On 20 January 2016 at 15:11, Aaron Meurer <asmeu...@gmail.com> wrote:
> SymPy has algorithms to find roots of quintics in radicals (when they
> exist). I don't recall if the algorithms work for symbolic inputs.
>
> One can take a general quintic (x**5 + a*x**4 + b*x**3 + c*x**2 + d*x + e)
> and shift it by y (replace x with x - y). Then expand and collect terms in
> x. The coefficient of x**3 is a quadratic in y. Hence, one can solve a
> quadratic in y in radicals terms of a and b, and shift the quintic by that.
> One then has a new quintic, with no cubic term, with the same roots shifted
> by some term which is expressible in radicals. Hence, the general quintic
> with no cubic term is not solvable in radicals, as a solution would give a
> solution to the general quintic (shift it back by the radical expression
> above, which would keep it in radicals).
The above argument extends to making any of the coefficients zero with
the exception of the constant coefficient e. If we could shift to x -
t so that we have a new polynomial with zero constant term then x = t
would be a root of the original polynomial which would imply being
able to find one root of (and hence all roots of) any quintic.
Otherwise though we can shift to make any of a, b, c or d zero.
However the OP asked about a less general case where the cubic and
quadratic coefficients are both zero. It's not possible in general to
shift a quintic so that both b and c are zero.
> Here is some SymPy code:
>
> a, b, c, d, e, x, y = symbols('a b c d e x y')
> q = x**5 + a*x**4 + b*x**3 + c*x**2 + d*x + e
> print(collect(q.subs(x, x - y).expand(), x, evaluate=False)[x**3])
> t = solve(collect(q.subs(x, x - y).expand(), x, evaluate=False)[x**3], y)[0]
> print(t)
> print(collect(q.subs(x, x - t).expand(), x, evaluate=False).get(x**3))
To make b and c zero we'd need to solve these two equations
simultaneously for one variable y:
In [5]: print(collect(q.subs(x, x - y).expand(), x, evaluate=False)[x**2])
6*a*y**2 - 3*b*y + c - 10*y**3
In [6]: print(collect(q.subs(x, x - y).expand(), x, evaluate=False)[x**3])
-4*a*y + b + 10*y**2
Clearly that will only work for some lucky values of a, b, and c. (One
that jumps out is the trivial solution that we could have y=0 if b and
c are both 0.)
So this is not a general quintic but neither sympy nor Wolfram can
solve it. I was wondering if it is possible to specifically verify
that it has an unsolvable Galois group but perhaps not.
--
Oscar
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