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The content is miles outside what I know about. So the question is a mixture of idle curiosity and maybe having this answered somewhere on the Internet. It is likely I will not be able to understand the answer.

How exactly does Wiles' proof of Fermat's Last Theorem fail for $n=2$?

anomaly
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JP McCarthy
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    The relevant result that links the modular theorem to Fermat for (prime) exponents $\ge5$ is [Ribet's theorem](https://en.wikipedia.org/wiki/Ribet%27s_theorem). – Sassatelli Giulio Jun 03 '22 at 10:19
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    @SassatelliGiulio that may be relevant (and ultimately so) but does not answer the question. – JP McCarthy Jun 03 '22 at 10:24
  • The Frey curve $y^2=x(x-a^p)(x+b^p)$ to $x^p+y^p=z^p$ is for an odd prime $p$. In the proof the properties of the modular form entailed by the form of Frey’s elliptic curve is used. – Dietrich Burde Jun 03 '22 at 11:05
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    The high-level overview of Wiles's proof is as follows: $(1)$ We assume that Fermat's Last Theorem is false i.e. there exists some $a,b,c > 0$ and prime $p > 2$ s.t. $a^p + b^p = c^p$. $(2)$ It follows from Ribet's Theorem that the elliptic curve $y^2 = x(x-a^p)(x + b^p)$ (known as the Frey Curve) has no associated modular form. $(3)$ However, it follows from the Modularity Theorem that every rational elliptic curve is modular. $(4)$ This is a contradiction. The reason that Fermat's Last Theorem doesn't hold for $p=2$ is that $(2)$ doesn't hold for $p = 2$. – Carefree Xplorer Jun 03 '22 at 11:08
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    @CarefreeXplorer so something like $y^2=x(x-3^2)(x-4^2)$ is modular? Suitably generalised to other parts of Pythagorean triples answers the question if you would like I will accept. – JP McCarthy Jun 03 '22 at 11:15
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    @JPMcCarthy I will refrain from answering as my knowledge on this topic is limited at best. I'm sure someone else can shed light on the details. – Carefree Xplorer Jun 03 '22 at 11:38
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    @CarefreeXplorer Do you know if Ribet's theorem covers $p=3$? On wikipedia, it only mentions $p\ge5$ (which also leaves out $n=4$). – Sassatelli Giulio Jun 03 '22 at 12:00
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    @SassatelliGiulio if the Wiles proof did not work for small $n>2$ of course there are alternative proofs for small $n$ and the question in the OP could also be directed at those small $n$. – JP McCarthy Jun 03 '22 at 12:03
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    Wiles used properties of (hyper-)elliptic curves. Curves of degree $2$ are not elliptic or hyperelliptic. – Peter Jun 03 '22 at 12:06
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    My comment above re $y^2=x(x-3^2)(x-4^2)$ should have been $y^2=x(x-3^2)(x+4^2)$. – JP McCarthy Jun 03 '22 at 12:13
  • @Peter I would have guessed that $y^2=x(x-a^2)(x+b^2)$ with $a^2+b^2=c^2$ is elliptic/degree two in the same way $y^2=x(x-a^p)(x+b^p)$ is... – JP McCarthy Jun 03 '22 at 12:20
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    The votes to close here are utterly bizarre. This is a natural question of wide interest which obviously should have an answer on this site. – Eric Wofsey Jun 03 '22 at 15:10
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    This is no different from asking "How does Wiles proof fail for n = 1?". It's fine to come up with such questions but it doesn't make it a **good** question unless there is some **specific** reason to **single out** n = 2, which this question does not. – user21820 Jun 03 '22 at 15:31
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    @user21820 There is an obvious way to single out $n=2$: it's the unique number for which there nontrivially exist $a,b,c$ with $a^n+b^n=c^n$. I really don't think something so obvious needs to be stated explicitly in the question. – Noah Schweber Jun 03 '22 at 15:37
  • @NoahSchweber: Uhh triviality is opinion-based, but that's up to you. And I'm not going to complain if people disagree with my opinion. – user21820 Jun 03 '22 at 15:44
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    A quote from meta: "Frankly, I think this is a prime example of where rigid policy is driving away from the type of non-homework content I wish we'd have more of." – Carefree Xplorer Jun 03 '22 at 15:45
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    @user21820 Actually I meant "nontrivial" in an entirely objective sense: *some but not all* pairs $(a,b)$ satisfy $\exists c(a^2+b^2=c^2)$, but all pairs $(a,b)$ satisfy $\exists c(a^1+b^1=c^1)$. – Noah Schweber Jun 03 '22 at 15:47
