Re: Brutal result, actually unanswerable

Rick Decker wrote:
> jstevh@xxxxxxx wrote:
>
> > Bill Dubuque wrote:
> >
> >>jstevh@xxxxxxx wrote:
> >>
> >>> a^3 + 3(-1+xf^2)a^2 - f^2(x^3 f^4 - 3x^2 f^2 + 3x) = 0
> >>>
> >>>integer solutions with non-zero nonunit integer f
> >>>and non-zero integer x coprime to f will either
> >>>have f itself as a factor, or be coprime to f
> >>
> >>As usual [1] there is an obvious and very elementary
> >
> >
> > Nope. You need to pay more attention.
> >
> >
> >>COUNTEREXAMPLE Specializing f = 2, x = 1 yields
> >>
> >> (a + 2) (a^2 + 7 a - 14) = 0
> >
> >
> > And as predicted by my theory, the rational root does indeed have 2 as
> > a factor.
>
> Actually, the fact that the integral root has 2 as a factor neither
> confirms nor denies your theory. I remind you that your purported
> result was that two of the roots will have f as a factor and the
> third will be coprime to f. Until you show all this, the fact that
> the root in this case is -2 should be regarded as nothing but trivia.
> BTW: as I've noted elsewhere, in this case *none* of the roots
> are coprime to 2, rather than the one you desire.
> >

Let me explain again, and I'm going to be objective and calm here.

If you have the possibility that two of the roots have f as a factor
while one is coprime to f, but only ONE of the roots is an integer
where you can see its factors, what are the possibilities?

I suggest to you Mr. Decker that the possibilities are that the integer
will have f as a factor or it will be coprime to f, so the integer
given DOES fit my theory.

A counterexample would be simple enough, as it would have some factor
of f as a factor, while not having f itself as a factor.

For instance, if f=32, and an integer root has 2 as a factor but is
coprime to 4, then my research would be proven false.

Or if f=10, and the integer was even but coprime to 5 that would prove
my research false.

Oh, one sidenote, if f has 3 as a factor then yeah, none of the roots
will be coprime to 3, as I did forget that condition. For some reason,
I usually do.


James Harris

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