PUMaC 2013 · 个人决赛（B 组） · 第 2 题

PUMaC 2013 — Individual Finals (Division B) — Problem 2

Find all pairs of positive integers ( a, b ) such that 3 a + 4 b 2 2 a + 2 b + 2 a b is a positive integer.

解析

Find all pairs of positive integers ( a, b ) such that 3 a + 4 b 2 2 a + 2 b + 2 a b is a positive integer. Solution When a = 1, we have b = 1 as a unique solution. When b = 1, we have 3 a + 4 ∈ Z 2 2 a + a + 2 and we can easily check that a = 1 , 2 are the only solutions. Assume from this point that a, b ≥ 2. 3 2 2 If 2 ≤ a ≤ b , we must have a ≤ a b and 4 b ≤ 2 b so the value is smaller than 1, thus cannot be an integer. So assume a > b . We have that 3 a + 4 b ∈ Z 2 2 a + 2 b + 2 a b 3 2 b ( a + 4 b ) ⇒ ∈ Z 2 2 a + 2 b + 2 a b 3 2 2 2 2 b ( a + 4 b ) 8 b − a − 2 ab 2 2 ⇒ − a ( a + 2 b + 2 a b ) = ∈ Z 2 2 2 2 a + 2 b + 2 a b a + 2 b + 2 a b 2 2 2 Now, for a ≥ 4, we have that 8 b − 2 ab − a < 0, thus 2 2 2 a + 2 ab − 8 b ∈ N 2 2 a + 2 b + 2 a b 2 2 2 2 However, we have ( a + 2 ab ) − ( b + 2 a b ) = ( a − b )( a + b − 2 ab ) < 0, so numerator is smaller than denominator. Contradiction. Thus we must have 4 > a > b ≥ 2: the only possible case is a = 3 , b = 2, and in that case the 35 value becomes which is not an integer. 3+12+36 Thus the only solutions are ( a, b ) = (2 , 1) , (1 , 1). 1