HMMT 二月 2025 · 团队赛 · 第 1 题

HMMT February 2025 — Team Round — Problem 1

[20] Let a , b , and c be pairwise distinct positive integers such that , , is an increasing arithmetic a b c sequence in that order. Prove that gcd( a, b ) > 1.

解析

[20] Let a , b , and c be pairwise distinct positive integers such that , , is an increasing arithmetic a b c sequence in that order. Prove that gcd( a, b ) > 1. Proposed by: Srinivas Arun 1 1 2 Solution 1: Observe that + = , so b ( a + c ) = 2 ac , and thus a | b ( a + c ). If we assume that a c b 1 1 gcd( a, b ) = 1, then we must have a | a + c , so a | c . However, < , so a > c , contradiction. Thus, a c gcd( a, b ) > 1, as desired. 2 1 1 Solution 2: Observe that − = , so (2 a − b ) c = ab and thus 2 a − b | ab . If we assume b a c that gcd( a, b ) = 1, then gcd(2 a − b, a ) = 1, so 2 a − b | b . Then 2 a − b | (2 a − b ) + b = 2 a , so 2 a − b | gcd(2 a, b ) ≤ 2. Thus 2 a − b ≤ 2. But a > b , contradiction. Thus, gcd( a, b ) > 1, as desired.