
Re: FLT Discussion: Simplifying
Posted:
Jan 16, 2001 11:09 AM


"Dik T. Winter" wrote: > > In article <9400ps$b5g$1@nnrp1.deja.com> jstevh@mydeja.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)(xsqrt(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. Offhand 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^21)z + sqrt((v^21).2xy)][(v^21)z  sqrt((v^21).2xy)] = > 0 (mod (x+y+vz)) when x^2 + y^2 = z^2, so > (v^21)z + sqrt((v^21).2xy) = 0 (mod (x+y+vz) > *or* > (v^21)z + sqrt((v^21).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^21)z + sqrt((v^21).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

