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The Wikipedia article on the proof of Fermat's Last Theorem has this sentence

If the link identified by Frey could be proven, then in turn, it would mean that a proof or disproof of either of Fermat's Last Theorem or the Taniyama–Shimura–Weil conjecture would simultaneously prove or disprove the other.

This suggests FLT and the modularity theorem are equivalent. While the fact that the modularity theorem implies FLT was a rather important part of Wiles' proof, I wasn't aware the reverse implication was true. Is it?

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    $\begingroup$ The modularity theorem is true so anything implies it. What implies the modularity of L-functions of rational elliptic curves should be essentially that for every $j(\tau) \in \mathbb{Q}$, some $j(\frac{a\tau+b}{d})$ appears in the invariants of the jacobian of some modular curve. $\endgroup$ – reuns Nov 15 '18 at 22:29
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This discussion seems to me a bit weird. Let us recall the facts:

1) The Shimura-Taniyama-Weil conjecture (now a theorem of Wiles et. al.) states that any elliptic curve $E$ defined over $\mathbf Q$ is "modular". This means roughly that the Hasse-Weil L-function $L_E(s)$ attached to $E$ comes from an L-function L$(f_E,s)$ attached to a certain modular form $f_E$. More precisely, let $f_E$ be the inverse Mellin transform of $(2\pi)^{-s}\Gamma(s)L_E(s)$; then $f_E\in S_2(\Gamma_0 (N))$, where $N$ is the conductor of $E$ and $f_E$ is a Hecke form.

2) Suppose that the Fermat equation $a^p+b^p=c^p$ admits a non trivial solution and consider the elliptic curve $E_{a,b,c}$ defined over $\mathbf Q$ by the equation $y^2=x(x-a^p)(x+b^p)$ (Hellegouarch, 1969). Frey (1986) conjectured that $E_{a,b,c}$ could not be modular (see the appendix). Then (Mazur and) Ribet (1990) proved a deep theorem on modular representations of $Gal(\bar {\mathbf Q}/\mathbf Q)$ which implied Frey's conjecture.

3) FLT follows from 1) and 2). Thus the modularity thm. implies FLT, but the converse does not hold, for purely logical but also for "intuitive" reasons.

Appendix. I take this opportunity to reproduce a short historical account which I gave as a comment to A question on FLT and Taniyama Shimura

Frey picked up this curve in Yves Hellegouarch's thesis *Courbes elliptiques et équation de Fermat", Besançon, 1972. Hell. himself has written a nice introductory book on the subject, "Invitation aux math. de Fermat-Wiles", Masson ed., 1997 (I don't know if there is an English translation). At the beginning of the 70s, Hell. had not the necessary coceptual tools to conjecture that his elliptic curve could not be modular, but he was aware that it possessed so many extraordinary properties that he had nicknamed it "Roland's mare" (Roland - Orlando in Italian - was presumably a nephew of emperor Charlemagne, a legendary hero in French mythology, and the central character of Tasso's epic poem "Orlando furioso"), a fabulous animal which had all the qualities in the world except... existence. Only in 1985, with all the theoretical progress made in more than 10 years, Frey was able to state his conjecture that "Roland's mare" could not be modular. See the historical appendix of Hell.'s book.

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Yes this reverse is also true

If there were a counterexample for FLT, it would also provide a counterexample of modularity by a specific construction.

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    $\begingroup$ The reverse implication of $A\rightarrow B$ is $B\rightarrow A$, not $\neg B\rightarrow \neg A$. $\endgroup$ – eyeballfrog Nov 15 '18 at 22:08
  • $\begingroup$ Yes, however this is actually an if and only if implication not just a one way. $\endgroup$ – Mathaddict Nov 15 '18 at 22:09
  • $\begingroup$ So you're saying that if there is any non-modular elliptic curve, it could be used to construct a counterexample to FLT? $\endgroup$ – eyeballfrog Nov 15 '18 at 22:11
  • $\begingroup$ No, that's what Wile's proof did (and then went on to prove that they don't exist), I'm saying that if there were a counterexample to FLT then it could be used to construct a non-modular semistable eliptic curve which would contradict modularity. $\endgroup$ – Mathaddict Nov 15 '18 at 22:15

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