To: NMBRTHRY@LISTSERV.NODAK.EDU

Dear moderator,

A week ago (6mar99) I submitted a posting, which uptill now has not
been passed, if I'm not mistaken. I would appreciate it to learn
why it was not posted yet: I'm anxious to have a discussion on the
proposed simple & direct FLT approach. You may notice, in the given
refs, that this material has already been presented at three major
conferences (Prague-96, Grenoble-96, Enschede-98), but somehow since
early 1995 got hardly any other attention from professionals.

Thank you for your time.

PS:  I attach the full text again: (flt-eds.txt, with corrected URL's)
-- Ciao, Nico Benschop.

1. =================================================== My posting after all:

Hi,

Although not a number theorist, I got in my (industrial) research
some experience on finite semigroups, as closures of Finite State
Machines (http://www.iae.nl/users/benschop/c-ranksm.dvi).
  While looking for a better way of  hardware design of binary
multipliers (the usual parallel array multipliers are easily seen
to be  *much* too powerful for their purpose), I discovered a special
property of the cubic roots of Unity: a^3 = 1 mod p^k (prime p=1 mod 6,
k>0. Namely: for k>1 they solve FLT mod p^k non-trivially, as Fermat's
Small Thm (FST) does (n^p = n mod p), so: (x+y)^p= x+y= x^p+y^p mod p^k
for k>0, with a+1 = -1/a.

For any solution in "core", which is the p-1 cycle subgroup of
the units group G_k coprime to p in semigroup(ring) Z(.) mod p^k, we
have: n^p = n md p^k, and (x+y)^p = x+y = x^p + y^p (mod p^k) any k>0,
which has interesting consequencies, re: a direct approach to FLT.

For any k such p-1 cycle exists, because there are p^{k-1} multiples
of p, hence |G_k| = p^k - p^{k-1} = (p-1)p^{k-1}, so G_k = A_k.B_k:
a direct product of core A_k of order p-1, and extension group B_k of
order p^{k-1}, having generator p+1.

It turned out that the cubic roots are a special case of the general
   "triplet" rootform solving x^p + y^p = z^p mod p^k (k>1):

  a+1 = -1/b,  b+1 = -1/c, c+1= -1/a, with abc = 1 (all mod p^k)  [1]

Here a,b,c are three residues mod p^k, not necessarily p-th powers
In fact [1] holds for all units (that is: in G_k): any residue is in
some triplet.
For p-th powers, the first occurrance of a triplet is at p=59 (in fact
two triplets), and they occur for both prime types +/- 1 mod 6. (see
table 2 in my paper */nfb0.dvi, for the 5 primes < 200 with triplets).

I've written this up late 1994, and presented it at several conferences
in Europe (on Semigroups in Prague'96, Logic Synthesis in Grenoble'96,
NL_Math_Congress in U-Twente'98),
  see "Triplets and symmetries of Arithmetic mod p^k"
     http://www.iae.nl/users/benschop/nfb0.dvi (full paper, 14 pgs)
     http://www.ams.org/preprints/11/199711/0index.html (abstract)

However it seems almost impossible to get it published in a math-journal
(ca. 15 submissions, without any useful feedback from referees/editors).
   Moreover, the same non-response occurred on sci.math, despite many
postings since oct95, trying to draw attention to this approach by
additive analysis of mult've semigroup Z(.) mod p^k  (with a special
way to break the Hensel lift ;-)

I would appreciate comments on the proof of the triplet structure of
the units group of residues mod p^k, and the corollary leading to a
 directproof of FLTcase1 (case2 is also using FST in a straightforward
way), *despite* the slightly inter- disciplinary approach (finite
semigroup structure applied to arithmetic) it is quite elementary.

