# Fermat’s Last Theorem: A one-operation proof

#### Victor Sorokine

Explanations

ramsey2879 said:
So much for a concise proof which does not require study of other documents which you did not refer to until now.
Anyways. Your logic is flawed. Statement 2 does not follow from statement 1 since u_{k+1} can be any number less than n.
Also while a,b,and c can each be multiplied by d to get u_{k+1} to equal n-1, how does this lead to statement 3 that u_{k+2}=2? On the contrary, u_{k+2} may equal n-1.
Similarly: statement 5 does not follow from statements 1-4
Statement 7 does not follow from statements 1-6.

About the counterexample. It still stands even though U'+U'' =/0. If the logic of (20°) in doc: Revista Foaie Matematică: www.fmatem.moldnet.md/1_(v_sor_05).htm
in pdf: http://fox.ivlim.ru/docs/sorokine/flt.pdf is correct it would make no difference whether or not U' + U'' = 0. The formula for U' and U'' should work for any a,b and c or your statement is not logical.
"Your logic is flawed. Statement 2 does not follow from statement 1 since u_{k+1} can be any number less than n."
1) u > 0, hence
(AFTER multiplication Fermat's equation by such d^n that (ud)_1 = n – 1, or "9")
2) u_{k+1} = "9" > 0.

" Also while a,b,and c can each be multiplied by d to get u_{k+1} to equal n-1, how does this lead to statement 3 that u_{k+2}=2?"
Let (u/u^k)_(2) = gn + (n – 1). Then there is such number f = 1 + xn that in the product [gn + (n – 1)](1 + xn) second digit is equal 2. Or [gn + (n – 1) – xn]_2 = 2, from here x = g – 2.

"The formula for U' and U'' should work for any a,b and c…"
Yes, but only in the Fermat's equation U' + U'' = 0, from here U'_{k+2} = U'' _{k+2} = 0 (since U'' _(k+2) = 0!).

V.S.

#### ramsey2879

Victor Sorokine said:
"Your logic is flawed. Statement 2 does not follow from statement 1 since u_{k+1} can be any number less than n."
1) u > 0, hence
(AFTER multiplication Fermat's equation by such d^n that (ud)_1 = n – 1, or "9")
2) u_{k+1} = "9" > 0.

" Also while a,b,and c can each be multiplied by d to get u_{k+1} to equal n-1, how does this lead to statement 3 that u_{k+2}=2?"
Let (u/u^k)_(2) = gn + (n – 1). Then there is such number f = 1 + xn that in the product [gn + (n – 1)](1 + xn) second digit is equal 2. Or [gn + (n – 1) – xn]_2 = 2, from here x = g – 2.

"The formula for U' and U'' should work for any a,b and c…"
Yes, but only in the Fermat's equation U' + U'' = 0, from here U'_{k+2} = U'' _{k+2} = 0 (since U'' _(k+2) = 0!).

V.S.
As to the first two paragraphs these statements belong in the proof to enable others to follow your logic. As to the counterexample, the counterexample is contrary to your admission that the formula for U' and U'' should work for any a,b,or c (whether or not a^n+b^n = c^n. I studied the text where you arrived at
"(4°) [a_(k)^n + b_(k)^n – c_(k)^n = U',
(5°) (n^(k+1))[a_{k+1} + b_{k+1} – c_{k+1}] + (n^(k+2))P = U'', including
Cf. (20°) in doc: Revista Foaie Matematică: www.fmatem.moldnet.md/1_(v_sor_05).htm
in pdf: http://fox.ivlim.ru/docs/sorokine/flt.pdf
but the text does not support the conclusion since these lines do not follow logically one line from another. As seen by my counterexample, U'' = U-U' =/ (n^(k+1))[a_{k+1} +b_{k+1}-c_{k+1}] +(n^(k+2))P as stated in (5°). If you still maintain that it is so, then you should start out with U'' = a^n + b^n -c^n - U' and line by line show logically how it can be so when my counterexample shows it not to be so.

#### Victor Sorokine

Ok!

ramsey2879 said:
but the text does not support the conclusion since these lines do not follow logically one line from another. As seen by my counterexample, U'' = U-U' =/ (n^(k+1))[a_{k+1} +b_{k+1}-c_{k+1}] +(n^(k+2))P as stated in (5°). If you still maintain that it is so, then you should start out with U'' = a^n + b^n -c^n - U' and line by line show logically how it can be so when my counterexample shows it not to be so.
OK:
1) U'' = a^n + b^n -c^n – U', or U'' + U' = = 0, where
2) U''_(k+2) = 0 (cf. 20°),
3) therefore U'_(k+2) = 0 (cf. 20°),
4) therefore (U'_{k+3} + U''_{k+3})_1 = 0 (cf. 20°).
5) therefore U'_{k+3} = – U''_{k+3} since U'_(k+2) = U''_(k+2) = 0 and U'' + U' = = 0.
6) Therefore U'' =/ - U' and a^n + b^n - c^n =/ 0.

