Fermat’s Last Theorem: A one-operation proof

[robert Ihnot]

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

Thank you very much for the complement, and the edition of the proof with the heading: "Page Math Journal" is a much improved version.

 

[Victor Sorokine]

For ramsey2879, Hurkyl, robert Ihnot and learningphysics

First how do you obtain -100 + 2, should not it be -100 + 50? And isn't this a contradiction that negative numbers are negative at every digit?

Dear ramsey2879, Hurkyl, robert Ihnot and learningphysics,
Here are our acquisitions:

1. Correction of (9°) in the proof:
(9°) u''_{k+2} = [v + (a_(k+1) + b_(k+1) – c_(k+1))_ {k+2}]_1.
Cause of the error: erroneous copying.

2. First digit in the negative number (in the example n = [13], k = 1, w = –80, (–80)_{3}).
It is logically:
The augmentation of the number w by n^(k+2) the digit w_{3} increases by 1 (if w_{3} = n – 1, then new w_{3} = 0). ALWAIS!
(2 + 50)_{3} = (300 – 80)_{3} = 2,
(1 + 50)_{3} = (200 – 80)_{3} = 1,
(0 + 50)_{3} = (100 – 80)_{3} = 0,
(–100 + 50)_{3} = (– 80)_{3} = –1.
Therefore:
If w_(2) < 0 and !w! < n^2, then w_{3} = –1. (–80)_{3} = –1, but 0!

To be continued

 

[ramsey2879]

Dear ramsey2879, Hurkyl, robert Ihnot and learningphysics,

1. Correction of (9°) in the proof:
(9°) u''_{k+2} = [v + (a_(k+1) + b_(k+1) – c_(k+1))_ {k+2}]_1.
Cause of the error: erroneous copying.
>snip
To be continued

That is not the only error, see 13°, "3" should be --5--. Also is it necessary to the proof that a, b, and c can be either negative or positive? For instance if a, b and c are each positive then "5° u'_k+1 = -1, 0 or 1" should be --u' = 0 or 1--. I think the proof can be more concise if a, b, c are made to be each positive. The proof is also too difficult to read. I suggest you look at the link at the top of this page on How to write Proofs.

 

[Victor Sorokine]

For all

The proof is also too difficult to read.

I would like my readers to take into account the following:
1. Since I don't master English very well, my replies could be a little be late.
2. Because of lack of time and being a little bit distracted, please be tolerant to typos in my texts.
3. Altough my wording style is not too detailed, it makes the reader to THINK and to generate a number of non-standard ideas. It may be more interesting that a simple reading.
4. The beginning of the discussion gives useful results. However, no basic flaw in the proof has been found so far. It would be a mistake to end the discussion prematurely.
Thank you all,
Victor

Other equivalent version of (9°):
(9°) /u''_{k+2} - v/ <= 1

That is not the only error, see 13°, "3" should be --5--.

Yes. Thank you very much! The digit "3" was in a version of the proof for n > 5.

There is a version of the proof for n > 2, but with other multipliers in (3°) and in (1*°). This version is shorter in 2 times, but it contains 1 supplement lemma.

 

[Hurkyl]

3. Altough my wording style is not too detailed, it makes the reader to THINK and to generate a number of non-standard ideas. It may be more interesting that a simple reading.

Such a methodology certainly has merit, but I don't think the situation is appropriate for it, for two reasons:

(1) Claiming a proof of the FLT is a very serious thing, and people won't take you seriously if you leave much of the detail to the reader.
(2) Working with individual digits is generally an exercise in technical detail, not enlightened learning. Your proof doesn't seem to carry any insight into how numbers work: it looks more like a complicated series of calculations that (alledgedly) leads to a contradiction.

That's how I begin looking at the argument: I simply wrote a program to do the calculations, and checked to see if things behaved as advertised. I haven't updated it to handle your new definition of "k-th digit" yet... could you state it precisely?

4. The beginning of the discussion gives useful results. However, no basic flaw in the proof has been found so far. It would be a mistake to end the discussion prematurely.

Someone suggested to me today what might be a fatal flaw: he points out that your argument, if valid, would not only prove there are no integer solutions to the FLT, but that there are no p-adic solutions either... but he thinks there's good reason to expect that there are p-adic solutions to the FLT.

 

[ramsey2879]

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.

