From his young friend Blaise Pascal in Paris he just got a letter, (say:) about the multiplicative structure (using factorial n!) of the coefficients in the expansion of (a+b)^n arranged in a triangle, skipping the a's and b's (clever trick to dispose of cluttering detail) although the coefficients are generated purely by addition, like a two-dimensional FiBonacci array, summing two neighbours repeatedly. - In fact, he just found that factorial structure himself, and he answered Blaise: . . is'nt it marvelous that truth is the same in Toulouse as in Paris ?! Quite different from theology, where truth depended on the last priest or cardinal you talked to.

No, that would never happen to his dear Mathematics, brought forward as the new science by Galileo (from Italy, in his 7-th year of house arrest, and two years before his death, for criticising the geo-centric, if not ego-centric, worldview pushed by the church), and by Descartes (who fled to Holland to save his skin, and escape the sad lot of Geordano Bruno at the stake in 1600; Galileo's students prepared an escape for him to Holland, but he felt too weak and old for the trip). --- Authority had no place in Math: everybody with a clear and open mind, possibly helped by some calculation device, that he sometimes borrowed from Blaise who designed one (photo)* , could check it out and find new laws, and test for the truth himself.

Later, Leibniz in Germany, even made a multiplier from that device, which sold quite well. Of course it worked decimal, that beautiful code introduced from Arabia to Europe some 400 years ago by FiBonacci (the son of Bonacci). Leibniz did give a fleeting thought to binary code, which was just coming from China, where it was used for the philosophy of balance (yin/yang) and in a 3+3 bit code for the 8 x 8 = 64 hexagrams of the I-Ching (book of Changes) [ic-59] since 2000 BC. But the number wheels in the calculator would be too many for quick carry propagation which was the bottleneck for speed, as well as requiring too much of the crank turning the wheels.

So Pierre sharpened his feather to do some doodling by hand in his notebook, using p-ary code in order to find some structure, and possibly find solutions to
[1] : a^p+b^p = c^p . . prime p>2 (composite powers he quickly found inessential). From Pascals' triangle he saw that p divides each number in the p-th row, except the 1's at the two ends, proven easily by their newly discovered factorial coefficient nature. And slowly his Small Theorem (FST) took shape: a^p+b^p=(a+b)^p for the last p-ary digit, or as it is now known: n^p = n mod p for all n (using the modulo notation of Gauss who thus formulated arithmetic 'closure' for residues mod m, in 1801). This was exciting because the exponent distributes over addition, most unusual, and probably leading to inequality for integers, rather than equality. Could FST also be possible mod p^k for k>1, and some special a,b ? That would at least provide a solution to his root equation [1] mod p^k.

To show what his Small Theorem FST is about : take any odd prime p (like 3, 5, 7, 11, 13, 17, 23, 29, 31 etc, having no proper divisor) and do residue arithmetic mod p. Normal number notation base p uses n = c.p + r, with carry 'c' and unique residue r from 0, to p-1. For residue arithmetic reset carry c=0. Fermat considered p-th powers g^p mod p, trying to find a generator g < p such that all its powers g = g^1, g^2, g^3, ..., g^(p-1) would be different. For instance mod p=7, then g=2 yields g^2= 4 and g^3= 8 = 1 mod 7 will not do since only half of the residues r < 7 are generated. But 3 does work : 3, 3^2 = 9 (base 10) = 1.7 + 2 = 2 mod 7, 3^3 = 3.3^2 = 3.2 = 6 = -1 mod 7. We can stop at -1 mod p , because the remaining h=(p-1)/2 powers mod p are the complement -g^i of those g^i (i=1..h) already found. Now 3 is called a primitive root of unity 1 (mod 7), since it generates all p-1 non-zero residues mod p. In short the generated p-1 cycle is: 3, 2, -1, -3, -2, 1 (mod 7). Now FST states that for every prime there is such primitive root g, generating a full p-1 cycle. It suffices (usually) to look for a primitive root among the divisors of p-1 (not proven), for instance p=191 has p-1 = 2.5.19 of which 19 is a primitive root (but not 2 or 5), its powers mod 191 generate a cycle of length 190.

