Solve the following system of equations:

$$
\begin{array}{r}
x+y+z=0 \\
x^{2}+y^{2}-z^{2}=20 \\
x^{4}+y^{4}-z^{4}=560
\end{array}
$$
In the given system of equations, $x$ and $y$ play the same role, while $z$ plays a slightly different one. It is therefore advisable to eliminate $z$. From (1), $z=-(x+y)$; substituting this into (2) and rearranging, we get:

$$
x y=-10
$$

Similarly, substituting the value of $z^2$ from (2) into (3),

$$
x^{4}+y^{4}-\left(x^{2}+y^{2}-20\right)^{2}=-2 x^{2} y^{2}+40\left(x^{2}+y^{2}\right)-400=560
$$

and from this, considering (4),

$$
x^{2}+y^{2}=29
$$

Thus, we have obtained a simple system of equations for $x, y$ in (4) and (5). From here,

$$
\begin{gathered}
\left(x^{2}+y^{2}\right)+2 x y=(x+y)^{2}=29-20=9, \quad \text { so } \quad x+y= \pm 3 \\
\left(x^{2}+y^{2}\right)-2 x y=(x-y)^{2}=29+20=49, \quad \text { so } \quad x-y= \pm 7
\end{gathered}
$$

and this, with the $2 \times 2$ possible choices of signs, gives four linear systems for $x, y$. For example,

$$
x+y=-3, \quad x-y=+7-\text { from which } \quad x=+2, \quad y=-5, \quad \text { and then by (1) } z=3 .
$$

This solution satisfies (2) and (3) as well. The same holds for the other three sign combinations, so the given system has four solutions:

$$
x, y, z=5,-2,-3 ; \quad-2,5,-3 ; \quad 2,-5,3 ; \quad-5,2,3
$$

Gergely János (Keszthely, Vajda J. g. III. o., t. )