The usual objection against a "direct" proof of the FLT inequality :
x^p + y^p <> z^p for integers, using residues is:
--- From equivalence mod p^k (occurring for some primes p>2)
you can never derive inequality for integers,
due to the Hensel lift :
--- a solution mod p^2 can be extended to equivalence mod p^k for all k>2.
There are (p-1)p^{k-1} residues mod p^k coprime to prime p,
forming a cyclic group of 'units' G_k = A_k B_k,
with necessarily
a cyclic 'Core' subgroup A_k of order p-1, where n^p = n (mod p^k).
...... So what about a solution 'in Core': . . . z = x+y mod p^k ?
...... Then (x+y)^p = x+y = x^p + y^p . . . mod p^k,
. . . where " Exponent p Distributes over a Sum" ( EDS-property ), yielding
. . . inequality (x+y)^p > x^p + y^p for the integers x,y < p^k (k>1) and x^p,y^p < p^{kp}
. . . corresponding to the residue equivalence, despite the known Hensel lift.
In fact, it can be shown that ALL solutions of FLT(case1) mod p^2 satisfy this
EDS condition.
Actually the cubic roots of 1 mod p^k are three p-th power
residues summing to 0 mod p^k,
solving FLT mod p^k. . .
As intro, consider the following:
For prime p>2 and x,y,z>0 coprime to p (FLT_case1):
..0.. FLT inequality x^p + y^p <> z^p follows from inequality mod p^2 (least signif 2 digits).
Remains the case of equivalence mod p^2. Then the Hensel lift:
. . . Solution (=) mod p^2 -----> (=) mod p^k (all k>1),
. . . seems to exclude derivation of inequality (<>) for integers.
Except if z = x+y mod p^k (with EDS-property), for instance in case of
--- the cubic roots of 1 mod p^k (p=1 mod 6, k>1), seen as follows:
..1.. There are p^{k-1} multiples of p mod p^k (all k>1).
..2.. So (p-1).p^{k-1} residues coprime to p, forming a group.
..3.. This group of units is cyclic G_k = A_k.B_k with:
. . . . . . cyclic "Core" subgroup A_k of order p-1 (all k>1).
..4.. Then n^p = n mod p^k, for the p-1 Core residues:
. . . . . . extending Fermat's Small Theorem (FST) mod p to mod p^k.
..5.. "Core theorem": . . Each Core-subgroup S>1 sums to 0 mod p^k.
------>(see refs. below; e.g: -1 in Core ---> even |S| yields pairwise 0 sum)
. . . . (only subgroups of extension group B_k have a non-zero sum = p^i, k>i)
..6.. Take: p=1 mod 6 (3 divides p-1) and a^3 = 1 mod p^k,
..7.. then: Core A_k (k>1) has subgroup S={a,a^2,a^3=1}
..8.. with sum(S) = 0 mod p^k. . . . by (5)
..9.. Hence: a + 1/a = -1 mod p^k, with each term in Core,
10.. so: (a + 1/a)^p = a^p + (1/a)^p mod p^k solves FLT mod p^k.
11.. ---> Exponent p Distributes over a Sum ("EDS" property).
12.. So Cubic root solution (9) of FLT mod p^k (k>1) yields:
FLT_case1 inequality for p-th powers of the corresponding
. . . . . integers < p^k (all k>1), despite the Hensel lift, due to (11).
Moreover, they are the only solution "in core" mod p^k for k>=3
. . . . . (but not for k=2: see the "triplets", for some p >= 59).
