# cubic diophantine equation

• Mar 23rd 2008, 01:58 PM
dioph
cubic diophantine equation
Please help me to find a source/reference for solving the equation (or similar) 4u^3 - v^2 = 3.

I am guessing that the only solutions are (1,1), (1,-1), (7,37), and (7,-37), but my efforts have not been successful.

Thanks,

Wayne
• Mar 24th 2008, 04:11 AM
Opalg
Quote:

Originally Posted by dioph
Please help me to find a source/reference for solving the equation (or similar) 4u^3 - v^2 = 3.

I am guessing that the only solutions are (1,1), (1,-1), (7,37), and (7,-37), but my efforts have not been successful.

There are infinitely many rational solutions, for example $u=\frac{73}{36},\ v=\pm\frac{595}{108}$. It seems unlikely that there are any more integer solutions, but I don't know how one might prove that.
• Mar 24th 2008, 10:00 AM
dioph
integer
Sorry - I meant to say integer solutions. Do you know of a source/reference for solving specifically equations of this form?

Thanks,

Wayne
• Mar 24th 2008, 10:34 AM
ThePerfectHacker
Yes. This looks like a problem with Elliptic curves (hence my avatar). A book on Elliptic curves might mention something about this equation. I have a book at home higher number theory, it mentions some stuff, but nothing like this.

Another idea:
If you can reduce the equation to $y^2 = x^3 - k$ where $k$ is square-free, then it might work. The first step is to note $(2u)^3 = 2v^2 - k$. And so $(2u)^3 = (\sqrt{2} v)^2 + 6$. And we are left with an equation, $x^3 = y^2 + 6$, but here $x,y$ are not integers. Perhaps you can take advantage of the fact that $\mathbb{Z}[\sqrt{-2}]$ has unique factorization and use it to prove that $x^3 = y^2 + 6$ has not only solutions in $\mathbb{Z}$ but more, no solutions in $\mathbb{Z}[\sqrt{-2}]$.