
1. Find all real solutions of the system

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


Solution. The left hand sides of the equations are square roots, i.e. non-negative, thus right hand sides are non-negative as well and $z \geqslant 1, x \geqslant 1, y \geqslant 1$.

We square the equations to get

$$
x^{2}-y=(z-1)^{2}, \quad y^{2}-z=(x-1)^{2}, \quad z^{2}-x=(y-1)^{2} \text {, }
$$

and sum up and simplify:

$$
\begin{aligned}
\left(x^{2}-y\right)+\left(y^{2}-z\right)+\left(z^{2}-x\right) & =(z-1)^{2}+(x-1)^{2}+(y-1)^{2}, \\
\left(x^{2}+y^{2}+z^{2}\right)-(x+y+z) & =\left(z^{2}+x^{2}+y^{2}\right)-2(z+x+y)+3, \\
x+y+z & =3 .
\end{aligned}
$$

On the other hand, summing up the inequalities $x \geqslant 1, y \geqslant 1$ a $z \geqslant 1$ we obtain $x+y+z \geqslant 3$, therefore $x=y=z=1$ is the only possibility. The triple $(x, y, z)=(1,1,1)$ is indeed a solution.

Conclusion: The problem has the unique solution $(x, y, z)=(1,1,1)$.
