HMMT 二月 2018 · 冲刺赛 · 第 29 题
HMMT February 2018 — Guts Round — Problem 29
题目详情
- [ 17 ] Let a , b , c be positive integers. All the roots of each of the quadratics 2 2 2 2 ax + bx + c, ax + bx − c, ax − bx + c, ax − bx − c are integers. Over all triples ( a, b, c ), find the triple with the third smallest value of a + b + c .
解析
- [ 17 ] Let a , b , c be positive integers. All the roots of each of the quadratics 2 2 2 2 ax + bx + c, ax + bx − c, ax − bx + c, ax − bx − c are integers. Over all triples ( a, b, c ), find the triple with the third smallest value of a + b + c . Proposed by: Henrik Boecken Answer: (1 , 10 , 24) √ 2 ± b ± b ± 4 ac The quadratic formula yields that the answers to these four quadratics are . Given that 2 a b all eight of these expressions are integers, we can add or subtract appropriate pairs to get that a √ 2 b ± 4 ac ′ b ′ 4 c ′ and are integers. Let b = and c = . We can rewrite the expressions to get that b a a a √ ′ ′ 2 ′ 2 ′ 2 ′ and b ± c are positive integers, which also tells us that c is a positive integer. Let b + c = n , ′ 2 ′ 2 b − c = m . ′ ′ c Notice that a + b + c = a (1 + b + ), so to find the third smallest value of a + b + c , we first find 4 ′ ′ ′ 2 ′ 2 2 small solutions to ( b , c ). To do this, we find triples ( m, b , n ) such that m , b , n form an arithmetic sequence. Because odd squares are 1 mod 4 and even squares are 0 mod 4, if any of these three terms is odd, then all three terms must be odd. By dividing these terms by the largest possible power of 2 ′ then applying the same logic, we can extend our result to conclude that v ( m ) = v ( b ) = v ( n ). Thus, 2 2 2 ′ we only need to look at ( m, b , n ) all odd, then multiply them by powers of 2 to get even solutions. ′ ′ We then plug in b = 3 , 5 , 7 , 9, and find that out of these options, only ( n, b , m ) = (1 , 5 , 7) works, ′ ′ ′ giving ( b , c ) = (5 , 24), a + b + c = 12 a . Multiplying by 2 yields that ( n, b , m ) = (2 , 10 , 14) also works, ′ ′ giving ( b , c ) = (10 , 96), a + b + c = 35 a . For 11 ≤ b ≤ 17, we can check that m = b + 2 fails to give ′ 2 2 c 15 − 11 ′ an integer n. For 11 ≤ b ≤ 17, m 6 = b + 2, a + b + c = a (1 + b + ) ≥ a (1 + 11 + ) = 38 a , 4 4 the smallest possible value of which is greater than 12 a with a = 1, 12 a with a = 2, and 35 a with a = 1. Thus, it cannot correspond to the solution with the third smallest a + b + c . For b ≥ 19, ′ 2 2 c 21 +19 ′ a + b + c = a (1 + b + ) ≥ a (1 + 19 + ) = 40 a , which, similar as before, can’t correspond to 4 4 the solution with the third smallest a + b + c . ′ ′ Thus the smallest solution is ( a, b , c ) = (1 , 5 , 24) , ( a, b, c ) = (1 , 5 , 6), the second smallest solution is ′ ′ ( a, b , c ) = (2 , 5 , 24) , ( a, b, c ) = (2 , 10 , 12), and the third smallest solution that the problem asks for is ′ ′ ( a, b , c ) = (1 , 10 , 96) , ( a, b, c ) = (1 , 10 , 24).