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

$$
\begin{aligned}
& \sqrt{x^{2}-y}=z-1, \\
& \sqrt{y^{2}-z}=x-1, \\
& \sqrt{z^{2}-x}=y-1 .
\end{aligned}
$$
SOLUTION. The left sides of the given equations have (as square roots) non-negative values, hence the inequalities $z \geqq 1, x \geqq 1$ and $y \geqq 1$ follow from the right sides.

We eliminate the square roots in the equations by squaring them:

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

We add the squared equations and simplify the result:

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

However, from the inequalities $x \geqq 1, y \geqq 1$ and $z \geqq 1$, it follows by addition that $x+y+z \geqq 3$, so the equality $x+y+z=3$ can only be satisfied if $x=y=z=1$. By substituting, we can verify that the triplet $(x, y, z)=(1,1,1)$ is indeed a solution (the only one, as follows from our procedure).

Let us add that if we do not notice the inequalities $x, y, z \geqq 1$ beforehand, but derive the relation $x+y+z=3$ after adding the squared equations, we can then use the determined value of the sum $x+y+z$ when adding the original (unsquared) equations, thus obtaining the equation $\sqrt{x^{2}-y}+\sqrt{y^{2}-z}+\sqrt{z^{2}-x}=0$ with the clear consequence: each of the square roots must be equal to zero.