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 easily checkable by
> > those of you with some expertise using math software as it covers
> > rationals as well as irrationals where with 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 nonunit integer f and
> > non-zero integer x coprime to f, those solutions will either have f
> > itself as a factor, or be coprime to f, for instance, if f = 81, and
> > you find a rational root that root would have either 81 as a factor or
> > be coprime to it.
>
> Can you please tell us which is the ring you're 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.
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