As intro I include the following recent posting:

----- From Fermat's Small Thm to his Large Thm ;-)  --------

His marginal note (around 1640?) in Bachet's translation of
              Diophantus' "Arithmetica":
>
> "Cubum autem in duos cubos, aut quadratoquadratum in duos
> quadratoquadratos, et generaliter nullam in infinitum ultra
> quadratum potestatem in dous ejusdem nominis fas est dividere:
> cujus rei demonstrationem mirabilem sane detexi.
> Hanc marginis exiguitas non caparet." -- Pierre de Fermat
>
> He had a jolly big margin.  -- Richard Carr.
---
( NB: Spacy enough for a modern version of the proof, based on
      an extension of his just discovered Small Thm... see below;-)
---
  One way of looking at Fermat's Large (cq Last) Thm inequality is
  via his own Small Thm: n^p = n mod p (prime p, any n) he just
  discovered (ca. 1637). It is easily expanded to mod p^k (k>1). The
  structure of Z(.) mod p is a p-1 cycle (containing 1) adjoined to 0.

  In Z(.) mod p^k there are p^{k-1} residues with divisor p,
  so p^k - p^{k-1} = (p-1)p^{k-1} residues coprime to p ("units"),
  which form a closure (group) which is cyclic for any k>0, which
  he may have guessed - if not proven.

And factor p-1 guarenteees a cycle ("core" subgroup) that extends to
n^p=n mod p^k (any k>1) for p-1 special "core" residues. He may not
have been able to formally prove these cyclic structures (hence did
not come out with them), but he had a fabulous intuition, and probably
had some clue how to go about proving FLT, at least for case1 (x,y,z
coprime to prime exponent p) occurring in this cyclic structure.

Surely he had not available the group-structure insights that came
later (>150 yrs) with Lagrange, Galois, Abel, &c. But his Small Thm
clearly showed the way to look at multiplication in a "closure", as
later (1800) developed formally by Gauss in his residue arithmetic.

Computing with k digits base p is possibly (probably?) what he did.
And why then not explore 2-digit arithmetic, for instance for base
p=3, 5, 7, just to see what happens?
Moreover the p-th power residues n^p mod p^2 form a p-1 cycle,
which is the core of units, with 'fixed-point' property
n^p=n mod p^2 for the terms of *any* solution of FLT mod p^2.

Mod 3^2 and mod 5^2, with 2- and 4- cyclic core, no solutions
exist of

(1)...  x^p + 1 = z^p mod p^2,

normed to one term=1, which is always possible since in a group
each element has an inverse - certainly in a cycle;-)
This proves the FLTcase1 inequality for p=3 and p=5 quite simply.

But (1) mod 7^2 *does* have a solution, in fact two, corresponding
to the two cubic roots <>1 of 1 mod 7^2.
The 6-cycle F_2 of 2-digit 7-th powers is generated by 43 (base 7):

   F_2 = {43, 42, 66, 24, 25, 01}  with a^3=01 for: {42, 24, 01}

The p-1 cycle mod p^k ("core" of units group G_k) exists for all k>0.
And it is obvious that each solution "in core", where n^p=n mod p^k,

satisfies: (x+y)^p = x+y = x^p + y^p  mod p^k.

     So the Exponent p Distributes over a Sum ("EDS")
     which for the corresponding integers X, Y, Z=X+Y < p^k
yields inequality, since (X+Y)^p > X^p + Y^p, with terms < p^{pk}.

The Cubic roots of Unity (see the equilateral yellow triangle in
    a blue circle, with Fermat's portrait in it, on the cover
    of Simon Singh's book on "Fermat's Last Theorem" ;-)
must have been fascinating for him, in those times where religion
played still such dominant role. (It is said by historians that
e.g. Newton wrote more on alchemy than on any other subject -
including what we now call 'science/math', see my homepg ref [e]).

Even IF he had a (partial FLTcase1) proof, based on the cubic roots
of 1 mod p^k (all p=1 mod 6, any k>1) then he might have been
reluctant to come out with it (re: Copernicus much earlier)
-- because of its "obvious" Holy Trinity associations...

