Q:

Solution to a linear combination of integers being zero

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 $\alpha/\beta \in \mathbb{Q}$?

A:

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
$$
\boldsymbol{(m, n) = (kp, kq)} \textbf{ for all } \boldsymbol{k \in \mathbb{Z}}.
$$