Notice that in any such full cycle we have g^(p-1) = 1 mod p, the end of the cycle before it repeats itself, and g^h = -1 halfway. In fact it is easy to see that for any non-negative residue r < p holds r^(p-1) = 1 mod p, or : r^p = r (mod p). Even an extension to higher precisions mod p^k (k digit base p arithmetic) holds :
For odd prime p and any precision k > 1 there is a primitive root g < p that generates a cycle of length (p-1).p^{k-1} of residues mod p^k, denoted extension FST_k. The reason is that the powers of p+1 generate mod p^k a residue cycle of length p^{k-1}. For instance the cycle mod 7^2 is 6.7 = 42 long (D.Adam's magic number), generated for instance by 03, given in 7 columns of 6 residues with two 7-ary digits each:

+03 +43 +13 +53 -44 -04 -34
+12 +62 +42 +22 -65 -15 -35
+36 +46 +56 +66 -61 -51 -41
+44 +04 +34 -03 -43 -13 -53
+65 +15 +35 -12 -62 -42 -22
+61 +51 +41 -36 -46 -56 -66 = +01

Notice (mod 7^2) : -56 = 11 = p+1 = [3^{p-1}]^{p-1} , and (11)^7 = 01 : a length p cycle.
The cubic roots a^3=1 are 42 and 24 = (-42)-1, hence 42 + 24 + 01 = 00 (mod 7^2),
. . . and they are p-th powers : 42 = 3^14 = (3^2)^7, 42 = 3^28 = (3^4)^7 (mod 7^2)

And indeed, after some more trials, at p=7 F. found something interesting mod 7^2 (using base 7 number code): 42+01=43 where a=42=a^7 and a+1=(a+1)^7, and another pair b, b+1 with similar properties: 24+01=25. Moreover b=1/a, and a+b= 66 = -1 (mod 7^2). So a^2+a+1=0, implying (a+1)^2=a and (a+1)^6=1 --> p-1 cycle mod p^2. He convinced himself that this holds for any prime p=6n+1, so 3 divides p-1, allowing a 3-cycle {a, a^2, a^3=1} in the p-1 cycle, to be called the 'core' of the cyclic group of roots of 1 mod p^k, for any k>=1 --- (now using the 'group' concepts of Abel and Galois, who factored Gauss' closure around 1830, into an uncoupled product resp. cascade- coupled product of cycles: solvable groups; the final chapter of what Nicolo Fontana, alias Tartaglia the stutterer from Brescia, started around 1540 when he found, just in time before a challenged competition, a formula for the roots of any cubic equation - subsequently stolen by Cardano and published in his "Ars Magna" in 1543, without reference to its source).

So this p-1 cycle would be his Medium Theorem (FMT) or 'Core theorem': n^p = n mod p^k (for odd prime p and any k>0, and p-1 special fixed-point core-residues coprime to p) as extension of FST to higher precisions k. It provides a solution to his k-digit Large Theorem (FLT/k): a^p+b^p= -1 mod p^k . . [2] for odd prime p, k>1 with {a,b} "in-core" where a^p=a, b^p=b and b=1/a, a^3=1 mod p^k, sothat a^2+a+1=0: the cubic root solution of his equation mod p^k (prime p=6n+1). Now the full FLT for positive integers that he was after, and which he suspected to yield inequality for all primes p>2, would follow if ... [2] were the only possible solution of FLT/k, apart from some scaling factor. Then the exponent distributes over a sum: a^p+b^p=(a+b)^p mod p^k, clearly impossible for integer p-th powers < p^{kp} since the 'mixed terms' with a^i.b^(p-i) in the expansion do not sum to zero (which they do mod p^k for the cubic root case).

In fact, notice the recursion (a+b)(a^n+b^n) = a^{n+1} + b^{n+1} + ab(a^{n-1} + b^{n-1}). Writing S(n) for a^n+b^n, and shifing n+1 --> n this becomes: S(n) = (a+b).S(n-1) - ab.S(n-2) . . [3]. Then it is not difficult to derive, given {a,b} in core, that: a+b = -1 --> ab=1, so actually a cubic root-pair is the only solution 'in-core' for k>2. In this case [3] becomes S(n)= -{S(n-1)+S(n-2)} which is like a negative FiBonacci sequence. Starting with -1, -1 it yields for increasing n the repeated sequence {-1,-1,+2}* with period 3 : precisely as expected for cubic roots!