Moreover, he may not have seen how to show solutions mod p^2
are _always_ "in core" -- which indeed they are, although not
necessarily mod p^k for k>2 (see "triplets" in homepg paper
ref[1]:      http://www.iae.nl/users/benschop/nfb0.dvi

With modern notation & concepts of residue arithmetic and
(commutative cyclic) groups in Z(.) mod p^k and its additive
structure of p-th power residues as "core-"subgroup -- a proof
of FLTcase1 could possibly fit in that margin of his! Maybe even
including case2, which _also_ is based on FST (thm4.2, ref[1]).

Plenty of "probable" possibilities re his marginal note, see:

  "On Fermat's marginal note: a suggestion"
       http://arXiv.org/abs/math.HO/0103051
       http://www.iae.nl/users/benschop/marg-flt.dvi

intro's:
         http://www.iae.nl/users/benschop/scimat98.htm
         http://www.iae.nl/users/benschop/campaign.htm
         http://www.iae.nl/users/benschop/func.htm
         http://www.iae.nl/users.benschop/ferm.htm

Relevant (historical & mathematical) comments are welcome.

Ciao, Nico Benschop  -- http://www.iae.nl/users/benschop

2. ================================================================ reply:

Date:         Mon, 15 Mar 1999 10:59:46 -0500
Sender:       Number Theory List 
From:         Alf van der Poorten 
Subject:      Re: The triplet structure of  Z(.) mod p^k

The writer complains that "it seems almost impossible to get it
published in a math-journal (ca. 15 submissions, without any useful
feedback from referees/editors).
Moreover, the same non-response occurred on sci.math, despite many
postings since oct95, ... ".

The writer plainly defines a response not to be useful (or not to
exist) if it advises him that he makes no contribution to our
understanding of Fermat's Last Theorem. He's right, I guess, in
thinking the response is not useful to him if he fails to understand
it. But he's wrong in insisting that there is some obligation on the
world to explain his blunders to him in a manner that he will find
convincing.

Shorn of its irrelevant remarks, he alleges, in effect, that
$a^p+b^b=c^p$ and $p$ does not divide $abc$ in the $p$-adics entails
that $c=a+b$ (note that these are $p$-adic integers). But, he now
alleges, viewed just, modulo $p^k$ --- so choosing residues $a$, $b$,
$c$ less than $p^k$ --- the conditions  $c=a+b$ and $c^p=a^p+b^p$ are
incompatible, hence FLT Case 1. Of course those conditiuons are not
incompatible modulo $p^k$ and, whatever, the writer cannot be showing
more than that.

Last time he wrote a multi-page letter to me and I replied: `I've
read your letter. You observe [in effect] that there are three
$p$-adic cube roots of unity in the ring of $p$-adic integers when
$p\equiv1\mod6$ and these roots of course satisfy $a^p+b^p+c^p=0$ (a
fortiori their truncations $\mod p^k$ do so $\mod p^k$. This
observation does not have much content, it has no originality, it has
nothing much to do with semi-groups, and it does not inform usefully
on Fermat's Last Theorem.'

I then got a multi-page reply which, of course, I disregarded. -- Alf v.d.P

3.  My reply (to nmbrthry.list, not posted by moderator, and to A.v.d.P)

From:  Nico Benschop ,  Sun, 21 Mar 1999 20:03:30 +0100
  To:  NMBRTHRY@LISTSERV.NODAK.EDU
  CC:  Alf van der Poorten 
--

Alf van der Poorten wrote:
>
> The writer complains that `it seems almost impossible to get it
> published in a math-journal (ca. 15 submissions, without any useful
> feedback from referees/editors).  Moreover, the same non-response
> occurred on sci.math, despite many postings since oct95, ... '.
>
> The writer plainly defines a response not to be useful (or not
> to exist) if it advises him that he makes no contribution to our
> understanding of Fermat's Last Theorem. He's right, I guess,
> in  thinking the response is not useful to him if he fails to
> understand it.

Not much to understand: no referee report(s) at all ;-)
With the remark from the editor, that the material is not
appropriate for his journal...

> But he's wrong in insisting that there is some obligation on the
> world to explain his blunders to him in a manner that he will find
> convincing.

