PUMaC 2021 · 团队赛 · 第 7 题

PUMaC 2021 — Team Round — Problem 7

The roots of the polynomial f ( x ) = x + x − x − x − x + x + 1 are all roots of unity. We say 2 iπr that a real number r ∈ [0 , 1) is nice if e = cos 2 πr + i sin 2 πr is a root of the polynomial f 2 iπr and if e has positive imaginary part. Let S be the sum of the values of nice real numbers p r . If S = for relatively prime positive integers p, q, find p + q. q

解析

The roots of the polynomial f ( x ) = x + x − x − x − x + x + 1 are all roots of unity. We say 2 iπr that a real number r ∈ [0 , 1) is nice if e = cos 2 πr + i sin 2 πr is a root of the polynomial f 2 iπr and if e has positive imaginary part. Let S be the sum of the values of nice real numbers p r . If S = for relatively prime positive integers p, q, find p + q. q Proposed by: Frank Lu Answer: 31 2 We try multiplying this polynomial by x − x + 1 , in an attempt to find which polynomial of n 10 5 the form x − 1 it divides. We see that this equals x − x + 1 , meaning that the roots of this polynomial are all 30th roots of unity. We now need to determine which of these divide into the polynomial given. 15 5 10 5 Notice that we have the equality ( x + 1) = ( x + 1)( x − x + 1) . Thus, we have the roots of 15 5 2 x + 1 , minus those that are roots of either x + 1 or x − x + 1 . But notice that the former are iπk/ 5 iπk/ 3 roots of the form e , where k is odd, and for the latter they have the form e , where k is odd. k This means that the possible values of r that we have (which we note are of the form for 30 1 7 11 13 17 19 23 29 some odd k ) are , , , , , , , . Of these, only the first four have positive imaginary 30 30 30 30 30 30 30 30 1 part (as positive imaginary part holds if and only if r lies between 0 and ). Our sum of the 2 32 16 first four is thus = , so our answer is 31 . 30 15