But what about other possible solutions not in core ? . . Without a clue, and Pascal's Calculator (PC) too slow and not geared for p-ary code to do some more experiments, he quit. Putting his notebook away, he let the problem rest. . .
If only he could have gone till p=59 (peanuts for our PC's) then he would have found a clue, and discovered a different kind of root which indeed is not 'in-core' with all three terms. With of course the next questions: if that would be the end of it - or maybe there would be still other types of FLT/k roots, and : would some EDS (Exponent Distributes over a Sum) type of argument hold for these new roots ?

"Triplets and symmetries of arithmetic mod p^k."

[12] "New Math (base p) applied to FLT . . . . (Fermat?) computing a^3=1 mod 7^2.
[margin.dvi] Expanded Version (21oct98) . . . . "On FLT and the Cubic roots of Unity." . . (abstract.htm)
[theory-edge] Example of triplets mod 59^2 (click2x) . . . Working with log code in G_k = g* (base g)
[sci.math-0] Re: Fermat . . extending Fermat's Small Thm. . . . . 10dec98
[sci.math-1] Re: FLT Question / and simple answer (case1) . . . 4feb99

[*] Two Pascaline foto's (of 5- and 8-digit calculators) ...222k ------------
Pierre de Fermat (1601-1665) biography
History of FLT
Re: JSH: Finally, an ending (FLT triplets: residue symm -n and 1/n)

       Author: benschop_nf  -- benschop_nf@my-dejanews.com
       Date:   1999/02/15
       Forum:  sci.math      " -- In FLT the Carry makes the difference --"
------------------------
Positive integer n = c.m + r
       ( modulus m, carry c, residue  r=0.. m-1 ) --> n = r mod m.
        p-th power:  n^p = (cm+r)^p = r^p  mod m.
        If n has k digits (base p), then n^p has upto kp digits.

FLT:     x^p+y^p = z^p ...[1]
     has no solution in integers x,y,z > 0; prime exponent p>2.
      " The sum of two p-th powers is not a p-th power" (anti-closure)
     Case1: x,y,z coprime to p,     Case2: p divides one of x,y,z.

Clues/reminders Case1:

1.  Represent integers over base p, and check first the
    necessary condition [1] mod p^k (k digit arithmetic base p)

2.  There are p^{k-1} residues mod p^k with divisor p (end on 0), so
    (p-1)p^{k-1} residues coprime to p, forming a cyclic group G_k.

3.  The p-th power residues form a subcycle F_k={n^p} of order
    |G_k|/p = (p-1)p^{k-2}, and for k=2 we have |F_2|= p-1, which all
    satisfy  n^p = n mod p^2 (n in "core" F_2). Compare FST: n^p=n mod p.

4.  Any solution of [1] mod p^k in F_k also satisfies [1] mod p^2, hence
    is "in core" F_2.  So that  (x+y)^p = x+y = x^p + y^p mod p^2  ....[2]
  --> So:  Exponent p Distributes over a Sum (EDS property), for residues.

5.  Any integer solution of [1] is in core F_2 mod p^2. For the corresponding
    2-digit integers X,Y: (X+Y)^p > X^p + Y^p; the three p-th powers < p^{2p}.

6.  Any extension by one or more digits: X'= X + u.p^2, Y'= Y + v.p^2
    (with terms < p^k) can be obtained (mod p^k, any k>2) from an assumed
    integer FLT equality by two operations (mpy: by factor s^p = 1 mod p^2,
    add: term t = 0 mod p^2) preserving the assumed integer equality,
    such that both X' and Y' are in core A_k. But such equivalence has
    no 0-extension to integers, so the assumption of an FLT equality for
    integers is false. (see for details: lem3.1 in ref[1] on homepage).

