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 Posted: Thu Oct 30, 2008 2:28 pm Post subject: Solvable Sextics x^6+ax+b = 0 |
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After an advice given in another post, it turns out it is quite easy
to find parametrizations for the _sextic_ analogue of the Bring
quintic, or,
x^6+ax+b = 0
as irreducible but solvable. We can decompose it as 1) two cubics with
coeffs determined by a quadratic, or 2) three quadratics with coeffs
det. by a cubic.
I. As two cubics:
Collect[Resultant[x^3 + v x^2 + (p v + q) x + (r v + s), v^2 + u, v],
x, Factor]
This has the complete soln:
x^6 + pq(2p^2+3q)x - (q/8)(4p^4-36p^2q+q^2) = 0
One nice example is p = 2, q = -2 giving:
x^6-8x+89 = 0
II. As three quadratics:
Collect[Resultant[x^2 + v x + (p v^2 + q v + r), v^3 + b v + c, v], x,
Factor]
This is trickier but, after eliminating the x^4, x^3 using q,r, making
the x^2 vanish entails solving the quadratic:
(4b^3+27c^2)p^2 -2(4b^3+27c^2)p +15c^2 = 0
To find rational p, we make its discriminant D a square,
D = (b^3+3c^2)(4b^3+27c^2) = y^2
which is an elliptic curve. One soln is given by setting b = c and
this becomes,
D = b^4(b+3)(4b+27) = y^2
which is easily solved.
P.S. Maybe I>ll take a look at x^8+ax+b = 0...
Tito |
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