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

$$
\begin{aligned}
& \sqrt{x-y^{2}}=z-1, \\
& \sqrt{y-z^{2}}=x-1, \\
& \sqrt{z-x^{2}}=y-1 .
\end{aligned}
$$
1. The values of square roots are always non-negative, and the values under the square roots are also non-negative, so the unknowns $x, y, z$ must satisfy the conditions $x, y, z \geq 1, x \geq y^{2}, y \geq z^{2}$, and $z \geq x^{2}$. From the last three inequalities, we have $\max \{x, y, z\} \geq \max \left\{y^{2}, z^{2}, x^{2}\right\}$. The opposite (non-strict) inequality holds because $t \leq t^{2}$ for every $t \geq 1$. Therefore, $\max \{x, y, z\} = \max \left\{y^{2}, z^{2}, x^{2}\right\} = 1$, so $x = y = z = 1$ and both sides of all three equations in the system are equal to zero (this is a check).

Alternative approach: Instead of reasoning about the maxima, we can proceed after identifying the solution from the first sentence as follows: $x \geq y^{2} \geq y \geq z^{2} \geq z \geq x^{2}$, the inequality between the extreme terms $x \geq x^{2}$ already implies $x = 1$, so the values of $y, z$ in the given chain of six terms are also equal to 1.

Conclusion. The system has a unique solution $x = y = z = 1$.

For a complete solution, 6 points are awarded, unless the check is not performed and the solution requires it, in which case 1 point is deducted. For incomplete solutions, award 1 point for writing all six inequalities from the first sentence of the solution, and 1 point to those who guess the desired triplet. Complete solutions based on raising the equations to a power (without substantial use of the inequality $t \leq t^{2}$ for every $t \geq 1$) are not known to the problem committee; if such solutions appear, please send them.

![](https://cdn.mathpix.com/cropped/2024_04_17_1ebf82ba476a6017f4f1g-2.jpg?height=496&width=625&top_left_y=1174&top_left_x=721)