\section*{Problem 1 - 071241}

Determine all ordered quadruples of real numbers \(\left(x_{1}, x_{2}, x_{3}, x_{4}\right)\) that satisfy the following system of equations:

\[
\begin{aligned}
& x_{1}+a x_{2}+x_{3}=b \\
& x_{2}+a x_{3}+x_{4}=b \\
& x_{3}+a x_{4}+x_{1}=b \\
& x_{4}+a x_{1}+x_{2}=b
\end{aligned}
\]

Here, \(a\) and \(b\) are real numbers (consider different cases!).
}

Case 1: \(a=0\). Then the system of equations is equivalent to \(x_{1}+x_{3}=b=x_{2}+x_{4}\), so all solution quadruples have the form \((s, t, b-s, b-t)\) with two real parameters \(s\) and \(t\). Substitution confirms that these are indeed solutions.

Case 2: \(a \neq 0\). Then equating equations (1) and (3) leads to \(a x_{2}=a x_{4}\) or \(x_{2}=x_{4}\). Similarly, from equations (2) and (4), we get the identity \(x_{1}=x_{3}\). Substitution now yields \(2 x_{1}+a x_{2}=b=2 x_{2}+a x_{1}\), thus \((2-a) x_{1}=(2-a) x_{2}\).

Case 2.1: \(a \neq 2\). Then it follows that \(x_{1}=x_{2}\), so all four variables are equal, and we obtain \((2+a) x_{1}=b\).

Case 2.1.1: \(a \neq -2\). Then the only solutions are the quadruples \(\left(\frac{b}{2+a}, \frac{b}{2+a}, \frac{b}{2+a}, \frac{b}{2+a}\right)\), which can be easily verified by substitution.

Case 2.1.2: \(a=-2\) and \(b=0\). Then the solution quadruples are given by \((t, t, t, t)\), where \(t\) runs through the real numbers. The verification confirms this result.

Case 2.1.3: \(a=-2\) and \(b \neq 0\). Then there is no solution because \((2+a) x_{1}=(2-2) x_{1}=0 \neq b\).

Case 2.2: \(a=2\). Then \(2 x_{1}+2 x_{2}=b\), so all solutions are given by \(\left(t, \frac{b}{2}-t, t, \frac{b}{2}-t\right)\), where the parameter \(t\) runs through the real numbers, and substitution confirms that these are indeed solutions.