\section*{Problem \(5-300935=321034\)}

Determine all triples \((x, y, z)\) of natural numbers for which the following holds:

\[
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{4}{5}
\]
}

O.B.d.A. let \(00\) and thus \(\frac{1}{x} \geq \frac{1}{3} \cdot \frac{4}{5}=\frac{4}{15}>\frac{4}{16}=\frac{1}{4}\), so \(x < 4\), because otherwise \(x=1\), and thus \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>\frac{1}{x}=1 \mathrm{im}\) would contradict the equation from the problem statement. There remain two possibilities for \(x\).

Case 1: \(x=2\). Then the following equation needs to be solved: \(\frac{1}{y}+\frac{1}{z}=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\). Since \(y \leq z\), \(\frac{1}{y} \geq \frac{1}{z}\), so \(\frac{1}{y} \geq \frac{1}{2} \cdot \frac{3}{10}=\frac{3}{20}>\frac{3}{21}=\frac{1}{7}\), which implies \(y \leq 6\). Since \(\frac{1}{3}>\frac{3}{10}\), \(y \geq 4\) as well, otherwise \(\frac{1}{y} \geq \frac{1}{3}\) would lead to a similar contradiction as above.

Case 1.1: \(y=4\). Then \(\frac{1}{z}=\frac{3}{10}-\frac{1}{4}=\frac{6-5}{20}=\frac{1}{20}\), so \(z=20\).

Case 1.2: \(y=5\). Then \(\frac{1}{z}=\frac{3}{10}-\frac{1}{5}=\frac{3-2}{10}=\frac{1}{10}\), so \(z=10\).

Case 1.3: \(y=6\). Then \(\frac{1}{z}=\frac{3}{10}-\frac{1}{6}=\frac{9-5}{30}=\frac{4}{30}=\frac{2}{15}\), so there is no solution here.

Case 2: \(x=3\). Then the following equation needs to be solved: \(\frac{1}{y}+\frac{1}{z}=\frac{4}{5}-\frac{1}{3}=\frac{12-5}{15}=\frac{7}{15}\). Since \(\frac{1}{y} \geq \frac{1}{z}\), \(\frac{1}{y} \geq \frac{1}{2} \cdot \frac{7}{15}=\frac{7}{30}>\frac{6}{30}=\frac{1}{5}\), so \(y \leq 4\). Since \(y \geq x\), \(y \geq 3\) as well.

Case 2.1: \(y=3\). Then \(\frac{1}{z}=\frac{7}{15}-\frac{1}{3}=\frac{7-5}{15}=\frac{2}{15}\), so there is no solution here.

Case 2.2: \(y=4\). Then \(\frac{1}{z}=\frac{7}{15}-\frac{1}{4}=\frac{28-15}{60}=\frac{13}{60}\), so there is no solution here either.

In summary, the triples

\((3,4,20),(3,20,4),(4,3,20),(4,20,3),(20,3,4),(20,4,3),(3,5,10),(3,10,5),(5,3,10),(5,10,3),(10,3,5),(10,5,3)\) of positive integers satisfy the equation.