HMMT 十一月 2008 · GEN1 赛 · 第 10 题

HMMT November 2008 — GEN1 Round — Problem 10

[ 8 ] Find the largest positive integer n such that n + 4 n − 15 n − 18 is the cube of an integer.

解析

[ 8 ] Find the largest positive integer n such that n + 4 n − 15 n − 18 is the cube of an integer. 3 3 3 2 Answer: 19 Note that the next cube after n is ( n + 1) = n + 3 n + 3 n + 1. After that, it is 3 3 2 3 3 3 2 ( n +2) = n +6 n +12 n +8. n +6 n +12 n +8 is definitely bigger than n +4 n − 15 n − 18, so the largest 3 2 3 3 2 cube that n + 4 n − 15 n − 18 could be is ( n + 1) . On the other hand, for n ≥ 4, n + 4 n − 15 n − 18 is 3 3 2 2 2 2 larger than ( n − 2) = n − 6 n +12 n − 8 (as 4 n − 15 n − 18 > − 6 n +12 n − 8 ⇐⇒ 10 n − 27 n − 10 > 0, which is true for n ≥ 4). 3 2 3 3 3 So, we will check for all solutions to n + 4 n − 15 n − 18 = ( n − 1) , n , ( n + 1) . The first case yields 3 2 3 2 2 n + 4 n − 15 n − 18 = n − 3 n + 3 n − 1 ⇐⇒ 7 n − 18 n − 17 = 0 which has no integer solutions. The second case yields 3 2 3 2 n + 4 n − 15 n − 18 = n ⇐⇒ 4 n − 15 n − 18 = 0 which also has no integer solutions. The final case yields 3 2 3 2 2 n + 4 n − 15 n − 18 = n + 3 n + 3 n + 1 ⇐⇒ n − 18 n − 19 = 0 which has integer solutions n = − 1 , 19. So, the largest possible n is 19. Remark: The easiest way to see that the first two polynomials have no integer solutions is using the n Rational Root Theorem , which states that the rational solutions of a polynomial ax + . . . + b are all ′ b ′ ′ of the form ± , where b divides b and a divides a . ′ a 2