1. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& x^{2}-y=z^{2} \\
& y^{2}-z=x^{2} \\
& z^{2}-x=y^{2}
\end{aligned}
$$
1. By adding all three equations after canceling the quadratic terms, we get

$$
x+y+z=0 \text {. }
$$

From this, we express $z=-x-y$ and substitute it into the first equation of the system. We obtain $x^{2}-y=(-x-y)^{2}$, which after simplification gives the equation $y(2 x+y+1)=0$. Therefore, we distinguish which of the two factors on its left side is zero.

In the case $y=0$, from equation (1) we get $z=-x$ and after substituting $y, z$ into the original system, we get for the unknown $x$ the single condition $x(x-1)=0$, which is satisfied only by $x=0$ and $x=1$. These correspond to the solutions $(x, y, z)$ of the form $(0,0,0)$ and $(1,0,-1)$.

In the case where $2 x+y+1=0$ or $y=-2 x-1$, from (1) we have $z=-x-y=x+1$. After substituting $y, z$ into the original system, we get for the unknown $x$ the single condition $x(x+1)=0$, which is satisfied only by $x=0$ and $x=-1$. These correspond to the solutions $(x, y, z)$ of the form $(0,-1,1)$ and $(-1,1,0)$.

Conclusion: The given system has exactly four solutions $(x, y, z)$: the triples $(0,0,0),(1,0,-1)$, $(0,-1,1)$, and $(-1,1,0)$.

Another solution. By adding the first two equations of the given system, we eliminate the unknown $x$ and get the equation $y^{2}-z^{2}=y+z$, which can be written in the product form

$$
(y+z)(y-z-1)=0 \text {. }
$$

Again, we distinguish which of the two factors in the last equation is zero.

In the case $y+z=0$, from the third equation of the given system, we get $x=0$ and from the first two equations after substituting $x=0$ and $z=-y$, we get for the unknown $y$ the single condition $y(y+1)=0$, i.e., $y=0$ or $y=-1$. The corresponding solutions $(x, y, z)$ are $(0,0,0)$ and $(0,-1,1)$.

In the case where $y-z-1=0$ or $z=y-1$, we get from the third equation of the system $x=z^{2}-y^{2}=(y-1)^{2}-y^{2}=1-2 y$. Substituting $x, z$ gives for the unknown $y$ the single condition $y(y-1)=0$, i.e., $y=0$ or $y=1$. The corresponding solutions $(x, y, z)$ are $(1,0,-1)$ and $(-1,1,0)$.

For a complete solution, award 6 points. The derivation of equation (1) or at least one of the three analogous equations in the product form (2) should be valued at 2 points.