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
8 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2020 at 20:12 | history | bounty ended | PMaynard | ||
| Dec 30, 2019 at 0:21 | comment | added | PMaynard | I have updated the question to add some context that might be related to your findings. This is where the Pell equation shows up. | |
| Dec 29, 2019 at 21:55 | vote | accept | PMaynard | ||
| Dec 29, 2019 at 21:55 | |||||
| Dec 27, 2019 at 22:32 | history | edited | Constantin-Nicolae Beli | CC BY-SA 4.0 |
Wrong indices. I replaced $x_k$ by $x_{2k}$
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| Dec 27, 2019 at 22:27 | comment | added | PMaynard | The pell equation above shows up in where this problem comes from. When I post the full background later this may help. So I'm wondering if maybe this form could also be a necessary condition for $c^2$ as well (although we still don't know if these are the only solutions). If this is the case it could make the problem easier to tackle. | |
| Dec 27, 2019 at 22:15 | comment | added | PMaynard | Very interesting. Unfortunately I haven't had time to study this so thank you for the insight. I want to believe that these are the only solutions because these are the only ones up to $10^7$, but then again there could always be more until shown otherwise. Honestly, there is just so much here to study so hopefully I can spend some more time on this as well. I was aware of the ratio being equivalent to the height. It is frustrating for $c^2$ because I can't apply any geometric meaning to that yet. What I'd also really like to know is if these solutions have any bearing on the $c^2$ case? | |
| Dec 27, 2019 at 21:27 | history | edited | Constantin-Nicolae Beli | CC BY-SA 4.0 |
a comma and some quotations signs
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| Dec 27, 2019 at 21:22 | history | answered | Constantin-Nicolae Beli | CC BY-SA 4.0 |