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    @CarefreeXplorer: Meta has recently been overrun by users who make a lot of noise with little substance, and I have no problem with having non-homework content that are **not** from cranks. Note that I am *not* saying that this thread here is at all crankery, but you cannot lump all non-homework content together and make a false dichotomy. – user21820 Jun 03 '22 at 15:49
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    @user21820 I'm honestly curious: how would *you* write this question? While short, I personally don't see any additional context that could be added that isn't completely trivial, and I don't see inherent value in padding out length for its own sake. What would you add here? – Noah Schweber Jun 03 '22 at 15:50
  • @NoahSchweber: That's **not objective** at all! Some but not all pairs (a,b) from ℤ satisfy ∃c ( a^7 + b^7 = c^7 ). You excluded negative and zero numbers? Well, that is up to you, but is not objective! To answer your meta-question, I would be happy with it if it had included exactly what I stated, namely something explaining why the asker is interested in "n = 2" rather than other n. That's all I need to see! – user21820 Jun 03 '22 at 15:50
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    @user21820 Are you seriously arguing that the context of $\mathbb{N}$ isn't implicit in a question about FLT? This is just pedantry for pedantry's sake. – Noah Schweber Jun 03 '22 at 15:53
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    @NoahSchweber: My ℕ includes zero. I thought yours did too. This isn't bad faith. I seriously think that ℕ is more reasonable than ℕ+ as a foundational structure, such as a model of PA. You should stop assuming people have bad faith, which is (um) bad faith? – user21820 Jun 03 '22 at 15:54
  • @NoahSchweber: I'll note that you said I was commenting in bad faith, and later edited it out of your comment. I answered all your questions sincerely, but it seems you're not really interested in understanding my view. That's alright; I'm not here to convince or fight with you. – user21820 Jun 03 '22 at 15:56
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    @user21820 Yes, I did edit that out, since I quickly thought better of it. – Noah Schweber Jun 03 '22 at 15:57
  • Ok no problem. I don't know whether you believe me or not, but I am truly being sincere when I say my ℕ includes zero. After all, this is also what all logicians do when axiomatizing ℕ using PA, and also what Friedman did when axiomatizing EFA, which is a natural background theory for Fermat's conjecture. I just don't find the quadratic case very non-trivial, because it's something that has an algebraic solution set. And I just don't see any truly objective way to cut out some possibilities in one case but not another. Excluding zero is at least 1 bit of information you're providing. – user21820 Jun 03 '22 at 16:01
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    @user21820 My $\mathbb{N}$ includes $0$ too. But FLT is a problem primarily in the provenance of number theory, which community broadly does *not* include $0$ as a natural in my experience (and certainly the standard "naive" phrasing of FLT is implicitly over the positive integers). My own foundational stance is far less important to me than being able to talk smoothly with other mathematicians. – Noah Schweber Jun 03 '22 at 16:23
  • @NoahSchweber: Alright! Your last comment made me curious whether there is any actual 'niceness' to quantifying over ℕ+ instead of ℕ in number theory. I had always thought it was a superficial difference, but if you know of any deep result that you feel *naturally* calls for ℕ+ over ℕ, I may change my mind! (Up till now, I had thought that even outside of logic, algebraically inclined people would also prefer the discrete semi-ring structure to one without an additive inverse, but...) – user21820 Jun 03 '22 at 16:27
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    @user21820: A little late to the game, but my feeling on this is that what counts as trivial is a matter of consensus—but it's pretty broadly held consensus. I see the aim as principally focusing in on "the interesting bits": Solutions to $a^n + b^n = c^n$ that include any of $a, b, c$ equaling zero are trivial not only because they're obvious, but also because they get in the way of more interesting solutions. I would liken it to the division between trivial and non-trivial for the zeros of the Riemann zeta function (and newcomers to it might not consider any of the zeros obvious!). – Brian Tung Jun 03 '22 at 19:54
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    @JPMcCarthy To address your specific question about the curve $E: y^2=x(x-3^2)(x-4^2)$: yes, this curve is modular. The map $(x,y) \mapsto (x+2, y)$ gives an isomorphism to $E': y^2=x^3+x^2-160x+308$. This is the elliptic curve with [LMFDB label 240.d5](https://www.lmfdb.org/EllipticCurve/Q/240/d/5). In the Related objects box on that page, you'll see the [modular form with LMFDB label 240.2.a.d](https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/240/2/a/d/). This means that this elliptic curve and modular form have the same $L$-function, which is one definition of $E'$ being modular. – Viktor Vaughn Jun 03 '22 at 23:10

