HMMT 二月 2023 · 团队赛 · 第 8 题
HMMT February 2023 — Team Round — Problem 8
题目详情
- [60] Find, with proof, all nonconstant polynomials P ( x ) with real coefficients such that, for all nonzero 1 real numbers z with P ( z ) 6 = 0 and P ( ) 6 = 0, we have z 1 1 1
- = z + . 1 P ( z ) z P ( ) z
解析
- [60] Find, with proof, all nonconstant polynomials P ( x ) with real coefficients such that, for all nonzero 1 real numbers z with P ( z ) 6 = 0 and P ( ) 6 = 0, we have z 1 1 1
- = z + . 1 P ( z ) z P ( ) z Proposed by: Luke Robitaille 4 k +2 4 k x ( x + 1) x (1 − x ) Answer: P ( x ) = or P ( x ) = 2 2 x + 1 x + 1 Solution: It is straightforward to plug in and verify the above answers. Hence, we focus on showing that these are all possible solutions. The key claim is the following. Claim: If r 6 = 0 is a root of P ( z ) with multiplicity n , then 1 /r is also a root of P ( z ) with multiplicity n . ′ ′ Proof 1 (Elementary). Let n be the multiplicity of 1 /r . It suffices to show that n ≤ n because we ′ can apply the same assertion on 1 /r to obtain that n ≤ n . n To that end, suppose that ( z − r ) divides P ( z ). From the equation, we have [ ( ) ] [( ) ( )] 1 1 1 N N z P + P ( z ) = z z + P ( z ) P , z z z N where N deg P + 1 to guarantee that both sides are polynomial. Notice that the factor z P ( z ) and ( ) n n N 1 the right-hand side is divisible by ( z − r ) , so ( z − r ) must also divide z P . This means that z ( ) N 1 n 1 there exists a polynomial Q ( z ) such that z P = ( z − r ) Q ( z ). Replacing z with , we get z z ( ) ( ) ( ) n P ( z ) 1 1 1 N − n n = − r Q = ⇒ P ( z ) = z (1 − rz ) Q , N z z z z n implying that P ( z ) is divisible by ( z − 1 /r ) . Proof 2 (Complex Analysis). Here is more advanced proof of the main claim. View both sides of the equations as meromorphic functions in the complex plane. Then, a root r with 1 multiplicity n of P ( z ) is a pole of of order n . Since the right-hand side is analytic around r , it P ( z ) 1 follows that the other term has a pole at r with order n as well. By replacing z with 1 /z , we P (1 /z ) 1 find that has a pole at 1 /r of order n . This finishes the claim. P ( z ) The claim implies that there exists an integer k and a constant such that ( ) 1 k P ( z ) = z P . z By replacing z with 1 /z , we get that ( ) 1 k z P = P ( z ) . z Therefore, = ± 1. Moreover, using the main equation, we get that k k 1 z 1 z (1 + z )
- = z + = ⇒ P ( z ) = . 2 P ( z ) P ( z ) z 1 + z This is a polynomial if and only if ( = 1 and k ≡ 2 (mod 4)) or ( = − 1 and k ≡ 0 (mod 4)), so we are done.