Timeline for The Diophantine equation $x^2 + bxy + cy^2 = p^z_1 \cdots p^{z_k}$
Current License: CC BY-SA 3.0
13 events
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| Nov 12, 2014 at 21:03 | history | edited | Jason Sawyer | CC BY-SA 3.0 |
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| Nov 12, 2014 at 20:58 | review | Close votes | |||
| Nov 13, 2014 at 10:36 | |||||
| Nov 12, 2014 at 20:57 | comment | added | Felipe Voloch | Why do you need linear forms in logs? Also if $b=0,c=-2$, there are infinitely many solutions with $z_i=0$, for example. | |
| Nov 12, 2014 at 20:54 | history | edited | Jason Sawyer | CC BY-SA 3.0 |
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| Nov 12, 2014 at 20:53 | comment | added | Jason Sawyer | @DanielLoughran : With the condition (x,y)=1, I am pretty sure the solution set is provably finite. (If the homogeneous polynomial on the left-hand side was irreducible of degree larger than 3, it's called a Thue-Mahler equation and it's well-known there is an effective algorithm to determine the finite number of solutions). | |
| Nov 12, 2014 at 20:52 | comment | added | Jason Sawyer | @FelipeVoloch That's true. And if you want to compute them you are looking at linear forms in logarithms. But I have a feeling that this should be simpler than the Thue-Mahler equation where the left-hand side is a homogeneous polynomial in x,y of degree at least 3. Perhaps it's just that you can use Davenport's lemma rather than linear forms in logs and LLL reduction. In any case, I can't find a reference and I'd like to. | |
| Nov 12, 2014 at 20:46 | comment | added | Daniel Loughran | It seems difficult to imagine an algorithm which could enumerate such a possibly infinite set. Certainly there is a very simple algorthim which will find all solutions given an infinite amount of time. | |
| Nov 12, 2014 at 20:25 | comment | added | Felipe Voloch | If the left-hand side is irreducible over the rationals and you go to the quadratic field where it factors, then you are looking at the $S$-units of that field where $S=\{p_1,\ldots,p_k\}$ and the generalization of the Dirichlet unit theorem to $S$-units describes all the solutions. | |
| Nov 12, 2014 at 19:54 | comment | added | Jason Sawyer | @eric : Naturally, I meant to find all the solutions. Not just one of them. The question has been edited to clarify this. | |
| Nov 12, 2014 at 19:52 | history | edited | Jason Sawyer | CC BY-SA 3.0 |
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| Nov 12, 2014 at 19:49 | comment | added | eric | Yes: $x=1$ and everything else is 0. | |
| Nov 12, 2014 at 19:41 | history | edited | Jason Sawyer | CC BY-SA 3.0 |
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| Nov 12, 2014 at 19:25 | history | asked | Jason Sawyer | CC BY-SA 3.0 |