\section*{Problem 3 - 031213}

Determine all real values of \(x_{1}, x_{2}, x_{3}\) that satisfy the equations

\[
\begin{aligned}
& x_{2}+x_{3}=p x_{1} \\
& x_{1}+x_{3}=p x_{2} \\
& x_{1}+x_{2}=p x_{3}
\end{aligned}
\]

and their dependence on the real number \(p\) (parameter)!
Assuming there is a solution \(\left(x_{1}, x_{2}, x_{3}\right)\) that satisfies equations (1), (2), (3), then this solution also satisfies the equivalent equations

\[
\begin{aligned}
x_{1}+x_{2}+x_{3} & =(p+1) x_{1} \\
x_{1}+x_{2}+x_{3} & =(p+1) x_{2} \\
x_{1}+x_{2}+x_{3} & =(p+1) x_{3}
\end{aligned}
\]

If \(p \neq -1\), then \(\left(x_{1}+x_{2}+x_{3}\right)(p+1)^{-1}=x_{1}=x_{2}=x_{3}=x\). From (1) to (3), it follows that \(2 x=p x\), so \(\left(x_{1}, x_{2}, x_{3}\right)=(0,0,0)\) is the only solution for \(p \notin\{-1,2\}\). For \(p=2\), all \(\left(x_{1}, x_{2}, x_{3}\right)=(x, x, x)\), \(x\) real, satisfy the equations.

If \(p=-1\), then equations (1) to (3) are equivalent to \(x_{1}+x_{2}+x_{3}=0\). Thus, all \(\left(x_{1}, x_{2}, x_{3}\right)=\left(x_{1}, x_{2},-x_{1}-x_{2}\right)\), \(x_{1}, x_{2}\) real, satisfy the equations.