Re: fermat's last theorem
- From: Gottfried Helms <helms@xxxxxxxxxxxxx>
- Date: Fri, 30 Dec 2005 15:14:18 +0100
Am 30.12.2005 11:59 schrieb Fermat:
> Richard, no offense, but you seem to have a limited understanding of
> mathematics. There is a emormous but fundamental difference between
> knowledge and understanding. You may know but you do not understand.
> Also "very likely in my opinion" is quite ambigous, please use complete
> sentences.
>
Well, recently I stumbled in such an error, which *could* be
happen, that also Fermat stumbled in. I'll report it here,
since the way of reasoning is very accurately along the famous
theorem, which Fermat dealt with and actually proved: his
little theorem.
Using the little theorem only one can easily find, that the
residues of f(n,a,b) = a^n - b^n (mod p) are cyclic, and the cyclicityness
puts some hard conditions on the primefactorization of f(n,a,b).
For example, use (a,b)=(2,1) and write only f(n).
Let ß(p) be the length of such a cycle for a prime p.
Now write
a
~ := 1 if b divides a, and 0 if it does not.
b
Write also {a,b} for the highest power of b dividing a, just
for a concise notation.
Then, from a short empirical table you would find a much
convincing result that, with a prime-exponent r the prime-
factorization of f(r) looks like
r r r
~ (1 + {r,3}) ~ (1 + {r,5}) ~ (1 + {r,7})
0 ß(3) ß(5) ß(7)
f(r) = 2 * 3 * 5 *7 *...
This shows in each exponent two rivaling conditions, which
bounds the exponent of each primefactor :
since gcd(ß(p),p) is 1 the ß(p)- and the {r,p}- term are in concurrence:
1) if r equals ß(p) the {r,p} is zero and the power of the primefactor p
is just 1*(1 + 0)
2) if r equals p, then we have 0(1 + 1)
For each combination of coprime (a,b) in f(r,a,b) this looks
completely the same, except that the factors of (a-b) must be
added to the rhs, if (a-b)>1.
But since a quick heuristic table suggests, that the primefactors
generally can only occur with an exponent 0 = 0(1+1) or 1=1*(1 + 0)
that difference (a-b) must provide primefactors q of powers
at least r-1 itself, with the consequence that not only a^r-b^r has
primefactors of the power of r, but even (a^1 - b^1) must be
completely composed of primefactors q to the power of r or of r-1,
like a-b = q^r * s^(r-1) ...
The next step even strengthening the assumption of having the
key of a proof, may then be, that with n=ß(p) *each* number n
occur at least once, namely the prime r, so whatever difference
a-b = q^r * s^(r-1) I choose, there will be -at least- one primefactor
p, whose ß(p) is exactly r, thus this primefactor has only the
exponent 1 in the above prime-factorization and must be contained
in (a-b) as (a-b) = q^r * p^(r-1) where p is such, that ß(p)=r
From all that follows, that the original equation, replacing "a" by
"b + q^r * p^(r-1)"
now looks like
(b +q^r*p^(r-1))^r - b^r =?= q^r * p^r * ...
and following that condideration one might come across some more
impossibilities and contradictions.
The empirical evidence for FLT from here is strong, -just look at
such tables- that there is in fact no solution, and such a reasoning
used only things, Fermat was not only aware of but even the
inventor.
Only that the table is in one point false; the formula cannot
generally use ones in the exponents, but must respect, that
there are effects like wieferich primes, where the "one" has to
be replaced by "two" ... and also by higher values. Say, µ, for
the "power of p at its first occurence ß(p)" or as implicite
functional definition:
f( ß(p) ,a,b) = p^µ * x
Then the corrected version is
r r r
~ (µ + {r,3}) ~ (µ + {r,5}) ~ (µ + {r,7})
0 ß(3) ß(5) ß(7)
f(r) = 2 * 3 * 5 *7 *...
(in fact here is a further correction omitted to keep the basic
approach readable)
The first occurence of µ=2 is at primefactor 1093 and if we keep
(a-b) on the lhs for this consideration, establishing the
cyclotomic version f(r,a,b)/(a-b) then µ>1 on the rhs are
extremely rare.
You need computer power to search for such numbers p systematically,
and for the pair (a,b) = (2,1) it was about 1900, that the two
occurences 1093 and 3511 were found, and until now it is checked,
that there are no others below ~10^12 (I think; see "wieferich primes")
So without the idea of the possibility, that µ could be greater
than 1 in the cyclotomic case, and is restricted by the primefactors
of (a-b) in the general case, the above observation may be
sufficient for a first shot, and a remark of "marvelous proof" -
because it is so simple, and relates practially only on the
little theorem, the cyclicityness-property and the contrary
restrictions given by r~ß(p) and {r,p}, seems "understandable".
Extending the empirical observations one may find some occurences
of a µ>1 and one would accept, that one must have overseen a
vital aspect. The conditions of divisibility of f(r,a,b)/(a-b) by
a square are very unclear, and - assumed this way of arguing would
sketch Fermats idea a bit at all (if actually he had one) - it is also
logical to simply withdraw such a -marvelous- idea of a proof and
try to find other ways, as he also did (the latter: to search).
Anyway, we shall not know, whether he had or had not a promising
key in his mind, and also not, whether the above reasoning is
related to his key-idea at all.
However, I think, it is an interesting way of looking at FLT in a very
elementary way, which still leads far into the matter of further
numbertheoretical questions and problems.
Gottfried Helms
.
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