SOLUTION. By squaring and subtracting the first two equations, we get $x^{2}-z^{2}=(z+1)^{2}-(x+1)^{2}$, which we can rearrange to $2\left(x^{2}-z^{2}\right)+2(x-z)=0$ or

$$
(x-z)(x+z+1)=0
$$

Similarly, we would get two more equations arising from (1) by cyclically permuting the unknowns $x \rightarrow y \rightarrow z$. Given this symmetry (the system remains unchanged even under any permutation of the unknowns), it suffices to consider only the following two cases:

If $x=y=z$, the original system reduces to a single equation $\sqrt{2 x^{2}}=x+1$, which has two solutions $x_{1,2}=1 \pm \sqrt{2}$. Each of the triples $(1 \pm \sqrt{2}, 1 \pm \sqrt{2}, 1 \pm \sqrt{2})$ is clearly a solution to the original system as well.

If, on the other hand, two of the numbers $x, y, z$ are different, for example, $x \neq z$, it follows from (1) that $x+z=-1$. Substituting $x+1=-z$ into the second equation of the system, we get $y=0$, and then from the third equation, we obtain $x^{2}+(x+1)^{2}=1$ or $x(x+1)=0$. The last equation thus has two solutions $x=0$ and $x=-1$, corresponding to $z=-1$ and $z=0$. It is easy to verify that both found triples $(0,0,-1)$ and $(-1,0,0)$ are solutions to the given system, as is the triple $(0,-1,0)$, which we get by permuting them.

The given system has a total of five solutions:

$$
(0,0,-1),(0,-1,0),(-1,0,0),(1+\sqrt{2}, 1+\sqrt{2}, 1+\sqrt{2}) \text { and }(1-\sqrt{2}, 1-\sqrt{2}, 1-\sqrt{2}) \text {. }
$$

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Solve the system in the set of real numbers

$$
\begin{aligned}
& x^{2}=y+z+1 \\
& y^{2}=z+x+1 \\
& z^{2}=x+y+1
\end{aligned}
$$

$[(0,0,-1),(0,-1,0),(-1,0,0),(1+\sqrt{2}, 1+\sqrt{2}, 1+\sqrt{2})$ and $(1-\sqrt{2}, 1-\sqrt{2}, 1-\sqrt{2})]$

N2. Solve the system in the set of real numbers

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

$[59-\mathrm{A}-\mathrm{S}-1]$

N3. Determine all triples $(x, y, z)$ of real numbers for which

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

$[58-\mathrm{B}-\mathrm{I}-2]$

N4. Solve the system of equations in the set of real numbers

$$
\begin{aligned}
x+y & =1 \\
x-y & =a \\
-4 a x+4 y & =z^{2}+4
\end{aligned}
$$

for unknowns $x, y, z$ and real parameter $a$. $[58-\mathrm{B}-\mathrm{II}-1]$

N5. Solve the system of equations in the set of real numbers

$$
\begin{aligned}
& x^{2}+2 y z=6(y+z-2), \\
& y^{2}+2 z x=6(z+x-2), \\
& z^{2}+2 x y=6(x+y-2) .
\end{aligned}
$$

$[\mathrm{A}-53-\mathrm{S}-3]$
1. In the field of real numbers, solve the system

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