Solve the following system of equations:

$$
\begin{aligned}
& x(x+y)+z(x-y)=a \\
& y(y+z)+x(y-z)=b \\
& z(z+x)+y(z-x)=c
\end{aligned}
$$
By adding our equations in pairs,

$$
(x+y)^{2}=a+b, \quad(y+z)^{2}=b+c, \quad(z+x)^{2}=c+a
$$

from here, two values come for the pairwise sums of the unknowns:

$$
x+y= \pm \sqrt{a+b}, \quad y+z= \pm \sqrt{b+c}, \quad z+x= \pm \sqrt{c+a}
$$

and the square root signs can be chosen independently in each case, resulting in a solvable system of linear equations. For the sake of clarity, we denote the 1 or -1 multiplier before each square root as $e_{1}$, $e_{2}$, and $e_{3}$.

The sum of the equations in (1) gives half the value of $x+y+z$, and by subtracting each of these equations from it, we always get the third unknown:

$$
\begin{aligned}
x & =\frac{1}{2}\left(e_{1} \sqrt{a+b}-e_{2} \sqrt{b+c}+e_{3} \sqrt{c+a}\right) \\
y & =\frac{1}{2}\left(e_{1} \sqrt{a+b}+e_{2} \sqrt{b+c}-e_{3} \sqrt{c+a}\right) \\
z & =\frac{1}{2}\left(-e_{1} \sqrt{a+b}+e_{2} \sqrt{b+c}+e_{3} \sqrt{c+a}\right)
\end{aligned}
$$

The choice of 3 signs is independent of each other, so the system can have $2^{3}=8$ solutions. Indeed, this is the number of solutions, if $a+b$, $b+c$, and $c+a$ are all positive. This number is halved every time a zero appears among the sums. However, if there is a negative number among the sums, the system has no real solutions.

Katona Viktor (Heves, Gymn., IV. grade). Somogyi Kornélia (Ajka, Bródy I. Gymn., III. grade).

Remark. It can already be seen from the original equations that if the system has a solution $x=\alpha, y=\beta, z=\gamma$, then the value triplet $x=-\alpha, y=-\beta, z=-\gamma$ is also a solution.