What is the blunder about noticing that integer arithmetic can be
decomposed in two components (with modulus m) n = c.m + r : residue r
and carry c (multiple of the modulus). And observing that ignoring the
carry, popular because of nice closure properties (formalized by Gauss
around 1800), leaves residue analysis as much less than half the story...

And: noticing the connection between the residue structure in Fermat's
Small theorem (n^p=n mod p, for prime p, all n) as a p-1 cycle coprime
to zero, and his Large Thm FLT: x^p + y^p <> z^p for p>2, with case1:
x,y,z coprime to p (prime p>2 suffices). Namely: extend FST mod p to
FST* mod p^2, and observe that there are p-1   p-th power residues in
precisely the same structure as his FST, but now for n^p mod p^2,
sothat again (n^p)^p = n^p mod p^2 "in core" (p-1 cycle mod p^2).

Any FLTcase1 integer solution of p-th powers solves the FLT eqn mod p^2,
however with the Exponent p Distributing over a Sum ("EDS"),
     since "in core": (x+y)^p = x+y = x^p + y^p  mod p^2.
Which implies for the corresponding 2-digit (base p) integers X,Y the
inequality of their p-th powers: (X+Y)^p > X^p + Y^p ...... < p^{2p}.
No extension X+a.p^2 and Y+b.p^2 can undo this inequality mod p^{2p}.

Frankly, no-one (including Alf) has shown this conclusion of inequality
for integer "solutions" to FLTcase1, to be invalid. And then there are
the much more interesting solutions: 'triplets' -- not only for p-th
power residues, but for ALL units (corpime to p). No reply on that either.

Frankly, whenever I received a reply like Alf's from someone familar
with number theory (and p-adics;-) I wondered wether the fact that
the p-th power of a k-digit integer (base p) has upto kp digits was
suppressed during their years of study (;-)

For digital hardware designers the concept of "carry" is a very
dominant part of life, and any ignorance about it is immediately
punished severly by dayly practice... So how come that the carry
plays no role (and in fact is 'strongly' ignored, as it were) in
the theory of numbers & arithmetic ?

Let me redo the simple example for p=7, from my previous posting, of
how to break the Hensel lift, in case there *is* a solution FLT mod p^2.
Core cycle F_2 = 43* = {43, 42, 66, 24, 25, 01} of 7-1=6 7-th power
residues has a 3-cycle subgroup {42, 24, 01} of cubes mod 7^2,
which obviously sum to zero: 42 + 24 = -1 mod 7^2,
where 42^7=42, 24^7=24 and (-1)^7 = -1, forming FLT mod p^2 solution:
              42^7 + 24^7 = 66^7 (mod 7^2)

The corresponding integer 7-th powers in 14-digit yield:

 1402646634642   = (42)^7
   21112533024   = (24)^7
============== +
 1424062500666              [LHS]

60262046400666   = (66) ^7  [RHS],  with LHS < RHS, and
^^^^^^^^^\
          \ inequality due to the carry(s).
               [preserved mod p^{2p} for extensions X+a.p^2, Y+b.p^2
               [see http://www.iae.nl/users/benschop/ferm.htm
               [at bottom, ref: "New Math (base p) applied to FLT"]

There is nothing like a counter example, is there? -- against the
mistaken & very popular opinion: "the Hensel lift cannot be broken".

> Shorn of its irrelevant remarks, he alleges, in effect, that
> $a^p+b^b=c^p$ and $p$ does not divide $abc$ in the $p$-adics entails
> that $c=a+b$ (note that these are $p$-adic integers). But, he now
> alleges, viewed just, modulo $p^k$ --- so choosing residues $a$, $b$,
> $c$ less than $p^k$ --- the conditions  $c=a+b$ and $c^p=a^p+b^p$ are
> incompatible, hence FLT Case 1. Of course those conditiuons are not
> incompatible modulo $p^k$ and, whatever, the writer cannot be showing
> more than that.

