Re: Fermat's Last Secret

Am 19.04.2006 13:40 schrieb quasi:
<big snip>

If we're stepping back to conjecturing, let me
repeat (and put it together in a different way)
what I have posted few times ago, -although, at
the end, I don't have the true connection to the
OP's idea...

-------------------------------

A pair of coprime x and y, raised to a power
of n behave similar to a cyclotomic function
resp. their primefactorization.
That means, (the primefactors separated into
3 sets q,r,s, whose characteristics I'll explain later):

f(n,x,y)
= x^n - y^n
= product q^k * product r^j * product s^m

The occurences of the sets of primefactors p,q,and r
are periodic with n, referring f()'s cyclic
group order based on x and y (mod primefactor q),
say "ord(q,x,y)".
I omit the "x,y"-notation, since the arguments in the
following focus the primefactors of a fixed pair of (x,y),
so let's write it simply "o(q)" to denote the
length of cyclic subgroup of "f(n,x,y) mod q"

The group orders of all primefactors are also a
divisor of phi(primefactor) = primefactor-1

So the above equation should be written as

f(n,x,y)
= x^n - y^n

= product q^k * product r^j * product s^m
o(q)=1 o(r)>1
o(r)|n

( where also all q,r,s are primes)

Now, in the FLT, n is an odd prime, say "p" for
readability, and o(q), o(r) and o(s) must divide p to allow
q,r or s being a primefactor of f(p,x,y). Thus o()
must be 1 or p itself to be a divisor of p.
Additionally the third set of s^m denotes now only the
primefactor p itself (if it occurs already in x - y):

f(p,x,y) x^p - y^p
= x^p - y^p = (x - y) * ----------
x - y

= product q^k * product r^j * product p^m
o(q)=1 o(r)=p o(p) = 1

"product r" refers to all primefactors, where
f(p,x,y) has group order p (mod r),

"product q" refers to all primefactors, where
f(p,x,y) has group order 1 (mod q), this means:
here all primefactors are collected, which already
occur in f(1,x,y) = x^1 -y^1

"product p" refers to the primefactor p itself,
if it occurs in f(1,x,y)= x - y already and
thus f(p,x,y) has order o(p,x,y)=1

---------------------


Now to have this a FLT-conformal equation, all
exponents on the rhs (k,j,m) must be the same
and must all equal p.

That this is an extreme hard condition becomes obvious,
if one notes, that with cyclotomic functions in most
cases the exponent at the first occurence of a
primefactor with order o(q,x,y)>1 is simply
1; the exceptions may be collected under the term
"generalized wieferich primes" (to a certain cyclo-
tomic function); for f(n,2,1) for instance we only
know two such exceptions where r = 1093 and r=3511
and their exponent at their first occurence (with
minimal n) being only 2. For f(n,3,1) and r=11 we
also have the exponent 11^2 for f(5,3,1) instead of
11^1. Let's call the value which exceeds the expected
exponent=1 the "wieferich excess", so that the
"wieferich excess" of 11 for f(n,3,1) "is" 1.

The exponent of a primefactor q grows also cyclically
with n, but not with the period of o(q,x,y) but just
with period of q itself.
That suggests a formula for the exponent of an
arbitrary primefactor:

to allow, for an integer n (prime or composite), that
q^k | f(n,x,y)

the following conditions must be met:
o(q,x,y) | n
q^(k-1)*o(q,x,y) | n

I put that into a formula with two new operands:
n
k = ~~~ ( a + w + {n,q})
o(q)

which means:

