What are the necessary conditions on integers $a, b$ for the sum of the fractions below to be reduced?

Question

1. $a/3+b/6$
2. $a/20+b/12$
3. $a/10+b/12$
   For 1., $2a+b$ would have to be a multiple of 2 or 3. $2a$ must be even, and for $2a+b$ to be divisible 3, then $b$ would have to be odd. I can show other obvious relations between $b$ and $a$, but I have trouble proving the properties of $b$ and $a$ specifically, as well as generalizing it to different cases as in the ones in 2 and 3.

Answer 1

Notate the first fraction's denominator as $c$ and the second fraction's denominator as $d$. Then set $g = \gcd(c, d)$. If $g > 1$ and, then if $\gcd(a + b, g) > 1$, you'll be able to reduce the sum of the two fractions.

To work through your second example: $g = \gcd(20, 12) = 4$. Then suppose $a = 9$ and $b = 7$. Then $$\frac{9}{20} + \frac{7}{12} = \frac{27}{60} + \frac{35}{60} = \frac{62}{60} = \frac{31}{30}.$$

Answer 2

Okay, so, given $d_a \neq d_b$, we have $$\frac{a}{d_a} + \frac{b}{d_b} = \frac{c}{d_c}.$$

How do we figure out what $c$ and $d_c$ are? $$\frac{a}{d_a} + \frac{b}{d_b} = \frac{m_a + m_b}{\textrm{lcm}(d_a, d_b)} = \frac{c}{d_c}$$ where $$m_a = \frac{\textrm{lcm}(d_a, d_b)}{d_a} a$$ and $$m_b = \frac{\textrm{lcm}(d_a, d_b)}{d_b} b.$$

Then it boils down to whether $\gcd(m_a + m_b, \textrm{lcm}(d_a, d_b)) > 1$ or not. I feel underwhelmed. Maybe there is a more exciting answer, but I don't see it.