Timeline for representation of integers as the product of linear forms in three variables
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2011 at 18:39 | comment | added | Dror Speiser | Oh yes! Thanks David. I've looked at too many quadratic imaginary fields lately that I forgot that units with negative norm can even exist... | |
| Jun 7, 2011 at 18:12 | vote | accept | Tom Hunt | ||
| Jun 7, 2011 at 15:45 | comment | added | David Loeffler | Dror: You have to restrict to norm 1 units, cf. my answer below, which crossed over with yours. | |
| Jun 7, 2011 at 15:43 | answer | added | David Loeffler | timeline score: 6 | |
| Jun 7, 2011 at 15:36 | history | edited | Tom Hunt | CC BY-SA 3.0 |
I changed my initial statement to reflect the comment of David.; added 2 characters in body
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| Jun 7, 2011 at 15:29 | comment | added | Dror Speiser | In your example $1,\theta,\theta^2$ generate the maximal order, so the equation is simply the norm equation whose solutions are exactly the units of the field generated by $\theta$. The field is totally real, hence has unit rank 2, and the group of units is generated by: {$-1,\theta+1,\theta^2-1$}. So the solutions are those such that $p+q\theta+r\theta^2=\pm (\theta+1)^a(\theta^2-1)^b$, for any integers $a,b$. (all computations were done using Sage) | |
| Jun 7, 2011 at 15:10 | comment | added | Tom Hunt | David, thank you for your reply. Indeed your restatement of the problem is correct. It may be less difficult than the problem he tries to solve. I looked at Mordell's particular solution as only a particular solution to my problem. | |
| Jun 7, 2011 at 14:54 | history | edited | Tom Hunt | CC BY-SA 3.0 |
fixed typo: changed indecies on the left hand side of particular solutions.
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| Jun 7, 2011 at 14:25 | comment | added | David Loeffler | So: you want to find all triples $(x, y, z)$ of integers such that $\prod_{\theta} (x + y\theta + z \theta^2) = 1$, where $\theta$ runs through the solutions to your cubic? (I think Mordell is trying to do something else much harder -- for him w is not part of the question; he's solving for the quadruple $(x, y, z, w)$.) | |
| Jun 7, 2011 at 12:39 | history | asked | Tom Hunt | CC BY-SA 3.0 |