This short (16 pgs) and direct proof of FLT is published by Univ. Bratislava
dept. of mathematics (Nov.2005), online:
direct FLT proof
and in book "Associative Digital Network Theory"
(Springer, May 2009)
In fact any FLT_case_1 solution of x^p + y^p = z^p mod p^k can be transformed
into one with the two lefthand terms in core A_k, by two transformations
(scale=mpy and shift=add), preserving the assumed integer FLT equality,
having a version of the EDS property that blocks an integer case_1 solution. #######
See --- "On Fermat's marginal note: a suggestion " ---
with abstract on http://piazza.iae.nl/users/benschop/marg-abs.htm ,
which refers to the full text in ref. [5] (.dvi) on my homepage:
. . . . . . http://piazza.iae.nl/users/benschop
For the general triplet rootform of FLT(case1) mod p^k, involving
three inverse-pairs (vs. one inverse-pair for the cubic rootform):
. . . a + 1/b = b + 1/c = c + 1/a = -1 mod p^k, with abc=1 mod p^k,
see ref. [1]: " Triplets and Symmetries of Arithmetic mod p^k "
" On State Machine decomposition I "
" On State Machine decomposition II "
" Example of triplets for p=59 "
--- Relevant comments are appreciated: n.benschop-at-chello.nl
Is FLT (case1) "simple" after all ?
Maybe Fermat knew about the cubic root solution (first at mod 7^2),
but noticed that an FLT_case1 proof requires them to be the only rootform.
.... Which is false: the triplets occur first at mod 59^2 (tough to find
without PC).
--- PS --- :
The "extension" subgroup B_k of units group G_k, consisting
of all p^{k-1} residues 1 mod p, has generator p+1.
I would be interested in a literature reference.
It is hard to believe that this is not known.
. . Thanks.. Nico Benschop, 23-jan-98.
---------------------------------------------------------------------
One more crank: applying New Math (base p) to FLT
------------------------------------------------------sci.math 14sep98
[** Bob S. wrote:
Note: I am not an expert on this subject. However, I have enough of a
background to be able (barely) to follow the explanations of others.
I have seen explanations of what I said above given by Larry
Washington, Karl Rubin, and J.P. Serre. If anyone thinks they are
out to discredit Mr. Benschop then I suggest that said person needs
a reality check.
A characteristic of cranks is that when professionals don't agree with
their "theories", then they start attacking their motives and accusing
them of bias. I am afraid Mr. Benschop has crossed that line.
I have seen Mr. Benschop's arguments. I am unconvinced by them.
(*) Hensel lifting a solution of the Fermat equation from
mod p^n to Q does not and CAN NOT work.
One can not prove FLT from local considerations.
-------------------------------------------------------------**]
NB:
Re(*): It occurred to me that the usual way to disqualify
a wide claim is simply by a counter example.
If I understand (*) correctly, the claim is: there is no way to derive
an inequality for integers from an equivalence mod p^k, for FLTcase1
(sometimes referred to as: "the Hensel lift cannot be broken").
How about this example, of "how to break the Hensel lift":
For p=7 there *is* equivalence x^p + y^p = z^p mod p^2 possible.
In fact the 7-th power residues form a 6 cycle, with an order 3
subcycle that sums to 0 mod 7^2 (trivial, as you should remark;-).
Each 7-pth power n^7 is "in core" mod 7^2, that is: in a 7-1 cycle,
since there are (p-1)p residues mod p^2 coprime to p, forming a cyclic
group as well, as is known.
Namely the 6_cycle generated by 3^7 = 31 mod 7^2, and in base7 code:
(43)* = { 43, 42, 66, 24, 25, 01 } mod 7^2 : the 6_cycle of 7_th
powers (called "core" A_2, where n^p = n mod p^2, in general:
core A_k with n^p=n mod p^k, and |A_k|= p-1 for all k>0).
The cubic roots a^3=01 are then
(base7): {42, 24, 01} sum 66+01=00 mod 7^2.
In the normal form a^p + b^p = -1 mod p^k this becomes (base 7):
(42)^7 + (24)^7 = (66)^7 mod 7^2 (in core: n^p=n mod p^k) ...[#]
This implies inequality for the "corresponding integers", namely
the same residues < p^2 interpreted as positive integers. Because
of the 'Exponent p Distributing over a Sum' (EDS) property for such
residues "in core", which does not hold for integers:
base7: (42)^7 + (24)^7 < (42 + 24)^7 ---- QED. [##]
Does not this example "break the Hensel lift", after all ?
