HMMT 十一月 2020 · 冲刺赛 · 第 25 题

HMMT November 2020 — Guts Round — Problem 25

[13] Let a , a , a , . . . be a sequence of positive integers where a = i ! and a + a is an odd perfect 1 2 3 1 i i +1 i =0 square for all i ≥ 1. Compute the smallest possible value of a . 1000

解析

[13] Let a , a , a , . . . be a sequence of positive integers where a = i ! and a + a is an odd 1 2 3 1 i i +1 i =0 perfect square for all i ≥ 1. Compute the smallest possible value of a . 1000 Proposed by: Sheldon Kieren Tan Answer: 7 Solution: Note that a ≡ 1 + 1 + 2 + 6 ≡ 2 (mod 8). Since a + a must be an odd perfect square, 1 1 2 we must have a + a ≡ 1 (mod 8) = ⇒ a ≡ 7 (mod 8). Similarly, since a + a is an odd perfect 1 2 2 2 3 square, we must have a ≡ 2 (mod 8). We can continue this to get a ≡ 2 (mod 8) and a ≡ 7 3 2 k − 1 2 k (mod 8), so in particular, we have a ≡ 7 (mod 8), so a ≥ 7. 1000 1000 2 2 Now, note that we can find some large enough odd perfect square t such that t − a ≥ 23. Let 1 2 a = t − a . Since a ≡ 7 (mod 8), we can let a − 7 = 8 k for some integer k ≥ 2. Now, since we have 2 1 2 2 2 2 2 (2 k + 1) − (2 k − 1) = 8 k , if we let a = (2 k − 1) − 7, then 3 2 2 2 2 a + a = a + ((2 k − 1) − 7) = (2 k − 1) + ( a − 7) = (2 k − 1) + 8 k = (2 k + 1) , 2 3 2 2 2 which is an odd perfect square. Now, we can let a = 7 and we will get a + a = (2 k − 1) . From 4 3 4 here, we can let 2 = a = a = a = · · · and 7 = a = a = a = · · · , which tells us that the least 5 7 9 4 6 8 possible value for a is 7. 1000