Timeline for How to solve this equation $a^2+3b^2c^2=7^c$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 27, 2022 at 11:20 | vote | accept | math110 | ||
| Dec 14, 2021 at 15:10 | comment | added | Fedor Petrov | @user914367 $2+i\sqrt{3}$ is an element, coprime residues modulo $p_1$ in $\mathbb{Z}[e^{2 \pi i/3}]$ are a multiplicative group | |
| Dec 14, 2021 at 14:01 | comment | added | Tashi | @FedorPetrov Why $p_1$ divides $(2\pm i \sqrt{3})^{p_1^2-1}-1$ follows from Lagrange theorem? What are the group and the element? | |
| Oct 28, 2021 at 9:48 | comment | added | Fedor Petrov | @user173628192 $x^{|G|}=e$ for every element $x$ of a finite group $G$, where $e$ is the unit element | |
| Oct 28, 2021 at 8:35 | comment | added | user434551 | @FedorPetrov what exactly does lagrange’s theorem state? | |
| Nov 17, 2020 at 11:43 | comment | added | Fedor Petrov | @user44191 ah, indeed, I used to think that the case $p_1=i\sqrt{3}$ was already excluded but it was not, now fixed | |
| Nov 17, 2020 at 11:42 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
added 214 characters in body
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| Nov 17, 2020 at 11:34 | comment | added | joro | You can get gcd(b,c)=2 with a,b,c=−137986556241619682249090303,9885082693522654065041378,64 | |
| Nov 17, 2020 at 11:31 | comment | added | user44191 | On a second look - I think you may need to address $c$ divisible by $3$ in the same way you addressed $c$ divisible by $2$, as your contradiction in 2.2 doesn't work when $p_1 = 3$. It should be solvable in much the same way as the even case; $3$ will divide $(x^{3^\alpha})^2 + (x^{3^\alpha})(y^{3^\alpha}) + (y^{3^\alpha})^2$ if it divides $x - y$, and so $9$ will divide $x^3 - y^3$, etc. | |
| Nov 17, 2020 at 10:55 | comment | added | user44191 | Ooh, the proofs in the second half are surprisingly pretty (I was working through the even case, and my proof there depended on the fact that $c \choose i$ was divisible by $2^m$ for $i$ odd, and I didn't have an answer for the odd case). I think it might be worth pointing out the exact reason why $gcd(p_1^2 - 1, c) = 1$ and writing the conclusion that $c$ has trivial prime factorization explicitly; a great answer! | |
| Nov 17, 2020 at 10:47 | history | answered | Fedor Petrov | CC BY-SA 4.0 |