Expert : . Hmm.., would that be a coincidence?

Rooky : . I don't know, you tell me! You are the Expert ;-)

Expert : . But what about that relation to FLTcase1?

Rooky : . See the many posts that I produced (sci.math) to wake some people up,
. . . . . . . . . (but somehow that did not quite succeed, since autumn '94 ;-()
. . . . Briefly: for residues of CR: (x^p)^3 = 1 mod p^k, y=1/x <>1, where p=1 mod 6,

Expert : Well, I'll have to check this ( with my colleague ...if I can find one ... who knows a bit more about finite semigroups, especially multiplicative semigroup Z(.) mod p^k and its additive properties ). It seems to me rather old_fashioned stuff, probably very old... Nowadays we are much further, applying semigroups (associative function composition algebra) as basis for much more abstract Category Theory ( thinking : all that fiddling in the finite is pretty boring... ).

Rooky : Ciao ( thinking : by my experience the past 3 years, I'm afraid I've seen the last of him;-( )

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Subject:      Re: proof of Goldbach's Conjecture
Author:       Nico Benschop 
Organization: Research
Date:         Fri, 3 Sep 1999 08:08:04 GMT

Tapio Hurme wrote:
>
> A tip - excuse me.
> FYI, Goldbach Conjecture (=GC) is equivalent to the conjecture:
> p = q+r +/- 1, where p, q and r are primes. p>q>r.
> Any prime is constructed as a sum of two smaller primes +/- 1,
> and proof that. Do not proof GC.
> Referring Lagrange p = a^2+b^2+c^2+d^2, as very well known,
> where a,b,c,d are integers - including zero values of a,b,c,or d.
>
> More easy p=a^2-b^2=(a+b)(a-b), where a-b=1.
> Further - obviously -  q and r equal to the sum of 4 squares , as above.
> Or alternatively: q or r= a difference of two squares (integers).
> Go 3D, and draw your conclusions referring infinite descending method
> by P.F.
>
> No mods, but have a fun - at least.  [*]
> Why to be complicated?               [*]  -- Tapio

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

http://www.iae.nl/users/benschop/ngb2.dvi
    "Arithmetic with squarefree modulus, its lattice of groups,
            a primeseive, and Goldbach's Conjecture"
http://www.ams.org/preprints/11/199805/0index.html (abstract)

http://www.iae.nl/users/benschop/carry.htm (intro: closure & generator)
    On the "carry" c in: n = cm + r.  Integer n, seen in terms of:
    Modulus m, Residue r, Carry c linking integers (n) to residues (r)

http://www.iae.nl/users/benschop/fewago.htm
    Intro: semigroups & arithmetic.
    On the use of finite semigroup (associative closure: function
    composition) for  arithmetic analysis: associative and commutative
    closure Z(.,+) Especially: additive structure of multiplicative Z(.),
    linking powersum- (Fermat, Waring) and primesum (Goldbach) problems.

Re[*]: I'm having fun all right ;-)
    Moreover, my initial aim is to gain insight WHY such well known
    problems are so difficult(*). [ my own hobby: FSM decomposition of
    Finite State Machines: "Digital Network Theory" if you like...]
    The reason(*) roughly is the
    "single-focus" of many specialists (which brought them so far in
    the first place, of course)... Still, there's something missing:

Dual Focus (at least) is better:
        study the relation between (.) and (+)
     -- and between (^) and (+) while you are at it.
   Residue arithmetic (Gauss, 1801) is only half the story:
   ---------- "the carry makes the difference" -------------------
         (in FLT, and in many other cases as well;-)
   http://www.iae.nl/users/benschop/ferm.htm (Cubic roots of 1 mod 7^2)
    The p-th power of a k-digit (base p) number has upto kp digits (!)

Unfortunately, it (the carry) is anti-closure type, namely 'generative'.
  So it does not quite fit nicely in a 'closed-theory', as it were.
  You'll see (my homepg "cup-a-soup" text) that 'exponentiation' (^)
  is of similar non-conformist nature ;-)
  Namely: not associative, not commutative, not distributing over (+)
     yet (^) is simply defined as repeated (.)
 Just as (.) is defined as repeated (+).
      with (+) and (.) both being associative & commutative and
      (.) distributing over (+).
 So how come: is (^) one bridge too far?
 Not quite, if you _really_ look at it
      (with modulo eyes, and a good nose for the carry: "follow your nose" :-)

Subject:      Re: JSH: Questions for Nico Benschop
Author:       Nico Benschop 
Date:         30 Sep 99 05:38:06 -0400 (EDT)    --- news:sci.math

