\section*{Problem 4 - 051244}

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

\[
\begin{aligned}
(*) & x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}+x_{4}=2 \\
(* *) & x_{1} x_{2}+x_{1} x_{4}+x_{2} x_{4}+x_{3}=2 \\
& x_{1} x_{3}+x_{1} x_{4}+x_{3} x_{4}+x_{2}=2 \\
& x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}+x_{1}=2
\end{aligned}
\]
Assuming \(\left(x_{1}, x_{2}, x_{3}, x_{4}\right)\) is a solution of the given system of equations, it follows from \(\left(^{*}\right)\) and \((* *)\) by subtraction

\[
x_{1} x_{3}-x_{1} x_{4}+x_{2} x_{3}-x_{2} x_{4}+x_{4}-x_{3}=0 \quad \text { thus } \quad\left(x_{3}-x_{4}\right)\left(x_{1}+x_{2}-1\right)=0
\]

Since the system of equations remains invariant under any cyclic permutation of the indices, the following equations are also obtained:

\[
\begin{aligned}
& \left(x_{1}-x_{2}\right)\left(x_{3}+x_{4}-1\right)=0 \\
& \left(x_{1}-x_{3}\right)\left(x_{2}+x_{4}-1\right)=0 \\
& \left(x_{1}-x_{4}\right)\left(x_{2}+x_{3}-1\right)=0 \\
& \left(x_{2}-x_{3}\right)\left(x_{1}+x_{4}-1\right)=0 \\
& \left(x_{2}-x_{4}\right)\left(x_{1}+x_{3}-1\right)=0 \\
& \left(x_{3}-x_{4}\right)\left(x_{1}+x_{2}-1\right)=0
\end{aligned}
\]

These equations are satisfied if at least one factor in each equation is zero. We distinguish the following cases:
1) \(x_{1}=x_{2}=x_{3}=x_{4}\)
2) \(x_{1}=x_{2}=x_{3} \neq x_{4}\)
3) \(x_{1}=x_{2} \neq x_{3} ; x_{1} \neq x_{4}\)
4) \(x_{i} \neq x_{k}\) for \(i \neq k\)

All other cases arise by cyclic permutation of the indices, as each quadruple derived from a solution by cyclic permutation of the indices is also a solution, and any quadruple can be transformed into one of the cases 1) to 4) by cyclic permutation.

1. In this case, from \((*)\):

\[
x_{1}^{2}+x_{1}^{2}+x_{1}^{2}+x_{1}=2 \Rightarrow x_{1}^{2}+\frac{x_{1}}{3}-\frac{2}{3}=0
\]

This equation is only satisfied if

\[
x_{1}=\frac{-1+5}{6}=\frac{2}{3} \quad \text { or } \quad x_{1}=\frac{-1-5}{6}=-1
\]

It is easy to verify that for

\[
x_{1}=x_{2}=x_{3}=x_{4}=\frac{2}{3} \quad \text { and } \quad x_{1}=x_{2}=x_{3}=x_{4}=-1
\]

the given system of equations is satisfied.

2. In this case, from (6) because \(x_{3} \neq x_{4}\):

\[
x_{1}+x_{2}=1 \quad ; \quad 2 x_{1}=1 \quad ; \quad x_{1}=x_{2}=x_{3}=\frac{1}{2}
\]

Furthermore, from \(\left(^{*}\right): \quad 3 x_{1}^{2}+x_{4}=2 \quad ; \quad x_{4}=2-\frac{3}{4}=\frac{5}{4}\)

It is easy to verify that for \(x_{1}=x_{2}=x_{3}=\frac{1}{2}, x_{4}=\frac{5}{4}\) the given system of equations is satisfied.

3. In this case, from (2), (3), (4), (5):

\[
x_{2}+x_{4}=x_{2}+x_{3}=x_{1}+x_{4}=x_{1}+x_{3}=1 \quad \text { thus } \quad x_{3}=x_{4}
\]

and from the relationship \(x_{3}=1-x_{1}\) derived from \(\left({ }^{*}\right)\):

\[
x_{1}^{2}+2 x_{1}\left(1-x_{1}\right)+1-x_{1}=2 \quad ; \quad x_{1}^{2}-x_{1}+1=0
\]

which is impossible because

\[
x_{1}^{2}-x_{1}+1=\left(x_{1}-\frac{1}{2}\right)^{2}+\frac{3}{4}>0
\]

Thus, there is no solution in this case.

4. In this case, from (1) and (2): \(x_{3}+x_{4}=x_{2}+x_{4}=1\), which is impossible because \(x_{1} \neq x_{2}\).

Therefore, the given system of equations is satisfied for the quadruples

\[
\left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}, \frac{2}{3}\right) ; \quad(-1,-1,-1,-1) ; \quad\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{5}{4}\right) ; \quad\left(\frac{1}{2}, \frac{1}{2}, \frac{5}{4}, \frac{1}{2}\right) ; \quad\left(\frac{1}{2}, \frac{5}{4}, \frac{1}{2}, \frac{1}{2}\right) ; \quad\left(\frac{5}{4}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)
\]

\section*{Adapted from \([5]\)}