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

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 [Broken] ; http://www.scientific.ru/dforum/altern - page 7.

Last edited by a moderator:

Related Linear and Abstract Algebra News on Phys.org
Hurkyl
Staff Emeritus
Gold Member
www.fmatem.moldnet.md/1_(v_sor_05).htm

This link does not work for me.

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

robert Ihnot said:
I can not follow the notation. What is the meaning of: Everywhere in the text
$$a1^10.$$ (Which is written in red.)?
In the all publications:
"Everywhere in the text a1 ≠ 0."
or: a_1 =/ 0 (the last digit of the number "a" =/ 0).
Thank,
vs

Hurkyl
Staff Emeritus
Gold Member
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 $u''_3 \neq [v + (a_2 + b_2 - c_2)_2]_1$

wow I have no clue what hurkyl just said =(

learningphysics
Homework Helper
What is meant by this in section 0.2:

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

Hurkyl said:
I agree, it is difficult to read, and not all statements are justified.

...you've asserted that ... the only assumption that's been invoked on a, b, and c is that the last digit of a + b - c is zero. ... 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
...
[/itex]
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.

learningphysics said:
What is meant by this in section 0.2:

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

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)?

learningphysics
Homework Helper
robert Ihnot said:
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)?
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

learningphysics said:
...Then it's possible to show that:
a_1 + b_1 + c_1=0
Thanks,
vs

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 [Broken] 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 by a moderator:
Hurkyl
Staff Emeritus
Gold Member
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.

learningphysics
Homework Helper
robert Ihnot said:
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
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.

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:
robert Ihnot said:
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.
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

Hurkyl said:
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.
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

learningphysics
Homework Helper
Victor Sorokine said:
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.
In the above 60Z-58Z = 50.

k = 1.
u' = 7 + 5 – Z = 0.
u'' = u – u' = 500 + 100 – 580 = 20.
So if k=1, then u''_(k+2) = 0.

u''_(k+2)= [v + (a_(k+1) + b_(k+1) – c_(k+1))_2]_1 = 1 + 0 = 1 = (–1, 0 or 1).
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.

Victor Sorokine: 3. (a^n_1 + b^n_1 – c^n_1)_1 = (a_1 + b_1 – c-1)_1 = u_1.

What are you saying by a^n_1, a raised to the last digit in n? I don't think so. Everyone assumes that a^n+b^n-c^n = 0, though you have never bothered to even state as much. What I think you mean by a^n_1 is the last digit in a^n; do you ever read any mathematical literature? Why can't you just follow convention?:

$$a^n+b^n-c^n=0 \equiv a+b-c\equiv0 Mod (n)$$

As for the final u_1, what on earth is that? Is it defined? I don't think so.

Last edited:
Hurkyl
Staff Emeritus
Gold Member
I don't know of any particular convention for denoting the k-th digit of a number, so :tongue2:.

u was defined to be a + b - c, so I thought it was pretty clear u_1 meant the last digit of u, anyways.

learningphysics said:
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.
Yes:
(9°) u''_{k+2} = [v + (a_(k+1) + b_(k+1) – c_(k+1))_k+2]1, where (a_(k+1) + b_(k+1) – c_(k+1))_ {k+2} = (–1, 0 or 1);

Let a = 507, b = 105, c = 58Z. Then:
u = 507 + 105 – 58Z = 60Z – 58Z = 50.
k = 1.
u' = a_1 + b_1 – c_1 = 7 + 5 – Z = 0.
u'' = u – u' = 500 + 100 – 580 = 50.
v = 5 + 1 – 5 = 1.
(a_(k+1) + b_(k+1) – c_(k+1))_3 = (07 + 05 – 8Z)_ 3 = (0Z – 8Z)_ 3 = 0.
u''_(k+2)= [v + (a_(k+1) + b_(k+1) – c_(k+1))_ 3]_1 = 1 + 0 = 1 = (–1, 0 or 1).
OK?
vs

robert Ihnot said:
Victor Sorokine: 3. (a^n_1 + b^n_1 – c^n_1)_1 = (a_1 + b_1 – c-1)_1 = u_1.
Certainly: 3. (a^n_1 + b^n_1 – c^n_1)_1 = (a_1 + b_1 – c-1)_1 = 0.

vs

Hurkyl said:
I don't know of any particular convention for denoting the k-th digit of a number, so :tongue2:.

u was defined to be a + b - c, so I thought it was pretty clear u_1 meant the last digit of u, anyways.
Yes. u_1 = 0.
vs