
1. Solve the system of equations

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

in the domain of real numbers.


Solution. Adding up all three equations and cancelling the quadratic terms gives

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

Thus $z=-x-y$ and substituting this into the first equation yields $x^{2}-y=(-x-y)^{2}$, or $y(2 x+y+1)=0$, hence either $y=0$ or $2 x+y+1=0$.

If $y=0$ then (1) implies $z=-x$ and upon substituting for $y, z$ into the original system we get for the unknown $x$ the single condition $x(x-1)=0$, which is satisfied only for $x=0$ or $x=1$. This corresponds to solutions $(x, y, z)$ of the form $(0,0,0)$ and $(1,0,-1)$.

If $2 x+y+1=0$, or $y=-2 x-1$, (1) implies $z=-x-y=x+1$. Substituting these $y, z$ into the original system yields for the unknown $x$ the single condition $x(x+1)=0$, which is fulfilled only for $x=0$ and $x=-1$. To these correspond the solutions of he form $(0,-1,1)$ and $(-1,1,0)$.

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