Timeline for Prove $\frac{\text{Area}_1}{c_1^2}+\frac{\text{Area}_2}{c_2^2}\neq \frac{\text{Area}_3}{c_3^2}$ for all primitive Pythagorean triples
Current License: CC BY-SA 4.0
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| Dec 27, 2019 at 6:17 | comment | added | Jonny Evans | @PMaynard What better way to learn some algebraic/Diophantine geometry than to have an example like this in mind. There's a fantastic book called "Ideals, varieties, algorithms" by Cox, Little and O'Shea which might be a good place to start if you're interested in concrete examples like this (rather than certain other Springer books called, say, "Algebraic geometry"). | |
| Dec 27, 2019 at 3:54 | comment | added | PMaynard | Thank you so much for pointing this out. I was a bit afraid of this and maybe I should have specified my background a bit more. I just got done with real analysis and am currently self studying abstract algebra so I'm afraid this is a bit out of my scope but I'll do my best. I'm going to try to pick up some algebraic geometry from the springer series in the next couple months or so. Hopefully I can pursue this route a bit further when I get there. Do you have any thoughts on the recursive nature of the solutions when c is not squared? It's pretty puzzling to me... | |
| Dec 27, 2019 at 3:02 | comment | added | user44191 | It may be easier to analyze the surface $\frac{(m_1^2 - n_1^2)m_1 n_1}{(m_1^2 + n_1^2)^2} + \frac{(m_2^2 - n_2^2)m_2 n_2}{(m_2^2 + n_2^2)^2} = \frac{(m_3^2 - n_3^2)m_3 n_3}{(m_3^2 + n_3)^2}$ as a subset of $(\mathbb{P}^1)^3$ (gotten using the usual parametrization of Pythagorean triples). | |
| Dec 27, 2019 at 2:28 | history | answered | Joe Silverman | CC BY-SA 4.0 |