HMMT 十一月 2023 · THM 赛 · 第 5 题

HMMT November 2023 — THM Round — Problem 5

解析

Compute the unique positive integer n such that is a perfect square. n Proposed by: Isaac Zhu Answer: 13 2 1989 Solution 1: We need n − to be a perfect square, so n | 1989. Also, this perfect square would be n 2 2 2 less than n , so it would be at most ( n − 1) = n − 2 n + 1. Thus, 1989 2 ≥ 2 n − 1 = ⇒ 1989 ≥ 2 n − n, n so n ≤ 31. Moreover, we need 1989 2 3 n ≥ = ⇒ n ≥ 1989 , n 2 so n ≥ 13. Factoring gives 1989 = 3 · 13 · 17, which means the only possible values of n are 13 and 17. 3 13 − 1989 2 Checking both gives that only n = 13 works. (In fact, = 4 .) 13 3 n − 1989 2 3 2 Solution 2: If = d then n − nd = 1989. Factorizing gives n 2 ( n − d ) n ( n + d ) = 3 × 13 × 17 . We can easily see that n = 13, d = 4 works since 1989 = 9 × 13 × 17.