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

$$
\begin{aligned}
& x y+1=z^{2}, \\
& y z+2=x^{2}, \\
& z x+3=y^{2} .
\end{aligned}
$$

(Tomáš Jurík)
SOLUTION. By subtracting the second equation from the first and making a minor adjustment, we get

$$
\begin{aligned}
y(x-z)-1 & =(z-x)(z+x), \\
-1 & =(z-x)(z+x+y) .
\end{aligned}
$$

Similarly, by manipulating the difference between the second and third equations, we obtain

$$
\begin{aligned}
z(y-x)-1 & =(x-y)(x+y) \\
-1 & =(x-y)(x+y+z) .
\end{aligned}
$$

From either of the two obtained equations, it follows that $x+y+z \neq 0$, so we can divide both by $x+y+z$ and then compare them to get

$$
z-x=\frac{-1}{x+y+z}=x-y
$$

Let $d=z-x=x-y$. Then $z=x+d$ and $y=x-d$, so all solutions to the original system are of the form $(x, y, z)=(x, x-d, x+d)$. Substitute such $y$ and $z$ into the individual equations.

Substituting into the second equation, we get

$$
(x-d)(x+d)+2=x^{2} \quad \text { or } \quad d^{2}=2, \quad \text { i.e., } \quad d= \pm \sqrt{2}
$$

Substituting into the first and third equations and using the result $d^{2}=2$, we can write the following in two columns side by side:

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

For $d=+\sqrt{2}$, we get $x=-\frac{1}{3 \sqrt{2}}=-\frac{1}{6} \sqrt{2}$, and for $d=-\sqrt{2}$, we get $x=\frac{1}{3 \sqrt{2}}=\frac{1}{6} \sqrt{2}$. By calculating the values of $y=x-d$ and $z=x+d$, we obtain the (only) two solutions to the problem, which are the triples

$$
(x, y, z) \in\left\{\left(-\frac{1 \sqrt{2}}{6},-\frac{7 \sqrt{2}}{6}, \frac{5 \sqrt{2}}{6}\right),\left(\frac{1 \sqrt{2}}{6}, \frac{7 \sqrt{2}}{6},-\frac{5 \sqrt{2}}{6}\right)\right\} .
$$

Note that in our approach, verification is not necessary, as we have ensured the satisfaction of all three original equations by the method of determining the values of $x, d$ for the derived triples $(x, x-d, x+d)$.

Remark. The same intermediate result $(x, y, z)=(x, x-d, x+d)$ can also be achieved by comparing the differences of other pairs of equations. For example, by comparing the equations $-1=(x-y)(x+y+z)$ and $-2=(z-y)(x+y+z)$, we get $z-y=2(x-y)$.