Task 2. Determine all triples $(x, y, z)$ of real numbers for which

$$
\begin{aligned}
& x^{2}-y z=|y-z|+1, \\
& y^{2}-z x=|z-x|+1, \\
& z^{2}-x y=|x-y|+1 .
\end{aligned}
$$
Solution. The system of equations is symmetric: if you swap $x$ and $y$, for example, the third equation remains the same and the first two equations switch places. We can thus assume without loss of generality that $x \geq y \geq z$. The system then becomes:

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

Subtracting the first two equations gives $x^{2}-y^{2}+z(x-y)=y-x$, or $(x-y)(x+y+z+1)=0$. This implies $x=y$ or $x+y+z=-1$. Subtracting the second and third equations gives $y^{2}-z^{2}+x(y-z)=y-z$, or $(y-z)(y+z+x-1)=0$. This implies $y=z$ or $x+y+z=1$.

We now consider two cases: $x=y$ and $x \neq y$. In the first case, $y \neq z$, because otherwise $x=y=z$ and this would give $0=1$ in the first equation, a contradiction. Thus, we have $x+y+z=1$, or $2x+z=1$. Substituting $y=x$ and $z=1-2x$ into the first equation gives $x^{2}-x(1-2x)=x-(1-2x)+1$, which simplifies to $3x^{2}-x=3x$, or $3x^{2}=4x$. This gives $x=0$ or $x=\frac{4}{3}$. With $x=0$, we get $y=0, z=1$, but this does not satisfy our assumption $x \geq y \geq z$. Thus, we are left with $x=\frac{4}{3}$, which gives the triplet $\left(\frac{4}{3}, \frac{4}{3},-\frac{5}{3}\right)$. Checking shows that this is a solution.

Now consider the case $x \neq y$. Then $x+y+z=-1$, so it cannot also be true that $x+y+z=1$, so we see that $y=z$. Now $x+y+z=-1$ gives $x+2z=-1$, so $x=-1-2z$. The first equation becomes $(-1-2z)^{2}-z^{2}=1$, which simplifies to $3z^{2}+4z=0$. This gives $z=0$ or $z=-\frac{4}{3}$. With $z=0$, we get $y=0, x=-1$, but this does not satisfy our assumption $x \geq y \geq z$. Thus, we are left with $z=-\frac{4}{3}$, which gives the triplet $\left(\frac{5}{3},-\frac{4}{3},-\frac{4}{3}\right)$. Checking shows that this is a solution.

By including the permutations of these two solutions, we find a total of six solutions: $\left(\frac{4}{3}, \frac{4}{3},-\frac{5}{3}\right),\left(\frac{4}{3},-\frac{5}{3}, \frac{4}{3}\right),\left(-\frac{5}{3}, \frac{4}{3}, \frac{4}{3}\right),\left(\frac{5}{3},-\frac{4}{3},-\frac{4}{3}\right),\left(-\frac{4}{3}, \frac{5}{3},-\frac{4}{3}\right)$, and $\left(-\frac{4}{3},-\frac{4}{3}, \frac{5}{3}\right)$.