PUMaC 2016 · 团队赛 · 第 14 题

PUMaC 2016 — Team Round — Problem 14

(12) Suppose P ( x ) = x + a x + . . . + a x + a satisfies 2015 1 0 P ( x ) P (2 x + 1) = P ( − x ) P ( − 2 x − 1) for all x ∈ R . Find the sum of all possible values of a . 2015 n

解析

Observe that the condition implies that the multisets of the roots of the two polynomials P ( x ) P (2 x + 1) and P ( − x ) P ( − 2 x − 1) are identical. 3 ′ Let R = ( r ∈ R , P ( r ) = 0) be the multiset of roots of P . Given r ∈ R , if r = − r ∈ R , then ′ we can cancel out the instances of r in the LHS with the instances of r in the RHS. We can ′ ′ ′ then let R = R/ ( r, r ). We repeat this removal of pairs ( r, r ) until no such pair remains, and call the resulting set S . Now, for any r ∈ S , r must be a root of P ( − 2 x − 1), so there exists s ∈ S such that − r − 1 − r − 1 n r = − 2 s − 1 ↔ s = . Taking an arbitrary r ∈ S , we inductively define r = for 0 n +1 2 2 n ≥ 0, and indeed, r ∈ S . This gives us for arbitrary k ≥ 0: n +1 k k 2 − ( − 1) k ( − 1) r − 1 3 r = k k 2 1 Since S is finite, there exists some n for which r = r , and solving yields r = − . Hence, S n 0 0 3 1 contains only roots equal to − . Then: 3 ∑ ∑ | S | a = − r = − r = 2015 3 r ∈ R r ∈ S Since | S | ∈ { 2016 , 2014 , . . . 0 } , the sum of all possible values of a is: 2015 ∑ 1 1008 · 1009 = · (2016 + 2014 + . . . + 0) = = 339024 . 3 3 Problem written by Bill Huang. n n n n