Timeline for Which Diophantine equations can be solved using continued fractions?
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| Oct 23, 2011 at 6:47 | comment | added | Samuel Hambleton | I think Dr. Jagy is saying that one can obtain all solutions of X^2 - D y^2 = 4 Z^3 by using representative binary quadratic forms of each of the classes of forms. For D = 229, there are three. This is correct, we can solve X^2 - D y^2 = 4 Z^3 by looking at classes of forms. Part of the proof is that the map from points of ( X^2 - D y^2 = 4 Z^3 ) to Cl^+(D)[3] is surjective. I am curious to know whether it can get computationally easier than this. Possibly not? | |
| Oct 19, 2011 at 23:24 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Oct 19, 2011 at 23:02 | history | answered | Will Jagy | CC BY-SA 3.0 |