n
~~~ = 1 if o(q) divides n, else = 0
o(q)

a is the power at q's first occurence (with minimal n)
except the wieferich-excess w
w is usually 0, and in the rare wieferich cases it can
be greater

{n,q} indicates the exponent to which q is a factor of n,
such that n = x* q^{n,q} and x doesn't contain q

------------------------------------------------------------

So for the non-wieferich primefactors q and the FLT-case,
where n is a prime p we have for the exponents k of any
primefactor q

p
k = ~~~~ ( a + w + {p,q})
o(q)

which reduces immediately for all q<>p, since q never divides
p and the {}-term vanishes:

p
k = ~~~~ ( a + w )
o(q)

and this means for all primefactors q, for which
f(n,x,y) has order p (mod q)

k = 1 + w for q<>p with o(q,x,y)=p
-------------------------------------------------

Now let's apply that considerations to the FLT-
primefactorization:

f(p,x,y) x^p - y^p
= x^p - y^p = (x - y) * ----------
x - y

= product q^k * product r^j * product p^m
o(q)=1 o(r)=p o(p) = 1

First let's consider "product q^k"

THis is the set of primefactors q, where f(p,x,y) has
order o(q)=1, that means, q is already factor of x - y.
(but note that q<>p, the primefactor p is dealt separately)
The exponent k is then
k= a + w; w is here 0 so "a" is simply the power of
q, to which q occurs already in x - y




Second, let's consider "product r^j"
The exponent j is
j = a + w = 1 + w
and only for generalized wieferich-primefactors (for
f(n,x,y)) w can be greater than zero.




Third, "product p^m" contains simply the prime-factor p
itself, and only, if it occurs already in x - y, such that

m = a + {p,p} = a + 1




---- FLT-condition -------------------------

So, all together, we have

f(p,x,y) x^p - y^p
= x^p - y^p = (x - y) * ----------
x - y

= product q^a * product r^(1+w) * product p^(a+1)
o(q)=1 o(r)=p o(p) = 1

and all exponents must equal p.
----------------------------------------------


That constitutes extremely hard conditions:

1)
in "product p^(a+1)"
a+1 = p
so x - y must
1a) either be non-divisible by p
1b) or x - y must contain p to the power of p-1

2)
in "product r^(1+w)"
f(p,x,y) must contain one (or more) wieferich-primefactors
with an wieferich-excess of exactly p-1

3)
in "Product q^a"
all primefactors q of (x-y) (except p itself, which is dealt
in 1)) must occur to the power of p already.

-------------------------------------------

Thus we have, that f(p,x,y) must have the following structure

f(p,x,y)
x^p - y^p
= x^p - y^p = (x - y)* ---------
x - y

4.1)

(x - y) = product q^p (4.1.a)
o(q)=1
or
= p^(p-1) * product q^p (4.1.b)
o(q)=1

4.2)

x^p - y^p
--------- = product r^(1+w) = product r^p (4.2.a)
x - y o(r) = p

or
= p * product r^(1+w) = p * product r^p (4.2.b)
o(r) = p

The term 4.1.a) means, that x-y must already be a perfect p'th power
4.1.b) allows exactly one additional cofactor of p^(p-1)

The term 4.2.a) means, that the factorization of the cyclotomic
part needs only primefactors, all having a wieferich excess,
and that all that primefactors must also have the same excess of
the value of p-1; the term in 4.2.b) allows one additional cofactor
of p in that term.

Wieferich-excesses (in my generalization here) are very rare
and usually small - although a functional relation to x and y
in f(n,x,y) is not known. The hard condition is, that the order
of f(p,x,y) (mod r) must also equal p-1, and perhaps the
most restrictive condition is, that looking at
x^p - y^p = z^p and x^p - z^p = y^p
simultaneoulsy, they must occur mutually to the same excess
in f(p,x,y) and f(p,x,z).

--------------------


Now let's look at the original problem of the prime-factorization of

g(p,x,y) = x^p + y^p

The things can be checked the same way, only that the group-
orders o(r) have to be dealt differently, for instance in 4.2)
o(r,x,y) must equal 2p.
However, this selects only different prime-factors, but doesn't
affect the consideration of the exponents, and only this is the
crucial point in the shown approach to the FLT-problem.

--------------------

I didn't look at the 2^k-exponent-idea of bassam king karzeddin
yet; if this is of importance, it should popup at any place of
my above arguments...

I'll look at it later.

Regards -

Gottfried

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