V.S.

#### ramsey2879

Victor Sorokine said:
OK:
1) U'' = a^n + b^n -c^n – U', or U'' + U' = = 0, where
2) U''_(k+2) = 0 (cf. 20°),
3) therefore U'_(k+2) = 0 (cf. 20°),
4) therefore (U'_{k+3} + U''_{k+3})_1 = 0 (cf. 20°).
5) therefore U'_{k+3} = – U''_{k+3} since U'_(k+2) = U''_(k+2) = 0 and U'' + U' = = 0.
6) Therefore U'' =/ - U' and a^n + b^n - c^n =/ 0.

V.S.
If a=5, b=4, and c=3, n=3 U''_(k+2)=200 =/ 0 so this means that 20° is not a true statement. What I meant by line by line is with the logic supporting each line so evident that no further study has to be made. Forget a reference to other work. Please show all work so it can be followed. I repeat here that so far you have not shown that U''_(k+2) must equal 0.

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#### Victor Sorokine

ramsey2879 said:
I repeat here that so far you have not shown that U''_(k+2) must equal 0.
- Cf. binominal theorem for PRIME n:
a^n = [a_(k+1) + n^(k+1)a_{k+2}]^n = a_(k+1)^n + n[a_(k+1)^(n – 1)][n^(k+1)a_{k+2} + n^(k+2)P] = a_(k+1)^n + n^(k+2)[a_{k+2}) + n^(k+3)P.
Therefore a^n_{k+2} is not a function from a_{k+2}
(and therefore U_{k+2} is not a function from v).

#### moshek

Dear Victor,

Maybe you know how to prove FLT in a very simple way by looking on base מ and the k+3 digit of a^n+b^n-c^n.

Your attitude is really beautiful !

But it look that there are many holes in the writing of the prove.
Ramsey work really hard for you you sould thank him , he discover mistakes and then you correct it - that's fine.

I want to ask you were is your responsibility to preset here the most update version of your work. If you can't do that I am afraid you will never receive the recognition that you want.

So please do arrange everyting together like you write it for the first time but with the new insight you got from us , and share your paper with us

Moshe

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#### Victor Sorokine

two holes

moshek said:
But it look that there are many holes in the writing of the prove.
Dear Moshe,
You are right: there are 2 holes:

After transformation u_{k+1} into n – 1 [with one-digit-multiplier g]
and u_{k+2} into 2 [with two-digit-multiplier 1 + ng]
we transform also
c_{k+1} into 0 [with multiplier 1 + (n^k)g; the digits u_{k+1} and u_{k+2} do not change!]
and c_{k+2} into 0 [with multiplier 1 + (n^(k+1))g; the digits u_{k+1} and u_{k+2} do not change!]
Now U' >> 0, U'' < 0, n > v > 0 [since c_{k+2} = 0].
And therefore U_{k+3} = (n - 2v)_1 =/0.

V.S.

#### moshek

Last theorem of Fermat..

Ok Victor thank you !

Still you have to write the all prove together.
Now Ramsey , do you think that we can declare
That FLT can be settle in a very simple and Organic way
as was done [ until now ] nicely by Victor ?

Thank you
Moshe

p/s : My own interest in mathematics is about the hidden connection
between mathematics an physics as you can read in my paper ]

