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: $$\begin{array}{rl}{x}^2-yz& =y-z+1,\\ y^2-zx& =x-z+1,\\ z^2-xy& =x-y+1.\end{array}$$ 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 $x\ne y$. In the first case, we have $y\ne z$, 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 $x\ne y$. 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 $(-\frac{4}{3},-\frac{4}{3},\frac{5}{3})$. $\square$