James Harris 
on Thu, 30 Sep 1999 12:30:53 -0400 wrote:

>Nico Benschop wrote in message
>> ... and I thought to engage in some math-discussion finally.
>>     But this looks more like some simpleminded chatbox ... ;-(
>>
>
> Sorry about that. Despite my warning, two people decided to lob
> verbal grenades. I felt a strong urge, which I decided not to deny,
> to toss their grenades back at them.
> But I did notice Robin Chapman had some comments that I'm sure you
> will attempt to address.
> I've made progress in reviewing your work. As I mentioned to you
> earlier, I'm testing your assertions with my favorite identity:
>
>  (x+y-z)^p = x^p + y^p - z^p + p(z-x)(z-y)(x+y)Q         ...[FI]
>
> Now, as I mentioned before, from it I can easily prove that if   >
there's an FLT counterexample x+y = z(mod p^2).

On your [FI]:
     Very similar to mine: (x+y)^p = x^p + y^p (mod p^2),
     which holds for all FLT_1 solutions (if they exist;-)
     Since they are 'in_core': the p-1 cycle of FST mod p^2.
      (each p-th power residue a=n^p satisfies a^p=a mod p^2)
      (as also each b=n^{p^2}, coprime to p, has b^p=b mod p^3, &c)

> However, this seems to be just the lower limit imposed by the
> identity.  As of yet, I see no indication that it must be higher.

There is no need: mod p^2 is necessary & sufficient to prove FLT_1,
 using the EDS property (Exponent p Distributes ove r a Sum)
 of *any* assumed integer solution of FLT_1.
 Although a=n^{p^[k-1]} satisfies a^p=a mod p^k for every k>1.
 Has to do with all residues mod p^k forming a cyclic group G_k
 of order (p-1)p^{k-1}.

This is _the_ big difference of 'thinking-frames' qua number theory:

 Many people think in NrTh of  numbers & equations.
 While I'm rather using (sub-)closures & generators,
         and especially the mult've subclosures of Z(.) mod p^k
         looking  for their _additive_ properties, such as: ...[+]

Like Fermat's Small Thm: n^p = n   mod p (prime p), for all n.
                     or: n^{p-1}=1 mod p (for n coprime to p)

It is much more intuitively appealing (to me;-) to think of
this property as a cyclic 'closure' of length p-1, generated
by one generator g as follows:
           ________________________<_______________________
          /                                                \
 G=g*:   g --> g^2 --> g^3 --> .... --> g^{p-2} --> g^{p-1}=1

where you 'see' immediately that g^{p-2} is the inverse g^{-1}=1/g
of g, since g.g^{p-2}=1 (in residues of course)

And: all p-th residues g^(mp), hence all p-th powers (since what
  holds for a generator *also* holgs for all, roughly speaking;-)
form a cycle of length |G|/p, so: G mod p^2, of cyclelength (p-1)p
has a subcycle of length p-1 (which I call its 'core' subgroup A_2),
as well as a subcycle of length p, called the 'extension' group B_2
of order p -- and generated by p+1 (for any prime p *that* is a
generator! -- although there are many more generators of B_2).

Thus  G_k = A_k.B_k    core |A_k| = p-1,
                   and extension group |B_k|= p^{k-1}
Re[+]:
where each subgroup S of core A in cyclic group G of units in Z(.)
     sums to zero (mod p^k):   sum(S)=0 mod p^k.
While any subgroupp T in B_k has:  sum(T)=p^t, for some t \leq  k-1.

Looking at NrTh this way, it is *much* easier to get a 'feeling' of
the whole (rather than local details), and come with fruitful ideas
and conjectures to prove.

-- In other words:  "nrs & eqns" is looking thru a microscope,

   while "closures & generators" is looking thru a telescope.

This is the dual focus I mentioned earlier: one should do *both*,
[ not at-the-same time of course: then you get the Ben Turpin effect,
  and a head_ache;-(  but alternatively, in a kind of 'bootstrap' ]

>
> Unfortunately, I didn't have time to do further research as I ended
> up having to do some work on my own webpage.  I'm still hoping on
> finishing up looking at your arguments tomorrow. -- James Harris

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

Subject:   Re: Why mathematicians had to be liars
   Date:   Mon, 01 Oct 2001
   From:   Nico Benschop <>
 Orga'n:   Digital Research
Newsgrp:   sci.math

James Harris wrote:
>
> [...] and my answer will be that you can't trust mathematicians.
> [...] --  James Harris

Well, I would'nt generalize like that, James!
I suggest to quantify: you can't trust *some* mathematicians...

