[**-- James Harris wrote:

I was thrilled just a few years ago by your proof of Fermat's "last theorem".
I've tended to be bullish about what human beings are capable of,
and besides I was also anxious to see a satisfying ending to a fascinating tale.

Unfortunately, I was soon put off by learning that I would be unable
to understand your proof without obstruse mathematics; and even if I
did understand the mathematical methods, I'd have to wade through up
to a hundred pages to look for the answer to a simply stated problem.
{. . . . . . . }

What I'm looking for is an ending to this story that the child in me
would be satisfied with. An ending that says that sometimes, the most
profound conclusions have simple reasons. And that the answers are
within everyone's reach. --**]

I doubt if you'll get an answer, James, but that of course is beside
your point. I agree, it would be more satisfying if such simple
statement as Fermat's Last Theorem . . FLT :

BTW: A short and direct proof of FLT (x^p + y^p =/= z^p) using
base (prime) p number representation -- first by residues mod p^k,
followed by carry extension and induction on precision k --
was published in the Acta Mathematica of Univ. Bratislava (Nov.2005)
online http://pc2.iam.fmph.uniba.sk/amuc/_vol74n2.html (pp 169-184)

To me the FLT problem seems to be at the core of arithmetic -- showing
two operations (+) and (^) that are apparently too far apart for comfort,
although (^) is repeated (.), which itself is repeated (+).
So (^) is only one step beyond (.), yet one step too much for closure!
. . . Notice that (^) is not associative, nor commutative,
nor does it distribute over (+) : (re: Pascal's triangle). It is a non-conformist
operation, despite its simple definition as repeated multiplication.

Alternatively, this anti-closure allows another more positive point of view:
apparently p-th powers are an excellent set of additive generators of
the naturals! (in fact, the sum of four of them, but as residues mod p^k, covers
ALL residues mod p^k, prime p>2, any k \geq K_p (some critical K_p < p);
see my homepg ref[6] -- a type of Waring_for_Residues result).

Notice, however, that Wiles' proof of FLT is only a corollary of a theory that
links elliptic curves (equation in two variables x,y: one cubic and
the other quadratic) to modular forms, related to function (a.x+b)/(c.x+d) of x.
If you saw Wiles' BBC interview, then you may have noticed his quoting Eichler:
--- " There are five basic arithmetic operations:
. . . . add, subtract, multiply, divide, and modular forms " ---
( rather than exponentiation ). This is the clue to the solution:
modular forms are the missing_link between (+) and (^), clearly due to
their structure as a balanced mix of (+) and (.).

Aside: 2 x 2 matrix filled with coefficients a,b in first row,
and c,d in second row, is another form of the same; matrices R,S,T
compose associatively: R(ST) = (RS)T just like functions f(x) do,
and a matrix is a balanced mix of (+) and (.) ...

I've been trying my hand at FLT, like yourself, but with a different approach,
namely using function composition, like f(g(x)) composing two functions
by first applying g(x) and using the result as argument in f(), yielding a
new function {fg}(x), in general different from {gf}(x).
Such algebra of function composition (just string_concatenation) is
associative: f(gh) = (fg)h = fgh -- hence no brackets necessary
(called "context-free") -- without any other restriction ( like commutative ),
called "semigroups".
Note: iteration a(aa) = (aa)a = aaa = a^3 is unique for an associative closure.

Function composition is a rather basic operation, which w.r.t. FLT also can
function as link between (+) and (^), similar to modular forms in Wiles' proof.
In fact, multiplication modulo some modulus m, denoted Z(.) mod m,
that is: residue arithmetic, is associative (and commutative as well: a.b=b.a)
and yields for summing p-th powers enough information to prove FLT, provided
modulus m= p^k (a prime power, p > 2) is taken. This should not be surprising,
since {n^p} form a cyclic subgroup of Z(.) mod p^k for every prime p > 2 ,
and any precision k > 0.

