Problem
NT-B1-M05-P022 A Product Divides a Sum
#22
★★★★☆ Level 4 of 5
Find all positive integers \(x,y\) such that \(xy\mid x+y+2\).
By symmetry assume \(x\le y\), then use \(xy\le x+y+2\).
Let \(x\le y\). Then \(xy\le x+y+2\). If \(x\ge3\), then \(xy\ge3y\), while \(x+y+2\le2y+2\), impossible for \(y\ge3\). Thus \(x=1\) or \(x=2\). For \(x=1\), \(y\mid y+3\), so \(y\mid3\), and \(y=1,3\). For \(x=2\), \(2y\mid y+4\), so \(y\le4\); checking \(y=2,3,4\) leaves only \(y=4\). By symmetry, the solutions are \((1,1)\), \((1,3)\), \((3,1)\), \((2,4)\), \((4,2)\).
Important as a first serious use of bounding in Diophantine equations.