PUMaC 2009 · 数论（B 组） · 第 4 题
PUMaC 2009 — Number Theory (Division B) — Problem 4
题目详情
- Find the number of ordered pairs ( a, b ) of positive integers that are solutions of the following equation: 2 2 a + b = ab ( a + b )
解析
- Find the number of ordered pairs ( a, b ) of positive integers that are solutions of the following equation: 2 2 a + b = ab ( a + b ) . 2 3 Solution. 1. Suppose at first that a = b . Then we get 2 a = 2 a , which yields a = 0 or a = 1. Since we must have positive integers, this yields a = b = 1 as a possible solution. Now a b if a 6 = b , then suppose WLOG that a > b . Then, the equation reduces to + = a + b . But, b a b 1 > , and a + b ≥ a + 1, because we have positive integers only. Then, we get a a a b
- 1 > + = a + b ≥ a + 1 b b a a i.e. > a , which means 1 > b , which is a contradiction. Note that we can divide throughout b freely by any of the two variables because they are positive integers.