Reminder: the p-th power of a k-digit number has upto kp digits.
For k digits equivalence (mod p^k):
     the FLT(case1) integer inequality is due to the (p-1)k carries.
     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

For more intro's & details:
  http://www.iae.nl/users/benschop/carry.htm (Sum & Carry: closure & generation)
  http://www.iae.nl/users/benschop/ferm.htm  (Fermat & Cubic roots of 1 mod 7^2)
  http://www.iae.nl/users/benschop/selmer.htm (is Wiles' proof by L-series non-residue?)
  http://www.iae.nl/users/benschop/fewago.htm (Semigroup Z(.) mod m_k
                                                  for Fermat, Waring & Goldbach)
  http://www.iae.nl/users/benschop/nfb0.dvi (FST & EDS --> FLT, also case2) ..[1]

Constructive comments are welcome. (15feb99)

"Re: JSH: Hidden proof revealed, FLT for p>3"   (sci.math 9mar99)

James Harris wrote:
>
> When I realized a couple of days ago that my latest attempt at a proof
> of FLT was still incomplete, I was a bit bummed.  It took me about an
> hour to notice something interesting, which was that I had proven
> Case 1. ( Not so easy to see when you're playing with p=3 all the
> time which doesn't have a Case 1.)
> [...skipped several pages of formula's...]
>

For goodness sakes, James, you call that a *short* proof for p=3 ?
FLTcase1 for p=3 is done in a few lines via residues mod 3^2 as follows:

There is no solution of x^3+y^3=z^3 mod 3^2 for int x,y,z coprime to 3.
Proof:
  There are (p-1)p = 6 residues mod 9 that are coprime to 3, forming a
  cyclic group G_2 with generator for instance 2: {2,4,-1,-2,-4,1}
  containing the p-1 subcycle F_2 = {-1,1} of 2 cubes <>0 mod 3.
  All pairsums yield {-2,2,0} none of which is a cube coprime to 3. ***

Corollary:
  With no solution mod 9, there is no case_1 solution for integers. QED.

BTW:
  For any prime p>3 the FLTcase_1 proof is not much longer, using core F_2
  and the "Exponent p Distributes over a Sum" (EDS) property of
  any solution in p-1 cycle core F_2 ={n^p mod p^2}:  ...FST extended.
               ^^           ^^^^                       {n^p=n mod p^2}
--
Ciao, Nico Benschop -- http://www.iae.nl/users/benschop/carry.htm

__________/ if stuck@closure(mod..), use the carry \________________
          \      FST --> FST* --> EDS --> FLT      /

James Harris wrote:
>
> My position has always been based on the belief that there's this
> simple proof of FLT out there that hasn't been found for myriad
> reasons. I've always believed it was so simple that it'd be
> taught in high school (and not in advanced math classes either).
> Strangely enough, I seem to have found it, but I now face
> a dilemma shared with a reader who knows my history.
> How do I really believe it's true?
> [...] -- James Harris

Well, with that essential doubt you're in good company.
       It was Pascal's most urgent question to himself:
   *How* does one convince others (and of course first oneself;-)

He had a clever way of arguing, for which the Church "hired" him:
to underpin the belief in God's existence with a "proof".
Which Spinoza, and later Clarke, thought to have found, using
Aristotle's type of logic reasoning. And for which Leibniz
sought a math-language to clear all doubt, like some form of
"calculus of logic reasoning". Eventually provided by George Boole
(1848/1854: "The Laws of Thought") in his binary & idempotent &
commutative arithmetic x^2=x, y^2=y, xy=yx (the basis of set theory).

The clue is every time in the mapping of your problem "faithfully"
(isomorphically) onto such calculus model (for you: arithmetic & the
algebra of polynomials). In Spinoza's case, Boole found the "proof"
to be circular, since the result (God exists) was in the premises
(axioma's)....

Of course it helps if others agree with one's arguments... but there
is no absolute guarantee there either: they can also be wrong after
all, especially in such notorious case as FLT & Goldbach, where
wellknown and respected mathematicians *did* think to have a proof
(Lame') but did make a mistake...

That's why one can hardly expect from a math-journal editor (& referee)
to accept a simple FLT proof that "looks good" (which he even might
think to be correct;-). Why should *he* put his reputation (and of
his journal) on the line? No way!

In other words, don't forget this golden rule of human behaviour:
  " Don't stick your neck out, unless *absolutely* necessary ! "
(and: progress in science & math is *not* absolutely necessary...)

