PUMaC 2009 · 代数（B 组） · 第 1 题

PUMaC 2009 — Algebra (Division B) — Problem 1

If φ is the Golden Ratio, we know that = φ − 1. Define a new positive real number, called φ √ 1 a + b φ , where = φ − d (so φ = φ ). Given that φ = , a, b, c positive integers, and the d d 1 2009 φ c d greatest common divisor of a and c is 1, find a + b + c .

解析

If φ is the Golden Ratio, we know that = φ − 1. I will define a new quantity, called φ , d φ √ 1 a + b where = φ − d (so φ = φ ). Given that φ = , a, b, c positive integers, and the d 1 2009 φ c d greatest common divisor of a and c is 1, find a + b + c . 1 2 Solution. 4038096. Let x = φ . The equation = x − 2009 is equivalent to 1 = x − 2009 x 2009 x √ 2 2 2009 ± 2009 +4 or also x − 2009 x − 1 = 0. This is a quadratic equation with solutions x = . 1 , 2 2 From the statement of the problem it follows that we need to consider the solution with the plus sign. Since GCD(2009 , 2)=1, the fraction is written in the appropriate way, hence 2 a + b + c = 2009 + 2009 + 4 + 2 = 4038096.