# How to solve $(xy)^2+a(xy)+bx+cy+d=0$ in integers?

Respected all.

We know that $x^2+y^2+2gx+2fy+c=0$ represents a circle and the parametric solution for it is $x=\cos t, y=\sin t$.

But I was wondering what would happened for the following equation $$(xy)^2+a(xy)+bx+cy+d=0$$ where $a,b,c,d,e\in \mathbb Z.$

Can anyone help me out how to solve this equation in integers ? I have no idea on it. please show me the path.

• That will be harder because it is a fourth degree polynomial. Quadratics are easier. – Thomas Andrews Jan 12 '15 at 6:17
• :-( so no way this time ? – Anjan3 Jan 12 '15 at 6:19
• I didn't say "no way." There might be a trick. Just that quadratics tend to be easier. – Thomas Andrews Jan 12 '15 at 6:20

## 2 Answers

Transform this equation using w=xy and eliminate y from the equation. This shows that you only have to look for the factors of c since x must divide c evenly. There are only finite factors in c. Otherwise this is a cubic equation and may be approached by trial and error ....x can only be a factor of c....starting from x=1...[very hard problem]

Assume that the curve is irreducible, then it has genus $\leq 1.$ In case of genus 1 you have to find a Weierstrass model and the (birational) map between them. Then it is (usually) easy to find the integer solutions (in the Weierstrass model) using a computer algebra system and go back to the original curve using the birational map. If the genus is zero, then you need a rational parametrization (here it may have infinitely many integer solutions, it depends on the "form" of the points at infinity). There are references for both the cases, if you google it. Hope that helps.