Timeline for System of diophantine equations related to Ozanam's problem
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 16, 2016 at 22:23 | comment | added | veg_nw | Actally I am interested in only $S_1$ because of the task I am currently working on. From this paper I have found the following solutions: $c=-{k}{a}( {a}^{8}+6\,{a}^{4}{b}^{4}-3\,{b}^{8}); d=-{k}{b}( 3\,{a}^{8}-6\,{a}^{4}{b}^{4}-{b}^{8}) $. But not sure if this is the general solution of this system. | |
| Apr 16, 2016 at 20:36 | comment | added | Tito Piezas III | It turns out this system $S_1$ is just a particular solution to the Mengoli Six-Square Problem (MSP). Are you interested only in $S_1$, or is your interest in the broader MSP? | |
| Apr 16, 2016 at 6:31 | comment | added | veg_nw | And also I have found the following article with some set of solutions, but not sure if this is a general solution. ac.els-cdn.com/S0315086084710056/… page 15. | |
| Apr 16, 2016 at 6:30 | comment | added | veg_nw | Yes. Sure. Obviously the set of solutions of $(a^4-b^4)(c^4-d^4)=X^2$ is larger. Would be interesting to get the general solution for this equation also. May be we can filter out the solutions of $(a^2\pm b^2) (c^2\pm d^2) = \square$ from there. | |
| Apr 16, 2016 at 1:07 | comment | added | Tito Piezas III | Not necessarily. A solution to $(a^4-b^4)(c^4-d^4) = x^2$ is $a,b,c,d =1,\,3,\,31,\,49$, but it does not solve $ (a^2\pm b^2) (c^2\pm d^2) = \square$. | |
| Apr 14, 2016 at 21:06 | comment | added | veg_nw | I think the field of solutions of this system should be similiar to $(a^4-b^4)(c^4-d^4)=X^2$ does this make sense? | |
| Apr 12, 2016 at 22:46 | history | asked | veg_nw | CC BY-SA 3.0 |