-- Breaking the Hensel lift : Carries make the difference in FLT mod p^k
Nico Benschop
FST (n^p=n mod p) and its (cyclic) "core"-extension to p-1 residues mod
p^k (any k>1) -- With any FLT mod p^k sol'n "in core" having:
(x+y)^p = x^p + y^p (mod p^k),
so: Exponent p Distributing over a Sum ("EDS") blocking a corresponding
integer solution {X^p,Y^p} < p^{kp}. After all: the p-th power of a
k-digit integer has kp digits, thus with (p-1)k "carries" beyond p^k.
And k=2 is nec&suf to settle FLTcase1; see:
http://www.iae.nl/users/benschop/nfb0.dvi (full 12pg text)
http://www.ams.org/preprints/11/199711/0index.html (intro)
http://www.ams.org/preprints/11/199712/0index.html (Fermat and the
Cubic roots of Unity)
"The Hensel lift":
The objection, as I learned after a while, is the next wrong opinion
against a direct FLT proof via residues (mod p^k). Namely the Hensel
extension lemma (in his p-adic number theory, early this century).
Which for FLTcase1 (x,y,z coprime to pime exponent p) has
the next consequence: a solution of x^p + y^p = z^p mod p^2 ...[1]
implies p solutions mod p^3, p^2 solutions mod p^4, .... and p^{k-2}
solutions mod p^k for *any* k>2 (hence equivalence for all k-->inf).
No inequality for integers was thus thought possible to follow....
This is simply because [1] can be "scaled" sothat one term is 1
(each x,y,z in units group G_k coprime to p has a unique inverse in
that cyclic group of order |G_k|=(p-1)p^{k-1} ).
In normalized 'scaled' version [1] becomes a^p + b^p=-1 (mod p^2), if
divided by -z^p. Now -1 mod p^k has code ...qqqqqqq (k digits q= p-1),
and *any* msd (more-significant-digit) extension of [1] with digitwise
sum (a^p)_i + (b^p)_i = p-1 will provide a proper solution (note that
for the same reason: *each* of the p^{k-2} possible msd extensions of
a 2-digit p-th power to a mod p^k residue is again a p-th power
residue;-)
Now you see the 'problem' of the infamous Hensel lift: a finite
integer solution requires *finite* k digits, and extended by leading
zero's. This cannot solve [1] extended to any k, because that does NOT
yield -1 (which has leading q's "forever" ;-)
You see: the Hensel lift (for k-->inf) was thought to imply no
possibility to derive integer inequality in FLTcase1.
However: a solution to [1] mod p^2 has one special property:
it occurs in "core" F_2 = {n^p} of p-th power residues coprime to p,
of which there are precisely p-1, within the group G_2 of (p-1)p
residues coprime to p (called "units").
This is most interesting, because Fermat's Small Theorem is *also*
about a p-1 cycle, namely n^p=n mod p --> n^{p-1}=1 mod p, for *each*
n<>0 mod p. So [1] is a solution "in core" mod p^2: the unique p-1
cycle of p-th power residues.
Then: (x+y)^ = x+y = x^p + y^p (mod p^2), while *each* FLT
solution of p-th powers *must* be in core F_2= {n^p} mod p^2, ...OK?
Conclusion: despite the Hensel lift: inequality for finite integers
(viz. with leading zero's) follows, due to the EDS property of *any*
solution mod p^2, with (X+Y)^p > X^p + Y^p in corresponding 2-digit
integers X,Y and 2p-digit p-th powers X^p, Y^p < p^2p.
Just check out a computation example for p=7 (with x,y the two non-1
cubic roots of 1 mod 7^2, as given an earlier sci.math reply to
bo...@rsa.com - copied in my homepage, at bottom: "New Math (base p)"
http://www.iae.nl/users/benschop/ferm.htm
Of course, it takes quite some courage to admit the oversight, that
the Hensel lift does *not* block deriving integer inequality, as was
always thought (this century, and before that as well, you don't need
Hensel's p-adics to see this). And moreover, this approach follows
*directly* by extending FST (Fermat's Own Small Thm), nota bene!
That is why no answer comes anymore: it is *so* ambarassingly simple,
that no math-journal, in the past 4 years & 15 submissions, had the
courage to publish it (despite three conferences where I presented it,
at different emphases): http://www.iae.nl/users/benschop/campaign.htm
Frankly, I'm about to give up on normal publication (maybe I'll try a
monograph, including Waring & Goldbach by the same method: add've
analysis of Z(.) mod m_k, with m_k = p^k for Fermat & Waring powersums,
and m_k = {\prod first k primes} for Goldbach).
Because publication in a math-journal would require not only a
(anonimous) referee to understand (easy;-) and agree (difficult),
but also the (non-anonimous) editor to agree (almost impossible;-(
Because: it *cannot* be so simple, can it...?...
I made a separate study of this phenomenon: look up the nine reasons
*why* this simple proof via FST & EDS --> FLT(case1) was overlooked,
in the Conclusions section of my special paper on Fermat & cubic roots
of 1 mod p^k: "On Fermat's marginal note: a suggestion"
http://www.iae.nl/users/benschop/marg-flt.htm
(rejected for publication in the full-text digest of the
Netherlands Math Congress NMC33, apr98, U-Twente, NL --
where I presented it last April - see abstr.digest p39)
I think it's more about human psychology than about math, really;-)
Ciao, Nico Benschop -- http://www.iae.nl/users/benschop
____ At equivalence FLT mod p^k, the Carry makes the difference _____
FST --> FST* --> --> EDS --> --> FLT