, 2025

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:

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.