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The sentence $S$ which Gödel in his proof of the incompleteness theorem proves to be be unprovable in the system of Peano arithmetic can be proved (as a true theorem of PA) outside PA (and necessarily only outside PA).

Compared to Fermat's last theorem $F$ which states that for $n>2, a, b, c \in \mathbb{N}$

$$a^n + b^n = c^n \rightarrow a = 0 \text{ or } b = 0$$

it's hard to write $S$ down as a sentence of PA and understand its meaning. Nevertheless it's a (true) theorem of pure number theory, perfectly expressible in the language of PA. And for both $S$ and $F$ no proof inside PA is known. (For $S$ there cannot be one.)

My question is: Can it be shown (and/or how can be shown) that Fermat's last theorem has no proof in the language of Peano arithmetic?

This would imply that one must leave the realm of Peano arithmetic to prove it. As a side question: How would one - in general terms - name the realms in which Gödel and Wiles performed their proofs? "Model theory" and "algebraic geometry"?

Hans-Peter Stricker
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    Surprisingly I cannot quickly find a duplicate, but a remark in this answer claims that "experts now believe that Wiles' proof can be converted mechanically into a proof over an incredibly weak system, even weaker than PA!" – hmakholm left over Monica Aug 20 '18 at 08:55
  • Gödel's sentence is undecidable in PA (but is true). Suppose you could similarly prove that $F$ is undecidable in PA. Then you would have shown it was true, as otherwise a counter-example would exist which contradicts undecidability. – user1729 Aug 20 '18 at 09:05
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    I don't understand the argument. How does it follow that $F$ is true from being undecidable? (This holds and is the meaning only of $S$.) – Hans-Peter Stricker Aug 20 '18 at 09:13
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    (Forget that we know that $F$ is true.) Suppose $F$ was both undecidable in PA and false. As it is false there exists integers $a$, $b$ and $c$, and an $n>2$ such that $a^n+b^n=c^n$. We can find these integers. Hence we can prove that $F$ is false within PA, contradicting undecidability. Hence, if you that $F$ is undecidable within PA it follows that $F$ is true. – user1729 Aug 20 '18 at 09:16
  • But how can we find them? They may be arbitrarily large. Doesn't miss this the essence of provability? Or did I miss something? – Hans-Peter Stricker Aug 20 '18 at 09:18
  • We can find them because we are assuming that they exist ("Suppose $F$ was...false."). – user1729 Aug 20 '18 at 09:22
  • Now I see the logic: If we knew that $F$ is undecidable, we know that it's true. So a proof of the undecidability of $F$ would be a proof of $F$ (just like with $S$ or any theorem $T$, right?) - but outside of PA. But from $F$ being true doesn't follow that $F$ is undecidable. So the undecidability of $F$ has not been proved (contrary to what you claim). Where lies the logical error? – Hans-Peter Stricker Aug 20 '18 at 09:45
  • The question has an implicit assumption that F would not be provable in PA, but there is no clear reason to make that assumption. The Gödel sentence is somewhat an exception - we generally expect that any $\Pi^0_1$ sentence which was first proved outside the field of logic will be provable in PA and in much weaker systems. So the questions would be better phrased as "How would we prove F in PA" rather than "How would we show F is not provable in PA". – Carl Mummert Aug 20 '18 at 11:43
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    @HansStricker I was just claiming "$F$ is undecidable in PA $\Rightarrow$ $F$ is true". Anything else I claimed was unintentional, and the opposite direction of this implication is almost certainly not true! – user1729 Aug 20 '18 at 11:51

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This paper shows that Fermat's Last Theorem (FLT) is not provable in weak arithmetics, including theories where exponentiation is not definable from addition and multiplication; in particular, FLT is not provable in Presburger artithmetic extended with a natural set of axioms for exponentials.

Concerning Peano arithmetic (PA), as far as I know, there is no negative result and the problem is still open. Actually, one of the main issues whether the proof of FLT can be formalized in PA is that it relies on higher-order structures and a priori there is no evidence that they can be reformulated in first-order language of PA. Moreover, even if all the necessary concepts can be stated in the first-order language of PA, Wiles' original proof of FLT uses set-theoretical assumptions unprovable in Zermelo-Fraenkel set theory with axiom of choice (ZFC, which is stronger than PA) and there is no a priori guarantee that we do not need axioms with greater strength than PA.

However, McLarty in this paper claims that in the proof of FLT:

[...] certainly much less than ZFC is used in principle, probably nothing beyond PA, and perhaps much less than that. [p. 359]

Macintyre in the appendix of his chapter of the book Kurt Gödel and the Foundations of Mathematics: Horizons of Truth proposed and sketched a project of formalizing Wiles's proof of FLT in Peano arithmetic. He says:

There is no possibility of giving a detailed account in a few pages. I hope nevertheless that the present account will convince all except professional skeptics that [the modularity theorem, which plays a key role in Wiles' proof] is really $\Pi^0_1$ [p. 15].

Taroccoesbrocco
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    How would you even state FLT in the language of Presburger arithmetic? – hmakholm left over Monica Aug 20 '18 at 10:13
  • @HenningMakholm - Formally you're right, I preferred a slightly ambiguous phrase instead of a more precise but heavier one (I don't even know the level of knowledge of the author of the OP). All technical details are well explained in the introduction of the first paper I cited. I summarize them in my comment below. – Taroccoesbrocco Aug 20 '18 at 10:40
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    @HenningMakholm - Roughly speaking, they deal with expansions $(\mathcal{B}, e)$ of models of arithmetical theories in the language $L = (0, 1, +, ·, \leq)$ by a binary function $e$ intended as an exponential ($e$ satisfies a certain natural set of axioms $Exp$). They construct a model $(\mathcal{B}, e) \models Th(\mathbb{N}) + Exp$ and a substructure $(\mathcal{A}, e)$ with $\mathcal{A} \models Pr$ (Presburger arithmetic) such that in both $(\mathcal{B}, e)$ and $(\mathcal{A}, e)$ FLT for $e$ is violated by cofinally many exponents. – Taroccoesbrocco Aug 20 '18 at 10:41