Intelligent life

Moshe Klein
‏2005–03–19

Planet Mars was associated twice during human history with the search for intelligent life. First was the effort to understand the retrograde movement of Mars on the background of the night sky. Second were the marks on Mars face that were at some point confused to be canals created by other cultures. Perhaps there is connection between Mars and the research of life in the area of mathematics?
Thousands of years ago people observed that Mars is moving sometimes in the opposite direction relative to the other stars. Mars is moving slowly on the background of the night sky so that you can distinguish the difference every day. From time to time it suddenly stops and then starts to move in the opposite direction. Then it stops again and continues its movement in the regular direction.
To explain this unusual movement the astronomers invented complex systems of wheels in which planet Earth was always in the center. The most complicated system was invented by Ptolemy in the second century AD, and included 12 wheels. But 1500 years later (about 400 years ago) Copernicus invented revolutionary idea to put the sun in the center of the world and not the planet Earth. In this model the explanation to the movement of Mars became very simple, bases on some interaction of movements between Mars and Earth. This is how the science revolution started, and it came with many hard struggles. The followers of Copernicus, Galileo and Newton, established the scientific way of thinking by developing new suitable mathematics.
The peak of this way of thinking brings in the 20 century the development of two central theories: Relativity and Quantum theory. These two theories changed the way man understood his place in the world completely. We are not passive observer of phenomena, but rather we have active part that is based on the interaction between us and the world.
Is there a way to establish similar principles in the rational field of mathematics? At first glance it seems there is no way to do it. The world of mathematics seems to be absolute and to exist independently of man. But on the other hand we know today that there is a need to develop a new view of mathematics that will bring better understanding of the relations between mathematics and the world of phenomena. This understanding was expresses in a lecture by Alain Connes, who developed 20 years ago a Non commutative Geometry, and is considered one of the leading mathematicians in the world today. "… We need today a new understanding of mathematics that is not necessarily founded on logic but rather on geometry." This observation ended the last lecture in the conference "100 to Hilbert" (free transcript from the speech). This conference was held in commemoration of another famous lecture – the lecture of Hilbert in Paris in 1900. Hilbert ended his lecture a hundred years ago with a vision of the discovery of an organic unity of mathematics.
A simple observation through these two important conferences as two eyes, one took place in Paris and the other in Los-angels with a difference of 100 years between them, raises the question if and how mathematics is moving, like Mars, on the background of the culture of mankind. This also raises the problem that has not been solved yet, of the inherent hidden connection between mathematics and the real world. But like the observation of Copernicus that allowed us to put the Sun in the center of the world and by doing this to understand much more simply the interactions between Earth and Mars, we can point to a third eye that will allow us to solve the problem of mathematics. The new common center of mathematics and physics is the discovery of the Organic unity of mathematics – Intelligent Life.

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#### ramsey2879

Victor Sorokine said:
- Cf. binominal theorem for PRIME n:
a^n = [a_(k+1) + n^(k+1)a_{k+2}]^n = a_(k+1)^n + n[a_(k+1)^(n – 1)][n^(k+1)a_{k+2} + n^(k+2)P] = a_(k+1)^n + n^(k+2)[a_{k+2}) + n^(k+3)P.
Therefore a^n_{k+2} is not a function from a_{k+2}
(and therefore U_{k+2} is not a function from v).
Standing alone these lines make no sense. For instance, in the counter example of Hurkyl, n=13, k = 2 and a=CCCCCCC so a^n = [CCC + n^(k+1)*C +N^(k+2)P]^n where P>0 =/ [a_(k+1) + n^(k+1)a_{k+2}]^n

#### ramsey2879

Victor Sorokine said:
Dear Moshe,
You are right: there are 2 holes:

After transformation u_{k+1} into n – 1 [with one-digit-multiplier g]
and u_{k+2} into 2 [with two-digit-multiplier 1 + ng]
we transform also
c_{k+1} into 0 [with multiplier 1 + (n^k)g; the digits u_{k+1} and u_{k+2} do not change!]
and c_{k+2} into 0 [with multiplier 1 + (n^(k+1))g; the digits u_{k+1} and u_{k+2} do not change!]
Now U' >> 0, U'' < 0, n > v > 0 [since c_{k+2} = 0].
And therefore U_{k+3} = (n - 2v)_1 =/0.

V.S.
I refuse to consider anymore a proof that is presented only piecemeal. Each time a hole in the proof is found you come up with additional detail that should have been presented originally. As requested by Moshe please submit a complete proof, line by line, with no detail missing, and which has been checked for typing errors. As a preliminary note, however, you should clearly state that each of a,b and c are multiplied by these multipliers, or your proof fails. However, it doesn't seem logical that the digit u_{k+2} can not change when a,b,c are multiplied by 1+(n^k)g to make c_{k+1} into 0.

#### arildno

Homework Helper
Gold Member
Dearly Missed
ramsey:
I've followed this for a while, and it is quite clear that Victor Sorokine is making up his proof as he goes along.
He doesn't have any proof, as he has claimed, only some muddled ideas as to how a proof might look like.

#### Victor Sorokine

+ physics

moshek said:
p/s : My own interest in mathematics is about the hidden connection
between mathematics an physics as you can read in my paper ]
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

arildno said:
ramsey:
I've followed this for a while, and it is quite clear that Victor Sorokine is making up his proof as he goes along.
He doesn't have any proof, as he has claimed, only some muddled ideas as to how a proof might look like.
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

Homework Helper
Gold Member
Dearly Missed
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

Victor Sorokine said:
Fantastic idea for my friends
Right contradiction: the number u is infinite
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…

V.S.

#### Hurkyl

Staff Emeritus
Gold Member
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

Gold Member
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".

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#### Chronos

Gold Member
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

Chronos said:
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.
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

For my Friends

Chronos said:
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.
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

ramsey2879 said:
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.
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/Resources/AoPS_R_A_HowWriteJustify.php [Broken]

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#### moshek

ramsey2879 said:
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/Resources/AoPS_R_A_HowWriteJustify.php [Broken]

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

from the begining to the end.

thank you
Moshe

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#### ComputerGeek

Hurkyl said:
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.