\section*{Problem 4 - 151224}

Determine all triples \((x, y, z)\) of real numbers that satisfy the system of equations

\[
\begin{array}{r}
x+y+z=a \\
x^{2}+y^{2}+z^{2}=a^{2} \\
x^{3}+y^{3}+z^{3}=a^{3}
\end{array}
\]

where \(a\) is a real number.
}

First, we have

\[
x y + x z + y z = \frac{1}{2} \left( (x + y + z)^{2} - \left( x^{2} + y^{2} + z^{2} \right) \right) = \frac{1}{2} \left( a^{2} - a^{2} \right) = 0
\]

From this, it immediately follows that

\[
\begin{aligned}
a^{3} & = (x + y + z)^{3} = x^{3} + y^{3} + z^{3} + 3 x(x y + x z) + 3 y(x y + y z) + 3 z(x z + y z) + 6 x y z = \\
& = a^{3} + 3 x(-y z) + 3 y(-x z) + 3 z(-x y) + 6 x y = a^{3} - 3 x y z
\end{aligned}
\]

and thus \(x y z = 0\), meaning that at least one of the three variables \(x, y, z\) must be 0.

Assume without loss of generality that \(z = 0\). From

\[
2 x y = (x + y)^{2} - \left( x^{2} + y^{2} \right) = a^{2} - a^{2} = 0
\]

it follows that either \(x\) or \(y\) must also be 0. If \(y = 0\), then \(x = a, y = 0, z = 0\) is indeed a solution, and all three solutions can be obtained by simply permuting the variables \(x, y, z\).