HMMT 十一月 2019 · 冲刺赛 · 第 3 题

HMMT November 2019 — Guts Round — Problem 3

[5] For how many positive integers a does the polynomial 2 x − ax + a have an integer root? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2019, November 9, 2019 — GUTS ROUND Organization Team Team ID#

解析

[5] For how many positive integers a does the polynomial 2 x − ax + a have an integer root? Proposed by: Krit Boonsiriseth Answer: 1 2 Let r, s be the roots of x − ax + a = 0. By Vieta’s, we have r + s = rs = a . Note that if one root is an integer, then both roots must be integers, as they sum to an integer a . Then, rs − ( r + s ) + 1 = 1 = ⇒ ( r − 1)( s − 1) = 1 . Because we require r, s to be both integers, we have r − 1 = s − 1 = ± 1, which yields r = s = 0 , 2. If r = 0 or s = 0, then a = 0, but we want a to be a positive integer. Therefore, our only possibility is when r = s = 2, which yields a = 4, so there is exactly 1 value of a (namely, a = 4) such that 2 x − ax − a has an integer root.