I would like my readers to take into account the following:
1. Since I don't master English very well, my replies could be a little be late.
2. Because of lack of time and being a little bit distracted, please be tolerant to typos in my texts.
3. Altough my wording style is not too detailed, it makes the reader to THINK and to generate a number of non-standard ideas. It may be more interesting that a simple reading.

Such a methodology certainly has merit, but I don't think the situation is appropriate for it, for two reasons:

(1) Claiming a proof of the FLT is a very serious thing, and people won't take you seriously if you leave much of the detail to the reader.
(2) Working with individual digits is generally an exercise in technical detail, not enlightened learning. Your proof doesn't seem to carry any insight into how numbers work: it looks more like a complicated series of calculations that (alledgedly) leads to a contradiction.

That's how I begin looking at the argument: I simply wrote a program to do the calculations, and checked to see if things behaved as advertised. I haven't updated it to handle your new definition of "k-th digit" yet... could you state it precisely?




Someone suggested to me today what might be a fatal flaw: he points out that your argument, if valid, would not only prove there are no integer solutions to the FLT, but that there are no p-adic solutions either... but he thinks there's good reason to expect that there are p-adic solutions to the FLT.

I think that Victor does not appreciate the lack of courtesy his methodology shows here. A proof is not supposed to make the reader full in the detail. It is assumed that the author has already covered all corners. It is the task of the author to communicate this to the reader. I again refer to the pages at http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php on how to write proofs. It begins with this opening paragraphs: "You've figured out the solution to the problem - fantastic! But you're not finished. Whether you are writing solutions for a competition, a journal, a message board, or just to show off for your friends, you must master the art of communicating your solution clearly. Brilliant ideas and innovative solutions to problems are pretty worthless if you can't communicate them. In this article, we explore many aspects of how to write a clear solution. Below is an index; each page of the article includes a sample 'How Not To' solution and 'How To' solution. One common theme you'll find throughout each point is that every time you make an experienced reader have to think to follow your solution, you lose.

As you read the 'How To' solutions, you may think some of them are overwritten. Indeed, some of them could be condensed. Some steps we chose to prove could probably be cited without proof. However, it is far better to prove too much too clearly than to prove too little. Rarely will a reader complain that a solution is too easy to understand or too easy on the eye"


As for my statement that to do otherwise demonstrates a lack of courtesy, I point out that if a simple proof of FLT is indeed at hand, it makes no sense to write the proof in such a manner that the reader has to redo much of the original work of the author. Most mathematicians already have a full schedule and don't appreciate having to give a lot of time to something that seems so dubious in the first place. If a simple proof is in fact known, then it should be simple to write it in a clear manner. As to the fact that English is a second language here, I don't see it as an excuse for omitting so much detail. While Victor's time may also be limited, he had the incentive to submit his proof to over seventy mathematicians for review and to post it in the internet. A little extra time spent on writing a clear proof in the first place thus could have saved untold volumns of time and would have been more rewarding for Victor's own interests, let alone the interests of others. This is not something that should be taken so lightly.

 

[Hurkyl]

Yes, there are (a great many) nontrivial solutions to x^p + y^p = z^p in the p-adics for all primes p.

If you can write down a solution to x^p + y^p = z^p modulo p^2, then that can be extended to a solution in the p-adics. (a great many, actually)


Here's an example for p = 13:

...3333^p + ...1217114C11^p = ...4444^p

which I generated from the base solution 3^13 + 1^13 = 4^13 (mod 13^2). (numbers here written base 10)

 

[ramsey2879]

Yes, there are (a great many) nontrivial solutions to x^p + y^p = z^p in the p-adics for all primes p.

If you can write down a solution to x^p + y^p = z^p modulo p^2, then that can be extended to a solution in the p-adics. (a great many, actually)


Here's an example for p = 13:

...3333^p + ...1217114C11^p = ...4444^p

which I generated from the base solution 3^13 + 1^13 = 4^13 (mod 13^2). (numbers here written base 10)