2 Answers2

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tl;dr version: A solution $a^\ell + b^\ell = c^\ell$ would generate a weird (i.e,. not modular) elliptic curve over $\mathbb{Q}$ for prime $\ell \geq 5$. Wiles proved that sufficiently weird elliptic curves don't exist, so that means Fermat's Last Theorem holds for $\ell \geq 5$.

There isn't an easy or concise answer to the question, simply because the number theory involved is so complicated. It had been known for a while before Wiles that Taniyama-Shimura conjecture would imply Fermat's Last Theorem, and Wiles proved it for a large enough class of curves to also prove that theorem. (The conjecture is a much deeper result than Fermat's Last Theorem, and Wiles' proof was also extended to the general case a bit later.)

The idea behind the conjecture is that for an elliptic curve $E/\mathbb{Q}$, you can reduce mod $p$ for each prime $p$ and count the number of points on $E$ over $\mathbb{F}_p$. Explicitly, given a Weierstass form $E:y^2 = x^3 + a x + b$, you can just count number of points $(x, y)\pmod{p}$ with $y^2 \equiv x^3 + ax + b\pmod{p}$. These can be tossed into something like an Euler product or a Dirichlet series to get to a resulting object called an $L$-series. (Sort of, anyway. There are corrections that have to be made for a finite number of primes $p$). If the Taniyama-Shimura conjecture holds, then we can do a bit of manipulation to get a nonzero function $f$ that is extraordinarily symmetric under the the group \begin{align*} \Gamma_0(N) = \left\{\begin{pmatrix} a & b \\ c & d\end{pmatrix}\in SL_2(\mathbb{Z}):\, c\equiv 0\!\!\pmod{N}\right\} \end{align*} acting on $\{z\in \mathbb{C}:\, \operatorname{Im} z > 0\}$ by $\begin{pmatrix} a & b \\ c & d\end{pmatrix}.z = (az + b)/(cz + d)$, where $N$ is a certain integer attached to $E$. (Specifically, $f$ is a modular form, and so the Taniyama-Shimura conjecture is also called the modularity theorem. Also, what I'm calling $N$ here isn't quite the conductor of $E$, but it divides it.)

This condition on the action $\Gamma_0(N)$ is pretty strict, and there aren't many such $f$. In particular, for $N = 2$, there isn't any nonzero function $f$ at all. But it turns out that given coprime (and nonzero) $a, b, c$ with $a^\ell + b^\ell = c^\ell$ for prime $\ell \geq 5$, the curve $E:y^2 = x(x - a^\ell)(x - c^\ell)$ happens to have $N = 2$. That contradiction implies that no such $a, b, c$ exist; that is, Fermat's Last Theorem holds for $\ell$. It's then clear that Fermat's Last Theorem holds for any exponent $n$ divisible by some prime $\ell \not = 2, 3$. But the cases $n = 3, 4$ are classical, and so the theorem holds for all $n > 2$.

