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Suppose we have an equation of the form $ax+by=0$ with $a,b,c \in \mathbb{Z}$. For simplicity, $a \neq 0, b \neq 0$. Then, a single solution to this equation is $(x_0, y_0)=(-a, b)$. My book states that all solutions are of the form $(x_0, y_0)=(-na,nb)$ with $n \in \mathbb{Z}$. But why? How we really know that this are all solutions. I tried this: Let $(x_1, y_1)$ be an arbitrary solution to this problem. Then $(x_0+x_1, y_0+y_1)$ is also a solution, so, "probably" $(x_0+nx_1, y_0 + ny_1)$ is a solution. Can someone help me out?

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First, it must be $\,(-b,a)\,$ is a solution to $\,ax+by=0\,\,,\,\,ab\neq 0\,$ , and not $\,(-a,b)\,$ .

Second, let us solve parametrically the equation:

$ax+by=0\Longleftrightarrow y=-\frac{a}{b}x\Longrightarrow \text{the general solution is}$

$\left(x\,,\,-\frac{a}{b}x\right)\Longleftrightarrow (-bx\,,\,ax)\,\,,\,\,x\in\Bbb Z$

Please do note the first option above may not be defined in the integers, since it could be $\,a/b\notin\Bbb Z\,$ , but the second one's already fine (if you insist in integer solutions. If you're fine with rational ones then the first form is ok), and it is of the required form (after the correction mentioned at the beginning of my answer)

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    So, now you are saying, $(x_0, y_0)$ is a solution $\Leftrightarrow (m x_0, m y_0)$ is a solution. This is not true. For example, if... Ah, I see :p. Thanks!2012-09-26