Fermat’s Last Theorem: A one-operation proof

Victor Sorokine

Thanks +
Another step to the perfection

Algorithm of the Proof (Case 1):
1) Transformation of u_{k+2} into 2 [or > 1, but < n] with multiplication by 1 + ng.
2) Transformation of c_3, c_4, … c_k, c_{k+1}, c_{k+2} into 0 with multiplication by 1 + gn^2, 1 + gn^3,… 1 + gn^(k+1).
Then:
3) v = (a_{k+2} + b_{k+2} – c_{k+2})_1 = (1; 2) > 0.
4) U' = a_(k+1)^n + b_(k+1)^n – c_(2)^n >> 0.
5) U" = U – U' < 0.
6) (– U")_{k+3} = v; U"_{k+3} = – v.
7) U'_{k+3} = – v.
8) U_{k+3}= (U'_{k+3} + U"_{k+3})_1 = – 2v =/0.
The proof is done.

Next topic: Case 2.

Victor

ramsey2879

All the more reason to insist that Victor present a complete proof in a single paper that is not lacking detail, i.e., such that each line is clearly correct from what has been just stated above. At this time, I don't think anyone can make out a proof from what Victor has given us. Does anyone disagree?

arildno

Frankly, I can't make head or tails out of this.
The least we should demand, is that V.S. shapes up his, at times, extremely confusing notation and presents a properly formatted proof for perusal (preferably LATEX).

As it is now, it is utterly impenetrable, at least to me.

Victor Sorokine

One step back: return to "Fantastic idea for my friends" (cf. Forum, 08.30.2005)

Heart of the proof:
After transformation of u_{k+1} into 1 the equation U = a_(k+1)^n + b_(k+1)^n – c_(k+1)^n = 0 has no solution since the number u = a_(k+1) + b_(k+1) – c_(k+1) is odd.

Attempt to correct this situation with help of the insertion in the numbers a_(k+1), b_(k+1), c_(k+1) following digits from a, b, c is doomed to failure:
transformation of U_s [=/ 0] into 0 requires to add an odd/even number to U_s and add en even/odd number to u. Therefore after this operation U = a_(s)^n + b_(s)^n – c_(s)^n =/ 0 and therefore there is other digit U_r =/ 0.

Attempt to correct this situation with help of the insertion in the numbers a_(s), b_(s), c_(s) following digits from a, b, c is doomed to failure…

AND SO AD INFINITUM!!!

Think yourself about that!

V.S.

Hurkyl

Ok, I'm going to face the fact that I'm just not finding the time to devise demonstrations of the flaws in the sequence of calculations.

Victor, you're going to have to start actually writing a proof or I'm going to close this. A proof involves writing out clear claims and a clear justification of them. A proof is not simply a list of equations, occasionaly with a couple words of explanation.

Chronos

Perhaps dissecting Wiles proof and pointing out the unnecessary steps would be instructive.

symplectic_manifold

You guys should change your forum policies...I wonder how you can stand that ignorance directed to you and spend your precious time with stupidness of this degree!...You should have closed this topic at the very beginning!
Only one look at the paper of this V.S. is enough to make clear for oneself what we are dealing with here: scientific incompetence, ignorance, insolance and the attempt to distinguish one's own "genius".

Chronos

I liked the VS conjecture. He gave the 'math' and asked fair questions. No harm in that. It was good science, and I enjoyed the discussion. I'd rather hear that kind of argument 100 times over than another 'logical' brief on why Einstein was wrong.

ramsey2879

True, there was no harm and Victor did present enough material to interest me as I learned something out of this thread. But it's time to insist that Victor present a proof that can be followed in the manner set forth by the forum. I would wait until Victor at least makes a reasonable attempt to do this.

Victor Sorokine

Thank! +:

Dear Friends!
Your criticism has help to me to finish the research.
While the proof is executed, I give any fact:

1) Obligatory transformation of the number u into 99…9900…00, which has s digits.
2) The contradiction is discovered in the digit… u_t, where t > s [EVRICA!]:
- or U'_"t" = a_(t)^n + b_(t)^n – c_(t)^n =/ 0,
- either the number of the digits in the number u is equal to t.
Attention: there are two specieses of digits u_t = 0:
(a_t + b_t – c_t)_1 = "9" and
a_t + b_t – c_t = – v (where 0 < v < n).

Thank + thanks
V.S.

Victor Sorokine

For interesting reflections

The digits and the endings in the proof + inequalities

Let s – the number of digits in the number u = 99…9900…00 и
t [t > s] – the least [and only one!] rank with an equality a_t + b_t – c_t = – v_"t" (where 0 < v_"t" < n). Then:

The numbers a_(s), b_(s), c_(s) is such:
*) at best: a_(s) = "9/2""9/2""9/2"…, b_(s) = "9/2""9/2""9/2"…, c_(s) = 000…;
*) at worst: a_(s) = 999…, b_(s) = 999…, c_(s) = 999… [Here "9" = n – 1].
But [a_(s) + b_(s) – c_(s)]_{s+1} = 1. Therefore:

If s < i < t, then for digits of the rank i there is htly the equality:
(a_i + b_i – c_i)_1 = 9, and again:
*) at best: a_(i) = "9/2""9/2""9/2"…, b_(i) = "9/2""9/2""9/2"…, c_(i) = 000…;
*) at worst: a_(i) = 999…, b_(i) = 999…, c_(i) = 999… [Here "9" = n – 1].

For the digits and endings of the rank t there is the equality:
**) (a_t + b_t – c_t)_1 = – 1 и [a_(t) + b_(t) – c_(t)]_{t+1} = 0

For rank i, where s =< i < t (cases *), there is the equality: a_(i)^n + b_(i)^n – c_(i)^n > 0.
For rank t (case **) there is the equality: a_(t)^n + b_(t)^n – c_(t)^n > < 0.

Therefore, there exist such u_r, where r > t, for which (U'_"r")_{r+2} = [a_(r)^n + b_(r)^n – c_(r)^n]_{r+2} =/0, а (U''_"r")_{r+2} = 0.

V.S.

ramsey2879

I am sorry Victor, but what you recently posted is not comprehensible, and is completely without merit given the warnings to avoid unjustified statements, i.e. which have no appearance of logical basis. We are not mind readers!! See http://www.artofproblemsolving.com/R...iteJustify.php

moshek

Hi ramsey2879

I appreciate very much your effort to glide victor with is very interesting intuition about FLT. Remember that Fermat did nor write any prove and A.Wiles did not answer if Fermat have the solution in his mind.
Therefore I think that this thread should be open.

Victor is certenly doing progress in his direction by our remarks, manly yours.

Moshe


Dear victor

Please write now all your paper on FLT
from the begining to the end.

thank you
Moshe

ComputerGeek

please oh please kill this thread.

moshek

Please explain to me why ?