So, what happens with $\ell = 2$? Well, the theorem obviously fails: There are plenty of points with $a^2 + b^2 = c^2$, and we're going to have to get a different value of $N$ for $\ell = 2$ in order for Taniyama-Shimura to still hold. There isn't an easy way to describe this part of the proof, but the basic idea is to compute the conductor of a certain elliptic curve, and there are a couple of cases describing its behavior when reduced mod $2$. If $\ell \geq 5$, we can eliminate all but a few easy cases and get an explicit value for $N$. That's probably not particularly useful, but see the caveat above that there isn't an easy or concise answer to this question. The major requirement of the proof is that $\ell$ be sufficiently large; the fact that $\ell \geq 5$ works is a happy coincidence.

If you're looking for a deeper reason why $\ell = 2$ fails, then the reason is that the Fermat curves $C:X^\ell + Y^\ell - Z^\ell$ are very different for $\ell = 2$ than for (prime) $\ell > 2$. The genus of $C$ is $g = \frac{1}{2}(\ell - 1)(\ell - 2) $, which is greater than $1$ for $\ell > 3$. Falting's Theorem states that such curves with $g > 1$ only have finitely many points where you can take the coordinates to all be rational. (The case $\ell = 3$ is different but easily handled classically.) For $\ell = 2$, this is clearly not the case; there are infinitely many Pythagorean triples. Now, "finitely many" is unfortunately a very long way from "zero," but that's the fundamental reason why $\ell = 2$ is different. The reason why Wiles' proof fails is that the Frey curve $E:y^2 = x(x - a^\ell)(x - c^\ell)$ gives you a different value of $N$ when $\ell = 2$, and we need $N = 2$ to get a violation of Taniyama-Shimura. As for why that value is different...well, it's a function of $\ell$, and it just happens to be different for $\ell = 2$ versus $\ell \geq 5$. I'm not sure there's anything really deep going on there.

I'm skipping over and simplifying a lot of the details here, but there's a more thorough but readable treatment in, e.g., the last chapter of Knapp's "Elliptic Curves."