--
Ciao, Nico Benschop.             | AHA: One is Always Halfway Anyway
http://www.iae.nl/users/benschop | Institute of Advanced Engineering
http://www.iae.nl/users/benschop/ferm.htm
http://www.iae.nl/users/benschop/func.htm

Subject: Re: Intuition in mathematics
   Date: Wed, 17 May 2000 10:52:15 GMT
   From: Nico Benschop 
Organization: Research
  Newsgroups: sci.math
  References: 1 -- 17
-----
Jan-Christoph Puchta (U-Freiburg.de) wrote:
>
> Nico Benschop wrote:
>
> (snip)
>
> > As explained in http://www.iae.nl/users/benschop/scimat98 ...
>
> The mistake in the document mentioned above is almost precisely in the
> middle of it. (I assume that the reader of this post knows about this
> document or looks at it right now)
>
> The following is claimed:
>    The fact that (42)^7 + (24)^7 = (66)^7 mod 7^2
>              and (42)^7 + (24)^7 < (42 + 24)^7
> imply a "break [of] the Hensel lift".
> However, (42)^7 + (49 + 24)^7 = (66)^7 mod 7^2 and
>          (42)^7 + (49 + 24)^7 > (42 + 24)^7 is also true,  ..[*]

Indeed, precisely my point: in fact ANY extension (your's is adding
1.7^2 to one operand, and 0.7^2 to the other, thus: beyond the modulus
of 7^2) of the given cubic root solution, yields integer inequality
(or here in [*]: mod 7^21, if you like;-)

> hence the EDS-property cannot be used to give information about
> magnitudes. -- JCP

However, it _does_ suffice to show the required inequality
for _any_ (integer) extension of a given solution mod p^2.
(See the scale/shift argument preceding lem3.1 in the MS)
--
Ciao, Nico Benschop -- http://www.iae.nl/users/benschop/nfb0.dvi
                       http://www.iae.nl/users/benschop/nf-abstr.htm

Subject: Re: Chapman ad nauseam (or: "Breaking the Hensel lift")
   Date: Wed, 24 May 2000 13:40:58 GMT
   From: Nico Benschop 
Organization: Research
  Newsgroups: sci.math
  References: 1 -- 17
---
John R Ramsden wrote:
>
> Nico Benschop  wrote:
> >
> > As Bob S. would say (his sig): "You can bring a horse to the water,
> > but you can't make him drink" (or something to that effect;-)
>
> That's probably what Robin thinks about you Nico, as you never seem
> to address the objections he raises (apart from in effect repeating
> your assertions).
>
> > Just have a look at http://www.iae.nl/users/benschop/nr-th.htm for
> > an interesting:  "On a non-discussion in a NumberTheory list",
> > with the same ignorance about the existence of the 'carry', beyond
> > residue arithmetic... and the same stubborn (and arrogant) refusal
> > to read, and learn the basic facts of arithmetic with residues &
> > carry:
> >      http://www.iae.nl/users/benschop/carry.htm
>
> I can't see how it makes sense to talk about a "carry" with residue
> arithmetic, provided one decides in advance the largest prime power
> one proposes to use as the modulus.
> This "carry" concept just seems superfluous and misleading.
> Can you explain it with simple examples, in isolation, ..[*]
> without launching into an elaborate schpiel about its
> application to your results?   Cheers -- John R Ramsden
>

Sure, let me try.
However re[*]: without the context of FLTcase_1 and the Hensel-lift
(which I'll explain) it is indeed not very useful to consider the carry
concept 'in isolation'...

Re: the concept of 'closure' as in residue arithmetic, versus that of
'generation' as represented by the 'carry', for instance in the p-th
power of a k-digit pos.integer (base p).

Denote any integer n = c.m + r, with some chosen finite mudulus m>1,
then with residue r in range 0 .. m-1, the 'carry' c = (n-r)/m follows,
the corresponding multiple of the modulus (possibly >1 digits base m).

Finite arithmetic Z(+,.) mod m is 'closed', yielding with every pair
of operands in [0,m-1] a result also in [0,m-1] ... considering the
'base m' representation.

Extending some result for residues mod m (say 'closure' under addition
or multiplication) to integers is done by extending the arithmetic to
mod m^k, by induction over k --> inf. Actually, k=2 suffices in the
first extension step, because for non-negative {a,b} < m we have
sum: 0 <= a+b < 2m < m^2 (carry= 0 or 1), and product: 0 <= a.b < m^2
as well (carry <= m-2).