Well, see above ;-)

> Last time he wrote a multi-page letter to me and I replied: `I've
> read your letter. You observe [in effect] that there are three
> $p$-adic cube roots of unity in the ring of $p$-adic integers when
> $p\equiv1\mod6$ and these roots of course satisfy $a^p+b^p+c^p=0$ (a
> fortiori their truncations $\mod p^k$ do so $\mod p^k$.
>   This observation does not have much content, it has no originality,
>   it has nothing much to do with semi-groups, and it does not inform
>   usefully on Fermat's Last Theorem.'
>
> I then got a multi-page reply which, of course, I disregarded.
                              (as you wil this reply, no doubt?)

Clearly, what can an engineer with that silly 'carry' idea
  possibly contribute to our insight in FLT ?-)

[...]
Let me include here, for him or anyone who *is* interested in my answer,
yet does not feel the need to read the full text, a six point sketch of
the argument of how to excape the Hensel lift (that is: for FLTcase1),
as part of the intro of that paper
     http://www.iae.nl/users/benschop/nfb0.dvi
      " On the triplet structure of Z(.) mod p^k "

  "In FLT: the Carry makes the difference"
   ---------------------------------------
Positive integer n = c.m + r
       ( modulus m, carry c, residue  r=0.. m-1 ) --> n = r mod m.
        p-th power:  n^p = (cm+r)^p = r^p  mod m.
        If n has k digits (base p), then n^p has upto kp digits.

FLT:     x^p+y^p = z^p ...[1]
     has no solution in integers x,y,z > 0; prime exponent p>2.
      " The sum of two p-th powers is not a p-th power" (anti-closure)
     Case1: x,y,z coprime to p,     Case2: p divides one of x,y,z.

Clues/reminders Case1:

1.  Represent integers over base p, and check first the
    necessary condition [1] mod p^k (k digit arithmetic base p)

2.  There are p^{k-1} residues mod p^k with divisor p (end on 0), so
    (p-1)p^{k-1} residues coprime to p, forming a cyclic group G_k.

3.  The p-th power residues form a subcycle F_k={n^p} of order
    |G_k|/p = (p-1)p^{k-2}, and for k=2 we have |F_2|= p-1, which all
    satisfy n^p=n mod p^2 (n in "core" F_2). Compare FST: n^p=n mod p.

4.  Any solution of [1] mod p^k in F_k also satisfies [1] mod p^2,
    hence is "in core" F_2.
        So that  (x+y)^p = x+y = x^p + y^p mod p^2  ......[2]
  So:  Exponent p Distributes over a Sum (EDS property), for residues.

5.  Any integer solution of [1] is in core F_2 mod p^2.
    For the corresponding 2-digit integers X,Y:
        (X+Y)^p > X^p + Y^p, with all p-th powers < p^{2p}.

6.  Any msd-extension by one or more digits: X'=X+a.p^2, Y'=Y+b.p^2
    clearly preserves the inequality mod p^{2p} if p>2
    (trivial core at p=2 ;-)

Note the clue: the p-th power of a k-digit number has upto kp digits.

I would be pleased to learn why for instance the cubic roots a^3 = 1
mod p^2 (prime p=1 mod 6), which solve a^p + 1 = (a+1)^p mod p^2 for
a != 1 mod p^2, should _not_ imply inequality for the corresponding
integers (base p) with p-th powers < p^{2p}.  Note that these are not
   the only possible solutions: also "triplets" (as described in the
   posting & in the paper) can occur for some primes p>=59, for which
   the same argument (escaping the Hensel lift) holds.

Or better still, since you claim this approach "does not have much
content, it has no originality, it has nothing much to do with semi-
groups, and it does not inform usefully on Fermat's Last Theorem",
show me exactly the error in the above argument, cq: a reference to
this approach in the literature in case it is not a new contribution.