anomaly
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    To the OP: I have no idea what kind of background in number theory you have, so the summary above might be more simplified or rough around the edges than what you're looking for. Knapp's a pretty good reference for this sort of thing and goes in a different direction than, say, Silverman. – anomaly Jun 03 '22 at 18:30
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    Also, I'm hedging a bit about why exactly the computation $N = 2$ fails for $\ell < 5$. There's not a great answer for that; the proximate reason is that the computation needs $32\ |c^\ell$. We can arrange for $c$ to be even, but we still need $\ell \geq 5$ to conclude that. – anomaly Jun 03 '22 at 18:44
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    thank you very much. I understand the FLT on a popular scientific basis so can follow the overview. Thanks for this answer but I am not sure does it answer the question. My reading is that you are saying the proof works for $\ell\geq5$. Why does the same proof not work for $\ell=2,3,4$ is the question. – JP McCarthy Jun 03 '22 at 20:16
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    I appreciate you have addressed precisely this in your second comment. I think unless someone else comes in yours will be accepted. – JP McCarthy Jun 03 '22 at 20:16
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    @JPMcCarthy: The answer is simply that what I'm calling $N$ above just turns out to be different for the curves $y^2 = x(x - a^\ell)(x - c^\ell)$ when $\ell = 2, 3$. I'm not sure there's any deep reason for that, but it has certainly _has_ to be different for $\ell = 2$; such $a, b, c$ do exist in that case, and they give a perfectly reasonable, modular elliptic curve (and for different values of $N$, there are such functions $f$). The computation of $N$ isn't particularly difficult, and Ribet's theorem gives it explicitly; it just requires some serious number theory. – anomaly Jun 03 '22 at 20:48
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    I must take your word here and if so as suspected the correct answer is beyond my own machinery. And I accept you pinpoint where the proof breaks down. Thank you. – JP McCarthy Jun 03 '22 at 20:50
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    (continued from above) I don't think it would be hard to prove a similar but less precise statement about $N$ for sufficiently large $\ell$; and with that, the fact that the technique actually works for $\ell$ as small as $5$ is just a nice coincidence. (FLT is known classically for some exponents $\ell \geq 5$, and more modern results give it for some but not all arbitrarily large primes.) The real cool bit, though, is that something as arbitrary and ad hoc as FLT ties in with a truly deep, important result like the modularity theorem. – anomaly Jun 03 '22 at 20:52
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    One last point that may be along the lines of what you have in mind: The Fermat curve $X^n + Y^n = Z^n$ is a conic for $n = 2$, an elliptic curve for $n = 3$, and a hyperelliptic curve for $n > 3$. (I'm ignoring some algebraic geometry issues; this is just a comment.) Conics aren't very interesting algebraically, but Falting's Theorem states that hyperelliptic curves have only finitely many rational (let alone integral) points. That's fundamentally the reason why $n = 2$ is different, but the reason Wiles' proof fails is what I mentioned above. But going from "finite" to "zero" is hard. – anomaly Jun 03 '22 at 21:04
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    @EricWofsey: Done. – anomaly Jun 03 '22 at 21:21
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    Well done @anomaly, good job – JP McCarthy Jun 03 '22 at 21:23
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    @anomaly I don't think you really meant "hyperelliptic" in your above comment. The Fermat curve is definitely not hyperelliptic for $n = 4$, and I don't think it is for any $n \geq 4$ (see [this answer](https://math.stackexchange.com/a/3494261)). But Faltings's Theorem only cares about the genus, so this doesn't really matter for the rest of your comment. – Viktor Vaughn Jun 03 '22 at 23:01
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    @ViktorVaughn: Shoot, you're right. It's too late to edit the comment now, but I'll fix the post. – anomaly Jun 03 '22 at 23:04
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The other "accepted" answer in my opinion more or less misses the point. So here is another answer.

Consider the elliptic curve

$$E: y^2 = x(x - a^p)(x + b^p).$$

The $p$-torsion points $E[p]$ of $E$ form a $2$-dimensional vector space over $\mathbf{F}_p$ which has an action of the Galois group, and this gives rise to a representation

$$\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F}_p).$$

The structure of the proof is based on the idea that the shape of $E$ implies that $\overline{\rho}$ has very special ramification properties. Assuming that $E$ and hence $\overline{\rho}$ is modular, Ribet's argument shows that either $\overline{\rho}$ should come from another modular form of a weight and level which doesn not exist (which is a contradiction, and this is the part which uses modularity) or $\overline{\rho}$ is not absolutely irreducible.

In particular, in order to complete the proof, one has to rule out the possibility that $E[p]$ is not absolutely irreducible. Assuming one can massage the choices of $a$ and $b$ to guarantee that $E$ is semistable, this turns out to be equivalent to showing that $E$ does not have a ratio nal point of order $p$. But since $E$ already has rational $2$-torsion, this is not possible for primes $p \ge 5$ by Mazur's theorem (https://en.wikipedia.org/wiki/Torsion_conjecture). Without Mazur's theorem (or something equivalent or stronger) you can't prove Fermat.

Returning to the case $p = 2$, the key reason the argument fails is that $\overline{\rho}$ can be (and indeed is) reducible, but knowing that $E[2]$ is reducible is not any contradiction to Mazur's theorem. In particular, you can write down $E: y^2 = x(x-9)(x+16)$ and then this is a perfectly good elliptic curve with rational $2$-torsion points.

The Phoenix
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