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 Posted: Wed Oct 29, 2008 12:44 pm Post subject: Solvable sextics and Fibonacci numbers |
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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 |
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Daniel Lichtblau
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| Posted: Wed Oct 29, 2008 3:51 pm Post subject: Re: Solvable sextics and Fibonacci numbers |
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On Oct 29, 7:44 am, tpie...@gmail.com wrote:
[quote]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
[/quote]
If you want to try by hand, you might look for cubic factors with
linear and quadratic terms that are complex conjugates of one another.
By computer, you can just blindly factor over the stated extension. A
few lines of Mathematica shows this explicitly.
In[24]:= l[n_] := GoldenRatio^(2*n) + GoldenRatio^(-2*n)
In[25]:= g[n_] := x^6 - l[n]*x^5 + l[n+1]*x - 1
In[26]:= InputForm[Factor[g[n], Extension->GoldenRatio]]
Out[26]//InputForm-((2^(-2 - 2*n)*((1 + Sqrt[5])^(1 + 2*n) + 2^(2*n)*(1 - Sqrt[5])*x +
2*(1 + Sqrt[5])^(2*n)*x^2 - 2^(1 + 2*n)*x^3)*(2^(2*n)*(-1 +
Sqrt[5]) +
(-1 - Sqrt[5])*(1 + Sqrt[5])^(2*n)*x - 2^(1 + 2*n)*x^2 +
2*(1 + Sqrt[5])^(2*n)*x^3))/(1 + Sqrt[5])^(2*n))
Since n is always a positive integer, the coefficients always live in
the desired extension of the rationals.
Daniel Lichtblau
Wolfram Research |
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I.M. Soloveichik
Guest
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| Posted: Wed Oct 29, 2008 5:43 pm Post subject: Re: Solvable sextics and Fibonacci numbers |
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[quote]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
[/quote]
It factors into 2 cubics over the field generated by sqrt(5). The sextic is the product of a cubic and its sqrt(5) conjugate. The cubic is (up to signs)
x^3+/- a^n*x^2-(1+sqrt(5))/2*x+/-b*a^n
where b=-2+sqrt(5) and a=(-3+sqrt(5))/2. |
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A N Niel
Guest
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| Posted: Wed Oct 29, 2008 6:32 pm Post subject: Re: Solvable sextics and Fibonacci numbers |
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In article
<14f30133-6db2-445b-88a4-ea1d880034b4@b2g2000prf.googlegroups.com>,
<tpiezas@gmail.com> wrote:
[quote]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
[/quote]
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). |
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Guest
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| Posted: Thu Oct 30, 2008 12:38 pm Post subject: Re: Solvable sextics and Fibonacci numbers |
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On Oct 29, 1:43 pm, "I.M. Soloveichik" <imsol...@yahoo.com> wrote:
[quote]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
It factors into 2 cubics over the field generated by sqrt(5). The sextic is the product of a cubic and its sqrt(5) conjugate. The cubic is (up to signs)
x^3+/- a^n*x^2-(1+sqrt(5))/2*x+/-b*a^n
where b=-2+sqrt(5) and a=(-3+sqrt(5))/2.- Hide quoted text -
- Show quoted text -
[/quote]
Quite a number of people found the proof. Thanks, all. It was then
easy to find the generalization.
Define: L_n = y1^(2n)+y2^(2n)
where y1 and y2 are the roots of y^2-ay-1 = 0. Then the sextic,
x^6-(L_n)x^5+(1-a)x^4-(1-a)x^2+(L_(n+1))x-1 = 0
is solvable in radicals.
Of course, when a=1, and the quadratic has the Golden Ratio as a root,
the sextic has a simpler form as its x^4 and x^2 terms vanishes.
P.S. I tried to extend this result to octics, but I haven>t found the
proper arrangement yet.
Tito |
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