...Namely those without an open mind, unwilling to even consider
other [direct] paths, because "the Hensel Lift cannot be broken"
- which in fact is false, in special cases: the solutions of
  case_1 FLT mod p^k, all equivalent via a linear tranformation to:
 'triplet'-structure a + 1/b == b + 1/c == c + 1/a == -1 mod p^k,
 for p-th powers a, b, c, with abc == 1 mod p^k.

After all, for a simple (yet very hard-to-prove) statement like:
 "the sum of two p-th powers is not a p-th power (for odd prime p)",
my next question would be: what *do* the sums of two p-th powers look
like? (for residues mod p^k: they cover half the group of units;-)
Note that 360 year passed since that conjecture (NOT theorem FLT!-)
was posed, so it is obvious that "something must be missing" in the
known & tried arithmetic approaches... but *what* ?

The clue, it seems to me, is that the statement x^p + y^p =/= z^p
has only two arithmetic operations: (^) and (+), of which (+) is
associative and commutative, while (^) lost both properties -
although defined (for integers;-) as repeated multiplication (*),
which itself is defined by repeated addition (+), with (*) also
associative and commutative.

So how come that making one step beyond (*) you get (^): loosing
all-of-a-sudden BOTH nice properties!?  Well, for residues mod p^k
(base p = the exponent in the statement) this is not quite true:
    there *are* solutions of FLT mod p^k for some primes p,
            a necessary condition for any integer solution!
E.g: all p=1 mod 6: namely the cubic roots of unity 1 mod p^k,
     and a few others for some p \geq 59: "triplet' solution type,
"breaking the Hensel Lift" (HL). The Hensel-Lift here means: if there
is a solution for k=2 (2-digit arithmetic: mod p^2) then it generates
p^{k-2} solutions mod p^k for any k>2, thus it *seems* no inequality
can be proven (since for k --> inf the equivalence stays valid: p-adics)

But: any solution of FLT mod p^k is equivalent to one with a cute
           'EDS' property: (x+y)^p == x^p + y^p (mod p^k),
in other words: Exponent p Distributes over a Sum (EDS).
-- Are you surprised that this (necessary) condition for any integer
solution to FLT (case_1) cannot be extended to hold for integers?!
Try for instance:      18^7 + 30^7 == -1 (mod 49),
in base 7 code:    (24)^7 + (42)^7 == (66)^7 mod 7^2. ..[2]
where 24^7 == 24, 42^7 == 42, 66^7 == 66 (mod 7^2).
See  http://de.arXiv.org/abs/math.HO/0103051
     on Fermat & the Cubic roots of Unity.

What's _missing_ is the insight that FLT is NOT a purely arithmetic
problem, but one requiring non-commutative FUNCTION composition:
of the two symmetries of arithmetic: complement -n for (+) and
inverse 1/n for (*), in residues mod p^k, with: a+1 == -1/a mod p^k.

-- NB  http://piazza.iae.nl/users/benschop/func.htm   (triplets mod p^k)
[2]..  http://piazza.iae.nl/users/benschop/selmer.htm (on a^3=1 mod 7^2)
       http://piazza.iae.nl/users/benschop/nr-th.htm  ("non-discussion")
       http://piazza.iae.nl/users/benschop/inertia.htm  (-;)
       http://piazza.iae.nl/users/benschop/anti-cl.htm  N(+) = {n^p}*
       http://piazza.iae.nl/users/benschop/carry.htm (Carry & Generation)
                                                 Residues --> Integers
                                                   Local  --> Global

Subject:  Re: To the mathematicians
   Date:  Tue, 02 Oct 2001 11:44:08 +0200
   From:  Nico Benschop <>
    Org:  Digital Research
Newsgrp:  sci.math
----------------
Jim Ferry wrote (re: James Harris):
>
> [...] However, you're not the only one who thinks that the math
> world is corrupt, that it stifles the ideas of outsiders to protect
> its privileged elite.  (Note that these two sentences make a
> proof of Harrisian rigor that James is not, in fact, crazy.)
> It's sad that you will not deign to work with them,    ...[*]
> that you'd  rather fight alone and lose than learn to work
> constructively with others.  [...]
> "Doing your best" would entail joining forces with those of like
> mind. You know who they are.  You know where to find them.
>
> > JSH:
> > But I'll draw on the confidence gained by being able to prove
> > Fermat's Last Theorem from scratch in a bit over six years.
> > It's just too sad that human beings would turn out to be an even
> > bigger problem, and sadder that I'm not surprised by that.

