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 $n=2$?

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    $\begingroup$ 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 $p = 2$. $\endgroup$
    – user905694
    Jun 3, 2022 at 11:08
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    $\begingroup$ @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. $\endgroup$ Jun 3, 2022 at 15:37
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    $\begingroup$ 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." $\endgroup$
    – user905694
    Jun 3, 2022 at 15:45
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    $\begingroup$ @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? $\endgroup$ Jun 3, 2022 at 15:50
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    $\begingroup$ @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. $\endgroup$ Jun 3, 2022 at 15:53

2 Answers 2


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."

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    $\begingroup$ 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. $\endgroup$
    – anomaly
    Jun 3, 2022 at 18:30
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    $\begingroup$ 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. $\endgroup$
    – anomaly
    Jun 3, 2022 at 18:44
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    $\begingroup$ 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. $\endgroup$ Jun 3, 2022 at 20:16
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    $\begingroup$ @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. $\endgroup$
    – anomaly
    Jun 3, 2022 at 20:48
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    $\begingroup$ @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). But Faltings's Theorem only cares about the genus, so this doesn't really matter for the rest of your comment. $\endgroup$ Jun 3, 2022 at 23:01

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.

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    $\begingroup$ Being pedantic, "rational $2$-torsion" should be "full rational $2$-torsion": there do exist elliptic curves with $10$ torsion. Otherwise great answer - and I agree with your assessment in the first line :) $\endgroup$
    – Mathmo123
    Jun 5, 2022 at 18:54

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