\section*{Problem 4 - 091034}

Determine all pairs of real numbers \(a\) and \(b(b<a)\), for which the sum of the two numbers, the product of the two numbers, and one of the differences of the squares of the two numbers are equal to each other.
Assume \(a\) and \(b\) are numbers of the required type. Then it follows that \(a \neq -b\); for if \(a = -b\), one would obtain from the problem statement \(0 = a + b = ab = -b^2\), thus \(b = 0, a = 0\), in contradiction to the requirement that \(b < a\).

According to the problem statement, either \(a + b = a^2 - b^2\) or \(a + b = b^2 - a^2\). The latter equation would, due to \(a + b \neq 0\), lead to \(1 = b - a\) and thus also to a contradiction with \(b < a\).

Therefore, only the possibility \(a + b = a^2 - b^2\) remains, from which, due to \(a + b \neq 0\), it follows that \(1 = a - b\), i.e., \(a = b + 1\). Substituting this into \(a + b = ab\) yields the equation \(2b + 1 = b^2 + b\), i.e., \(b^2 - b - 1 = 0\), which has the solutions

\[
b_1 = \frac{1 + \sqrt{5}}{2} \quad ; \quad b_2 = \frac{1 - \sqrt{5}}{2}
\]

and only these. The corresponding values for \(a\) are

\[
a_1 = \frac{3 + \sqrt{5}}{2} \quad ; \quad a_2 = \frac{3 - \sqrt{5}}{2}
\]

Thus, at most the pairs \((a_1, b_1)\) and \((a_2, b_2)\) can be solutions to the problem, which is confirmed by verification. Solution adopted from \([5]\)