For 2 digits equivalence:
--->  the FLT(case1) inequality is due to the 2(p-1) carries. <---

Ciao, Nico Benschop.     -- http://www.iae.nl/users/benschop

Subj:  Re: The triplet structure of Z(.) mod p^k
Date: Tue, 30 Mar 1999 23:00:29 -0500
From: "Victor S. Miller" 
  To: Nico Benschop 
  CC: alf@macadam.mpce.mq.edu.au (Alf van der Poorten)

Mr. Benschop, I'm sorry, but I'm not going to forward your response to
the list.  I believe that I erred in sending your original post and
the reply to the list, and I don't wish to compound this mistake.  I
fully understand and applaud your interest in FLT -- you are certainly
not alone there.  However, it is clear that you are either unable or
unwilling to take legitimate criticism of your efforts.  The Number
Theory net list is not the appropriate place for long technical
discussions and/or arguments.  Instead, it is a place for technical
queries, or announcement -- which can include the availability of
technical papers.  Anyone on the list who wishes more details has been
adequately notified and will get in touch with you if interested.  I
might suggest, that if you want your writings to get a fair hearing
you need to show that you've spent a little effort in finding out what
the mathematical community has done in this regard, and to point out
how your method is different.  Mathematicians are busy people, and
need to have something to tantalize them sufficiently to make the
investment of their time and effort worthwhile.
-- Victor Miller -- moderator Number Theory Net

Subj:  Re: The triplet structure of Z(.) mod p^k
Date:  Wed, 31 Mar 1999 21:09:30 +0100
From:  Nico Benschop 
  To:  "Victor S. Miller" 
--

Victor S. Miller wrote:
>
> Mr. Benschop, I'm sorry, but I'm not going to forward your response to
> the list.  I believe that I erred in sending your original post and
> the reply to the list, and I don't wish to compound this mistake.  I
> fully understand and applaud your interest in FLT -- you are certainly
> not alone there.  However, it is clear that you are either unable or
> unwilling to take legitimate criticism of your efforts.
                   ^^^^^^^^^^^^^^^^^^^^^^^

Frankly, *if* you read Alf v.d.Poorten's reply, you can hardly qualify
that as "legitimate criticism" -- it is rather in the style of
crank-bashing, *not* answering my suggestion of how to break the Hensel
lift (via the 'exponent p distributes over a sum' property of each
solution to FLT mod p^2, as extension of the FST: n^p = n mod p).

> The Number Theory net list is not the appropriate place for long
> technical discussions and/or arguments.  Instead, it is a place for
> technical queries, or announcement -- which can include the availability
> of technical papers.  Anyone on the list who wishes more details has
> been adequately notified and will get in touch with you if interested.
> I might suggest, that if you want your writings to get a fair hearing
> you need to show that you've spent a little effort in finding out what
> the mathematical community has done in this regard, and to point out
> how your method is different.

Since begin 1995 I've written many essays and a few papers on this
approach, and had it accepted for presentation at three conferences
(two International: "Semigroups & Applications" Prague -jul96, and
 "Logic & Architecture Synthesis" Grenoble -dec96, and one National
Math-Congress, U-Twente apr98). I think I have done my share. ;-)

However, in trying to get it published, the cold sholder (at best)
is turned & the submission rejected - no referee report is returned,
no motivation other than: 'it does not fit our journal' or some
similarly weak excuse. Even the example of the cubic roots of 1 mod 7^2
is, as you and v.d.Poorten do, ignored. I am really disappointed at the
lack of professionalism apparent from number theorists, as well as from
semigroup specialists (the intersection of these sets seems to me empty,
probably one cause of embarrasment at such a simple approach to FLTcase1
- starting from Fermat's Small Thm via the semigroup/ring Z(.) mod p^k)

> Mathematicians are busy people, and
> need to have something to tantalize them sufficiently to make the
> investment of their time and effort worthwhile.
>
>                 Victor Miller -- moderator Number Theory Net

If this simple approach, breaking the Hensel lift -- which has always
been thought to block any direct approach to FLT via residues (which
indeed need one extra non-residue step, as shown) -- is not tantalizing,
then I'm sure there is something very wrong with the profession at large.

