HMMT 二月 2026 · 冲刺赛 · 第 31 题
HMMT February 2026 — Guts Round — Problem 31
题目详情
- [16] Let ζ = cos + i sin . It is given that the polynomial x + 16 x + 3 x − 229 has three distinct real 19 19 roots, and its largest root can be uniquely written in the form 2 18 a ζ + a ζ + · · · + a ζ 1 2 18 2 2 2 for some rational numbers a , . . . , a . Compute a + a + · · · + a . 1 18 1 2 18
解析
- [16] Let ζ = cos + i sin . It is given that the polynomial x + 16 x + 3 x − 229 has three distinct 19 19 real roots, and its largest root can be uniquely written in the form 2 18 a ζ + a ζ + · · · + a ζ 1 2 18 2 2 2 for some rational numbers a , . . . , a . Compute a + a + · · · + a . 1 18 1 2 18 Proposed by: Pitchayut Saengrungkongka Answer: 564 Solution: Let the three roots be p, q, r , and the root in question is p . The following claim recovers the remaining two roots. × Claim 1. For any t ∈ F , the expression 19 18 ∑ tj R = a ζ t j j =1 × is still a root of the same cubic. Furthermore, as t ranges across F , each root of the cubic appears in 19 R exactly 6 times. t Proof. Consider the polynomial 18 ∑ j 3 2 f ( x ) = a x , g ( x ) = f ( x ) + 16 f ( x ) + 3 f ( x ) − 229 . j j =1 Then we have that f ( x ) (and hence g ( x ) ) has integer coefficients. Moreover, g ( ζ ) = 0 . Since the 18 2 18 2 minimal polynomial of ζ is x + · · · + x + x + 1 , it follows that x + · · · + x + x + 1 divides g ( x ) , t which means that g ( ζ ) = 0 . To show that each root appear 6 times, we note that the polynomial ∏ ( x − R ) t × t ∈ F 19 3 2 has integer coefficients by the theory of symmetric polynomial. However, since the cubic x + 16 x + 3 x − 229 is irreducible (one only needs to check that ± 1 or ± 229 are not a root), this polynomial must ∏ 3 2 3 2 6 × be a power of x + 16 x + 3 x − 229 , so we have that ( x − R ) = ( x + 16 x + 3 x − 229) , implying t t ∈ F 19 that each root appears exactly 6 times. ©2026 HMMT × Thus, summing R across all t ∈ F gives t 19 − ( a + · · · + a ) = 6( p + q + r ) = ⇒ a + · · · + a = 96 . 1 18 1 18 Next, observe that for any t , we have ∑ 2 2 t ( j − k ) R = | R | = a a ζ t j k t 1 ≤ j,k ≤ 18 × Summing this across all t ∈ F gives 19 ∑ 2 2 2 2 2 6( p + q + r ) = 18( a + · · · + a ) − 2 a a j k 1 18 1 ≤ j<k ≤ 18 2 2 2 = 19( a + · · · + a ) − ( a + · · · + a ) . 1 18 1 18 2 2 2 2 By Vieta, p + q + r = 16 − 2 · 3 = 250 , which implies that 2 6 · 250 + 96 2 2 a + · · · + a = = 564 . 1 18 19 Remark. If 1 7 8 11 12 18 A = ζ + ζ + ζ + ζ + ζ + ζ 2 14 16 3 5 17 B = ζ + ζ + ζ + ζ + ζ + ζ 4 9 13 6 10 15 C = ζ + ζ + ζ + ζ + ζ + ζ , then it is straightforward to check that 3 A + 6 B + 7 C , 3 B + 6 C + 7 A , and 3 C + 6 A + 7 B are the roots of this cubic. Note also that A , B , and C are clearly real. It is interesting to observe that the exponents in A are the cubic residues modulo 19 .