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Does a heuristic explanation exist why $x^n+y^n=z^n \;\forall x,y,n,z\in \mathbb{Z},n>2$ doesn't have any solutions?

I'm not asking for an elementary proof or for an explanation of Wile's proof but maybe there is some kind of intuitive reasoning why the proposition should be true. Why did Fermat and many other mathematicians think that it was true? Did they just try a lot of different values?

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    $\begingroup$ Possible duplicate of Intuituive reason why Fermats last theorem holds $\endgroup$ – Robert Israel Mar 28 '17 at 17:47
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    $\begingroup$ Fermat thought he had a proof. The driving reason for early exploration was likely as much because they trusted Fermat's note that he had a proof, rather than heuristics. $\endgroup$ – Thomas Andrews Mar 28 '17 at 17:48
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    $\begingroup$ There's a heuristic argument suggesting that there are finitely many solutions for fixed $n$ (which can be proven using Faltings' theorem), and an argument suggesting that there are finitely many solutions overall using the abc conjecture, but I'm not aware of a heuristic argument for strengthening "finitely many" to "none." $\endgroup$ – Qiaochu Yuan Mar 28 '17 at 17:58
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    $\begingroup$ @user254665 Even Fermat was a human being that can have made an elementary error. If I am informed right, he claimed that he has found an easy proof. This is very unlikely because it took very long until a proof was found. But who knows, perhaps Fermat found a very difficult proof which seemed easy to him ... $\endgroup$ – Peter Mar 28 '17 at 20:41
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    $\begingroup$ Ironical his conjecture that every number of the form $2^{2^n}+1$ is prime. Today, there are good reasons to believe that there is no Fermat-prime besides the primes Fermat already knew. This would mean that the opposite of Fermat'c conjecture would be true. $\endgroup$ – Peter Mar 28 '17 at 20:46
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One explanation for why $x^n+y^n=z^n$ should not hold for $xyz\neq 0$ and $n>2$ comes from the abc-conjecture, in terms of the radical of an integer. Basically it says that if $a+b=c$ and $a$ and $b$ are high powers of an integer, then $c$ cannot be a high power of an integer. For example, if $a=2^{13}$ and $b=5^{13}$, then $$ a+b=2^{13}+5^{13}=1220711317=7\cdot 53\cdot 131\cdot 25117=c, $$ and $c$ is far from being a high power.

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  • $\begingroup$ Hmm, this should not have been a heuristic for people like Fermat himself, since the statement of the abc-conjecture is not so old, and I think also the whole idea is not so easy to grasp. $\endgroup$ – Gottfried Helms Mar 28 '17 at 18:59
  • $\begingroup$ No, not for Fermat, but he OP also asks for an intuitive reason for the present time. $\endgroup$ – Dietrich Burde Mar 28 '17 at 19:08
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I don't know whether the early pioneers have thought this way, but they might have.

If you write $ x^n - z^n = w $ and look at the prime-factorization of $w$ then you can use Fermat's little theorem to deduce that the exponents of the primefactors of $w$ grow only logarithmically with $n$. So it is very unlikely, that for larger $n$ we could have $w = p^n \cdot q^n \cdot \ldots \cdot u^n $ where $p \ldots u$ are the primefactors of $w$ - and this might have been very obvious to Fermat himself and also to Euler, who had analyzed this so-called "cyclotomic expression"s on the lhs and had generalized Fermat's little theorem to his "totient"-formula.

So ... this might have been a heuristic evidence for them pioneers ...

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  • $\begingroup$ How exactly do we apply Fermat's little theorem here? $\endgroup$ – Jack M Mar 28 '17 at 21:40
  • $\begingroup$ @Jack : by consideration of primefactors $p_k$ in $w$, and then we find, that their occurence in $x^n-z^n$ is periodic when increasing $n$ : for a primefactor $p$ the lhs $x^{p-1}-y^{p-1}$ is divisible by $p$ (this is little Fermat) and also $x^{(p-1)a}-y^{(p-1)a} \qquad a \in \mathbb N$ and so on. Using Euler's totient this is even more fluently expressible. Another keyword "LTE" or "Lifting the exponent" (google) $\endgroup$ – Gottfried Helms Mar 29 '17 at 0:24
  • $\begingroup$ @Jack : I've made an essay focusing that primefactorization of the cyclotomic expression. The very last chapter might illustrate this more explicite: go.helms-net.de/math/expdioph/CyclicSubgroups_work.pdf $\endgroup$ – Gottfried Helms Mar 29 '17 at 0:29
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For $n=3$ and $n=4$ it can be proved by infinite descent, as done by Euler and Fermat.

Lamé proposed a similar infinite descent approach when the cyclotomic integers have unique factorization. Alas, this does not always happen.

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