[seqfan] Re: Difficult problem

[Robert Israel]

The answer is no.

The equation of your plane says that "a" is a root of a polynomial of 
degree <= 4 over the field generated by x,y,z.  If x,y,z were expressible 
in terms of radicals, so would a (since quartics can be solved in terms of 
radicals), except in the case where the polynomial is trivial (which
requires x=y=z=0).

Cheers,
Robert Israel

On Sun, 27 Dec 2009, Artur wrote:

> Dear Seqfans,
>
> We have real plane in 3D Cartesian space
> x(1-a)+(a^2-a)y+(2-a^4)z=0
> where a is single one real root of quintic polynomial a^5-a-1=0
> Does exist on this plane rational points (different as od x=y=z=0) ?
> If not does existed points expressible by radicals (polynomial a^5-a-1
> have not solvable Galois group S5 over rationals)
>
> Best wishes
> Artur
>
>
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