Diophantine equation with division

Question

How can I find all the cases where y is positive integer in the next equation:

$$\frac{ax + b}{c-x} = y$$

• $a,b,c,x$ are not negative integers
• $a,b,x < c$
• $ax + b = 0$ is a trivial solution

Answer 1

In accordance with the usual convention, we avoid caps, which make equations harder to solve.

We are looking for positive integer solutions $(x,y)$ of $$\frac{ax+b}{c-x}=y,$$ where $a$, $b$, $c$ are non-negative and $x\lt c$.

Equivalently, we want, together with the bound on $x$, $ax+b=cy-xy$, that is, $xy+ax-cy=-b$, that is, $(x-c)(y+a)=-ac-b$, which looks better if written as $$(c-x)(y+a)=ac+b.$$ To find all solutions, we factor $ac+b$ as $ac+b=uv$, where $1\le u\le c-1$ and $v\gt a$.

Finding non-trivial such factors (if there are any) may be computationally very difficult if $ac+b$ is large.

Answer 2

$Y > 0$ if $AX > B$ is the only case if you want to avoid trivial solutions