Timeline for Which Diophantine equations can be solved using continued fractions?
Current License: CC BY-SA 3.0
17 events
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Jan 7, 2014 at 13:41 | comment | added | Alexey Ustinov | Linear equations are forgotten: $ax+by=1$, $ax+by+cz=N$, see [Frobenius number for three numbers.][1] [1]: mathoverflow.net/questions/23153/… | |
Feb 25, 2012 at 11:58 | answer | added | John R Ramsden | timeline score: 2 | |
Feb 24, 2012 at 19:35 | answer | added | duje | timeline score: 3 | |
Oct 19, 2011 at 23:02 | answer | added | Will Jagy | timeline score: 1 | |
Oct 19, 2011 at 11:28 | history | edited | Samuel Hambleton | CC BY-SA 3.0 |
added 315 characters in body
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Oct 19, 2011 at 8:29 | answer | added | Will Jagy | timeline score: 2 | |
Oct 19, 2011 at 6:50 | comment | added | Samuel Hambleton | Thank you very much Dr. Jagy. I apologise for the crime and the inconvenience. There are only two accounts. | |
Oct 19, 2011 at 5:59 | comment | added | Will Jagy | I have asked them to merge the two registered accounts of yours that I know about, 17053 and 18487. | |
Oct 19, 2011 at 2:50 | comment | added | Samuel Hambleton | The points should read $(11, 1, -3)$, $(-3, 5, 1)$, and ideal $(-3, 2 + (1 + \sqrt{229})/2)$. I would like to know of more methods for solving Diophantine equations, especially surfaces. Professor Elkies'methods look promising. Joro's question 70913, and particularly Schoof's article linked there shows that there may be some good reasons to want an easy method for finding non-principal ideals. I am particularly keen to learn methods for solving $18 x y +x^2 y^2 -4 x^3 -4 y^3 -27 = D z^2$. | |
Oct 19, 2011 at 2:39 | comment | added | Samuel Hambleton | ... done as: T = Union[DeleteCases[Partition[Flatten[Table[If[IntegerQ[Sqrt[d y^2 + 4 z^3]] && GCD[Sqrt[d y^2 + 4 z^3], z] == 1, {Sqrt[d y^2 + 4 z^3], y, z}, {w, w, w}], {y, -100, 100}, {z, -100, 100}]], 3], {w, w, w}]]; There are points of the set $T$ not in $S$, for example : $(11, 1, 3)$. The other form of the Pell surface is $B^2 + B C - 57 C^2 = A^3$. With point $(11, 1, 3)$ corresponding to $(A, B, C) = (3, 5, 1)$, which should map to the ideal $(3, 2 + (1 + \sqrt{\Delta })/2)$. I suspect that the "parametrization" leads to principal ideals. | |
Oct 19, 2011 at 2:27 | comment | added | Samuel Hambleton | Let $\Delta = 229$ and $K = \mathbb{Q}(\sqrt{\Delta })$. Then $\text{Cl}^+(K)[3] \simeq (\mathbb{Z}/ \3 )$. Using the "parametrization" in Mathematica5, d = 229; S = Union[DeleteCases[Partition[Flatten[Table[P = {(t^3 + 3 d t u^2)/4,(3 t^2 u + d u^3)/4,(t^2 - d u^2)/4};If[IntegerQ[P[[1]]] && IntegerQ[P[[2]]] && IntegerQ[P[[3]]] && GCD[P[[1]], P[[3]]] == 1, P, {w, w, w}], {t, -100, 100}, {u, -100, 100}]], 3], {w, w, w}]]; gives the Points of the "Pell surface" $x^2 - d y^2 = 4 z^3$ from the "parametrization". On the other hand, a brute force search for points satisfying this Eqn. can be ... | |
Oct 19, 2011 at 2:13 | comment | added | Samuel Hambleton | Thank you Dr. Jagy for comment, interest in the question, and for your help with question 71727 ! I've also seen that you're interested in ternary quadratic forms, which I would like to learn about too. My response will require quite the maximum characters, so I'll stop chatting and get to it. | |
Oct 17, 2011 at 0:03 | comment | added | Will Jagy | Samuel, could you be a little more specific about your comment about solving $x^2 - D y^2 = 4 z^3,$ especially why you are not satisfied with your parametrized solution? I printed out your paper with Franz, "Arithmetic of Pell Surfaces," where I think equation (1.1) with $n=3$ is what you are discussing below. Also Buell's book, especially pages 147-157, discussing Nagell 1922 and Yamamoto 1970. | |
Oct 13, 2011 at 2:48 | vote | accept | Samuel Hambleton | ||
Oct 13, 2011 at 2:42 | answer | added | Noam D. Elkies | timeline score: 25 | |
Oct 13, 2011 at 2:36 | history | edited | Yemon Choi |
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Oct 13, 2011 at 2:29 | history | asked | Samuel Hambleton | CC BY-SA 3.0 |