Hi, Vladimir asked me to give an example of how my semigroup
approach to FLT (case1: x^p + y^p = z^p mdo p^k; x,y,z coprime to p)
works.

Prime p=59 is a nice example.
Have a look at my paper (homepg,ref[1]: */nfb0.dvi) table.2
(last page #14). You will see there the list of solutions of
the normed eqv: a^p + b^p = -1 mod p^2 (for prime p<200)...(2)..pg4.

At p=59 you see that, taking generator 2 we have G = 2* mod 59^2,
with order of units group |G| = (p-1)p = 58.59, and
the p-th powers form a subcycle F = { n^p } of order p-1 = 58.

The two p-ary components of each (a^p, b^p) pair summing to -1
mod 59^2, are listed for p=59: there are 6 solutions, in fact two
triplets (because solutions other than cucbic roots "C3" occur only
three-fold, as proven in thm3.2).

They are, if you look only at their mod p values (like the ones that
Palmerston mentions):
(15,43) (11,47) (54,04) and: (44,14) (38,20) (03,55)

Precisely these triples belong together, because they satisfy the
equivalence of a triplet as given in thm3.1, namely:

a + 1/b = b + 1/c = c + 1/a = -1 mod p^2, with abc=1 mod p^2

(where a,b,c are actually p-th power residues).

Now how did I find that? Very simple: I knew that the cubic roots
form an inverse pair (first discovered for p=7, see table 1, pg13):
a + 1/a = -1 because (.a): a^2 + 1 = -a mod p^k (p=1 mod 6)
thus: a^2 + a + 1 = 0, is one factor of a^3-1=(a-1)(a^2 +a+1)=0.

These solutions at p=59 cannot be a cubic root solution, because
p = -1 mod 6, so 3 does not divide p-1: there is NO subcycle of
order 3 with zero sum. To see if these solutions had anything to
do with inverse pairs in some mysterious way, I needed a way to
recognize inverses easily, which can be done by encoding each
p-th power residue by the exponent of a generator of the p-1 cycle
F = { n^p } they are in, in this case G = 2*, so F = (2^{p-1})*,
right? I dont even know now what number that generator of F is,
but I do know the exponents of the solutions pairwise in this
"log" (base f=2^{p-1}, F = f*) terms:

[-2,-25] [25,23] [-23,2] and: [27,19] [-19,8] [-8,-27]

Do you see the pattern? A negative sign in the exponent means an
inverse, does'nt it? And you see precisely the "chain" of inverses
as they are in the three equivalences of a triplet as in thm3.1

The next question: *why* always three? For the proof of that, the
function composition appraoch pays off: see in a + 1 = -1/a the
two main symmetries occur: Complement C(n)= -n for (+), and Inverse
I(n) = 1/n for (.) togther with Successor S(n)=n+1. The cubic root
solution connects the three together into a solution of FLT mod p^k.
This is the only place where non-commutativity (of function
composition) plays an essential role: the general "rootform" of
solutions of FLT mof p^k is the triplet.!. In fact not only for
-th powers, but also for ALL units in group G...(see section 3.1)

You see *why* I find FLT to be "at the core" of arithmetic?
(pun intended: I later chose the name "core" A_k, after first "axis"
as the subgroup of order p-1 in units group G_k for every k>0).

The cubic roots of 1 can *not* occur here at p=59, but a variant of
it apparently *does*: a+1= 1/b, b+1=1/c, c+1=1/a, with abc=1 mod p^k.
And NO other solutions can exist, see thm3.2...

Furthermore: *why* can't such solutions in residues (mod p^2)
be extended to a solution of FLT for integers?

The reason is that, in fact *independent* of the rootform (cubic
root solution, or triplet) each solution of FLT mod p^2 "in core"

A_2 = { n^p = n mod p^2} ('units" subgroup of order p-1)
satisfies: (x+y)^p = x+y = x^p + y^p mod p^2.

Interpreting these residues as the 2 lsd's of integer p-th powers
(of 2p digits base p) you have inequality (x+y)^p > x^p + y^p.

And F_2 = A_2, so 2-digit p-th power residues are the "core" itself,
easily seen by its order: |F| = |G|/p = (p-1).p^{k-1}/p = (p-1).p^{k-2}
which for k=2 equals p-1: the core order for *every* k>0 (FST: k=1).

Notice the typical method of research/discovery: use the computer
to do some (guided) simple experiments, in a print-format that
exposes the feature you might be looking for, and possibly displaying
some new 'pattern', that then must be algebraically explained, if
you guess it is not a Coincidence - but Necessity.

I hope this helps. Comments are welcome.
--
Ciao, Nico Benschop.
http://www.iae.nl/users/benschop

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