PUMaC 2012 · 代数（A 组） · 第 5 题

PUMaC 2012 — Algebra (Division A) — Problem 5

[ 5 ] What is the smallest natural number n greater than 2012 such that the polynomial f ( x ) = 6 4 n 4 n 6 4 2 ( x + x ) − x − x is divisible by g ( x ) = x + x + 1?

解析

[ 5 ] What is the smallest natural number n greater than 2012 such that the polynomial f ( x ) = 6 4 n 4 n 6 4 2 ( x + x ) − x − x is divisible by g ( x ) = x + x + 1? 4 2 2 2 Solution: Let g = x + x + 1 = ( x + x + 1)( x − x + 1). If a is solution to the polynomial 2 2 g = x + x + 1 and b is solution to g = x − x + 1, then f is divisible by g ⇐⇒ a and b are 1 2 solutions to f , as well, so f ( a ) = f ( b ) = 0. 2 3 2 On the other hand, if a + a + 1 = 0, then a = 1; also, since b − b + 1 = 0, it follows that 3 b = − 1. 6 4 n 4 n 6 n n f ( a ) = ( a + a ) − a − a = (1 + a ) − a − 1 2 n n = ( − a ) − a − 1 = 0 2 3 Noting that we can substitute every a with 1, f ( a ) = 0 ⇐⇒ n ∈ { 6 k + 1 , 6 k + 5 } Similarly, 6 4 n 4 n 6 n n n f ( b ) = ( b + b ) − b − b = (1 − b ) + ( − 1) b − 1 n 2 n n n = ( − 1) b − ( − 1) b − 1 = 0 3 Noting that we can substitute every b with − 1, f ( b ) = 0 ⇐⇒ n ∈ { 6 k + 1 , 6 k + 5 } So the smallest instance of n larger than 2012 is 2015. This is the final answer. Answer: 2015 Thanks to Calvin Deng for pointing out the error in the original solutions with f ( b ).