I am weak in this area so I did an internet search on p-adics. One site of interest was http://mathforum.org/library/drmath/view/65286.html Another is http://www.math.niu.edu/~rusin/known-math/93_back/aaargh The last site has a section that reads:
"FLT is Fermat's Last Theorem, and it is a falsity. Because when you accept that the Naturals are the Infinite Integers, or p-adics and that Finite Integer was a ill-defined concept, then , FLT has solutions in all exponents. FLT is a patently false conjecture.
FLT is easy to prove in p-adics as a false conjecture. However,
Pierre Fermat never had the p-adics in his day.
And all attempts at proving FLT were so difficult not because the
problem is a difficult one for anyone who knows p-adics can solve it in
an hour. FLT was difficult for 350 years because it was false in the
first place and those that tried to prove it were using a fake concept
of Finite Integer. Imagine today a Quantum Physicist trying to use
Newtonian spinning spheres for electrons. Electrons exist but they are
never little spinning Newtonian spheres. P-adics exist, but "finite" in
Finite Integers is a illusion, a mirage"
OK there are many solutions of FLT in p-adics. Now, how does that show Victor's proof to have a fatal error?

 

[Victor Sorokine]

A proof is not supposed to make the reader full in the detail. ... This is not something that should be taken so lightly.

About the difficult of the proof:
1. Pierre Fermat had had more difficulties: it was necessary to make all calculation mentally, in his mind.
2. In the September I chall illustrate all operation with numerical calculations. But I am ready to explain in every detail any statement in my proof.

 

[Hurkyl]

Victor's proof is all about looking at the digits of various numbers, and only the right-most few: it never looks at any of the digits to the left of the (k+3)-th digit.

So, there are two ways of looking at it...

(1) Anything Victor states ought to be true about p-adic values as well as integer values, because the proof doesn't care what's happening to the left of the (k+3)-th digit.

(2) A p-adic solution can be used to generate digits of a "problematic" set of integers a, b, and c, which do satisfy a^p + b^p = c^p, as long as you're only looking at the rightmost (k+3) digits.


If you look at the p-adic solution I generated, we could use it to generate the "problem" set:

a = 3333333
b = 7114C11
c = 4444444

for which u = 6003B00. So, we need to multiply everything by 4 so the rightmost digit is a 5...

a = 0CCCCCCC
b = 22454944
c = 144444443

and we have u = B010500, so that k = 2, and also that the rightmost 5 digits of a^13 + b^13 = c^13 agree, so all the assumptions Victor actually uses in his proof ought to be satisfied, so he shouldn't be able to derive a contradiction.

 

[ramsey2879]

Victor's proof is all about looking at the digits of various numbers, and only the right-most few: it never looks at any of the digits to the left of the (k+3)-th digit.

So, there are two ways of looking at it...

(1) Anything Victor states ought to be true about p-adic values as well as integer values, because the proof doesn't care what's happening to the left of the (k+3)-th digit.

(2) A p-adic solution can be used to generate digits of a "problematic" set of integers a, b, and c, which do satisfy a^p + b^p = c^p, as long as you're only looking at the rightmost (k+3) digits.


If you look at the p-adic solution I generated, we could use it to generate the "problem" set:

a = 3333333
b = 7114C11
c = 4444444

for which u = 6003B00. So, we need to multiply everything by 4 so the rightmost digit is a 5...

a = 0CCCCCCC
b = 22454944
c = 144444443

and we have u = B010500, so that k = 2, and also that the rightmost 5 digits of a^13 + b^13 = c^13 agree, so all the assumptions Victor actually uses in his proof ought to be satisfied, so he shouldn't be able to derive a contradiction.

I think you are assuming that if a^p + b^p -c^p = 0 , i.e. mod 13^7 where a b and c are p-adic numbers then (a*k)^p + (b*k)^p -(c*k)^p also equals zero mod 13^7. I did a check on this using your numbers and according to my calculations, it is not true. I must add that p-adic numbers are new to me, but my calculations cause me to wonder about your statement that "the rightmost 5 digits of the new p-adic numbers agree with the formula a^13+b^13=c^13. Am I off base here?

 

[Hurkyl]

I could have the wrong numbers: I generated them at work, and I haven't programmed a way to check them here at home.

