Timeline for About the solutions of $ \dfrac{x^p - y^p}{x - y} = a^2+pb^2 $
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 24 at 16:32 | vote | accept | user967210 | ||
| Jun 29 at 1:44 | comment | added | Will Jagy | it's in the Disquisitiones, articles 124 and 357. In my translation pages 82 and 439-440. | |
| Jun 29 at 1:11 | comment | added | Will Jagy | Andre Weil, in Number Theory: An approach through History, on pages 336 and 337, says that Gauss proved this. Calling $F= \frac{x^p - y^p}{x-y}$ splits into two factors in $\mathbb Q(\sqrt {\pm p}) ,$ leading to polynomials $P,Q$ with half-integer coefficients and $F = P^2 - p Q^2$ when $p \equiv 1 \pmod 4,$ but $F = P^2 + p Q^2$ when $p \equiv 3 \pmod 4,$ | |
| Jun 27 at 15:06 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor formatting
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| Jun 27 at 14:39 | answer | added | Max Alekseyev | timeline score: 7 | |
| Jun 27 at 8:08 | comment | added | user967210 | @WillJagy Thank you for pointing that out.I just intended that because the primitive prime factors have the form $ (2k_i p+1) $, when you take their $ \bmod{4p} $ you have just 2 possible outcomes $q_i= \{1, 2p+1\} $ and in the case $ p ≡ 3\bmod4 $, for both it holds $ \left(\frac{-p}{q_i}\right) = 1 $ (it's just a way to prove that $ \left(\frac{-p}{2k_i p+1}\right) = 1 $ alway hold, but maybe there is a simpler one). Do you think that the reason underlying the close form cases is that they are Heegner numbers? It would be possible to find something similar for $p=43$ and others? Thank you | |
| Jun 27 at 3:01 | answer | added | Will Jagy | timeline score: 5 | |
| Jun 26 at 18:06 | comment | added | Will Jagy | @DanielAsimov it is an older usage of the word ever, now largely obsolete. I looked up the etymology, it all makes sense... I would have written "always," or perhaps just omitted the word ever. | |
| Jun 26 at 18:01 | comment | added | Daniel Asimov | What does "ever admits solutions" mean? (Sorry, I don't want to have to read the proof to find out.) | |
| Jun 26 at 17:59 | comment | added | Will Jagy | The thing about 7, 11, 19 is class number one. en.wikipedia.org/wiki/Heegner_number | |
| Jun 26 at 16:58 | comment | added | Will Jagy | as I had indicated, the fact that a prime $q \equiv 1 \pmod p$ is no guarantee of a representation as $q = u^2 + p v^2$ or as $q = s^2 + st + \frac{1+p}{4} t^2,$ the parent form. So I don't see your last paragraph as a proof of your conjecture. For $p=23$ the two binary forms agree for this purpose. I've got a few dozen primes expressible as your $q =\frac{x^{23} - y^{23}}{x-y}.$ By factoring $x^3 - x + 1 \pmod q$ I can tell whether $q = u^2 + 23 v^2.$ So far, all succeed. See zakuski.math.utsa.edu/~jagy/Hudson_Williams_1991.pdf | |
| Jun 26 at 9:12 | history | asked | user967210 | CC BY-SA 4.0 |