\section*{Problem 5 - 071225}

All ordered pairs of real numbers \((x, y)\) are to be determined for which the system of equations

\[
\begin{aligned}
x \cdot\left(a x^{2}+b y^{2}-a\right) & =0 \\
y \cdot\left(a x^{2}+b y^{2}-b\right) & =0
\end{aligned}
\]

is satisfied. Here, \(a\) and \(b\) are real numbers with \(a \neq 0, b \neq 0\) and \(a \neq b\).
}

We perform a case distinction:

1. Case: \(x=0\).

Then the second equation becomes \(y \cdot b \cdot\left(y^{2}-1\right)=0\), which, due to \(b \neq 0\), leads to \(y=0\) or \(y= \pm 1\). For all three elements \((x, y) \in\{(0,-1),(0,0),(0,1)\}\), verification confirms that they are indeed solutions to the system of equations.

2. Case: \(x \neq 0\).

Then it follows from the first equation that \(a x^{2}+b y^{2}-a=0\), and thus, due to \(a \neq b\), \(a x^{2}+b y^{2}-b \neq 0\). Therefore, from the second equation, we directly get \(y=0\). Substituting this into the equation just obtained, we get \(a x^{2}-a=0\) or \(x= \pm 1\). Here, too, all elements of the set \(\{(-1,0),(1,0)\}\) are solutions to the system of equations, as the verification confirms.

Thus, the given system of equations has a total of five solutions, which are noted in the two cases.