Timeline for Question related to Fermat curve: Does the equation $A x^n + By^n = C z^n$ have any solution in $\mathbb{N}$?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 30, 2016 at 3:03 | review | Close votes | |||
| Jan 30, 2016 at 20:59 | |||||
| Jan 25, 2016 at 17:22 | comment | added | Geoff Robinson | Yes,thanks. I saw Igor Rivin's answer too. | |
| Jan 25, 2016 at 15:15 | comment | added | Matt Young | @GeoffRobinson One can choose positive $x,y,z$ first, and then standard number theory gives $A,B,C \in \mathbb{Z}$. Then it seems to me that moving terms from one side of the equation to the other as necessary can get to an equivalent equation with all positive terms. | |
| Jan 23, 2016 at 22:53 | comment | added | Vesselin Dimitrov | In the case $A = B = 1, C= 2^r$, this has been proved in work of Ribet and Darmon-Merel. | |
| Jan 23, 2016 at 22:44 | comment | added | Vesselin Dimitrov | The right question seems to be whether, for any given $A,B,C \in \mathbb{Z}$ (not all zero), the equation $Ax^n + By^n = Cz^n$ has only finitely many solutions $(x,y,z;n)$ with $\mathrm{gcd}(x,y,z) = 1$ and $n > 3$. This follows from the $abc$-conjecture. | |
| Jan 23, 2016 at 22:35 | review | Close votes | |||
| Jan 24, 2016 at 5:43 | |||||
| Jan 23, 2016 at 22:31 | comment | added | Pietro Majer | Maybe the question is not perfectly stated, but I think it is clear that it asks about what is known about other diophantine equations $Ax^n+By^n=Cz^n$ where $A, B, C, n$ are given , and $x,y,z$ are the unknowns. | |
| Jan 23, 2016 at 22:13 | answer | added | Igor Rivin | timeline score: 3 | |
| Jan 23, 2016 at 21:04 | comment | added | Matt Young | You can choose any $x,y,z$ so that $(x,y,z) = 1$, and then pick $A,B,C$ so that $Ax^n + By^n = Cz^n$. | |
| Jan 23, 2016 at 21:01 | comment | added | Johnny T. | Thank you. I fixed the question to avoid these degenerate cases. | |
| Jan 23, 2016 at 21:00 | history | edited | Johnny T. | CC BY-SA 3.0 |
added 20 characters in body
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| Jan 23, 2016 at 20:57 | comment | added | Seva | ... or $z=1$ and $C=Ax^n+By^n$ ... | |
| Jan 23, 2016 at 20:56 | comment | added | Geoff Robinson | You need to exclude some other degenerate cases to make the question interesting: for example, if $x = y = z$ and $A+B = C$, there will be a solution. | |
| Jan 23, 2016 at 20:52 | history | asked | Johnny T. | CC BY-SA 3.0 |