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

1. Jul 18, 2005

### Victor Sorokine

Eight months have passed since the first time when an elementary proof of FLT was found, laid down and proposed to the attention of 70 mathematicians (they have asked for it; most of them are specialized in numbers theory). About 1,000 individuals visited internet sites with the text of the proof. However, no positive nor negative opinions have been expressed so far.

The idea of the elementary proof of the Fermat’s Last Theorem (FLT) proposed to the reader, is extremely simple:
After splitting the numbers a, b, c into pairs of sums, then grouping them in two sums U' and U'' and multiplying the equation a^n + b^n – c^n = 0 by 11^n (i.e. 11 power n, and the numbers a, b, c by 11), the (k+3)-th digit from the right in the number a^n + b^n – c^n (where k is the number of zeroes at the end of the number a + b – c) is not equal to 0 (the numbers U' and U'' are “multiplied” in a different way!). In order to understand the proof, you only need to know the Newton Binomial, the simplest formulation of the Fermat’s Little Theorem (included here), the definition of the prime number, how to add numbers and multiplication of a two-digit number by 11. That’s ALL! The main (and the toughest) task is not to mess up with a dozen of numbers symbolized by letters.
The redaction of the text dates of June 1, 2005 (after a discussion on the Faculty of Mathematics of Moscow University site).

The texts of the proof can be found on following sites:
English version of the demonstration (4kb): Revista Foaie Matematică: www.fmatem.moldnet.md/1_(v_sor_05).htm

Russin version in pdf : http://fox.ivlim.ru/docs/sorokine/vtf.pdf

FORUMS (Russian-language):
http://lib.mexmat.ru/forum/viewforum.php?f=1&sid=3fefd56c6fe2fa0e361464672ea92292 ;
http://forum.dubinushka.ru/index.php?showforum=40 ; http://www.scientific.ru/dforum/altern - page 7.

2. Jul 18, 2005

### Hurkyl

Staff Emeritus
www.fmatem.moldnet.md/1_(v_sor_05).htm

This link does not work for me.

3. Jul 19, 2005

### Victor Sorokine

4. Jul 19, 2005

### robert Ihnot

I can not follow the notation. What is the meaning of: Everywhere in the text
$$a1^10.$$ (Which is written in red.)?

5. Jul 20, 2005

### Victor Sorokine

In the all publications:
"Everywhere in the text a1 ≠ 0."
or: a_1 =/ 0 (the last digit of the number "a" =/ 0).
Thank,
vs

6. Jul 20, 2005

### Hurkyl

Staff Emeritus
I agree, it is difficult to read, and not all statements are justified.

Statement 9, in particular: you've asserted that $u''_{k+2} = [v + (a_{k+1} + b_{k+1} - c_{k+1})_2]_1$, and the only assumption that's been invoked on a, b, and c is that the last digit of a + b - c is zero, and that the last nonzero digit is 5. You've given no proof of that assertion, and I can provide a counterexample:

In this example, n = 13. All numbers are written in base 13.

a = 400
b = 206
c = 286

so that

u = a + b - c = 350
u' = 0
u'' = 350
v = 4
$u''_3 = 3$
$a_2 + b_2 - c_2 = -8$
$(a_2 + b_2 - c_2) = 0$
$v + (a_2 + b_2 - c_2)_2 = 4$
$[v + (a_2 + b_2 - c_2)_2]_1 = 4$
So that $u''_3 \neq [v + (a_2 + b_2 - c_2)_2]_1$

7. Jul 20, 2005

### Artermis

wow I have no clue what hurkyl just said =(

8. Jul 21, 2005

### learningphysics

What is meant by this in section 0.2:

[Digits in negative numbers are <<negative>>]

9. Jul 21, 2005

### Victor Sorokine

Dear Hurkyl,

1. From a^n + b^n – c^n = 0 and from Fermat’s Little Theorem for (a_1, b_1, c_1) =/ 0 we have: (a_1 + b_1 – c_1)_1 = 0 (= u_1).

a = 400, b = 206, c = 286 and (abc)_1 = 0
is not Case 1, where (abc)_1 =/ 0.

a = 400, b = 206, c = 286, k =2, u = c-b and c-b = 286 – 206 = 80 has 1 zero in the end
is not Case 2, where u = c-b, c-b has (cf. §1) kn-1 = 2x13 – 1 = 25 zeros in the end.

Respectfully yours,
vs.

10. Jul 21, 2005

### Victor Sorokine

Digits in negative numbers are <<negative>> =
= all digits (=/ 0) in negative numbers are negative.
vs

11. Jul 21, 2005

### robert Ihnot

None of this makes any sense to me: Victor Sorokine: From a^n + b^n – c^n = 0 and from Fermat’s Little Theorem for (a_1, b_1, c_1) =/ 0 we have: (a_1 + b_1 – c_1)_1 = 0 (= u_1).

What is (a_1, b_1, c_1) ? or the next statement: (a_1 + b_1 – c_1)_1 = 0 (= u_1)?

