# Solution to a linear combination of integers being zero

by Sahand Tabatabaei   Last Updated October 09, 2019 21:20 PM

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

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