Re: FLTMA: A little group theory


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.

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

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.

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

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).

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

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

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.

--c

.

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