All positive integers m,n such that $an+b=cm$? Given positive integers a,b,c, how to find all positive integers m,n such that $an+b=cm$?
Is there always infinitely many m,n for all a,b,c?
If $(n_0, m_0)$ is the smallest solution, are all other solutions of the form $(n_0+ct, m_0+at)$ for some positive integer t?
 A: How: For how to do it, look for the Extended Euclidean Algorithm.  It is not particularly hard to carry out, but a proper description would be quite lengthy. The Extended Euclidean Algorithm is also described in most books on Elementary Number Theory.  There are many computer implementations, but for $a$ and $c$ of modst size, the procedure can be easily carried out by hand. 
If $d$ is the greatest common divisor of $a$ and $c$, the Extended euclidean algorithm produces integers $x$ and $y$ such that $ax+d=cy$. But from that one can solve your more general problem.
Existence of solution: Certainly there do not always exist such $m$ and $n$. For example, let $a=2$, $b=1$, $c=2$. Then $an+b$ (that is, $2n+1$) is always odd, and $cm$ (that is, 2m$) is always even, so they can never be equal.
In general, if the greatest common divisor of $a$ and $c$ divides $b$, there always is a solution. If the greatest common divisor of $a$ and $c$ does not divide $b$, there is no solution.
Infinitely many solutions: If there is a solution $(n_0,m_0)$, then there are infinitely many solutions. For suppose that $an_0+b=cm_0$. Then for any integer $t$, we have
$$a(n_0 +ct)+b=c(n_0+at).$$
(Just expand. There will be a term $act$ on the left, and a term $cat$ on the right, and they cancel.)
So if $(n_0,m_0)$ is a solution, then so is $(n_0+ct, m_0+at)$ for any positive integer $t$. 
