Re: FLTMA: A little group theory

Hello, all. Hello, Chip.

Chip Eastham wrote:
Doug Goncz wrote:
It seems rather that you should be saying:

(x^p + y^p) = 0 mod z, and
(z^p - x^p) = 0 mod y, and
(z^p - y^p) = 0 mod x

But if we look at sequences with (x,y,z) = 1, one of {x,y,z} even, and
x<y<z<(x+y)

(x^n + y^n) mod z + (z^n - x^n) mod y + (z^n - y^n) mod x or
(x^n + y^n) mod z * (z^n - x^n) mod y * (z^n - y^n) mod x

by hand waving arguments (call them a lemma), we find that n is always
2 or composite, never an odd prime.
So FLT is proved.

I think I'd spend a bit of time trying to search out
a counterexample. You're conjecturing something
stronger than FLT, which implies (though you may
not have realized it) that the above can only occur
if n is a power of 2.

Yes,

(x^n + y^n) == 0 mod z &
(z^n - x^n) == 0 mod y &
(z^n - y^n) == 0 mod x

is the more standard way to say this.

Now, why or how would this imply that this can only occur if n = 2^m?
Group theory?

I have solutions with n = 2 and n composite but never n odd prime.

Yes, I realize this is stronger than FLT.
You'd agree if one could prove this stronger statement, then FLT would
follow, right?
And that sometimes it is easier to prove the stronger of two
statements, yes?

I wonder if there is another way to say "one of {x,y.z} even" or
another similar conclusion that can be drawn.

Rewriting for easier discussion:

From
(0) a^n + b^n = c^n in Z,
we have, in Z+,
(1) p prime
(2) gcd(x,y,z) = 1 or (x,y) = (y,z) = (x,z) = 1, a relation I still
misunderstand....
(3) x^p + y^p = z^p
(4) one of {x,y,z} even
(5) x<y<z<(x+y)
(6) (x^p + y^p) == 0 mod z
(7) (z^p - x^p) == 0 mod y
(8) (z^p - y^p) == 0 mod x

I note that in my tests, leaving out (2), (3), (4), or (5) results in
violations of (1), which does not prove that (2), (3), (4), and (5)
together imply (1), just as (6), (7), and (8) do not imply (0), but are
only necessary for (0).

I have "Finite Group Behavior 3.0" software to play with. :)

We are talking here about all cyclic multiplicative groups of integers
with order phi(n), n in Z+, aren't we?

With x, y, and z pairwise coprime, is
phi(z) / gcd( phi(z), x ) ever equal to
phi(z) / gcd( phi(z), y )? ( I can start some checks )

Mantra for today:

"If I had wanted to smoke outside my home, I'd have brought my
cigarettes.
Nicotrol will just have to do until I get back."

Maybe cold turkey just doesn't work for me and I can quit by stages. ;)


Doug

.

Relevant Pages

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