Re: Solvable sextics and Fibonacci numbers
- From: A N Niel <anniel@xxxxxxxxxxxxxxxxxxxxx>
- Date: Wed, 29 Oct 2008 09:32:47 -0400
In article
<14f30133-6db2-445b-88a4-ea1d880034b4@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<tpiezas@xxxxxxxxx> wrote:
While doing some research on solvable sextics, I stumbled upon this
unusual connection between it and Fibonacci/Lucas numbers.
Define the sequence, starting with n=0:
L_n = {2, 3, 7, 18, 47, 123, 322,...}
This is A005248 in the OEIS (Online Encyclopedia of Integer Sequences)
and has formula:
L_n = phi^(2n) + (1/phi)^(2n)
where phi is the golden ratio = (1+Sqrt[5])/2.
Conjecture: "The irreducible sextic x^6 - ( L_n)x^5 + (L(n+1))x - 1 =
0 is solvable in radicals and factors over the extension Sqrt[5]."
Thus, x^6-123x^5+322x-1 = 0 is solvable, and so on.
Anybody knows how to prove the conjecture?
Tito
from Maple:
x^6-3*x^5+7*x-1 =
-(1/4)*(2*x^3-3*x^2+x^2*5^(1/2)-x-x*5^(1/2)-4+2*5^(1/2))*(-2*x^3+3*x^2+x
^2*5^(1/2)+x-x*5^(1/2)+4+2*5^(1/2))
x^6-7*x^5+18*x-1 =
-(1/4)*(-2*x^3+7*x^2+3*x^2*5^(1/2)+x-x*5^(1/2)+11+5*5^(1/2))*(2*x^3-7*x^
2+3*x^2*5^(1/2)-x-x*5^(1/2)-11+5*5^(1/2))
x^6-18*x^5+47*x-1 =
-(1/4)*(-2*x^3+18*x^2+8*x^2*5^(1/2)+x-x*5^(1/2)+29+13*5^(1/2))*(2*x^3-18
*x^2+8*x^2*5^(1/2)-x-x*5^(1/2)-29+13*5^(1/2))
x^6-47*x^5+123*x-1 =
-(1/4)*(-2*x^3+47*x^2+21*x^2*5^(1/2)+x-x*5^(1/2)+76+34*5^(1/2))*(2*x^3-4
7*x^2+21*x^2*5^(1/2)-x-x*5^(1/2)-76+34*5^(1/2))
x^6-123*x^5+322*x-1 =
-(1/4)*(2*x^3-123*x^2+55*x^2*5^(1/2)-x-x*5^(1/2)-199+89*5^(1/2))*(-2*x^3
+123*x^2+55*x^2*5^(1/2)+x-x*5^(1/2)+199+89*5^(1/2))
Guess the pattern (in terms of Lucas numbers or Fibonacci numbers,
perhaps).
.
- References:
- Solvable sextics and Fibonacci numbers
- From: tpiezas
- Solvable sextics and Fibonacci numbers
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