Re: Solvable sextics and Fibonacci numbers
- From: Daniel Lichtblau <danl@xxxxxxxxxxx>
- Date: Wed, 29 Oct 2008 08:51:13 -0700 (PDT)
On Oct 29, 7:44 am, tpie...@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
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
.
- References:
- Solvable sextics and Fibonacci numbers
- From: tpiezas
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