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Gauss-Wantzel theorem asserts that a polygon with $n$ sides is constructible if and only if $n$ is a product of a power of $2$ and distinct prime Fermat numbers, where the Fermat number of index $k$ is $F_{k}=2^{2^k}+1$. Conjecturally, the only prime Fermat numbers are those with index $k\leq 4$.

Kinda coincidentally, those values of $k$ are also those for which the symmetric group of index $k$ and order $k!$ is solvable. My question is thus: is it really a coincidence? Or can we exhibit a rather natural structural similarity between a sequence of constructible polygons the order (as number of sides) of each dividing the order of the next one and a sequence of subgroups of $S_4$ to show that there is no prime Fermat number of index at least $5$? In other words, can we reasonably expect to prove in a Galois theoretic framework that $F_{k}$ for $k$ positive is prime if and only if the general polynomial equation of degree $k$ is solvable by radicals?

Thanks in advance.

Edit: As $S_{4}$ is solvable, there exists a normal series with abelian factor groups $\{e\}=G_{0}<...<G_{4}=S_{4}$. The idea is to map $G_{i}$ to a product $N_{i}$ of $n+1$ Fermat primes such that $N_{i}\mid N_{i+1}$.

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closed as unclear what you're asking by Lucia, GH from MO, paul garrett, Yemon Choi, Chris Godsil Aug 16 '15 at 18:41

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    $\begingroup$ "Is it really a coincidence?" Yes. $\endgroup$ – user9072 Aug 16 '15 at 16:37
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    $\begingroup$ Can you please tell me why? $\endgroup$ – Sylvain JULIEN Aug 16 '15 at 16:42
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    $\begingroup$ I'm voting to close this question as off-topic because it seems speculative without any basis $\endgroup$ – Yemon Choi Aug 16 '15 at 17:57
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    $\begingroup$ Other symmetric groups are not solvable by well understood reasons which have nothing common with the reasons why Fermat's numbers $F_5$, $F_6$, $\dots$, appear to be compisite. To best of my knowledge the reasons to believe that other Fermat's numbers are composite are probabilistic, not algebraic. $\endgroup$ – Fedor Petrov Aug 16 '15 at 18:34
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There are five Fermat primes but only four solvable symmetric groups, so I'm afraid there is not even a coincidence here.

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    $\begingroup$ But what are the chances that two small integers would differ by exactly $1$? :) $\endgroup$ – paul garrett Aug 16 '15 at 18:26
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    $\begingroup$ It depends on whether count $S_0$ or not (this if exists should have $0!$ elements). $\endgroup$ – Fedor Petrov Aug 16 '15 at 18:36
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    $\begingroup$ @FedorPetrov Of course it exists! It is the group of self-bijections of the empty set. $\endgroup$ – მამუკა ჯიბლაძე Aug 16 '15 at 21:24

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