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It is by now extremely well known that Sir Andrew Wiles proved the Taniyama-Shimura conjecture, and therefore, through the Frey-Hellegouarch curve, that for $n \geq 5$ the only integer solutions to the equation $x^n + y^n = z^n$ satisfy $xyz = 0$. The latter statement, of course, is the famous Fermat's Last "Theorem".

I have never studied Wiles' proof in depth. Throughout the years I've heard tidbits of the proof in various talks, many of them quite excellent, and now have a vague understanding of the general idea. However, it is still unclear to me exactly what prevents Wiles' proof from applying to general Fermat curves of the shape

$$\displaystyle ax^n + by^n + cz^n = 0, a,b,c \in \mathbb{Z} \setminus \{0\}.$$

I believe that the problem lies in the construction of the Frey-Hellegouarch curve, or the lack of a construction. Is this the main (only) problem?

As a related question, is there a thorough study of exactly the sets of integers $\{a,b,c\}$ for which the Wiles method will find all rational points on the corresponding Fermat curve?

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    $\begingroup$ 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!) $\endgroup$
    – Alison Miller
    Commented Dec 30, 2023 at 2:32
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    $\begingroup$ 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 $2\lvert abc\rvert$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 30, 2023 at 2:58
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    $\begingroup$ 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 $1+2=3$. $\endgroup$
    – Mike Bennett
    Commented Dec 31, 2023 at 2:36
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    $\begingroup$ 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. $\endgroup$
    – KConrad
    Commented Dec 31, 2023 at 3:35

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