Re: FLTMA: A little group theory


Chip Eastham wrote:
The Dougster wrote:
Chip Eastham wrote:

(much snipped)

Generally the elementary approach to FLT fixes an
exponent n > 2 and tries to prove that no solutions
x,y,z (xyz > 0) exist. In your case, the approach is
more a matter of fixing x,y,z > 0 such that:

x < y < z < x+y, (x,y,z) = 1, and xyz is even

and showing that no odd exponent n satisfies:

z^n = y^n mod x
z^n = x^n mod y
x^n = -y^n mod z


Er, no odd prime exponent, right? Otherwise, yes, a proof by
contradiction.

I thought I remembered your original claim as being
that n couldn't be odd, which of course is more than
enough to establish FLT. For that no odd prime n
would suffice.

No, I've seen n = 15 and other composite odds. It seems to fall out as
either 2 or composite, but there may be a pattern I *don't* see, not
yet. The point of my argument was that n was either degenerate (=2) or
composite. Composite n proves FLT, if it is in every case. I will try
to develop that argument after working the exercise.


From the above we can deduce some of the
other results you want:

x > 1

And x < y < z < z+y by Mahanobolis.

x,y,z are pairwise coprime

Thus the set of exponents n for which:

z^n = y^n mod x

will form a cyclic subgroup of Z/phi(x)Z,
because the above can be rewritten:

(z/y)^n = 1 mod x

In what algebra, no, in what arithmetic, is a fraction composed of
coprime numbers to any power equal to an integer? I don't get this
part, although it seems to follow.

The point is that since y is coprime to x, we
have y is invertible in Z/xZ. In fact by the
usual iteration of the Euclidean algorithm,
one gets integers a,b such that:

ax + by = 1

Oh! y is indeed invertible, and this is a multiplicative group. OK.

Er, is that Euclid's algorithm for GCD?

You draw x by y and mark off x^2, leaving y-x by x, then repeat....like
working with the golden ratio, only y/x isn't golden, so it resolves to
a square representing GCD(x,y)^2?

Or something else? It must be something else.


Hence by = 1 mod x, or in other words
b = 1/y mod and b is the multiplicative
inverse of y in Z/xZ. In this ring dividing
by y iis equivalent to multiplying by b.

Got it.

b is also coprime to x, ie. b in Z/xZ*
and thus z/y = bz mod x is in Z/xZ*.

OK. But that is something we didn't learn. If (x,y,z) = 1, then (x,
y^(-1) mod z) = 1 also? I can use that. Give me a few days.

Thus the exponents n which satisfy
this are simply the integer multiples
of the multiplicative order of (z/y) in
Z/xZ*. We know this order divides
phi(x), and we can rule out this order
being 1 (because 0 < z - y < x, so
z/y is not already 1 mod x).

I'd write z*(y^-1) but that's just me.
I follow.

Yeppers on that. But how? We haven't
learned what the multiplicative order of
z/y is in Z/xZ*.

Once you have z/y = bz mod x, the order
of this element in the multiplicative group
Z/xZ* is defined in the usual way: the
least positive integer exponent k so that:

(z/y)^k = 1 mod x

Sure, <z/y> mod x = k | (z/y)^k == 1 mod x & 0 < j < k ==> (z.y)^j =/=
1 mod x

Consider x = 7, y = 10, z = 13
and determine the exponents
which respectively satisfy each
constraint:

(z/y)^n = 1 mod x
(z/x)^n = 1 mod y
(x/y)^n = -1 mod z
Also
(y/x)^n = -1 mod z, right?

(7, 10, 13):

10^(-1) mod 7 = ?
7^(-1) mod 13 = ?
10^(-1) mod 13 = ?
7^(-1) mod 13 = ?

I guess I need to write the tables. More tomorrow.


and show that collectively the
exponents which satisfy all
three constraints are the odd
multiples of 6.

This is one of my favorite examples and I will reply soon.

I am at my Pookie's house now, enjoying XM radio, a (for me) rare
Molson, and have been off tobacco completely for some 24 hours now,
after starting my quit 2 October. Breaking the neck of the monkey on my
back, one slow twist at a time....

Rootin' for ya.

29 hours! Then a slip... :( no big deal, keep at it....

Doug

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