HMMT 二月 2023 · 团队赛 · 第 2 题

HMMT February 2023 — Team Round — Problem 2

[30] Prove that there do not exist pairwise distinct complex numbers a , b , c , and d such that 3 3 3 3 a − bcd = b − cda = c − dab = d − abc. ◦

解析

[30] Prove that there do not exist pairwise distinct complex numbers a , b , c , and d such that 3 3 3 3 a − bcd = b − cda = c − dab = d − abc. Proposed by: Rishabh Das Solution 1: First suppose none of them are 0. Let the common value of the four expressions be k , and let abcd = P . Then for x ∈ { a, b, c, d } , P 3 4 x − = k = ⇒ x − kx − P = 0 . x However, Vieta’s tells us abcd = − P , meaning P = − P , so P = 0, a contradiction. 3 3 3 2 Now if a = 0, then − bcd = b = c = d . Then without loss of generality b = x , c = xω , and d = xω . 3 3 But then − bcd = − x 6 = x , a contradiction. Thus, there do not exist distinct complex numbers satisfying the equation. 2 2 Solution 2: Subtracting the first two equations and dividing by a − b gives a + b + ab + cd = 0. 2 2 2 2 2 2 2 2 2 2 2 2 Similarly, c + d + ab + cd = 0. So, a + b = c + d . Similarly, a + c = b + d . So, b = c . 2 2 2 2 Similarly, a = b = c = d . Now by Pigeonhole, two of these 4 must be the same. ◦