12. Jul 21, 2005

### learningphysics

robert,
a_1 refers to the last digit of a (in base n)

Fermat's little theorem says that if a number n is prime then: a^n - a is divisible by n (a can be any integer I think). So we have a^n-a divisible by n... b^n-b divisible by n... c^n-c divisible by n... so their sum is divisible by n...

(a^n - a) + (b^n-b) - (c^n-c) is divisible by n so...
(a^n + b^n - c^n) - (a + b - c) is divisible by n...
0 - (a+b-c) is divisible by n...
so
(a + b - c) is divisible by n...
(a + b - c)_1 (last digit of a+b-c) = 0... since the number is in base n and divisible by n the last digit must be zero...

Then it's possible to show that:
a_1 + b_1 + c_1=0

13. Jul 21, 2005

### Victor Sorokine

Thanks,
vs

14. Jul 21, 2005

### robert Ihnot

Victor Sorokine: Then it's possible to show that:
a_1 + b_1 + c_1=0

WEll, I am not sure if base 11 has something to do with this, and, very possibly, Hurkyl is arguing about: You've given no proof of that assertion, and I can provide a counterexamplle: In this example, n = 13. All numbers are written in base 13.

I am reminded of Able's submission to the French Academy, and after two years Able wondered why he had not heard. http://www.shu.edu/projects/reals/history/abel.html Unfortunately, the Academy picked Legendre and Cauchy as referees to judge it. The former, who was in his seventies, claimed that he could not read the handwriting and left all the work to the latter. The latter, who was much more interested in his own work and possibly just a bit jealous, brought the work home and promptly "misplaced" it.

learning physics: (a + b - c)_1 (last digit of a+b-c) = 0... since the number is in base n and divisible by n the last digit must be zero... Who said anything about writing it in base n, how can that change the fundamental properties? Anyway that does not effect a+b+c, since if a+b-c==a+b+c ==0 Mod n, then for n not 2, c is divisible by n, is that what you are saying? If so no one for centuries thought that obvious, let alone in only a few short steps; if, I have this right, what you mean is to assert that n, the power of the exponent, a prime, must divide one of the terms, particularly c to boot!

You mention 70 mathematicians and 1000 internet hits, possibly if the paper was rewritten and the notation easier to understand, you might get a better response.

Last edited: Jul 21, 2005
15. Jul 21, 2005

### Hurkyl

Staff Emeritus
Your PDF says that case 1 is that $(bc)_1 \neq 0$.

a = 507
b = 105
c = 58C (C in base 13 = 12 in base 10)

My calculations again show statement 9 to be in error.

16. Jul 21, 2005

### learningphysics

That's part of his paper... writing it in base n... Anyway this is what I meant:

(a+b-c)_1 =0

where (a+b-c)_1 is the last digit of a+b-c... I didn't mean multiplication by the last digit of a+b-c. Sorry about that.

17. Jul 21, 2005

### robert Ihnot

learningphysics: That's part of his paper... writing it in base n... Anyway this is what I meant: (a+b-c)_1 =0

Well if you look at definition 1, 1* let us assume that $$a_n+b_n-c_n$$= 0.

From what is above it in the paper, it would mean that we are talking about the nth digit of an expression related by the equation (never even defined as) $$a^n+b^n=c^n$$, where n is a prime conventionally expressed as p. (This assumes we have n digits to speak of, or maybe n is just a dummy variable that runs from n=1, n=?)

HOWEVER, he might be talking about modulo n, which has never, as usual, been defined as the meaning. I somehow think it is the second meaning, possibly.

Last edited: Jul 21, 2005
18. Jul 22, 2005

### Victor Sorokine

Where is a mstake:
1. (a^(n-1))_1 = 1 (Little FT).
2. [(a^(n-1))_1 x a_1]_1 = a_1 = a^n_1.
3. (a^n_1 + b^n_1 – c^n_1)_1 = (a_1 + b_1 – c-1)_1 = u_1.
vs

19. Jul 22, 2005

### Victor Sorokine

Let the digits in base 13: 1, 2, 3, 4, 5, 6, 7, 8, 9, 9+1 = X [=10 in base 10], X+1 = Y [=11], Y+1 = Z [=12];
Z+1 = 10 [=13], 10+1 = 11 [=14]…
Let a = 507, b = 105, c = 58Z. Then:
u = 507 + 105 – 58Z = 60Z – 58Z = 20.
k = 1.
u' = 7 + 5 – Z = 0.
u'' = u – u' = 500 + 100 – 580 = 20.
v = 5 + 1 – 5 = 1.(a_(k+1) + b_(k+1) – c_(k+1))_2 = (0 + 0 – 8)_2 = 0.
u''_(k+2)= [v + (a_(k+1) + b_(k+1) – c_(k+1))_2]_1 = 1 + 0 = 1 = (–1, 0 or 1).
OK?
vs

20. Jul 22, 2005

### learningphysics

In the above 60Z-58Z = 50.

So if k=1, then u''_(k+2) = 0.

u''_(k+2) = 0, but [v + (a_(k+1) + b_(k+1) – c_(k+1))_2]_1 = 1, so the equation is false it seems.