NT-B2-M07-P020 Another Three-Fraction Sum

First prove \(x\le 4\). Then treat \(x=2,3,4\) separately.

From \(x\le y\le z\), we have \(\frac{3}{x}\ge \frac{2}{3}\), hence \(x\le 4\). The value \(x=1\) is impossible because \(\frac{1}{x}=1>\frac{2}{3}\).

For \(x=2\): \(\frac{1}{y}+\frac{1}{z}=\frac{1}{6}\), so \((y-6)(z-6)=36\). With \(y\le z\), this gives \((y,z)=(7,42),(8,24),(9,18),(10,15),(12,12)\).

For \(x=3\): \(\frac{1}{y}+\frac{1}{z}=\frac{1}{3}\), so \((y-3)(z-3)=9\). We get \((y,z)=(4,12),(6,6)\).

For \(x=4\): \(\frac{1}{y}+\frac{1}{z}=\frac{5}{12}\). Multiplying and completing a product gives \((5y-12)(5z-12)=144\). We need \(y\ge 4\), \(y\le z\), hence \(5y-12\le 12\), and this divisor must be congruent to \(3\) modulo \(5\). Among divisors of \(144\) not exceeding \(12\), the possibilities are \(3\) and \(8\); \(3\) gives \(y=3\), invalid, while \(8\) gives \(y=4,z=6\). The answer is \((2,7,42),(2,8,24),(2,9,18),(2,10,15),(2,12,12),(3,4,12),(3,6,6),(4,4,6)\).

The difficulty is the careful handling of \(\frac{5}{12}\), where the factorisation is \((5y-12)(5z-12)\).