Pete Agoras :  3^2 + 4^2 = 5^2       ---->   (2 D rectangle)   (n+1)^i
   and Pascal : (n+1)^3 = n^3 + 3.n^2 + 3.n +1 (1331=11^3)  Pascal_Triangle
          and : (n-1)(n+1) = n^2 -1            (1+-1)              1 = 11^0
        imply : 3^3 + 4^3 + 5^3 = 6^3  <----   (!)               1 1 = 11^1
                ---------------------                          1 2 1 = 11^2
Because:                  |---5^2---|  |--4^2--|             1 3 3 1 = 11^3
 6^3 = (5+1)^3 = 5^3 + 3.(3^2 + 4^2) + 3.5 +1              1 4 6 4 1 = 11^4
               = 5^3 + 3^3 + (3+1).4^2
  So: 6^3      = 5^3 + 3^3 + 4^3        ---?---> (3 D)    'Coincidence' ?

Come to think of it, going back to P.Agoras, use (121=11^2) and (1+-1) then:
                          (3-1)(3+1)
     5^2 = (4+1)^2 = 4^2 +   2.4    +1 = 4^2 + 3^2 :   Simple comme bonjour.

And it stops at 6^3 :  the next step is too much (and it gets worse beyond):
^^^^^^^^^^^^^^^^^^^
  7^4 = (6+1)^4 = 6^4 + 4.6^3 + 6.6^2 + 4.6 +1 = 6^4 + 5.(5+1)^3 + 5^2 =

      = 6^4 +5.(5^3 +3.5^2 +3.5 +1) +5^2 = 6^4 +5^4 + 5.[(3.5^2 +3.5 +1) +5]
                                                      |--------------------|
                    where:   4^4 + 3^4 = 256+81 = 337 < 5.[91+5]= 480.

Moreover, generalizing P.Agoras' equality to a sum of two n-th powers,
you get inequality for n>2: x^n + y^n <> z^n     (Fermat Large Theorem FLT).
This follows from :  x^p = x  mod p (prime p>2)  (Fermat Small Theorem FST),
and it's extension:  x^p = x  mod p^k (k>1) for p  x-values  (Core Theorem):
     FLT mod p^k (k>=1) has solutions: trivial for k=1 (FST) & all primes p,
     for k>=2 at primes p=1 mod 6  and some p>=59: x^p + y^p = z^p  mod p^k.
----->   They  have  z = x+y mod p^2,  sothat  x^p + y^p = (x+y)^p  mod p^2,
    which of course cannot hold for positive integers,  proving FLT (case 1)

Fermat's equality (~1640): a^p + b^p = c^p (prime p>2) can have solutions in residues (mod p^k), but only if c = a+b (and p=1 mod 6). . [1], so exponent p distributes over a sum (or special triplet solutions for some p>=59 with a similar property). Binomial expansion implies inequality a^p + b^p < (a+b)^p for integers: ergo FLT (case1).

And XLT (extra large theorem) : did you know that every residue mod p^k (prime p, any k>0) is the sum of at most 4 p-th power residues [4] . . (re: for 'mod-free' integers : Euler & Lagrange's 4 squares, and Waring's power-sums, both around 1770). Following the question: if the sum of two p-th powers is not a p-th power, what is it then ? - The answer is : mod p^k the pairsums of core-residues are all distinct (upto commutation); the rest follows, see [4]. A positive and constructive result (what can be done) is more interesting than a negative answer (what cannot be done).

<^> Cubic roots of 1. . . Triplets and the solutions of FLT mod p^k. [1]

sci.math  (2feb99)       Re: "Final simple FLT post, answers"
NB -> Frank Manus

You're welcome.  A simple FLT proof *is* possible.  Just two more hints:
   Let FLT be  x^p+y^p=z^p has no integer solution x,y,z>0 for p>2 ...[1]
then:
1. It suffices to show this for prime exponents, since (^) distributes over (.)
2. FLT has two cases. Case1: p not divides x,y,z. Case2: p divides one of x,y,z.