_____________________________________________________________endcopy
[-- 13nov99:
In fact for prime p=6m+1: residue cubic roots a^3 = 1 mod p suffice,
with a^2 = 1/a mod p, so: a + a^2 = np-1, hence a(a+1)+1 = n.p (prime):
(6m+1=p) p.n = a(a+1)+1 a a^2 a^3
m=1 p=7 .1 = 2.3 + 1 2 4 8 = 1 mod 7
2 13 .1 = 3.4 + 1 3 9 27 = 1 mod 13
3 19 .3 = 7.8 + 1 7 49 343 = 1 mod 19
5 31 .1 = 5.6 + 1 5 25 125 = 1 mod 31
6 37 .3 = 10.11+ 1 10 100 1000 = 1 mod 37
7 43 .1 = 6.7 + 1 6 36 216 = 1 mod 43
10 61 .3 = 13.14+ 1 13 169 2197 = 1 mod 61
For p=7: 2^3 = 1 mod 7, so cubic root 2, with inverse 4 (mod 7)
and: 2^7 = 128 (decimal)= 2.7^2 + 4.7 + 2 = 242 (base 7)
4^7 = 128^2 (dec)= (242)^2 (base 7) = 65524 (base 7)
--------- +
66066 (base 7)
while rhs: (p-1)^p = (7-1)^7 = ^^^
= 7^7 -7.7^6 +21.7^5 -35.7^4 +35.7^3 -21.7^2 +7.7 -1
= [3,-5, 5,-3, 1, 0,-1] (base 7)
= [2, 2, 4, 4, 0, 6, 6] = 2244066 (7^2 -1 mod 7^3)
^^^
So: 2^7 + 4^7 = 6^7 mod 7^3, but != mod 7^4.
In general let a^3=1 mod p (a < p): cubic root of 1 mod p (p=1 mod 6),
then a^p + (a^2)^p = (p-1)^p mod p^3, but != mod p^4.
--]
NB:
Just in case you don't believe inequality [##] for some reason;-)
check it out with the skills acquired with new Math: working in
another base different from ten, in this case base 7.
In general (base p) the p-th power of a k-digit number has at most kp
digits. Because the largest k-digit number is p^k -1,
and (p^k -1)^p yields, by the Pascal/Newton binomial expansion:
p^{kp} - p.p^{k(p-1)} ... (+/- smaller terms, down to -1) < p^{kp}.
Working out the 7-th powers in [##] base 7 we get, by longhand MPY:
42 24
42 24
--- x --- x
114 132
231. 51.
---- + ---- +
2424 642
42 24
---- x ---- x
5151 3531
13332. 1614.
------- + ----- +
141501 = (42)^3 23001 = (24)^3
141501 23001
------ x ------ x
141501 23001
1101405.. 102003...
141501... 46002....
626604.... --------- +
141501..... 562046001 = (24)^6 1
------------ + 24 1 1
24031253301 = (42)^6 --------- x 1 2 1
42 3241253004 1 3 3 1
----------- x 1454125002. 1 4 6 4 1
51062536602 ----------- + 1 5 10 10 5 1
132155406504. 21112533024 = (24)^7 1 6 15 20 15 6 1
------------- + 1 7 21 35 35 21 7 1
1402646634642 = (42)^7 |
21112533024 = (24)^7 |
============== + v
1424062500666 (LHS), while RHS = (66)^7 = (7^2 -1)^7 =
| | |
7^10 d5 d0 = 7^14 -7.7^12 = 7^14 - 7^13
+21.7^10 -35.7^8 +3.7^11 -5.7^9
+35.7^6 -21.7^4 +5.7^7 -3.7^5
| | | + 7.7^2 -1 + 7^3 -1
60262046400666 (RHS) <-- = <------------------/.............../
^^^^^
Notice that the integer equality only holds for the 5 lsd, so mod 7^5,
and INequality is due to the msd's (more significant digits: carries)
that are ignored in residue arithmetic (formalized by Gauss, 1800)
Furthermore, the 7-th powers of core residues in A_k yield residues
in core A_{k+1}, which here is A_3 = {643, 642, 666, 024, 025, 001}
cubic roots: |^^ |^^ |^^
again summing to zero: 642 + 024 + 001 = 000.
Could'nt Fermat easily have done such calculations by hand?
