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

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

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

This link does not work for me.

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

Victor Sorokine
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

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 $u''_3 \neq [v + (a_2 + b_2 - c_2)_2]_1$

Artermis
wow I have no clue what hurkyl just said =(

Homework Helper
What is meant by this in section 0.2:

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

Victor Sorokine
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.

Victor Sorokine
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

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)?

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

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

Thanks,
vs

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 [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.

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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.

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.

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.

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Victor Sorokine
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

Victor Sorokine
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

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.

robert Ihnot
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.

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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.

Victor Sorokine
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

Victor Sorokine
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

Victor Sorokine
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

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

Well, if you want to guess about it, but more conventionally the writer could begin by introducing u as: Let u represent... A person might have thought it was defined somewhere else.

Homework Helper
Victor Sorokine said:
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.

So u''_{k+2} = 0, where k=1.

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).

u''_(k+2) = 0... but [v + (a_(k+1) + b_(k+1) – c_(k+1))_ 3]_1 = 1 so the equation still seems false.

Is u''_(k+2) = 0 when k=1 and u''=50? I'm just looking at the third digit from the right of u'', which is 0.

Staff Emeritus
Gold Member
Well, if you want to guess about it, but more conventionally the writer could begin by introducing u as: Let u represent... A person might have thought it was defined somewhere else.

Statement 2° in his PDF:

Let u = a + b - c ...

robert Ihnot
Hurkyl: Statement 2° in his PDF:
Let u = a + b - c ...

I see that is correct.

ramsey2879
Victor Sorokine said:
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);
(9°) u''_{k+2} = [v + (a_(k+1) + b_(k+1) – c_(k+1))_2]_1, where (a_(k+1)
+ b_(k+1) – c_(k+1))_ 2 = (–1, 0 or 1);
Victor Sorokine said:
Yes:
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.
But (a_(k+1) + b_(k+1) + c_(k+1))_2 = (a_2 + b_2 - c_2)_2 = (0+0-8)_2
= -8_2 = 0?

Thus
u''_{k+2} = 50_3 = 0 <> (1+0)_1 = 1?

This seems to be a contradiction at first, but -8 is equivalent to -10 + 5 thus this must be that occasion where (a_2 + b_2 - c_2)_2 = -1. I can think of no other logical reason that -1 could result.

Thus -8_2 in my mind should equal -1 which then gives u''_3 = 0 which is correct!

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ramsey2879
Hurkyl said:
Your PDF says that case 1 is that $(bc)_1 \neq 0$.
In introductory notes, Victor also rules out a_1 as equaling 0. This does not cause a problem as at least one of a_1 and b_1 <> 0.
Hurkyl said:

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

My calculations again show statement 9 to be in error.

But look at my analysis of this in my last post which I edited just previously. I think there is no contradiction of statement 9 here!

Staff Emeritus
Gold Member
There is by the way Victor defines his terms... the second digit of -8 is not -1, it's 0.

I think I'd prefer some n's complement type thing, where -80 would actually be the left-infinite triskadecimal1 number ...CCCC50. (Adding 80 to this would yield 0)

In any case, if you would rather the second digit of -8 to be -1, then that should require that the first digit be a 5... does that mess anything up?

In any case, we'll have to wait and see if Victor likes one of these alterations, and sees if it works out the kinks in his logic. (And I still maintain it would be nice to see proofs of all steps)

1: I don't know if this is the standard term, or if there is even a standard.

ramsey2879
Hurkyl said:
In any case, we'll have to wait and see if Victor likes one of these alterations, and sees if it works out the kinks in his logic. (And I still maintain it would be nice to see proofs of all steps)

I also would like Victor to explain this more as his last post is a contradiction. It seems that his statements are so brief, that it is difficult for even the author to fully understand what is exactly meant upon reading them. I think the failure to fully set out the logical steps that lead to each statement adds to the confusion here.

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Victor Sorokine
For Hurkyl, robert Ihnot, learningphysics and ramsey2879

learningphysics said:
So u''_{k+2} = 0, where k=1.

u''_(k+2) = 0... but [v + (a_(k+1) + b_(k+1) – c_(k+1))_ 3]_1 = 1 so the equation still seems false.

Is u''_(k+2) = 0 when k=1 and u''=50? I'm just looking at the third digit from the right of u'', which is 0.

Dear Hurkyl, robert Ihnot, learningphysics and ramsey2879,
thank you very much for your help! Your criticism is polite and constructive.
The example of Hurkyl is exellent!

Accurate difinition (for a = 507, b = 105, c = 58Z, Z = 7+5, k =1):
u' = 7 + 5 – Z = 0, u'' = u – u' = 500 + 100 – 580 = 50; u''_{3} = 0;
v = (a_{3} + b_{3} – c_{3})_1 = 1 (cf 4°);
(a_(2) + b_(2) – c_(2))_{3} = (–1, 0 or 1) = (00 + 00 – 80)_{3} = (–100 + 2)_{3} = –1;

From (9°):
u''_{3} = [v + (a_(2) + b_(2) – c_(2))_3]_1 = (1 – 1) = 0.

ramsey2879
Victor Sorokine said:
Dear Hurkyl, robert Ihnot, learningphysics and ramsey2879,
thank you very much for your help! Your criticism is polite and constructive.
The example of Hurkyl is exellent!