Case1 is quite easy to prove, by expanding Farmat's Small Thm: n^p=n mod p,
for prime p, and all n. Namely consider arithmetic mod p^2 (2-digits base p),
then there are (p-1)p residues coprime to p (case1 !), and in general mod
p^k: there are p^k - p^{k-1} = (p-1)p^{k-1}. Any product of two nrs coprime
to p is also coprime to p (closure). They form a "group" G, and it is known
that this group is cyclic --> generated by the powers of just one nr, say
G ={g^i} for i=1..|G| (order |G| of the cyclic group). So mod p^2 the
p-th powers, that is: each p-th iteration of g, form a cycle of length p-1
---> Just like FST: n^p=n mod p, meaning: it's a p-1 cycle, which Fermat
discovered around 1637, just when he made his marginal note (!). But this
p-1 cycle mod p^2 is for 2-digit residues, and x^p=x, y^p=y, z^p=z mod p^2
for any x,y,z in this "core" cycle.  Clearly, for any solution "in core":
         (x+y)^p = x+y = x^p + y^p mod p^2....[2]

Notice that *every* solution of p-th power integers is "in core" mod p^2,
and that in [2] the Exponent p Distributes over a Sum (EDS property, as
trivially holds for FST mod p) --> Hence such solution in Core mod p^2
cannot be extended to integer p-th powers, of which the 2 lsd's necessarily
solve [2],because of inequality (X+Y)^p > X^p + Y^p for the corresponding
integer p-th powers < p^{2p}, where X,Y,Z are 2-digit integers solving [2]
mod p^2. ...QED (FLTcase1).

For details  http://www.iae.nl/users/benschop/nfb0.dvi  (full text)
http://www.iae.nl/users/benschop/scimat98.htm           (intro)
http://www.iae.nl/users/benschop/sgrp-flt.htm  (functions>arithm)
http://www.iae.nl/users/benschop/selmer.htm    (Hensel lift escape)
http://www.iae.nl/users/benschop/campaign.htm  (Hensel lift escape)
http://www.ams.org/preprints/11/199711/0index.html (abstract)

   and let me know what you tink of _that_ "simple" solution (for case1).

I bet you: simpler cannot (by direct extension of Fermat's own Small Thm ;-)

       Ciao xxxxxxxxxxxxxxxxxxxxxxxx1.1xxxxxxxxxxxxxxxxxxxxxxxx Nico
       FLT--------- AHA: One is Always Halfway Anyway -----------EDS

Subject:  Re: What is the definition of a "simple" proof of FLT?
   Date:  Sun, 09 Sep 2001 15:31:00 GMT
   From:  Nico 
  Org'n:  Digital Research
Newsgrp:  sci.math

"Robert J. Kolker" wrote:
>
> Wayne McDermott wrote:
>
> > I have been following the recent discussions regarding FLT with
> > some interest though no pretense of beginning to understand the
> > mathematics. From what I know of the history of the impossibility
> > theorem, Fermat claimed to have found a simple proof but didn't
> > have quite enough room in the margin to write it out. Is this
> > "simple" proof still being sought? The correspondence
> > to date appears to be using tools that I'm sure Pierre Fermat did
> > not have at his disposal, and as the Wiles-Taylor proof apparently
> > ran to 130 pages (I haven't read it!) it could hardly be described
> > as "simple".
> > Was Fermat just pulling everyones leg? Am I wildly off-topic here?
> > -- Kind regards
>
> Fermat was a contemporary of Descartes so ask what kind of
> math did he know?  Answer: algebra and geometry. Calculus
> had not been invented, nor group theory. So his simple proof was
> most likely some kind of induction proof involving elementary
> algebra with perhaps some geometricly related construct.
> Given the amount of brain power that has been expended
> on FLT and the fact that the best mathematicians in the
> world have had to resort to methods "way outside the box" my
> guess is the Fermat thought he had a proof but it was wrong.
> --  Bob Kolker

I guess a 'simple' proof could be understood by math-undergrads,
using basic methods taught in college. If such proof is possible,
say to start with the easier case_1 of FLT: x^p + y^p =/= z^p
for x,y,z coprime to prime p > 2, then something obvious must have
been overlooked, which of course is highly unlikely -- but not
impossible. Start with trying to find solutions for residues mod m,
where modulus m should relate to the problem at hand:
   how to test inequlality of integers?  By using a unique number
   representation and comparison. For instance here: p-ary code
   (that is: representation base p) -- and for k digits is m = p^k.

