PUMaC 2021 · 代数（B 组） · 第 8 题

PUMaC 2021 — Algebra (Division B) — Problem 8

Let f be a polynomial. We say that a complex number p is a double attractor if there exists a 2 polynomial h ( x ) so that f ( x ) − f ( p ) = h ( x )( x − p ) for all x ∈ R . Now, consider the polynomial 5 4 3 2 f ( x ) = 12 x − 15 x − 40 x + 540 x − 2160 x + 1 , n P and suppose that it’s double attractors are a , a , . . . , a . If the sum | a | can be written as 1 2 n i i =1 √ √ a + b , where a, b are positive integers, find a + b . 1

解析

Let f be a polynomial. We say that a complex number p is a double attractor if there exists a 2 polynomial h ( x ) so that f ( x ) − f ( p ) = h ( x )( x − p ) for all x ∈ R . Now, consider the polynomial 5 4 3 2 f ( x ) = 12 x − 15 x − 40 x + 540 x − 2160 x + 1 , n P and suppose that it’s double attractors are a , a , . . . , a . If the sum | a | can be written as 1 2 n i i =1 √ √ a + b , where a, b are positive integers, find a + b . Proposed by: Frank Lu Answer: 49 5 4 3 2 2 Let f ( x ) = 12 x − 15 x − 40 x + 540 x − 2160 x + 1 . Notice that that f ( x ) − f ( p ) = h ( x )( x − p ) 2 if and only if f ( x + p ) − f ( p ) = h ( x + p ) x . But notice that f ( x + p ) − f ( p ) is also a polynomial 2 in x that equals 0 when x = 0 . This is divisible by x if and only if the coefficient of x is 0 . But notice that, for p to satisfy this, we see that the x coefficient of f ( x + p ) − f ( p ) is the same as 4 3 2 that of f ( x + p ) , which is just 60 p − 60 p − 120 p + 1080 p − 2160 . We will now figure out when 4 3 2 this is 0 . We re-write this, factoring out the 60 , as p − p − 2 p + 18 p − 36 . Notice that this has 2 integer solutions 2 , − 3 , and factoring yields ( p − 2)( p + 3)( p − 2 p + 6) . This yields two more √ √ √ √ √ √ solutions 1 + i 5 , 1 − i 5 . Our final answer is thus just 2 + 3 + 2 6 = 5 + 2 6 = 25 + 24 . Hence, we see that yields 49 , as desired. 3