NT-B2-M07-P013 Three Unit Fractions

First prove \(x\le 3\). Then solve a two-fraction equation for each possible \(x\).

Since \(x\le y\le z\), \(1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le \frac{3}{x}\), so \(x\le 3\). If \(x=1\), the remaining sum is \(0\), impossible. If \(x=2\), then \(\frac{1}{y}+\frac{1}{z}=\frac{1}{2}\), so \((y-2)(z-2)=4\). With \(y\le z\), this gives \((y,z)=(3,6),(4,4)\). If \(x=3\), then \(\frac{1}{y}+\frac{1}{z}=\frac{2}{3}\). Since \(y\ge 3\), the maximum possible left side is \(\frac{2}{3}\), so \(y=z=3\). The answer is \((2,3,6),(2,4,4),(3,3,3)\).

A classic combination of bounding and factorisation; the ordering \(x\le y\le z\) is essential.