Now there are 'trivial' solutions mod p^k for k=1: Fermat's Small
Theorem: n^p == n mod p for all naturals n < p,  which he discovered
around 1637 (about that time he wrote his 'marginal note' on FLT;-)
By FST:  (x+y)^p == x+y == x^p + y^p (mod p)  x,y coprime to p,
for instance all x,y,z < p with z=x+y will do. No big deal of course.

So he might have looked at equivalence mod p^2 (last 2 digits).
And by Jove: there *are* solutions mod p^2, starting at p=7.
For instance if prime exponent p=1 mod 6, then the cubic roots
of 1 mod p^k, for any k > 1, will solve FLT_1:
   a^3==1 mod p^k --> a + 1/a == -1 mod p^k (where 1/a == a^2).
A generalization of this form is 3-fold (called 'triplet'):
  a + 1/b == b + 1/c == c + 1/a == -1 mod p^k, where abc==1 mod p^k.
This in fact is the general rootform for FLT case_1, not difficult
to prove, see [1]. Intro at http://www.iae.nl/users/benschop/func.htm

Now the next question is: if there is a solution (case_1) of the FLT
equation, thus the sum of two p-th powers *does* yield a p-th power,
then for residues mod p^k this solution must also hold for all k > 0.
A triplet-form solution mod p^k (k > 1) is a non-trivial necessary
condition for a  integer solution of FLT_1.

So it must be a triplet form (with cubic root form as special case).
Then: can any triplet rootform (necessary condition for an integer
solution) be extended to integers? The answer is NO, because via a
simple additive & multiplicative (viz. linear) transformation any
triplet solution can be brought into a form where x^p==x and y^p==y
mod p^k, and (x+y)^p == x+y (viz.  a solution "in Core", which core
is the unique and always existing subgroup of units of order p-1:
this is where undergrad stuff comes into play;-)
    sothat (x+y)^p == x+y == x^p + y^p (mod p^k)
(more precisely: two terms can be brought into Core, for a non-cube
solution, being sufficient, see 'core-increment' form in the paper)

Notice here the crucial fact that for *each* solution mod p^k this
equivalent form can be reached (by a linear equivalence preserving
transformation). And this form has Exponent p Distributing
over a Sum (in short: the EDS property holds for each solution in
Core, and: every solution of FLT_1 can be transformed to 'in Core'
or at least two terms 'in Core').

Now since for integer arithmetic (^) does NOT distribute over (+),
the 'direct' proof of FLT_1 *can* be completed without much trouble,
see [1]..  http://de.arXiv.org/abs/math.GM/0103014

And for a suggestion about Fermat's marginal note (the cubic root
solution mod p^2 for p=1 mod 6, starting at p=7):
see   http://de.arXiv.org/abs/math.HO/0103051

Recently this direct proof of FLT was published in the Nov.2005
issue of the Acta Mathematica of the Univ. of Bratislava, see :
http://pc2.iam.fmph.uniba.sk/amuc/_vol74n2.html , *and*

published as chapter 8 of book "Associative Digital Network Theory":
http://www.springer.com/computer/communications/book/978-1-4020-9828-4

With a similar proof (mod m_k = \prod. first k primes) of Goldbach
Conjecture (chapter 9) of only 10 pgs. This residue-and-carry method
(mod m_k) is quite general, e.g ch.10 shows a Waring-for-residues
result: each residue mod p^k (odd prime p, any k>0) is the sum of
at most 4 p-th power residues.

-- NB --
Intro to FLT: http://piazza.iae.nl/users/benschop/scimat98.htm

PS: you should critically consider R.Chapman's unsupported
  negative opinion: shall we say he's just jealous, or in denial?-)
He follows "His Master's Voice" (=A.v.d.P) as reported in the
non-discussion:  http://piazza.iae.nl/users/benschop/nr-th.htm
And about a formula of the probability to get a straight direct
proof of FLT accepted, (equiv. to 'breaking the Hensel Lift')
-- or a 'simple' proof of any other well known hard problem, see:
         http://piazza.iae.nl/users/benschop/inertia.htm