Re: FLTMA: Finitely Generated Abelian Groups


The Dougster wrote:
Tonico wrote:
The Dougster wrote:
The Dougster (I) wrote, in an early-morning fog:
The Dougster (I) wrote:
We had from x^p + y^p = z^p with gcd(x,y,z) = 1, that
(x/y)^ = -1 mod z and I wonder if, since
-( (x/y)^p) = 1 mod z, and p is odd for a counterexample, that
( (-(x/y))^p) = 1 mod z is a contradiction from a countexample. Hm.

I will program and see.

No, ( ( -(x/y) )^p ) == 1 mod z is no contradiction, but I don't think
it follows.
( ( -(x/y) )^p ) = z^q is a contradiction, but it's a stretch. Hm. It's
only true for
p = 2.

No, that's wrong.

( ( -( x * ( y^(-1) mod z)))^p) / z^q = 0 isn't right, either.

Doug
*******************************************************
Man, you seem to be conducting one hell of a thread convo with
yourself: poor you! You must be lonely as hell. Here, I will
participate in your thread so that you won't feel so lonely. You're
welcome!
Regards
Tonio

Thanks for joining me, Tonio!

It is only my little thread here in sci.math that is lonely as hell.

I now am wondering if we compute k = the multiplicative inverse of y
modulus z, that

(-(xk))^p = qz + 1, making this statement true for all even p

for 2, the only even prime

and false
for all odd p, with the restriction that q is positive, but
conventionally q can be negative.

Let's see. With x = 3, y = 4, z = 5, inverse y mod z is 4. Then (-12)^p
= qz - 1 for p = 2 and q = 29, but I don't see a solution for qz + 1.
Clerical error?

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

Doug

.

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