HMMT 二月 2007 · GEN1 赛 · 第 8 题

HMMT February 2007 — GEN1 Round — Problem 8

[ 5 ] Determine the largest positive integer n such that there exist positive integers x, y, z so that 2 2 2 2 n = x + y + z + 2 xy + 2 yz + 2 zx + 3 x + 3 y + 3 z − 6

解析

[ 5 ] Determine the largest positive integer n such that there exist positive integers x, y, z so that 2 2 2 2 n = x + y + z + 2 xy + 2 yz + 2 zx + 3 x + 3 y + 3 z − 6 2 2 Answer: 8 . The given equation rewrites as n = ( x + y + z + 1) + ( x + y + z + 1) − 8. Writing 2 2 r = x + y + z + 1, we have n = r + r − 8. Clearly, one possibility is n = r = 8, which is realized by 2 2 2 x = y = 1 , z = 6. On the other hand, for r > 8, we have r < r + r − 8 < ( r + 1) . 1