Moreover, mod p^k is relevant because Fermat's Small Theorem (FST):
. . . n^p = n mod p (for prime p, and all n), which he discovered around 1637,
can be extended to mod p^k for p-1 values of n, forming a p-1 cycle in Z(.) mod p^k
(for each prime p > 2, and each precision k > 0).

These properies are sufficient to tackle the FLTcase1 problem (in case_1,
of two cases: x,y,z are copime to p), although for a long time it was thought
impossible to derive from an equivalence x^p+y^p = z^p mod p^k the required
inequality for integers. -- But there is one special sub_case of case_1,
with z=x+y mod p^k, namely where x^3=1 mod p^k (so x is a cubic root of unity,
as they say): allowing to derive the inequality by a special property that is
as basic as the FLT statement: exponent p does distribute over a sum in
that sub_case, which in fact it (trivially) does in FST mod p -- but FST' mod p^k
is less trivial. . . In fact ALL solutions of FLT mod p^k (case1) have
such property, or a similar one (for the generalization of cubic root solution,
namely the "triplets" - see my homepg ref[1]) with the same effect:
inequality for integers. Notice that cubes play an essential role here,
as they do in elliptic curves (no coincidence!).

So you see, whether taking modular forms as go-between, or function composition:
it boils down to getting to grips with (^) related to (+), their "distance" is too large
"beyond Phytagoras" (where the sum of two squares sometimes does yield
a square, such as: 3^2+4^2 = 5^2).

So don't let anyone tell you that FLT is a nice & tough problem,
but rather 'esoteric' and of little use for normal arithmetic:
I hope you see that it in fact is "at_the_core" of integer arthmetic.

And solving the riddle in a way that improves insight into the basic operations
and structure of arithmetic is a must for the further development of
discrete math, as it raced, with (finite) computers, into the present information age.

That's why I think Wiles' proof comes short of your, and many other's,
expectations; and why I hope finite semigroups (function composition)
will be taken a "bit" more seriously. --- They have, frankly, been neglected
since Category Theory took over the lead into the abstract development
of semigroups, since the sixties. Rather than using their finite structure
(as model for FSM's - finite state machines : computers) to develop a detailed
structure theory of for instance Digital Networks ( FSM ), comparable
with what Shannon did in 1938 - seeing Boolean Algebra (an idempotent
and commutative form of arithmetic -- see G.Boole: "Laws of Thought", 1854)
as isomorphic, same form, to binary logic circuits.

Hopefully you, and others, will understand & agree with my analysis of the
crucial importance of FLT in general, and of linking (+) and (^) in
particular --- either by modular forms, or by function composition.

Subject: Algebra in grad school
   Date: 11 May 1998 (sci.math)    -- "Multiply mod 10 (semigroup)" --
NB:
Modern Algebra can be spoiled, especially if kept purely abstract.
  That is roughly: syntax only, hence no object representation -
- which is absolutely necessary for examples, and understanding.

For instance, an eye opener for me was the structure of multiplication
mod 10, as a (commutative) semigroup, with generators 2, 7 and 5;
drawn on paper as two 4-cycles 2*, 7* and a 2 x 2  'boolean lattice'
of idempotents 1, 5, 6, 0:
 { idempotent x.x = x mod m,  and ordering (transitive/asym/reflexive)
  of commut've idt's: x >= y <-> x.y= y.x= y, so x is identity for y }
     <---------------
   7 --> 9 --> 3 --> 1            Notice additive isomorphism "+5":
                    / \
  <--------------  /   \           { 2^i +5 = 7^i, and 0 +5 = 5 }
2 --> 4 --> 8 --> 6     5           and back again +5 : 7* --> 2*
                   \   /                                5  --> 0
                    \ /            Hence "+5" is a "symmetry"
                     0              (automorphism of order 2)
                                    = "inner-projection" 1-1 into)

   Do this for several moduli, say upto m=16, and you can learn more
about boolean algebra, set-theory, lattices, groups, fields, ordering
& equivalence, semigroups, divisors of zero, etc... than most courses
and/or books will tell you. For non-commutativity, take the symmetries
of an equilateral triangle, yielding smallest non-comm've group D3,
which can *also* be drawn nicely, with three 2-cycles and one 3-cycle.

