Problem
ALG-B1-M03-P024 Positive Integer Solutions with Product
#24
★★★★★ Level 5 of 5
Find all positive integers \(x,y,z\) such that \(xyz=x+y+z+2\).
Assume \(x\le y\le z\) and bound \(x\).
Without loss of generality, let \(x\le y\le z\). If \(x\ge3\), then \(xyz\ge3yz\), while the right side is at most \(3z+2\); for \(y\ge3\), \(z\ge3\), this is impossible. Hence \(x=1\) or \(x=2\).
If \(x=1\), then \(yz=y+z+3\), so \((y-1)(z-1)=4\). This gives \((y,z)=(2,5),(3,3),(5,2)\).
If \(x=2\), then \(2yz=y+z+4\). For \(y=1\), \(z=5\); for \(y=2\), \(z=2\); for \(y\ge3\), the left side is too large: \(2yz-y-z-4\ge6z-3-z-4=5z-7>0\). Taking permutations into account, all solutions are permutations of \((1,2,5)\), \((1,3,3)\), and \((2,2,2)\).
A higher-level problem: factorisation is combined with bounding.