I'll tell you how they were generated so you can play with them yourself:

You start with a (nonzero) solution to a^p + b^p = c^p (mod p^2)

For example, when p = 13, I believe that 1^p + 3^p = 4^p. In any case, you could always use 1^p + 0^p = 1^p, or (p-1)^p + 1^p = 0^p


Now, you add digits one at a time, allowing you to increase that exponent. Here's the derivation and proof:

You're given:

(1) a^p + b^p - c^p = 0 (mod p^n)

with at least 1 of the a, b, c nonzero, and you want to find digits x, y, z, such that:

a' = a + x p^(n-1)
b' = b + y p^(n-1)
c' = c + z p^(n-1)
(2) a'^p + b'^p - c'^p = 0 (mod p^(n+1))

We know from (1) that
a^p + b^p - c^p = q * p^n (mod p^(n+1))

So, we calculate (mod p^(n+1))
a'^p + b'^p - c'^p =
a^p + p a^(p-1) x p^(n-1) + 0 + b^p + p b^(p-1) y p^(n-1) + 0 - c^p - p c^(p-1) z p^(n-1) - 0
(The higher order terms from the binomial theorem all have a p^(2n-1) factor, and since n >= 2, 2n-1 >= n+1)
= (a^p + b^p - c^p) + p^n (a^(p-1) x + b^(p-1) y - c^(p-1) z)
= q p^n + p^n (a^(p-1) x + b^(p-1) y - c^(p-1) z)
= p^n (q + a^(p-1) x + b^(p-1) y - c^(p-1) z)

So all we need to do is to solve the equation
(q + a^(p-1) x + b^(p-1) y - c^(p-1) z) = 0 (mod p)

for x, y, and z. This is a severely underdetermined equation: it will have lots of solutions, unless the coefficients of x, y, and z are all zero... but that can't happen because at least one of a, b, and c are nonzero.

So, this yields an algorithm for generating a solution:

Set a = p - 1, b = 0, c = 1, n = 2. (Or, pick your favorite solution with nonzero a)

We see that a^p + b^p = c^p (mod p^n)

Now, find the q such that a^p + b^p - c^p = q p^n (mod p^(n+1))
Pick any value for y and z, then solve for x in

(q + a^(p-1) x + b^(p-1) y - c^(p-1) z) = 0 (mod p)

Now, set:
a <-- a + x p^(n-1)
b <-- b + y p^(n-1)
c <-- c + z p^(n-1)
n <-- n + 1

And we still have that a^p + b^p = c^p (mod n)

 

[ramsey2879]

Dear Hurkyl, Your calculations appear correct and work out, you just mistyped b = 22456944. I also would like to delete my two messages on this error. I tried to delete the first, without success as of now.

 

[Victor Sorokine]

All number-endings () and digits {},

Here are all number-endings () and digits {}, which are taken into account in the proof of the FLT (Case 1):

(1°) a^n + b^n – c^n :: [= 0]
(2°) u :: [= a + b – c > 0]
(2°) u_{k} :: [= 0; if a, b, c have not common factors then k >1]
(2°) u_{k+1} :: [=/ 0]
(4°) u' :: [= a_(k) + b_(k) – c_(k); !(a_(k) + b_(k) – c_(k))_ {k+1}! <= 1]
(3°, 3a°) u_{k+1} :: [= (u'_{k+1} + u''_{k+1})_1 = 5]
(5°) u'_{k+1} :: [!u'_{k+1}! <= 1]
(7°) u'_{k+2} :: [= 0 always]
(4°) u'' :: [= u – u' = a – a_(k) + b – b_(k) – c + c_(k)]
(6°) u''_{k+1} :: [= (u_{k+1} – u'_{k+1})_1 = (4, 5 or 6)]
(8°) u''_{k+2} :: [= u{k+2} always]
(4°) v :: [= (a_{k+2} + b_{k+2} – c_{k+2})_1; (vn^(k+1) + a_(k+1) + b_(k+1) – c_(k+1))_(k+2) = u_(k+2)]
(10°) a_(k+1) + b_(k+1) – c_(k+1) :: [!(a_(k+1) + b_(k+1) – c_(k+1))_{k+2}! <= 1]
(9°) u''_{k+2} :: [(v – 1)_1 <= u''_{k+2} <= (v + 1)_1]
(9°) v :: [(u_{k+2} – 1)_1 <= v <= (u_{k+2} + 1)_1]

