Timeline for Is surface $x^2+z^2=2\cdot y^2$ something of a Möbius strip?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 6, 2014 at 7:57 | comment | added | S. Carnahan♦ | -1 What is your question? | |
| Aug 6, 2014 at 7:06 | comment | added | individ | I would like to say a few words about the parameterization of the equation: $$a^2+b^2=qc^2$$ If you can imagine a factor as the sum of two squares. $q=t^2+k^2$ Then the solutions are. $$a=-tp^2+2kps+ts^2$$ $$b=kp^2+2tps-ks^2$$ $$c=p^2+s^2$$ Interestingly, all known formulas of Pythagorean triples. Are a special case of these formulas. | |
| Jul 1, 2013 at 7:05 | comment | added | Włodzimierz Holsztyński | Thank you Will Jagy and Jérémy Blanc for wiki link and information. | |
| Jul 1, 2013 at 5:03 | comment | added | Jérémy Blanc | Your surface is a quadratic surface so it is rational (project from a smooth point). By the way, the Möbius strip can be embeddded into $\mathbb{P}^2$, so it is also rational. Hence, both admit parametrisations. | |
| Jul 1, 2013 at 3:38 | comment | added | Will Jagy | Also, if you put positive (primitive) Pythagorean triples $a^2 + b^2 = c^2$ with $a$ odd, then $$ (a,b,c) \mapsto (|a-b|, a+b, c) $$ is a bijection from the positive Pythagorean triples to your triples, thus giving you a perfectly good tree. | |
| Jul 1, 2013 at 0:59 | comment | added | Will Jagy | There should be no trouble isolating the positive solutions by the method here: en.wikipedia.org/wiki/Tree_of_Pythagorean_triples | |
| Jul 1, 2013 at 0:06 | answer | added | Joseph O'Rourke | timeline score: 5 | |
| Jun 30, 2013 at 23:59 | comment | added | Włodzimierz Holsztyński | @Will Savin, thank you. (I'll still look at this surface, some things about it still feel to me interestingly odd). | |
| Jun 30, 2013 at 23:04 | comment | added | Will Sawin | The positive solutions are just the ones where $\sqrt{2}-1 < \alpha/\beta < \sqrt{2}+1$. They therefore form a triangular slice of the solutions overall. In general, one should never expect to be able to extract positive solutions, except by the use of inequalities - since being positive is, after all, an inequality. | |
| Jun 30, 2013 at 23:01 | history | edited | Włodzimierz Holsztyński |
wrong tag replaced by the intended one
|
|
| Jun 30, 2013 at 22:53 | history | asked | Włodzimierz Holsztyński | CC BY-SA 3.0 |