PUMaC 2017 · 个人决赛（B 组） · 第 2 题

PUMaC 2017 — Individual Finals (Division B) — Problem 2

Let a ( x ), a ( x ), and a ( x ) be three polynomials with integer coefficients such that every 1 2 3 polynomial with integer coefficients can be written in the form p ( x ) a ( x ) + p ( x ) a ( x ) + p ( x ) a ( x ) 1 1 2 2 3 3 for some polynomials p ( x ), p ( x ), p ( x ) with integer coefficients. Show that every polynomial 1 2 3 is of the form 2 2 2 p ( x ) a ( x ) + p ( x ) a ( x ) + p ( x ) a ( x ) 1 1 2 2 3 3 for some polynomials p ( x ), p ( x ), p ( x ) with integer coefficients. 1 2 3

解析

Let Z [ x ] denote the set of polynomials with integer coefficients. Given b ( x ) , . . . , b ( x ) ∈ Z [ x ], 1 k let ( b ( x ) , . . . , b ( x )) = { p ( x ) b ( x ) + · · · + p ( x ) b ( x ) | p ( x ) , . . . , p ( x ) ∈ Z [ x ] } . 1 k 1 1 k k 1 k We are given that ( a ( x ) , a ( x ) , a ( x )) = Z [ x ], so there exists p ( x ) ∈ Z [ x ] and q ( x ) ∈ 1 2 3 1 ( a ( x ) , a ( x )) such that 2 3 p ( x ) a ( x ) + q ( x ) = 1 . 1 1 Then, we have 2 2 p ( x ) a ( x ) + ( p ( x ) a ( x ) + 1) q ( x ) = 1 , 1 1 1 1 2 so 1 ∈ ( a ( x ) , a ( x ) , a ( x )). In fact, given any c ( x ) ∈ Z [ x ], we have 1 2 3 2 2 c ( x ) p ( x ) a ( x ) + c ( x )( p ( x ) a ( x ) + 1) q ( x ) = c ( x ) , 1 1 1 1 2 so ( a ( x ) , a ( x ) , a ( x )) = Z [ x ]. 1 2 3 2 2 Now, note that ( a ( x ) , a ( x ) , a ( x )) = ( a ( x ) , a ( x ) , a ( x )), so we may repeat this argument 1 2 3 2 1 3 2 2 to find that ( a ( x ) , a ( x ) , a ( x )) = Z [ x ]. If we repeat this argument one more time, we have 2 1 3 2 2 2 ( a ( x ) , a ( x ) , a ( x ) ) = Z [ x ], as desired. 1 2 3 Problem written by Matt Tyler