# Does the function Solve determine if a polynomial is NOT solvable by radicals?

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?

• But Solve did return an exact solution set. Root objects are exact solutions. – m_goldberg Jan 1 '16 at 14:19
• Have a look at Quartics in the documentations. – user36273 Jan 1 '16 at 14:28
• 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"? – Geoffrey Critzer Jan 1 '16 at 14:58
• 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. – murray Jan 1 '16 at 18:00
• @murray. Yes, thank you. This is the answer that I expected, along with the documentation to substantiate it. – Geoffrey Critzer Jan 1 '16 at 18:45

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