HMMT 二月 2010 · TEAM1 赛 · 第 5 题

HMMT February 2010 — TEAM1 Round — Problem 5

[ 20 ] Show that, for every positive integer n , there exists a monic polynomial of degree n with integer coefficients such that the coefficients are decreasing and the roots of the polynomial are all integers.

解析

[ 20 ] Show that, for every positive integer n , there exists a monic polynomial of degree n with integer coefficients such that the coefficients are decreasing and the roots of the polynomial are all integers. n Solution: We claim we can find values a and b such that p ( x ) = ( x − a )( x + b ) is a polynomial of degree n + 1 that satisfies these constraints. We show that its coefficients are decreasing by finding a k general formula for the coefficient of x . Team Round A ( ) ( ) n n k k k − 1 n The coefficient of x is b − ab , which can be seen by expanding out ( x + b ) and then k k − 1 multiplying by ( x − a ). Then we must prove that ( ) ( ) ( ) ( ) n n n n k +1 k k k − 1 b − ab < b − ab , k + 1 k k k − 1 or ( ( ) ( )) ( ( ) ( )) n n n n k − 1 k ab b − > b b − . k k − 1 k + 1 k ( ) n ( ) k Choose b > max in order to make sure the right-hand term in each product on each side of n ( ) k − 1 the inequality sign is positive (we’ll be dividing by it, so this makes things much easier), and choose ( ) n n b b − ( ( ) ( )) k +1 k a > max to make sure the inequality always holds. Since there are only finite values n n b − ( ) ( ) k k − 1 that k can take on given a fixed n (namely, integers between 0 and n inclusive), we can always find values of a and b that satisfy these constraints.