And the "right-copy" ab=b semigroup of any order, say order 2 for a
model of the common memory element in computers: the FlipFlop alias
"set-reset FF" where a='0' and b='1' (for engineers). --- Guess how
to model the if-the-else, alias "mux" (multiplexer) alias "branch"
element? (see my homepage ref[2] on state machines, and the 5 basic
state machine types, for practical purposes): It is ab=a, and ba=b,
the left-copier semigroup, represented over three states
(one initial state that's left forever).

Moreover, this way you have the "explorative do-it-yourself exitement".

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

____AHA principle of Dynamic Balance: One is Always Halfway Anyway___


Subject:  Re: more NFB garbage
          [was: Can a^p + b^p == c^p mod p^k (odd prime p, 0 < a,b,c < p) ]
   Date:  Tue, 30 Apr 2002 17:21:01 GMT
   From:  Nico Benschop 
    Org:  Digital Research : Finite & Associative
Newsgrp:  sci.math
-------
Pertti Lounesto wrote:
>
> Nico Benschop wrote:
>
> >Pertti Lounesto wrote:
> >
> >>Is Chapman's example a counterexample to Benschop's lemma?
> >>Does it satisfy all the assumptions of Benschop's lemma, without
> >>the conclusions being valid?  Has anybody verified the validity of
> >>Chapman's counterexample?  Does Benschop accept the validity?
> >>Does Benschop continue his useless and erroneous explorations?
> >>Has Chapman managed to instigate cognitive growth in Benschop? [*]
> >>
> >>Robin Chapman wrote:
> >>
> >>>[...]
> >>>
> >     RC ridiculing my changing lemma 2.2 after I admitted a mistake,
> >        namely with q = |B|=p^{k-1}:  n^{q-1} == 1 mod p^{k-1}
> >        and not:  n^{q-1} == 1 mod p^k. Moreover, this is about
> >        p-th power residues, which require to start mod p^2 (k=2)
> >        rather than mod p (k=1). That's all. I would'nt call it a
> >        typo, but it is a minor change, a rather simple correction.
> >
> >>> -- Robin Chapman
> >>>
> >
> > So RC sneers at me for taking him seriously, and correcting an error!
> >
> >Re[*]: Sure, and he does not appear to like it...
> >     strange fellow, that RC.  But thanks anyway.     -- NB
>
> Good.  You are not a boor, while you thank Chapman for
> pointing a flaw in your reasoning.  But, as an exercise, in
> order to learn to write better thank returning, rewrite all
> the above without any negativity about Chapman. I understand
> that after Chapman's insults, this does require some effort.

I think I went far enough, Pertti. I do appreciate someone's time
to read my MS and comment on it. That I have to filter out insulting
language and/or subject title change, I take into the bargain (if
that's the way it is done on sci.math;-)

> If there are errors in a text, it is often difficult to know what
> the author tries to say.  I once wrote a referees report of a
> paper, which had three (!) unrelated sign errors in one line.
> The topic was new to me (and to everybody), and it took me
> some effort to rectify the thing.  It was only expectable
> that the author behaved boorish, and did not thank me.

Well, now that preliminary and form_ mistakes have been ironed out,
why not yourself have a look at it. It is straightforward elementary
residue arithmetic, however with a new aspect to it: the carry is
revived - since long being ignored, I guess roughly since Gauss [1801].
Of course this is necessary, in order to move from FLT mod p^k to FLT
for integers (which was deemed 'impossible' due to the Hensel lift;
but then, if you insist on ignoring the carry, that's what you get:
blocking a simple and direct proof;-(

-- NB - http://piazza.iae.nl/users/benschop/nf-abstr.htm   Intro
        http://piazza.iae.nl/users/benschop/carry.htm       ,,
        http://piazza.iae.nl/users/benschop/nfb0.dvi       MS

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