Determine all triples $(x, y, z)$ of non-negative real numbers that satisfy the system of equations

$$
\begin{aligned}
& x^{2}-y=(z-1)^{2} \\
& y^{2}-z=(x-1)^{2} \\
& z^{2}-x=(y-1)^{2}
\end{aligned}
$$
Expanding the brackets and adding everything together gives

$$
x^{2}+y^{2}+z^{2}-(x+y+z)=x^{2}+y^{2}+z^{2}-2(x+y+z)+3,
$$

so $x+y+z=3$. Without loss of generality, we assume that $x \leq y, z$. Then $0 \leq x \leq 1$. Therefore, $x^{2} \leq x$, so $x^{2}-y \leq x-y \leq 0$. On the other hand, $x^{2}-y=(z-1)^{2} \geq 0$. Thus, equality must hold in both $x^{2} \leq x$ and $x-y \leq 0$. From the first equality, we get $x=0$ or $x=1$, and from the second, $x=y$. Suppose $x=y=0$, then from $x+y+z=3$ we get $z=3$. But then $x^{2}-y=(z-1)^{2}$ does not hold, a contradiction. We are left with the case that $x=y=1$. Then $z=3-1-1=1$. This triplet indeed satisfies the conditions and is therefore the only solution.