I missed A=0 on wolfram, sorry for confusion

On Wed, Jan 20, 2016, 9:38 AM Aaron Meurer <[email protected]> wrote:

> On Wed, Jan 20, 2016 at 10:30 AM, Oscar Benjamin <
> [email protected]> wrote:
>
>> On 20 January 2016 at 15:11, Aaron Meurer <[email protected]> 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.
>>
>
> Oh I didn't notice that :)
>
>
>> > 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.
>>
>
> SymPy is not able to do that (I checked and the roots_quintic function in
> the polys only works for rational coefficients).
>
> If WolframAlpha can compute Galois groups I wasn't able to get it to work.
> Probably GAP can do it, if someone wants to check that.
>
> Aaron Meurer
>
>
>>
>> --
>> Oscar
>>
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