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I am working on a program that uses a combination of heuristics and brute-force search to try to find solutions of polynomials of degree $>4$ in radicals. Although not all such polynomials are solvable, it turns out that many polynomials naturally occurring in the theory of special functions are actually solvable, and my program was able to find those solutions. See, for example, https://math.stackexchange.com/a/3345906/19661 and https://math.stackexchange.com/a/3353303/19661.

Could you please provide some other naturally occurring polynomials that you suspect might be solvable in radicals, so that I could test and improve my program. My intention is to open-source it once it is stable enough.

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  • $\begingroup$ I can understand a vote to close, but I could not find a better way and audience to solicit this information. $\endgroup$
    – Vladimir Reshetnikov
    Commented Sep 24, 2019 at 18:45
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    $\begingroup$ "Many polynomials naturally occurring in the theory of special functions are actually solvable" you are quite wrong, it is because you are looking at polynomials coming from special values of modular functions and modular forms with algebraic coefficients, all of those split completely in abelian extensions of $\Bbb{Q}(\sqrt{-d})$ (it is the theory of elliptic curves with complex multiplication) $\endgroup$
    – reuns
    Commented Sep 24, 2019 at 18:47
  • $\begingroup$ @reuns Thanks. Yes, I understand that this is specific to certain classes of special functions. I am looking for examples where polynomials are suspected to be solvable in radicals, but no such solution has been found yet. $\endgroup$
    – Vladimir Reshetnikov
    Commented Sep 24, 2019 at 20:16
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    $\begingroup$ For arbitrary solvable polynomials, the naive algorithm is to compute the normal closure (by factoring the polynomial repeatedly in extensions obtained by adding its roots) and its Galois group, finding a sequence of quotients making it solvable, thus obtaining a tower of cyclic extensions having radical closed-forms. When the polynomial comes from modular forms this procedure has a concrete meaning in term of the ideal class group of an order in $O_{\Bbb{Q}(\sqrt{-d})}$. $\endgroup$
    – reuns
    Commented Sep 24, 2019 at 20:34

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I don't know about "naturally occurring", but you might try to generate some randomly.

For example, the following Maple code produced $3\,{x}^{6}-3\,{x}^{5}-4\,{x}^{4}-6\,{x}^{3}+3\,{x}^{2}+3$.

with(GroupTheory):
for i from 1 do
  p:= randpoly(x,degree=6,coeffs=rand(-10..10));
  if degree(p,x) < 6 then next fi;
  if not irreduc(p) then next fi;
  if IsSoluble(GaloisGroup(p,x)) then print(p); break fi
od:

EDIT: Another approach is to start with an expression in radicals, and find its minimal polynomial over the rationals. Maple can do this using evala(Norm(expression - x)).

q:= convert(sqrt(3^(1/2)+5^(1/3))+sqrt(-3^(1/2)+5^(1/3)),RootOf):
factor(evala(Norm(q-z)));

$$ \left( {z}^{12}+36\,{z}^{8}-320\,{z}^{6}+432\,{z}^{4}+1728 \right) ^{ 4} $$ So the minimal polynomial is $z^{12} + 36 z^8 - 320 z^6 + 432 z^4 + 1728$.

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  • $\begingroup$ And solvable polynomials are as easy to produce as towers of radical extensions $\endgroup$
    – reuns
    Commented Sep 24, 2019 at 18:58
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    $\begingroup$ The smallest real root is $x_1=\frac{1}{18} \left(3-\sqrt{57}+\sqrt[3]{12\,\alpha}+\sqrt[3]{2/\alpha} \left(11\cdot \sqrt[3]9-3\,\sqrt[6]{3} \cdot\sqrt{19}\right)\right),$ where $\alpha=288-7 \sqrt{57}+9 \sqrt{999-42 \sqrt{57}}.$ I believe there is a known systematic way to solve all solvable polynomials up to degree $7$. $\endgroup$
    – Vladimir Reshetnikov
    Commented Sep 24, 2019 at 19:06
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    $\begingroup$ Quintics, sextics, septics. BTW, do you know if any of these algorithms are implemented in Maple? $\endgroup$
    – Vladimir Reshetnikov
    Commented Sep 24, 2019 at 19:25
  • $\begingroup$ In general Maple's solve command doesn't seem to go higher than quartics (or polynomials that are easily reducible to quartics). $\endgroup$
    – Robert Israel
    Commented Sep 24, 2019 at 21:21
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    $\begingroup$ For an octic try ${x}^{8}+4\,{x}^{7}+8\,{x}^{6}+7\,{x}^{4}-4\,x-7$. $\endgroup$
    – Robert Israel
    Commented Sep 24, 2019 at 21:25

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