PUMaC 2016 · 代数（A 组） · 第 7 题

PUMaC 2016 — Algebra (Division A) — Problem 7

Let S be the set of all polynomials P with complex coefficients, such that P ( x ) = P ( x ) P ( x − P

解析

We first determine all solutions to the equation P ( x ) = P ( x ) P ( x − 1). Let P ( x ) = ( x − α ) i ∏ ∏ ∏ 2 where α are the roots of P (including multiplicity). Then ( x − α ) = ( x − α ) ( x − i i i √ ( α + 1)). Thus, considering the sets of roots of both sides of this equation, we get {± α } = i i ∣ ∣ √ 2 ∣ ∣ { α } ∪ { α + 1 } . Now we note that if max α = M > 1 then max | α | = M > M , i i i i ∣ ∣ √ 2 ∣ ∣ contradiction. Similarly, if min α = N < 1 then min | α | = N < N , contradiction. i i (Unless N = 0. But this is impossible, since the multiplicity of zero as a root on both sides of the equation cannot be equal.) Therefore all roots α must satisfy | α | = | α + 1 | = 1, so i i i ( ) √ − 1+ 3 i α = ± . Now counting the distinct roots, the two different roots must have the i 2 2 n same multiplicity, so the solutions are P ( x ) = ( x + x + 1) for some positive integer n , or n 2 2 P ( x ) = 0 or P ( x ) = 1 (constant solutions). It follows that P (1) = 3 , so P ( x ) = ( x + x +1) . 0 0 2 Therefore P (10) = 111 = 12321 . 0 Problem written by Mel Shu.