Re: Brutal result, actually unanswerable

> In article <43g0kgF1mdiveU2@xxxxxxxxxxxxxx>
> =?ISO-8859-1?Q?Jos=E9_Carlos_Santos?=
> <jcsantos@xxxxxxxx> writes:
> > jstevh@xxxxxxx wrote:
> > > But the result that follows from the argument is
> s easily checkable by
> > > those of you with some expertise using math
> h software as it covers
> > > rationals as well as irrationals where with
> h rational solutions with the
> > > equation
> > >
> > > a^3 + 3(-1+xf^2)a^2 - f^2(x^3 f^4 - 3x^2 f^2 +
> + 3x) = 0
> > >
> > > when you find integer solutions with non-zero
> o nonunit integer f and
> > > non-zero integer x coprime to f, those solutions
> s will either have f
> > > itself as a factor, or be coprime to f, for
> r instance, if f = 81, and
> > > you find a rational root that root would have
> e either 81 as a factor or
> > > be coprime to it.
> >
> > Can you please tell us which is the ring you're
> e working with?
>
> Let's see. He states that f and x are integer, so we
> have a cubic with
> integer coefficients. We know that if such a cubic
> has an rational
> root, that root is also integer, and so the cubic is
> reducible over
> the integers. (James still does not grasp the
> difference between
> reducible polynomials and irreducible polynomials.)
> But I think the
> number of cases in which that polynomial has integer
> roots is
> extremely small, if there are any at all.
> --
Hi,
Could we check how many possibilities there are for
integers? Let in very beginning rewrite:
a^3 +3Ba^2 -f^2 C = 0 ....................(1)
where B=xf^2 -1 .............................(2)
C=x(x^2 f^4 -3x f^2 +3)................(3)
now from eq.(1) should be C or f/a
1) let C=ca then eq.(1):
a^2 +3Ba -f^2 c =0
D=9B^2 +4f^2 c = d^2
what offers d^2 - (3B)^2 = 4f^2 c
with d = f^2 + c; 3B = f^2 -c
or more generally for f=mk:
d = m^2 +ck; 3B = m^2 -ck
and so on we could try to find some proper
integers but involving them to:
a^3 +3a^2 -2 =0 is just some
incorrect and somehow trivial...
Ro-bin

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