Timeline for Summary of why Wiles' method does not work for general Fermat curves
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 3 at 2:40 | history | protected | Noah Schweber | ||
| Dec 31, 2023 at 3:35 | comment | added | KConrad | There is a paper by Darmon and Merel that treats $(a,b,c) = (1,1,2)$: math.mcgill.ca/darmon/pub/Articles/Research/18.Merel/paper.pdf. See the Main Theorem on page 2. | |
| Dec 31, 2023 at 2:36 | comment | added | Mike Bennett | Then Frey-Hellegouarch curves are easy to construct. The problem that arises is that essentially any nontrivial solution to the $S$-unit equation $x+y=1$, where $S$ is the set of primes dividing $abc$, leads to an obstruction to the method. By way of example, one can't solve $x^n+y^n=6z^n$ because $1+2=3$. | |
| Dec 30, 2023 at 2:58 | comment | added | R. van Dobben de Bruyn | My (very limited) understanding is that there are generalisations of level lowering for more general Frey curves, but it doesn't get you all the way down to level 2 (where there are no nonzero cusp forms, which is what ultimately gives the contradiction for FLT). I believe that level lowering in this case produces a cusp form of level $2\lvert abc\rvert$. | |
| Dec 30, 2023 at 2:32 | comment | added | Alison Miller | These curves are referred to as "twisted Fermat curves" in the literature. My rough impression is that one can still try to study the elliptic curve Y^2 = X (X + ax^n) (X - b y^n) which is the analogue of the Fermat curve, but that deriving a contradiction from the modularity of that curve is harder (as it should be, since sometimes solutions exist!) | |
| S Dec 30, 2023 at 2:31 | history | suggested | mathworker21 | CC BY-SA 4.0 |
fixed English
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| Dec 30, 2023 at 2:31 | review | Suggested edits | |||
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| Dec 30, 2023 at 2:08 | history | asked | Stanley Yao Xiao | CC BY-SA 4.0 |