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?


  • 3
    $\begingroup$ Isolate $y$ and use long division. $\endgroup$ Feb 24 '12 at 23:08
  • 2
    $\begingroup$ @Qiaochu, I think that qualifies as an answer. $\endgroup$ Feb 24 '12 at 23:16
  • $\begingroup$ @Gerry: well, the details are a bit annoying if $D$ isn't equal to $\pm 1$... $\endgroup$ 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.

  • $\begingroup$ Thanks. This was extremely helpful. $\endgroup$
    – SPSmith
    Mar 6 '12 at 13:08

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