Timeline for About the solutions of $ \dfrac{x^p - y^p}{x - y} = a^2+pb^2 $
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 24 at 16:32 | vote | accept | user967210 | ||
| Jun 30 at 5:34 | comment | added | Max Alekseyev | @user967210: I've updated the code in my answer to handle all odd primes. | |
| Jun 30 at 5:33 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Jun 30 at 0:32 | comment | added | Will Jagy | @Max I put two expositions in my answer, a chapter by Cox and a survey by Savitt, which appears to give a complete proof of the $a^2 \pm p b^2$ thing. | |
| Jun 29 at 22:39 | comment | added | user967210 | @MaxAlekseyev Thank you! Can I just ask you one last thing? How do I need to modify your Sage code to show the cases $ p \equiv 1 \pmod 4 $ for $ \frac{x^p - y^p}{x-y} = P^2 - p Q^2 $? I tried something, but I'm new to sage. Thanks | |
| Jun 29 at 22:05 | comment | added | Max Alekseyev | @user967210: I think it was just by chance. The factors of $b$, besides $x-1$, are generally palindromic but do not necessarily divide $\Phi_p$. | |
| Jun 29 at 21:35 | comment | added | Max Alekseyev | @WillJagy: Nice find! I suspected that this subject should be well studied. | |
| Jun 29 at 8:39 | comment | added | user967210 | @MaxAlekseyev Thank you so much! I tried with a few other cases, but it seems that $b$ is not factorizable in terms of cyclotomic polynomials, like I showed for the cases $p=7, 11, 19$, that just happened by chance or is there an underlying reason? Thank you! | |
| Jun 29 at 1:28 | comment | added | Will Jagy | in my English translation of the Disquisitiones Gauss announces the result in article 124, page 82, then gives detail in article 357, pages 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 14:59 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Jun 27 at 14:52 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Jun 27 at 14:39 | history | answered | Max Alekseyev | CC BY-SA 4.0 |