# Solution to cubic Diophantine equation in two variables

Does anyone know how to solve the following cubic Diophantine equation in two variables:

$$Ax^3 + Bx^2 + Cx + Dxy + Ey = 0$$

where A, B, C, D, E are known integers and $x$, $y$ are unknown integers to be solved?

Thanks,

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Isolate $y$ and use long division. –  Qiaochu Yuan Feb 24 '12 at 23:08
@Qiaochu, I think that qualifies as an answer. –  Gerry Myerson Feb 24 '12 at 23:16
@Gerry: well, the details are a bit annoying if $D$ isn't equal to $\pm 1$... –  Qiaochu Yuan Feb 24 '12 at 23:37

EDIT, 4:23 pm: I did not originally notice that, with fixed integers $A,B,C$ I was looking at a quadratic form value in setting $A E^2 - B DE + C D^2 = 0.$ This is impossible in nonzero $D,E$ unless $B^2 -4 AC = N^2$ for some integer $N.$ Or, put another way, $A x^2 + B x + C$ factors over $\mathbb Q$ and has two rational roots. Indeed, in this case $$A x^2 + B x + C = (D x + E)\left( \frac{Ax}{D} + \frac{C}{E} \right).$$ Well, unless $B^2 -4 AC = N^2,$ the number of solutions is finite. Once a particular $D,E$ are chosen, the number of solutions is probably finite anyway...
ORIGINAL: If $$BDE = AE^2 + C D^2$$ then there infinitely many solutions, all $x$ such that $$\frac{ADx^2 +(BD-AE)x}{D^2}$$ is an integer, which includes $x$ any multiple of $D^2$ but may have others. What is true is that if you take a successful value $x$ and then try $x + D^2,$ you get another success. So it is a set of values $\pmod {D^2}.$
If $$BDE - AE^2 - C D^2 \neq 0,$$ the set of solutions is finite. The first requirement is that $$(Dx+E) \; | \; (BDE^2 - A E^3 - C D^2 E)$$ which gives a finite set of $x$ values to check when $D \neq 0,$ a reasonable restriction here. For some of these $x$ values, $y$ may be integral.