Re: FLTMA: A little group theory

Hm.

From x^p + y^p = z^p, p an odd prime, we have

G sub z = { w | gcd(w,z) = 1}
(x/y)^p == -1 mod z
(x/y)^(kp) == (-1)^(kp) mod z
(x/y)^(2p) == 1 mod z
2 divides |<x/y> in Z sub z|

and we want to show p is not an odd prime, contradicting p = an odd
prime from the primitive counterexample, to show there are no candidate
counterexamples and so no primitive counterexamples or counterexamples
to FLT.

It will suffice to show p = qr ,the product of primes q and r, or p =
2. We can call these cases case 1 and case 2

We have
(z/x)^p == 1 mod y
(z/y)^p == 1 mod x

and I wonder about

G sub xy { w | gcd(w,xy) = 1}, the group of number relatively prime to
xy.

gcd(x,y) ==> phi(xy) = phi(x)phi(y).

I think it's true that
(zx)^p == 1 mod xy and
(zy)^p == 1 mod xy.

I am not sure.

Doug

.

Relevant Pages

  • Re: FLTMA: Using FLT as a lemma
    ... The Dougster wrote: ... then there would be a primitive counterexample, ... p an odd prime. ... Now, since FLT is true, we know the zeros of this sequence will never ...
    (sci.math)
  • Re: FLTMA: A little group theory
    ... divides |in Z sub z| ... and we want to show p is not an odd prime, contradicting p = an odd ... prime from the primitive counterexample, to show there are no candidate ...
    (sci.math)
  • Re: FLTMA: Using FLT as a lemma
    ... If n below is never odd prime, it is always comppsite or 2. ... then there would be a primitive counterexample, ... Now, since FLT is true, we know the zeros of this sequence will never ...
    (sci.math.research)