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 Posted: Thu Oct 30, 2008 2:32 pm Post subject: Solvable Octics x^8+ax+b = 0 |
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Turns out there is an _octic_ analogue to the Bring quintic as well --
and we can give a parametrization for this:
Collect[Resultant[x^4 + v x^3 + (p v + q) x^2 + (r v + a1)x + (a2 v +
a3), v^2 + a4, v], x, Factor]
It is easy to find a1, a2, a3, a4 to eliminate the x^6, x^5, x^4, x^3
terms. To make the x^2 term vanish, one has to solve the quadratic:
r^2 - 2(p^2+q) + (q/2)(-2p^2+q) = 0
To find rational r, we have to make its discriminant D a square:
D = 4p^4+12p^2q+2q^2 = y^2
If we set p = 1, this is just a quadratic polynomial to be made a
square, the _complete_ soln of which is q = (4mn)/(m^2-6mn+7n^2).
(The explicit expression for r is rather messy, so it suffices to give
p,q.)
Two nice examples are:
x^8-256x+832 = 0
x^8-384x+624 = 0
The first has {p,q,r} = {1,-8,-4} and the second as {1,-6,-4}. The
first factors over Sqrt[-1], the second, over Sqrt[-3]. In general,
this family of octics factors over Sqrt[2q].
P.S. If we go higher, solvable decics x^10+ax+b = 0 are now tricky,
since even if it factors over a Sqrt extension, there is no guarantee
the two quintic factors are solvable.
So are there solvable decics of form x^10+ax+b = 0?
Tito |
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