HMMT February, 2025

Number theory Difficulty 5.0 AIME Prove it United States

Problem:

Let $a$ , $b$ , and $c$ be pairwise distinct positive integers such that $\frac{1}{a}$ , $\frac{1}{b}$ , $\frac{1}{c}$ is an increasing arithmetic sequence in that order. Prove that $\gcd(a, b) > 1$ .

Solutions — 2

Solution 1

Solution:

Observe that $\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$ , so $b(a + c) = 2ac$ , and thus $a \mid b(a + c)$ . If we assume that $\gcd(a, b) = 1$ , then we must have $a \mid a + c$ , so $a \mid c$ . However, $\frac{1}{a} < \frac{1}{c}$ , so $a > c$ , contradiction. Thus, $\gcd(a, b) > 1$ , as desired.

Solution 2

Solution:

Observe that $\frac{2}{b} - \frac{1}{a} = \frac{1}{c}$ , so $(2a - b)c = ab$ and thus $2a - b \mid ab$ . If we assume that $\gcd(a, b) = 1$ , then $\gcd(2a - b, a) = 1$ , so $2a - b \mid b$ . Then $2a - b \mid (2a - b) + b = 2a$ , so $2a - b \mid \gcd(2a, b) \leq 2$ . Thus $2a - b \leq 2$ . But $a > b$ , contradiction. Thus, $\gcd(a, b) > 1$ , as desired.

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