HMMT 十一月 2023 · 冲刺赛 · 第 28 题
HMMT November 2023 — Guts Round — Problem 28
题目详情
- [15] There is a unique quadruple of positive integers ( a, b, c, k ) such that c is not a perfect square and p √ 4 3 2 a + b + c is a root of the polynomial x − 20 x + 108 x − kx + 9. Compute c . √
解析
- [15] There is a unique quadruple of positive integers ( a, b, c, k ) such that c is not a perfect square and p √ 4 3 2 a + b + c is a root of the polynomial x − 20 x + 108 x − kx + 9. Compute c . Proposed by: Pitchayut Saengrungkongka Answer: 7 Solution: There are many ways to do this, including bashing it out directly. p √ The four roots are a ± b ± c , so the sum of roots is 20, so a = 5. Next, we compute the sum of squares of roots: q q 2 2 √ √ √ 2 a + b ± c + a − b ± c = 2 a + 2 b ± 2 c, 2 2 so the sum of squares of roots is 4 a + 4 b . However, from Vieta, it is 20 − 2 · 108 = 184, so 100 + 4 b = 184 = ⇒ b = 21. Finally, the product of roots is √ √ 2 2 2 2 ( a − ( b + c ))( a − ( b − c )) = ( a − b ) − c = 16 − c, so we have c = 7 . 4 3 Remark. Here we provide a justification that there is a unique quadruple. Let P ( x ) = x − 20 x + p √ 2 108 x − kx + 9 and r = a + b + c . Then, note that √ • r cannot be an integer because if not, then b + c must be an integer, so c must be a perfect square. • r cannot be a root of an irreducible cubic. One can check this directly because the remaining factor must be linear, so it must be x ± 1, x ± 3, or x ± 9. We can also argue using the theory of field extensions: [ Q ( r ) : Q ] = 3. However, we have the tower, √ deg 2 Q ( r ) ⊃ Q ( b + c ) ⊃ Q , √ and by multiplicativity of degree of field extensions, this forces [ Q ( r ) : Q ( b + c )] to not be an integer. 2 • r cannot be a root of quadratic. If not, then P factors into two quadratic polynomials ( x − mx + 2 p )( x − nx + q ). We then have m + n = 20, pq = 9, and mn + p + q = 108. By AM-GM, we have mn ≤ 100, so p + q ≥ 8, which forces ( p, q ) = (1 , 9), but this makes mn = 98, which is impossible. Thus, the minimal polynomial of r must be the quartic P ( x ). This means that all roots of P are √ a ± b ± c , and we can proceed as in the solution. √