Repeated multiplication, for instance n^p (prime p) over base p,
yields two idempotents mod p: 1^2 == 1 mod p, and 0^2 == 0 mod p.
(re Fermat's Small Thm FST: n^p==n mod p, for all n, and any prime p).

But mod p^2 notice that (p+1)^p == 1 mod p^2, producing a carry =0
"beyond mod p", denoted (11)^p = 01 (mod p^2), although p+1 (code 11)
is idempotent mod p (single digit). And any n
1. Those p-1 residues with n^p == n mod p^k
(there are precisely p-1 such, coprime to p) form a p-1 order cyclic
group ('cycle') which I call the core A_k, of order p-1 for every k>0,
extending FST from k=1 to all k>1. Actually, any k-digit residue (k>1)
can be recognized as a p-th power residue by only its two lsd (less
significant digits): ANY msd extension of a proper p-th power
residue mod p^2 is _also_ a p-th power residue. ..[&]

In the context of FLTcase_1 the major (& correct) objection against
a direct proof (via ONLY residues mod p^k) of the required inequality
for integers is the "Hensel lift". Meaning that given an equivalence
x^p + y^p == z^p mod p^2 (for some prime p>2, and some x,y,z integers
coprime to p), a solution can be obtained mod p^k for _every_ k>2,
which are _still_ residues: thus 'never' become integers, no matter how
large k becomes, since finite integers require inf.# of leading zero's
beyond the msd (most significant digit) of weight p^{k-1}.

Hensel's infinite extension ("lift") is easily seen by normalizing to
a^p + b^p == -1 mod p^k, obtained upon multiplying by -z{-p} in units
group mod p^k. Notice that "-1" is represented as a string of k digits
all equal to p-1. Then ALL p^{k-2} extensions of a^p + b^p == -1 mod p^2
yield equivalence, freely chosing the next msd (a^p)_i in [0..p-1], and
corresponding (b^p)_i =(p-1)-a.
Each such extension is by [&] a p-th power residue mod p^k (k>2).
So k=2 is necessary & sufficient for the Hensel lift of p-th powers
[which form a p-1 cyclic subgroup of all (p-1)p units mod p^2].

So to prove inquality for integers one requires:
 ------- something beyond residues, in general: the 'carry' -----
showing that the p-th powers a^p and b^p of residues {a,b} < p^k, which
_as integers_ have upto kp digits, must necessarily yield inequality
of digits with weights p^m (m >= k), and that for EVERY finite k>0.

And that this holds for _every_ finite k that you please to take.
Compare it with a 'bootstrap' principle: look beyond the k digits
of the residue equivalence you start with (and that DOES exist for some
primes p>6 and some x,y,z residues mod p^2: I derive the general
'triplet' form of any solution of FLTcase_1 for residues mod p^k, k>1:
 a+1 == -1/b,  b+1 == -1/c, c+1 == -1/a, and abc == 1 mod p^k,
not only for p-th power residues a,b,c, but for ALL units).

The clue of this 'bootstrap' inequality is that exponent p distributes
over a sum mod p^2 (EDS property, or a similar property for non-cubic
roots of 1 in residues): (a+b)^p == a^p + b^p mod p^k, where the
required inequality for p-th power integers < p^{kp} is caused by the
(p-1)k carries, for ANY finite k>1 that you please to take.
Since (a+b)^p > a^p + b^p for the corresponding integers a,b < p^k,
the diffence being equal to the sum of the 'mixed_terms' in the binomial
expansion, which sum vanishes mod p^k, however NOT mod p^{kp}.
*** So: for p-th power integers the carries make the difference. ***
[for details see lemma 3.1, and preceeding eqns (4,5,5') in the MS]
 --
Ciao, Nico Benschop -- http://home.iae.nl/users/benschop/nfb0505.pdf (*)
                        http://home.iae.nl/users/benschop/campaign.htm

(*) published in Acta Methematica of the Univ. Bratislava (nov 2005):
    as: "Additive structure of the group of units mod p^k,
     with Core and Carry concepts for extension to integers"
    http://pc2.iam.fmph.uniba.sk/amuc/_vol74n2.html (pg 169 - 184)