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Re: FLT Discussion: Simplifying
Posted:
Jan 16, 2001 11:09 AM
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"Dik T. Winter" wrote:
>
> In article <9400ps$b5g$1@nnrp1.deja.com> jstevh@my-deja.com writes:
> > Given x^2 + y^2 = 0, x and y nonzero integers, show that no solution
> > exists.
> >
> > Proof by contradiction:
> >
> > (x+sqrt(-1)y)(x-sqrt(-1)y) = x^2 + y^2 = 0, so
> >
> > x = sqrt(-1)y *or* x = -sqrt(-1)y.
>
> James, you have to *prove* that last step. You cannot rely on the standard
> proof because it uses: x^2 + y^2 <> 0 whenever x != 0 or y != 0.
>
> > First off, it's worth noting that I'm treating the sqrt(-1) as an
> > *operation*. I think some of you live under the false notion that
> > something like sqrt(2) is the actual number.
>
> I must be dense, but what the heck do you mean? If it is not a number
> you have to define how to do multiplication of operations with numbers.
>
> So, James, how do I multiply a number with an operation, and what do I
> get?
>
Rest assured, Dik, it is not you who are dense.
> > [(v^5 + 1) z^2 -(5v^3 + sqrt(5v^6 - 20v))xy/2] [(v^5 + 1) z^2 -(5v^3 -
> > sqrt(5v^6 - 20v))xy/2] = 0(mod (x+y+vz)/h), when x^5 + y^5 = z^5, and
> > for the simplest case h = (x+y)^{1/5}.
>
> Yup, except that this is necessarily not valid when z is a multiple of 5.
>
> > It might not suprise you that now I go ahead and say that
> > [(v^5 + 1) z^2 -(5v^3 + sqrt(5v^6 - 20v))xy/2] = 0(mod (x+y+vz)/h)
> > *or*
> > [(v^5 + 1) z^2 +(5v^3 + sqrt(5v^6 - 20v))xy/2] = 0(mod (x+y+vz)/h.
>
> Like above, you have to *prove* this step. Off-hand I can say already
> that the step is unproven, because whenever x^5 + y^5 != z^5 this is
> clearly false. (The initial congruence above is not valid when that
> is the case.)
>
> > The equivalent step with easy polynomials would be
> > [x^2 + y^2][x^2 - y^2] = 0(mod (x+y)), so
> > x^2 + y^2 = 0 (mod (x+y))
> > *or*
> > x^2 - y^2 = 0(mod (x+y))
> > (which is actually a bit misleading but I won't go into why at this
> > point to try and lessen the confusion level).
>
> Nope, it is utterly misleading because you do not have a condition in the
> first congruence. A simple equivalent step is:
> [(v^2-1)z + sqrt((v^2-1).2xy)][(v^2-1)z - sqrt((v^2-1).2xy)] =
> 0 (mod (x+y+vz)) when x^2 + y^2 = z^2, so
> (v^2-1)z + sqrt((v^2-1).2xy) = 0 (mod (x+y+vz)
> *or*
> (v^2-1)z + sqrt((v^2-1).2xy) = 0 (mod (x+y+vz)
> Which is, eh, false (note that the first conditional congruence is
> *indeed* true).
>
> > It has to do with that modulus (x+y+vz)/h.
>
> Wrong at the first try. It has to do with the condition in the first
> congruence.
>
> > Can you imagine some factor that can multiply times (x+y+vz)/h to give
> > something like
> > [(v^5 + 1) z^2 -(5v^3 + sqrt(5v^6 - 20v))xy/2] ?
>
> Can you imagine some factor that can multiply times (x+y+vz) to give
> something like [(v^2-1)z + sqrt((v^2-1).2xy)] ?
Alas, Dik, it is possible for Mr. Harris to imagine just about
anything that suits his fancy.
> --
> dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
> home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
--Chuck Giffen
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