
1. Find all real solutions of the system

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


Solution. The square roots and their arguments have to be positive, therefore $x, y, z \geqslant 1, x \geqslant y^{2}, y \geqslant z^{2}$, and $z \geqslant x^{2}$. The last three inequalities imply $x \geqslant$ $y^{2} \geqslant y \geqslant z^{2} \geqslant z \geqslant x^{2}$, and since $x \geqslant 1$, the inequality $x \geqslant x^{2}$ forces $x=1$, and all the values in the given chain of inequalities are equal, especially $x=y=z=1$, which is indeed a solution.

Conclusion. There is a unique solution $x=y=z=1$.
