solve this simple equation:$ax^2+byx+c=0$

Question

I need help solving the diophantine equation:$$ax^2+bxy+c=0$$ The quadratic formula does not seem to help much. Please help.

Answer 1

Note that $x$ must be a factor of $c$ so write $c=dx$ then $ax+by+d=0$ (or $x=0$ which can occur only if $c=0$). Now $x$ is fixed by $d$ and $y=-\frac {ax+d}b$. You need to test whether this is an integer.

Answer 2

Usually we solve Diophantine equations for positive integer solutions. Therefore I think you need positive integer solutions.

Consider the equation $ax^2+bxy+c=0$ as a quadratic equation of $x.$ For integer values of $x$ discriminant of this should be a perfect square. $$△_x=(by)^2-4ac=z^2$$ for some integer $z.$ Since $(by+z)(by-z)=4ac,$ this can have only finitely many integer solution for $y$ and $z.$
For each of these values, you have two rational solution for $x$ given by $\dfrac{-by+z}{2a}$ and $\dfrac{-by-z}{2a}.$
Pick positive integer solutions.