\section*{Problem 3B - 251043B}

Given are real numbers \(a_{1}, a_{2}, a_{3}, a_{4}\).

Determine for each possible case of these \(a_{1}, \ldots, a_{4}\) all triples \(\left(b_{1}, b_{2}, b_{3}\right)\) of real numbers (or prove, if applicable, that no such triples exist) for which the system of equations

\[
\begin{aligned}
a_{1} x_{1}^{2}+a_{2} x_{2}^{2}+x_{3}^{2} & =b_{1} \\
x_{2}^{2}+a_{3} x_{3}^{2} & =b_{2} \\
x_{2}^{2}+a_{4} x_{3}^{2} & =b_{3}
\end{aligned}
\]

has exactly one triple \(\left(x_{1}, x_{2}, x_{3}\right)\) of real numbers as a solution.
}

The system of equations is linear in \(x_{1}^{2}, x_{2}^{2}\) and \(x_{3}^{2}\), so with a solution triplet \(\left(x_{1}, x_{2}, x_{3}\right)\), any triplet of the form \(\left( \pm x_{1}, \pm x_{2}, \pm x_{3}\right)\), where the signs can be chosen independently, is also a solution. For there to be a unique solution, it must therefore hold that \(x_{1}=x_{2}=x_{3}=0\).

For the triplet \((0,0,0)\) to be a solution at all, \(b_{1}=b_{2}=b_{3}=0\) must hold, because otherwise, for some \(b_{i} \neq 0\), the solution triplet \(\left(x_{1}, x_{2}, x_{3}\right)=(0,0,0)\) would not satisfy the \(i\)-th equation.

Therefore, the system of equations in the problem has exactly one solution if and only if \(b_{1}=b_{2}=b_{3}=0\) and the linear system

\[
\begin{aligned}
a_{1} y_{1}+a_{2} y_{2}+y_{3} & =0 \\
y_{2}+a_{3} y_{3} & =0 \\
y_{2}+a_{4} y_{3} & =0
\end{aligned}
\]

has only the trivial solution \(y_{1}=y_{2}=y_{3}=0\), meaning the three equations are linearly independent. For this, \(a_{3} \neq a_{4}\) must hold, because otherwise the second and third equations would be identical. If \(a_{3} \neq a_{4}\), it follows directly from these two equations that \(y_{2}=y_{3}=0\). It then follows that \(a_{1} \neq 0\) must hold, because otherwise \(\left(y_{1}, y_{2}, y_{3}\right)=(1,0,0)\) would also be a solution to the system of equations. If \(a_{1} \neq 0\), it follows again directly with \(y_{2}=y_{3}=0\) that \(y_{1}=0\).

In summary, the following holds:

If \(a_{1}=0\) or \(a_{3}=a_{4}\), there is no triplet \(\left(b_{1}, b_{2}, b_{3}\right)\) such that the system of equations in the problem has exactly one solution \((x_{1}, x_{2}, x_{3})\). If, however, both \(a_{1} \neq 0\) and \(a_{3} \neq a_{4}\), this is the case exactly for \(\left(b_{1}, b_{2}, b_{3}\right)=(0,0,0)\). (The solution is then \(\left(x_{1}, x_{2}, x_{3}\right)=(0,0,0)\).)