5. Find all real numbers $x, y, z$ such that $x^{2}+y^{2}+z^{2}=x-z=2$.
SolUTION
First, we claim that $y=0$.
Proof: if $y \neq 0$ then $y^{2}>0$, so we have $x^{2}+z^{2}=2-y^{2}<2$.
Since $x=(2+z)$, we have:
$$
\begin{aligned}
(2+z)^{2}+z^{2} & <2 \\
4+4 z+2 z^{2} & <2 \\
z^{2}+2 z+1 & <0 \\
(z+1)^{2} & <0
\end{aligned}
$$

Which is not possible, so $\boldsymbol{y}=\mathbf{0}$.
Given that $y=0$, we have $x^{2}+z^{2}=2$. We can proceed in a similar way to above, substituting $x=(2+z)$ :
$$
\begin{aligned}
(2+z)^{2}+z^{2} & =2 \\
4+4 z+2 z^{2} & =2 \\
z^{2}+2 z+1 & =0 \\
(z+1)^{2} & =0
\end{aligned}
$$

Hence $(z+1)=0$ so $z=\mathbf{1}$, and from this we deduce $\boldsymbol{x}=\mathbf{1}$.
It is straightforward to check that these values satisfy the original equations, so the only solution is $(x, y, z)=(\mathbf{1}, \mathbf{0}, \mathbf{- 1})$.