PUMaC 2019 · 代数（B 组） · 第 4 题

PUMaC 2019 — Algebra (Division B) — Problem 4

Let f ( x ) = x + 4 x + 2. Let r be the difference between the largest and smallest real solutions p q of the equation f ( f ( f ( f ( x )))) = 0. Then r = a for some positive integers a, p, q so a is square-free and p, q are relatively prime positive integers. Compute a + p + q. 100 i

解析

Let f ( x ) = x + 4 x + 2. Let r be the difference between the largest and smallest real solutions p q of the equation f ( f ( f ( f ( x )))) = 0. Then r = a for some positive integers a, p, q so a is square-free and p, q are relatively prime positive integers. Compute a + p + q. Answer: 35 Proposed by: Kevin Feng 2 2 2 2 2 4 Note that f ( x ) = x +4 x +2 = ( x +2) − 2. Then f ( x ) = (( x +2) − 2)+2) − 2 = ( x +2) − 2. n 4 n 2 4 2 It is easy to see by induction that f ( x ) = ( x + 2) − 2, so f ( x ) = ( x + 2) − 2. √ √ 16 16 4 Then the real solutions to f ( x ) = 0 are at x + 2 = ± 2, or x = − 2 ± 2. Hence, the √ 17 16 16 difference between the two of them are 2 2 = 2 , which gives us an answer of 2 + 16 + 17 = 35 . 100 i