Re[*]:  But, in an indirect way, he *does* cooperate: the attention
    he keeps drawing (for some unfathomable reason;-) is very useful!
I admit of 'misusing' (sp?) it to occasionally drop a line, drawing
attention to my approach via extending FST (Fermat's Small Thm) to
mod p^k for k>2 (odd prime p), combined with the fact that each
solution of FLT mod p^k has exponent p distributing over a sum,
or is by a linear transformation equivalent to such. And the rest is:
remembering that for integers this does not hold (Pascal's binomial
expansion;-) - with inequality upon 'quadratic analysis' mod p^{3k+1}
combined with the equivalence mod ^p^k (assuming some FLT solution for
integers < p^{kp}: knowing there are (p-1)k 'carries' for p-th
powers of k-digit (base p) integers, 'making the difference',
see e.g. http://de.arXiv.org/abs/math/0103067  for all divisors r of
p \pm 1  having distinct r^{p-1} mod p^3 in subgroup B_3 == (p+1)^*
mod p^3 of units.

Interesting observation: I have counters on several of my homepages, and
whenever I refer to them in a posting of mine in one of JSH's threads,
BINGO: these counter jump up, with vistits from various wellknown
institutions like Harvard.edu, UCLA.edu, Princeton.edu, Rutgers.edu,
MIT.edu, Penn.edu, Cornell.edu, Oxford.ac.uk, Boeing.com, Ford.com,
HP.com, uni-freiburg.de, rug.nl, cwi.nl, utu.fi, etcetera...
[ PS: some industrial labs do realize that FLT is at the core of
  arithmetic, http://de.arXiv.org/abs/math.GM/0105029 on log-arithmetic]
which would not occur in threads started by myself (I too have
some sci.math experience since 1995, plus several math-conference
experiences  'whereby hangs a tale';-)

So I don't belittle James' efforts in sci.math. On the contrary,
I value his remarkable attention drawing capacity - which is of
good use (eventually, I hope...  "Persistance furthers", I-Ching;-)

Moreover, his 'gut feeling', or intuition, I do share. Namely that
FLT *is* approachable with elementary means (although slightly more
is required than JSH uses: "breaking the Hensel Lift" with the cubic
roots of Unity being the main clue;-)

In due time the hesitating attitude of reputation-conscious editors /
referees of math-journals will bend. The Mathematical Intelligencer's
opinion on  http://de.arXiv.org/abs/math.HO/0103051 (Fermat & the cubic
roots of unity) may shift from "not convincing historically" to
something less evasive, and more positive. Or Annals of Mathematics'
reply recently on http://de.arXiv.org/abs/math.GM/0103014 (the Triplet
additive structure of the units group mod p^k) from a rather neutral
non-informative "we only publish the very best", to a bit more
specific info on the paper's content, at least...

-- NB -- http://piazza.iae.nl/users/benschop/nr-th.htm


 Subj: Re: Reasons to keep an open mind ;-)
 Date: Wed, 03 Oct 2001 11:28:57 +0200
 From: Nico Benschop
Org'n: Digital Research
Newsgrp: sci.math

"David C. Ullrich" wrote:
>
> On Tue, 02 Oct 2001 09:53:28 +0200, Nico Benschop wrote:
>
> >For some conventional wisdoms:
> >
> >"Heavier-than-air flying machines are impossible."
> >-- Lord Kelvin, president, Royal Society, 1895.
> >
> >"Who the hell wants to hear actors talk?"
> >-- H.M. Warner, Warner Brothers, 1927.
> >
> >"I think there is a world market for maybe five computers."
> >-- Thomas Watson, chairman of IBM, 1943.
> >
> >"There is no reason anyone would want a computer in their home."
> >-- Ken Olson, president/founder of Digital Equipment Corp, 1977.
> >
> >and other jewels, see:
> >
> >   http://www.ufomind.com/area51/list/1997/nov/a29-001.shtml
> >
> >"The Hensel Lift cannot be broken"
> >-- Most mathematicians,
>
> Really? Of course I don't know most mathematicians,
> but I know a few, and I've never heard any of them say this.

It is also referred to as (e.g. Bob Silverman, sci.math, 11may98
 bobs@rsa.com wrote in : "One more crank" ):
  "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."
  -- http://piazza.iae.nl/users/benschop/selmer.htm

> What's a specific example of someone who's said this?