In engineering such attitude ('cover-up' and 'damage control') would,
if at all apparent, not last very long, and be exposed critically by the
profession itself. Of course, the vested interest of v.d.Poorten, as
author of a book on FLT, is clearly obscuring his view on this proposed
way of 'breaking the Hensel lift'. That he would go as far as he did in
ignoring the facts, and not responding to my example mod 7^2 (I sent the
posting by email to him: no reply) is sad for him.

Luckily, one side result of this research is a new way of binary
multiplication (log-code using 3 as semi-primitive root of 1 mod 2^k).
It was patented recently, so the real interest will *not* come from math
but from computer engineering: FLT is at the core of arithmetic, and
especially binary arithmetic (with an extra sign 2-cycle group factor).
  I hope to have made it clear that not I, but you and your list-members
are loosing in this case. BTW: why don't you pass my submission to prof.
Andrew Wiles, he is at U-Princeton as well, isn't he? (not that I expect
an answer, mind you;-)

-- Ciao, Nico Benschop.
http://www.iae.nl/users/benschop
http://www.iae.nl/users/benschop/campaign.htm (intro: cubic roots of 1)
http://www.iae.nl/users/benschop/scimat98.htm (sci.math.research)
http://www.iea.nl/users/benschop/nf-abstr.htm (abstract + refs)

Subject:   Re: No Proof of the Goldbach Conjecture  (sci.math 16 oct'02)
   From:   Nico Benschop 
    Org:   Digital Research : Finite Associative Networks
--------
Robin Chapman wrote:
> 
> [...] 
> But anyway, he has not answered my earlier questions.

Your point on the precise form of the induction hypothesis 
I did acknowledge as being useful, and I'm working on it. 
The whole concept of "induction over k" - as suggested by the
described Euclidean prime sieve (= multi prime number code,
over m_k : the product of the first k primes) - may not be
necessary at all: just the range of k values relevant to
any given 2n may suffice. 

> Mr Benschop, are you going to withdraw your MS
> "On Z mod (product of the first k primes), a Boolean 
>  lattice of 2^k groups and Goldbach's conjecture"
> now that you know that the main "proof" is unsound?

I don't see the need for that, unless you convince me that
my approach is qua *method* essentially insufficient, which
you have not done, by a long shot. Your remarks are rather at
the level of nitpicking on details of form, or slightly above 
that level. In other words: useful but hardly of essence.

> Also which journal did you submit it to? -- Robin Chapman

As mentioned before: none of your business, and I really don't
see the relevance of your repeated question on this. Any serious
editor/reviewer allows corrections to an interesting approach,
unlike you - who apparently does not even know such concept,
viz. the difference between mainline and form_detail.

As for your earlier request of me providing a 'written statement'
from prof. A.Cohen (TU-Eindhoven) regarding him using the phrase
"the Hensel lift cannot be broken" (meaning that FLT cannot be 
proven by residue analysis mod p^k alone - which I agree with:
one extra step is required, namely realizing that exponent p
distributes over a sum for any solution of FLT_1 mod p^2): 

Such request is ridiculous, mr. Chapman, you'll just have to take 
my word for it. And if you don't believe me and urgently require 
such info (heaven knows why) - just ask him yourself.

It happened either late 1994 or early 1995, and he referred me to 
a book on Arithmetic by J-P. Serre regarding Hensel and p-adics, 
which I did study, to find the source of this mis-understanding 
of the Hensel lift blocking a direct FLT proof, see my findings 
       in http://home.iae.nl/users/benschop/selmer.htm .

Your request is so strange and devoid of reality, that I fear you 
are already sliding 'down the chute of ridicule' - of which I
warned you to be aware of, regarding your zealous and misdirected
efforts as self-appointed saviour of the Holy Grail (Wiles' FLT
proof, and Goldbach's Conjecture, &c.)    

Luckily one critical remark on your strange behaviour just came
today in this thread (from Charlie-Boo) - so you hear it now also 
from another: well meant critique to save you from yourself and your
overzealous ways. Again, your remarks on math content are appreciated,
but the demeaning language you use is despicable and unworthy of
your profession. 

-- NB - http://home.iae.nl/users/benschop

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