gives the triple $(\frac{4}{3}, \frac{4}{3}, -\frac{5}{3})$. We verify that this is indeed a solution.

Now consider the case $x \neq y$. Then we have $x + y + z = -1$, hence we cannot have $x + y + z = 1$, and we see that $y = z$. Now $x + y + z = -1$ yields $x + 2z = -1$, hence $x = -1 - 2z$. Now the first equality becomes $(-1 - 2z)^2 - z^2 = 1$, which can be simplified to $3z^2 + 4z = 0$. From this, we conclude that $z = 0$ or $z = -\frac{4}{3}$. With $z = 0$, we find $y = 0$, $x = -1$, which does not satisfy our assumption $x \ge y \ge z$. Hence, the only remaining possibility is $z = -\frac{4}{3}$, and this gives rise to the triple $(\frac{5}{3}, -\frac{4}{3}, -\frac{4}{3})$. We verify that this is indeed a solution.

By also considering the permutations of these two solutions, we find all six solutions: $(\frac{4}{3}, \frac{4}{3}, -\frac{5}{3})$, $(\frac{4}{3}, -\frac{5}{3}, \frac{4}{3})$, $(-\frac{5}{3}, \frac{4}{3}, \frac{4}{3})$, $(\frac{5}{3}, -\frac{4}{3}, -\frac{4}{3})$, $(-\frac{4}{3}, \frac{5}{3}, -\frac{4}{3})$, and $(-\frac{4}{3}, -\frac{4}{3}, \frac{5}{3})$. $\square$