When is the sum of three reduced rationals equal to an integer?
This may be a duplicate of Question 1550437 but even if it is, there is no answer associated with this other question.
Given three moduli, $m1$, $m2$, and $m3$, not necessarily distinct or coprime, but $m_i > 1$ $\forall m_i$, when does the sum of rationals: \begin{equation} \left( \frac{r_1}{m_1} + \frac{r_2}{m_2} + \frac{r_3}{m_3} \right) \end{equation} admit integer solutions? The remainders, $r_1$, $r_2$, and $r_3$, are all coprime to the corresponding modulus; i.e. $(r_1, m_1) = (r_2, m_2) = (r_3, m_3) = 1$. The rationals in the sum are in reduced form: i.e. $r_i < m_i$ $\forall r_i$
The question can be restated for what constellation of modulii do coprime remainders exists such that: \begin{equation} \left( \frac{r_1}{m_1} + \frac{r_2}{m_2} + \frac{r_3}{m_3} \right) \in \mathbb{Z} \end{equation} is true for at least one constellation of remainders?
I have found some solutions: \begin{align} \left( m_1, m2 \right) = \left( m_1, m3 \right) = \left( m_2, m3 \right) = 1 &\implies \nexists r_1, r_2, r_3 \text{ $\mid$ } \left( \frac{r_1}{m_1} + \frac{r_2}{m_2} + \frac{r_3}{m_3} \right) \in \mathbb{Z} \\ \left( m_1, m2 \right) = 1 \text{ and } m_3 \ne m_1 m_2 &\implies \nexists r_1, r_2, r_3 \text{ $\mid$ } \left( \frac{r_1}{m_1} + \frac{r_2}{m_2} + \frac{r_3}{m_3} \right) \in \mathbb{Z} \\ \left( m_1, m2 \right) = 1 \text{ and } m_3 = m_1 m_2 &\implies \exists r_1, r_2, r_3 \text{ $\mid$ } \left( \frac{r_1}{m_1} + \frac{r_2}{m_2} + \frac{r_3}{m_3} \right) \in \mathbb{Z} \end{align}
But this is too piece-meal for my tastes.
There must be some application of the Chinese remainder theorem that can tell me when: \begin{equation} m_2 m_3 r_1 + m_1 m_3 r_2 + m_1 m_2 r_3 \equiv 0 \text{ mod } \text{ lcm}\left( m_1, m_2, m3 \right) \end{equation} does or does not admit a solution?
Can anyone point me to an article or text which discusses under what conditions the above congruence does or does not have solutions in the three variables: $\left(r_1, r_2, r_3\right)$?
Do I solve for the an unrestricted solution for $\left(r_1, r_2, r_3\right)$ and then see if the co-prime condition is met? Or can I use the co-primality of the $r_i$'s to my advantage?