Timeline for Finding Generators of O( Z^3,x^2 + xy + y^2 - z^2) and integer solutions
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2011 at 5:15 | comment | added | Will Jagy | en.wikipedia.org/wiki/… suggests four possible articles, # ^ Gilder, J., Integer-sided triangles with an angle of 60°," Mathematical Gazette 66, December 1982, 261 266 # ^ a b Burn, Bob, "Triangles with a 60° angle and sides of integer length," Mathematical Gazette 87, March 2003, 148–153. # ^ a b Read, Emrys, "On integer-sided triangles containing angles of 120° or 60°", Mathematical Gazette 90, July 2006, 299-305. # ^ Selkirk, K., "Integer-sided triangles with an angle of 120°," Mathematical Gazette 67, December 1983, 251–255 | |
| Oct 8, 2011 at 4:58 | comment | added | Will Jagy | John, evidently your version is due to Berggren (1934), anyway see en.wikipedia.org/wiki/Pythagorean_triple | |
| Oct 8, 2011 at 4:23 | comment | added | john mangual | Somewhere, earlier this summer I read a note deriving the pythagorean triple tree from Euclidean geometry. Towards the end it showed there were 5 matrices generating solutions to $x^2+xy+y^2-z^2=0$ probably as consequence of Law of Cosines. Now I can't find it. | |
| Oct 8, 2011 at 4:21 | comment | added | john mangual | @Keith, I should fix the typo. I guess the correct statement is "the orbits of (3,4,5) and (4,3,5) by multiplication of these 3 matrices generate all positive pythagorean triples" You need a few more to get the entire orthogonal group but these are normal subgroups or something, switching halves of the "light-cone". | |
| Oct 7, 2011 at 3:52 | comment | added | Ian Agol | This might be in Allcock's paper "The reflective Lorentzian lattices of rank 3". ma.utexas.edu/users/allcock | |
| Oct 6, 2011 at 3:50 | comment | added | KConrad | @John, there are two problems with your description of the orthogonal group of $x^2 + y^2 - z^2$. First, the vector [5,4,3] isn't even a zero of that quadratic form. Typo? Second, why do you say these three matrices generate the orthogonal group? I think you're forgetting to look at stabilizer subgroups too. For example, the diagonal matrix diag(-1,1,1) is in the orthogonal group. Can you write it in terms of your three matrices? I know five generators of that orthogonal group, not three. | |
| Oct 6, 2011 at 2:39 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Oct 5, 2011 at 22:05 | comment | added | Will Jagy | John, as in my answer, the Gram matrix $G$ for the original Pythagorean triples would be the diagonal matrix with diagonal entries $ (1,1,-1).$ You should check, for your three matrices above, that $$A^T G A = G, \; B^T G B = G, \; C^T G C = G,$$ which is what you want if you are writing your triples as column vectors, as I do. If it turns out you are using row vectors, similar check with the transposes of your three matrices instead. | |
| Oct 5, 2011 at 19:58 | answer | added | Will Jagy | timeline score: 7 | |
| Oct 5, 2011 at 14:38 | history | edited | john mangual | CC BY-SA 3.0 |
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| Oct 5, 2011 at 14:35 | comment | added | john mangual | how might I find the lattice corresponding to a quadratic form $x^2+xy+y^2-z^2$ ? What is the lattice for $x^2+y^2-z^2$ ? | |
| Oct 5, 2011 at 14:31 | history | edited | john mangual | CC BY-SA 3.0 |
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| Oct 5, 2011 at 6:37 | comment | added | Will Jagy | I see, you need severa automorphs of $x^2 + x y + y^2 - z^2.$ Meanwhile, rather than primes $ \equiv 1 \pmod 4,$ For this problem you want primes $z \equiv 1 \pmod 3.$ So your first triple would be $(3,5,7)^T$ | |
| Oct 5, 2011 at 6:22 | history | asked | john mangual | CC BY-SA 3.0 |