Prof. A.M.Cohen (Technical Univ. Eindhoven, and CWI.nl Amsterdam)
In fact, when mentioning this, he advised me to read some more
on arithmetic from J-P.Serre, which I did. That's how I understood
this objection against a direct proof of FLT via residues mod p^k,
having to do with the Hensel Lift (and p-adics as asymptotic
residues for k --> inf), in this simple & special FLT case_1:
 the normed :  a^p + b^p + 1 == 0 mod p^k (of units)  ...[1]

Any solution of this equivalence mod p^2 can be extended as follows:
   Let a_i and b_i be the i-th digit (base p) of a^p resp. b^p,
then by [1]:
              a_i + b_i = p-1 for all i \geq 0, ..[2]

              Since -1 mod p^k == \sum (p-1)p^i (i=0, .. , k-1)

And given such solution mod p^2 (i=0,1) it suffices to extend for i>1
such that [2] is satisfied: yielding free choice out of p digits for
one of {a_i, b_i} at each extra extension digit. So the initial
solution mod p^2 yields p^{k-2} extension solutions mod p^k for k>2.

Now the Hensel Lift, applied to this case, might suggest that there
is NO way you can prove the FLT case1 inequality for ("mod-free")
integers. Since the Hensel Lift says FOR RESIDUES mod p^k:
equivalence can be obtained for ANY k, no matter how large k is.

This is certainly true, BUT  the structure of solutions of [1]
   implies that exponent p distributes over a sum, roughly:
if   a^p + b^p == c^p mod p^k (k>1), then necessarily c == a+b
(or a variant thereof, for the more general 'triplet' solution
a + 1/b == b + 1/c == c + 1/a, with abc==1 mod p^k for some p \geq 59).

See http://de.arXiv.org/abs/math.GM/0103014 for this EDS property
(for solutions "in core" subgroup of order p-1 of units, resp. with
at least two p-th power terms in core) can be generalized to all FLT
case_1 type of solutions, and necessarily to INequality for finite
integers (with the given residue equivalence).

I repeat: indeed I agree that from a ('local') residue equivalence
one cannot IN GENERAL derive a result for integers (like the FLT
inequality) -- but in the special case of the cubic roots of 1 mod p^k,
and its generalized 'triplet' rootform for FLT case_1, the FLT
inequality *does* follow, due to that special 'Exponent Distributes
over a Sum" (EDS) property of any FLT mod p^k solution (which is a
necessary condition for any integer FLT case_1 solution).

Put another way: for an FLT solution a^p + b^p == c^p mod p^k,
for ANY k>1: assuming it represents the residues of some finite
(integer) sol'n of p-th power integers < p^{kp}, the corresponding
k-digit (base p) integers A, B, C yield INequality already mod p^{3k+1},
hence certainly mod p^{kp} for prime p>3.

You can see this as a 'bootstrap': eqv mod p^k --> ineqv mod p^{3k+1},
referred to as 'quadratic' analysis (see lemma 3.2 in the paper).
The extra precision +1 coming from taking the p-th power, the 3k
from matching maximally 3 coefficients (of a quadratic polynomial)
to satisfy equivalence, which cannot be exceeded.

> >
> >-- NB -- http://piazza.iae.nl/users/benschop/quotes.htm
> >         ..... "An open mind is a joy forever" .....
> >        http://piazza.iae.nl/users/benschop/inertia.htm


Subject:   Re: Reasons to keep an open mind ;-)
   Date:   Thu, 04 Oct 2001 11:28:33 +0200
   From:   Nico Benschop
  Org'n:   Digital Research
 Newsgr:   sci.math

Lee Rudolph wrote:
>
> "Robin Chapman"  writes:
>
> >how the f**k
>
> Man, it's been years since I've seen that.
> Doesn't everyone write "how the f^k" these days? -- Lee Rudolph

Well, one is never too old to learn;-)
And from R.Chapman I learned to write "losing it" with one 'o',
(after all, he *is* a specialist, so he should know;-)

And re:   a^p + b^p == c^p mod p^k ..[1]   for k>1
          (replacing f by p [for prime] is no problem I hope,
           although it *does* require _some_ flexibility...)

Consider  A^p + B^p = C^p ..[2] with k-digit non-negative integers
A,B,C < p^k of the same representation (base p) as residues a,b,c resp.
For instance in the normed form:  c == -1 mod p^k, and C = p^k -1.

Then [2] is false, no matter how large k is taken. The eqv [1] can
always be normalized by a linear transformation (multiply and add
   with units mod p^k) to a form with two terms "in core" A_k,
  [the subgroup of units with order p-1, where a^p == a, b^p == b].

Assuming [1] to be residues of  some (finite) integer solution
(coprime to p: case_1), there is a corresponding finite precision k,
and this leads necessarily to inequality for [2] ...(or mod p^{kp}
if you like;-) and in fact 'already' with inequivalence mod p^{3k+1}.

-- NB -- http://de.arXiv.org/abs/math.GM/0103014