From FST to FLT via the Cubic roots of 1

All p-ary k-digit residues that are relative prime to p, hence no divisor p, form a cyclic group (one generator, say g).

Cyclic group G= g* (mod p^k),
with |G|= (p-1).p^(k-1) = A.B
is a direct product of two co-prime cycles A and B :

Each n in cyclic Core A of order |A| = p-1 satisfies: n^p = n mod p^k that is Fermat's Small Theorem (FST) extended to mod p^k (k>1). Every core-subgroup S >1 sums to zero (re: -1 in S for even |S| ).

Cyclic extension group B = (p+1)* of order p^(k-1), is generated by the powers of p+1 (mod p^k).

Plot all residues in G counter- clockwise along a circle, +1 right and -1 left (as usual in the 'complex' plane, only this is not complex).

All p-th powers in G form a cyclic subgroup ('subcycle') F = {n^p} of G, of order |G|/p, so |F| = (p-1).p^(k-2), and in fact F=A mod p^2. Cubic root 'a' of 1, hence a^3 = 1 (mod p^k), is in Core cycle A of G , and is a p-th power mod p^k (k>1) since it is in F. Notice that any solution to Fermat's Last Theorem (FLT) equation x^p + y^p = z^p must also hold mod p^k (prime p>2, k digits) for all k>1. . . Normalize to: a^p + b^p = -1 mod p^k.

Now a^3 -1 = (a-1)(a^2 +a +1) = 0 --> a^2 +a +1 = 0 (if a<>1) is a sum of three p-th powers with sum 0. . . Hence a counter example to the FLT inequality, but only for residues mod p^k (for any prime p=1 mod 6, sothat 3 divides p-1, and any k>1).

And a^2 = 1/a, sothat a +1 = -1/a for cubic root 'a' of 1 (mod p^k)..... (a<>1), while all three terms are in Core, thus a^p=a, 1^p=1 and (1/a)^p=1/a, and consequently : (a+1)^p = a^p + 1 (mod p^k),
where Exponent p Distributes over a Sum (EDS), which clearly cannot hold for integers.

In preprint [1] it is shown that all normed FLT mod p^k solutions (case 1 : x,y,z relative prime to p) have the next triplet form : a +1/b = b + 1/c = c + 1/a = -1, with a.b.c=1 (mod p^k), of which the cubic root solution is a special case (a=b=c). For p-th powers in F < G they occur for some p>=59, and all have this EDS property, sothat FLT (case 1) has no solution for integers: the sum of two p-th powers is not a p-th power (p>2).

Notice that there are just two symmetries of aritmetic: for addition and for multiplication (mod p^k):

Complement -n with n+(-n)= 0, and Inverse 1/n with n.(1/n)= 1 in G . . . linked in the G circle by a diagonal resp. a vertical . . (mirror about the horizontal axis).

Notice the cubic FLT_k root is a horizontal link : A+1 = -1/A . . ( A @ 120', 1/A=A^2 @ 240' )

Subject:  Re: The thirst for Challenge
   Date:  Tue, 18 Sep 2001 10:52:06 +0200
   From:  Nico Benschop 
  Org'n:  Amspade Research (Digital)
Newsgrp:  sci.math

Chan-Ho Suh wrote:
>
> "Nico Benschop"  wrote:
> [...]
> I must say the deck seems stacked against you, but I will certainly
> analyze your papers more thoroughly after having learned the relevant
> mathematics (which I expect to do this year); I am a geometer so [*]
> cannot comment with great confidence on the veracity of your results.

Re[*]: then you are excellently prepared, with an unprejudiced and
      open mind, to follow this approach:
[1] http://de.arXiv.org/abs/math.HO/0103051 
     "On Fermat's marginal note: a suggestion."

[2] http://home.iae.nl/users/benschop/nfb0306.pdf 
 "Triplets as additive structure of the Units group mod p^k ..."

Intro's at http://home.iae.nl/users/benschop/scimat98.htm [3]
           http://home.iae.nl/users/benschop/carry.htm
           http://home.iae.nl/users/benschop/cubic.htm
           http://home.iae.nl/users/benschop/ferm.htm

[1-3]: On "Fermat and the cubic roots of unity (1 mod p^k)"
   Clue: if someone shows you (x+y)^p == x^p + y^p  mod p^k,
  (prime p>2, k>1) for  x, y, x+y  'in Core' mod p^k, where
  'Core' = order (p-1) cyclic subgroup of units mod p^k, with
   n^p == n mod p^k for all n in Core, would you be very surprised
   if it can be shown that there is no corresponding integer
   ("mod-free") equation of p-th power integers < p^{kp} ?
  (with integers X, Y, X+Y such that x==X, y==Y, x+y==X+Y mod p^k )
  (Hint: since Pascal/Newton/Fermat & binomial expansion of (x+y)^p:
            exponent p does not distribute over a sum, for integers;-)
-- NB

-- (c) N.F.Benschop (Amspade Research) -- sep'97 -- may'03