Solve the following system of equations:

$$
\begin{aligned}
& x(x+y+z)=a^{2}, \\
& y(x+y+z)=b^{2}, \\
& z(x+y+z)=c^{2} .
\end{aligned}
$$
I. Solution: Add the three equations and factor out the common $(x+y+z)$ factor.

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

from which

$$
x+y+z= \pm \sqrt{a^{2}+b^{2}+c^{2}}
$$

Substituting this value of $x+y+z$ into the original equations, and assuming that $a^{2}+b^{2}+c^{2} \neq 0$, or at least one of $a, b, c$ is not 0, we obtain the following two systems of roots:

$$
\begin{aligned}
& x_{12}= \pm \frac{a^{2}}{\sqrt{a^{2}+b^{2}+c^{2}}} \\
& y_{12}= \pm \frac{b^{2}}{\sqrt{a^{2}+b^{2}+c^{2}}} \\
& z_{12}= \pm \frac{c^{2}}{\sqrt{a^{2}+b^{2}+c^{2}}}
\end{aligned}
$$

(The same sign must be taken within one system of roots.)

István Csákvári (Bp. III., Árpád g. II. o. t.)

Note: The above solution is correct if one or two of the quantities $a, b$, and $c$ are 0. If $a=b=c=0$, then our system of equations turns into the indeterminate equation

$$
x+y+z=0
$$

II. Solution: We have seen that at least one of $a, b, c$ must differ from 0. Suppose $a \neq 0$, then it is free to divide (2) and (3) by (1). Thus

$$
\begin{array}{llll}
\frac{y}{x}=\frac{b^{2}}{a^{2}}, & \text { from which } & (4) & y=\frac{b^{2}}{a^{2}} x \\
\frac{z}{x}=\frac{c^{2}}{a^{2}}, & \text { from which } & (5) & z=\frac{c^{2}}{a^{2}} x
\end{array}
$$

Substituting the values thus obtained into equation (1)

$$
x\left(x+\frac{b^{2}}{a^{2}} x+\frac{c^{2}}{a^{2}} x\right)=x^{2} \frac{a^{2}+b^{2}+c^{2}}{a^{2}}=a^{2}
$$

from which

$$
x= \pm \frac{a^{2}}{\sqrt{a^{2}+b^{2}+c^{2}}}
$$

Substituting this value of $x$ into (4) and (5), we obtain the same system of roots as in the first solution.