7. In the field of real numbers, solve the system

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

(Tomáš Jurík)
Solution. The given system of equations is symmetric, so it suffices to look for solutions that satisfy the inequalities $x \geqq y \geqq z$. Under this assumption, we can eliminate the absolute values, and thus we get

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

By successively subtracting equations (1) and (2), and (2) and (3), we obtain after simple manipulations the equations

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

From this, it follows that all three numbers $x, y, z$ cannot be mutually distinct, but they cannot all be the same either, because then we would get $0=1$ in the original equations. Exactly two of them are therefore different, so either $x=y>z$ and $x+y+z=1$, or $x>y=z$ and $x+y+z=-1$.

Notice that a triple $(x, y, z)$ satisfies the original system if and only if the "opposite" triple $(-z,-y,-x)$ does. The transition to the opposite triple does not change the established order of the numbers in the triple and converts the second case from the previous paragraph to the first one.

Therefore, it suffices to solve the first case, where $x=y>z$ and $x+y+z=1$. This gives us $z=1-2 x$. Substituting, for example, into equation (1) we get after rearrangement $x(3 x-4)=0$, so $x=0$ or $x=\frac{4}{3}$. This corresponds to the triples $(x, y, z)$ equal to $(0,0,1)$ and $\left(\frac{4}{3}, \frac{4}{3},-\frac{5}{3}\right)$. The first triple clearly does not satisfy the assumption $x \geqq y \geqq z$, so the original system is satisfied only by the second triple (verification in the given process is not necessary).

With regard to changes in the order of the unknowns and transitions to opposite triples, the system has 6 solutions:

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