Solution to a linear combination of integers being zero

Question

I'm not very familiar with number theory so forgive me if this is a basic question.

Consider the equation: $$\alpha \ n + \beta \ m = 0,$$ with $n,m$ being integers. One can easily see that the above equation can only have a solution when $\alpha/\beta \in \mathbb{Q}$. My question is:

Is there a general solution $(n,m)\in \mathbb{Z}^2$ for the above equation in terms of an arbitrary $\alpha,\beta$ satisfying $\alpha/\beta \in \mathbb{Q}$?

@Bill Dubuque I don't consider this a duplicate. This question asks when alpha and beta are not necessarily rational. — Caleb Stanford, Oct 10 '19 at 16:43
@6005 It trivially reduces to the linked dupe once we choose a (reduced) rational rep of $\ \alpha/\beta.\ \ \ $ — Bill Dubuque, Oct 10 '19 at 17:26

score 1 · Accepted Answer · answered Oct 09 '19 at 21:16

Yes, there is a general solution of the form you ask. Given $\frac{\alpha}{\beta} \in \mathbb{Q}$, then write $\frac{\alpha}{\beta} = \frac{p}{q}$, where this fraction is in reduced form ($p$ and $q$ are relatively prime).

Then the equation rearranges to $$ m = - \frac{\alpha}{\beta} n \qquad \text{or equivalently:} \quad m = -\frac{p}{q} n, $$ and this means $n$ must be a multiple of $q$. So the general solution to this is $$ \boldsymbol{(m, n) = (kp, kq)} \textbf{ for all } \boldsymbol{k \in \mathbb{Z}}. $$

Solution to a linear combination of integers being zero

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