2
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If I input:

Solve[x^5 - x - 1 == 0, x]
(* {{x -> Root[-1 - #1 + #1^5 &, 1]}, {x -> 
   Root[-1 - #1 + #1^5 &, 2]}, {x -> Root[-1 - #1 + #1^5 &, 3]}, {x ->
    Root[-1 - #1 + #1^5 &, 4]}, {x -> Root[-1 - #1 + #1^5 &, 5]}} *)

$x^5 - x - 1=0$ is NOT solvable by radicals (according to various sources).

My question is:

When Mathematica does not return an exact solution to a polynomial equation (as in the above example), does this imply that the polynomial is not solvable by radicals?

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  • 1
    $\begingroup$ But Solve did return an exact solution set. Root objects are exact solutions. $\endgroup$ – m_goldberg Jan 1 '16 at 14:19
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    $\begingroup$ Have a look at Quartics in the documentations. $\endgroup$ – user36273 Jan 1 '16 at 14:28
  • $\begingroup$ Maybe a better way to state my question is this: Does there exists an algorithm that can determine if a polynomial equation is "solvable by radicals"? $\endgroup$ – Geoffrey Critzer Jan 1 '16 at 14:58
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    $\begingroup$ The direct answer to your question is no, and this is documented in tutorial/AlgebraicNumbers. It is indicated there that Solve[x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23 == 0, x] returns a list of Root objects but that 2^(1/3) + 3^(1/2) is a solution. $\endgroup$ – murray Jan 1 '16 at 18:00
  • $\begingroup$ @murray. Yes, thank you. This is the answer that I expected, along with the documentation to substantiate it. $\endgroup$ – Geoffrey Critzer Jan 1 '16 at 18:45
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This is to put murray's comment on record, so it can be accepted by the OP.

The direct answer to your question is no, and this is documented in tutorial/AlgebraicNumbers. It is indicated there that

Solve[x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23 == 0, x]

returns a list of root objects, even though 2^(1/3) + 3^(1/2) is a solution

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