(using the New Math method, with base 7 ;-)
And algebraically concluded that for each prime p=6m+1 (so 3 div p-1)
such cubic solution of FLTcase1 mod p^k (at least necessary for
integer equality) *does* yield INequality for the corresponding
integer p-th powers, for *every* k > 1.
(let me know if I made an error in the calculations).
Which leaves the question if there might be still *another* type of
solution (non-cubic), which indeed occurs for the first time at p=59:
the triplets. A bit too much for experimantation, even if he borrowed
Pascal's Calculator (PC), ... or maybe after all? And FLTcase2...
See my homepage - Intro at http://home.ise.nl/users/benschop/scimat98.htm
and:
ref[1] "Triplets and symmetries of arithmetic mod p^k " */nfb0306.pdf
ref[5] "On Fermat's marginal note: a suggestion" */marg-flt.pdf
Ciao, Nico Benschop.
FLT-----------------if stuck@closure, use the carry------------EDS
Subject: Re: FLT
Date: Tue, 10 Jul 2001 20:01:05 GMT
From: Nico Benschop
Org: Digital Research
Newsgrp: sci.math
------------
"W. Dale Hall" wrote:
>
> MAppell917 wrote:
> >
> > > I am an amateur at mathematics. I am curious about Fermat's Last
> > > Theorem, however. Since it has been proved, after about 350 years
> > > with something like 200 pages of mathematics, do you think that
> > > Fermat had a proof as he claimed?
> >
> > Yes, I do think he had a proof
> > and I am in the minority who think he did.
> ... (stuff deleted) ...
> > Mike
>
> Do you have any basis (aside from a vague notion that simply-stated
> things can be proven simply) for this belief? Do you have any expertise
> that suggests that such a proof is possible?
>
> Aside from a certain perennial that sprouts regularly on sci.math and
> spouts all manner of FLT-related jibberish (together with great loads of
> abuse for anyone who cares to question his correctness), and a small
> trickle of once-or-twicers on the same topic, I haven't heard of
> *anyone* with a serious involvement in mathematics who holds to the
> belief that Fermat actually had a proof.
>
> I'm curious and wonder who cares to disabuse us all of the notion that
> Fermat-believers are like Leprechaun-hunters. -- Dale.
I think Fermat, just after discovering his Small Thm (FST: n^p == n mod p)
could have found a clue, possibly the cubic roots of 1 mod p^2 ...[2]
re: an extension of FST to mod p^k for k>1, which he later found not
to be sufficient for a complete proof.
The disbelief in a direct approach, via residues mod p^k, to FLT (case_1)
stems from Hensel's lemma (on p-adics), which implies that any solution
of normalized equivalence: x^p + y^p == -1 mod p^k (any k > 1) ...[1]
is an extension of such solution mod p^2.
This is taken to mean (wrongly;-) that no integer inequality can be
derived from the nec.condition [1], since k can be ANY positive integer,
also for limit k --> inf. ('asymptotic' p-adic solution).
Given any residue solution [1] for some k>1 (a necessary condition for
an integer solution), no p-th power integer solution, with terms < p^{kp}
can be derived from it.
In fact, taking such k-digit solution terms as integers < p^k,
their p-th powers imply necessarily inequivalence mod p^{3k+1},
thus at "triple precision" (for any precision k > 1).
This is a bootstrap type of phenomenon, 'breaking the Hensel lift';-)
The cubic roots of 1 mod p^k (p=1 mod 6) play a critical role here:
[notice: just as Wiles' elliptic curves being of degree three]
a^3 == 1, with a + 1/a = -1 (mod p^k)
and its generalization to the 'triplet' rootform of FLT(case_1) mod p^k:
a + 1/b == b + 1/c == c + 1/a == -1, with abc == 1 (mod p^k).
Which btw is the general structure for ALL units in the unit-group [3]
(not only p-th power residues mod p^k).
-- NB - http://piazza.iae.nl/users/benschop/scimat98.htm
"On Fermat's marginal note: a suggestion". . . . . [2]
"Triplets and Symmetries of the Units group mod p^k ..." [3]