
1. Find all triplets $(x, y, z)$ of real numbers satisfying

$$
\begin{aligned}
& x^{2}-y z=|y-z|+1, \\
& y^{2}-z x=|z-x|+1, \\
& z^{2}-x y=|x-y|+1 .
\end{aligned}
$$

(Tomáš Jurík)


Solution. The system is symmetric, hence without loss of generality we can assume $x \geqslant y \geqslant z$. Removing the absolute values we get

$$
\begin{aligned}
& x^{2}-y z=y-z+1, \\
& y^{2}-z x=x-z+1, \\
& z^{2}-x y=x-y+1 .
\end{aligned}
$$

Subtracting (1) and (2), resp. (2) and (3) and rewriting we obtain

$$
\begin{aligned}
& (x-y)(x+y+z+1)=0 \\
& (y-z)(x+y+z-1)=0 .
\end{aligned}
$$

This implies that $x, y, z$ can not be all mutually different. However, they can't be all equal either as this would yield $0=1$ in the original system. Thus precisely two of them are equal and we get two cases: Either $x=y>z$ and $x+y+z=1$, or $x>y=z$ and $x+y+z=-1$.

Observe that $(x, y, z)$ is a solution if and only if $(-z,-y,-x)$ is a solution, thus it suffices to solve the first case $x=y>z$ and $x+y+z=1$. Then we have $z=1-2 x$. Plugging this into (1) we obtain $x(3 x-4)=0$, hence $x=0$ or $x=\frac{4}{3}$. This corresponds to triples $(x, y, z)$ equal to $(0,0,1)$ and $\left(\frac{4}{3}, \frac{4}{3},-\frac{5}{3}\right)$. The first triplet violates the assumption $x \geqslant y \geqslant z$, the other one indeed is a solution to the original system.

Together with the symmetries we discarded along the way, this gives 6 solutions:

$$
\begin{gathered}
\left(\frac{4}{3}, \frac{4}{3},-\frac{5}{3}\right),\left(-\frac{5}{3}, \frac{4}{3}, \frac{4}{3}\right),\left(\frac{4}{3},-\frac{5}{3}, \frac{4}{3}\right), \\
\left(\frac{5}{3},-\frac{4}{3},-\frac{4}{3}\right),\left(-\frac{4}{3},-\frac{4}{3}, \frac{5}{3}\right),\left(-\frac{4}{3}, \frac{5}{3},-\frac{4}{3}\right) .
\end{gathered}
$$
