\section*{Problem 4 - 151044}

Determine all unordered pairs \((x, y)\) of two natural numbers \(x, y\) with \(x \neq y\), for which the following holds!

The arithmetic mean of \(x\) and \(y\) is a two-digit number. If one swaps the digits of this number, one obtains the geometric mean of \(x\) and \(y\) (which is the number \(\sqrt{x y}\)).
}

Let \(\frac{x+y}{2}=u\) and \(\sqrt{x y}=v\). By assumption, there exist \(a, b \in\{0,1, \ldots, 9\}\) such that \(u=10 a+b\) and \(v=10 b+a\).

From \(x+y=2 u\) and \(x y=v^{2}\), it follows by Vieta that \(x, y\) are the solutions of the quadratic equation \(z^{2}-2 u z+v^{2}=0\). These are given by \(z=u \pm \sqrt{u^{2}-v^{2}}\).

Thus, \(u^{2}-v^{2}=(u+v)(u-v)=11(a+b) \cdot 9(a-b)\) must be a perfect square.

Since \(a-b\) is not divisible by 11 (because \(a \neq b\) and \(x \neq y\)), \(a+b\) must be divisible by 11. This is only possible if \(a+b=11\). It follows that \(a-b\) must also be a perfect square, so \(a-b \in\{1,4,9\}\). Since \(a+b=11\) is odd, \(a-b\) must also be odd, so \(a-b \neq 4\). If \(a-b=9\), then \(a=9\) and \(b=0\) would hold, which contradicts \(a+b=11\). Therefore, \(a-b=1\).

Thus, we get \(a=6\) and \(b=5\), so \(u=65\) and \(v=56\), from which we finally obtain

\[
\{x, y\}=\left\{u \pm \sqrt{u^{2}-v^{2}}\right\}=\{65 \pm 11 \cdot 3\}=\{32,98\}
\]