HMMT 二月 2021 · 团队赛 · 第 3 题

HMMT February 2021 — Team Round — Problem 3

[50] Let m be a positive integer. Show that there exists a positive integer n such that each of the 2 m + 1 integers n n n n 2 − m, 2 − ( m − 1) , . . . , 2 + ( m − 1) , 2 + m is positive and composite.

解析

[50] Let m be a positive integer. Show that there exists a positive integer n such that each of the 2 m + 1 integers n n n n 2 − m, 2 − ( m − 1) , . . . , 2 + ( m − 1) , 2 + m is positive and composite. Proposed by: Michael Ren Solution: Let P be the set of prime divisors of the 2 m + 1 numbers m +1 m +1 m +1 2 − m, 2 − m + 1 , . . . , 2 + m. We claim that ∏ n = m + 1 + ( p − 1) p ∈ P m +1 works. To check this, let k be any integer with | k | ≤ m . We can take some prime q | 2 + k , as ∏ m +1 m +1 2 + k ≥ 2 − m ≥ 3. Let ( p − 1) = ( q − 1) α . Then, applying Fermat’s little theorem, we q p ∈ P have n m +1 q − 1 α m +1 q 2 ≡ 2 · (2 ) ≡ 2 ≡ − k (mod q ) n Thus, 2 + k is divisible by q and is bigger than q , so it is positive and composite. This works for each of the required k , so we are done.