PUMaC 2020 · 数论（A 组） · 第 2 题
PUMaC 2020 — Number Theory (Division A) — Problem 2
题目详情
- How many ordered triples of nonzero integers ( a, b, c ) satisfy 2 abc = a + b + c + 4?
解析
- How many ordered triples of nonzero integers ( a, b, c ) satisfy 2 abc = a + b + c + 4? Proposed by: Austen Mazenko Answer: 6 a + b +4 Since 2 ab − 1 6 = 0 for integers a, b , we need c = to be an integer. If | a | , | b | ≥ 2 then 2 ab − 1 8 | 2 ab − 1 | > | a + b + 4 | unless a = b = 2, so c = . Thus, one of a, b is in {− 1 , 1 } . If a = 1, 7 then (2 b − 1) | ( b + 5) and b = 1 , 6, giving (1 , 1 , 6) and cyclic permutations. If a = − 1, then (2 b + 1) | ( b + 3), so b = − 1 or b = 2. In either case, we get ( − 1 , − 1 , 2) and cyclic permutations. This exhausts all possible cases, so our answer is 6.