Re: FLTMA: A little group theory

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.

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.

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*.

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....

Doug

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