(1*°) u* :: [= 11u = 11a + 11b – 11c = a* + b* – c*]
[Let u/n^k = pn + q; then (pn + q)11_(2) = ((pn + q)(n + 1))_(2) = (qn + pn + q)_(2);
from here (u/n^k)_1 = q, (u/n^k)_2 = (p + q)_1; example: 230 x 11 = …530]
(4°) u*' :: [= a*_(k) + b*_(k) – c*_(k); !u*'_{k+1}! <= 1]
(9°) a*_(k+1) + b*_(k+1) – c*_(k+1) :: [!(a*_(k+1) + b*_(k+1) – c*_(k+1))_{k+2}! <= 1]
(4°) u*'' :: [= u* – u*']
(4°) 11u'
(4°) 11u''
(11°) u*_{k+1} :: [= u_{k+1} = 5]
(12°) u*'_{k+1} :: [= u'_{k+1}]
(13°) u*''_{k+1} :: [= (u*_{k+1} – u*'_{k+1) }_1 = (5 – u*'_{k+1})_1 = (4, 5 or 6)]
(14°) (11u')_{k+2} :: [= (u'_{k+2} + u'_{k+1})_1; (11u')_{k+2} * 0]
(15°) (11u'')_{k+2} :: [= (u''_{k+2} + u''_{k+1})_1]
(16°) u*_{k+2} :: [= (u_{k+2} + u_{k+1})_1 = (u''_{k+2} + u_{k+1})_1 = (u''_{k+2} + 5)_1]
(16а°) u*'_{k+2} :: [= 0]
(17°) u*''_{k+2} :: [= (u*_{k+2} + (–1, 0 or 1))_1 = (u''_{k+2} + (4, 5 or 6))_1]
(18°) v* :: [= (u*_{k+2} – (–1, 0 or 1)) _{k+2})_1 = ((u_{k+2} + u_{k+1})_1 – (–1, 0 or 1))_1 ]

(19°) U' :: [= (a_(k+1))^n + (b_(k+1))^n – (c_(k+1))^n, U'' = (a^n + b^n – c^n) – U', U = U' + U'']
(19°) U*' :: [= (a*_(k+1)) ^n + (b*_(k+1)) ^n – (c*_(k+1)) ^n,
(19°) U*'' :: [= (a*^n + b*^n – c*^n) – U*'
(19°) U* :: [= U*' + U*'']
(19а°) U'_(k+1) :: [= U*'_(k+1) = 0]
(20°) U(k+2) = U'(k+2) = U''(k+2) = U*(k+2) = U*'(k+2) = U*''(k+2) = 0 [always!].
(20b°) U''_{k+3} :: [= (a_{k+2} + b_{k+2} – c_{k+2})_1 = v]
(22°) (11^nU') _{k+3} :: [= U'_{k+3} = (U*'_{k+3} + (11u') _{k+2})1 = (U*'_{k+3} + u'_{k+1})_1]
(23°) U*'_{k+3} :: [= U'_{k+3} – u'_{k+1}]
(24°) U*''_{k+3} :: [= v* = (u_{k+2} + u_{k+1})1 – (–1, 0 or 1)]

(25°) (U*'_{k+3} + U*''_{k+3})_1 :: [= (U*'_{k+3} + U*''_{k+3} – U'_{k+3} – U''_{k+3})_1 =
= (U*'_{k+3} – U'_{k+3} + U*''_{k+3} – U''_{k+3})1 = (– u'_{k+1} + v* – v) =
= (– u'_{k+1} + [u_{k+2} + u_{k+1} – (–1, 0 or 1)] – [u_{k+2} – (–1, 0 or 1)])_1 = (– u'_{k+1} + u_{k+1} + (–2, –1, 0, 1, or 2))_1 = ( u''_{k+1} + (–2, –1, 0, 1, or 2))_1 = (2, 3, 4, 5, 6, 7 or 8) =/ 0, therefore the expression 1° is an inequality]

 

[Hurkyl]

I don't understand your notation at all.

 

[Victor Sorokine]

Considerable Simplification Of The Proof

I don't understand your notation at all.

I listed all digits and numbers used in the proof. Square brackets are used to write down their values and also the link to other digits or numbers. I am ready to illustrate any number in an example with the base 13. Which number or its value is unclear to you?

Example:
(1°) a^n + b^n – c^n :: [= 0]
Let a = …507, b = …105, c = …58Z, Z = 7 + 5 and
…507^n + …105^n – …58Z^n = 0.

(2°) u :: [= a + b – c > 0]
u = …507 + …105 – …58Z = ...050; and u > 0.

(2°) u_{k} :: [= 0]
u_(1) = 0, but u_(2) =/ 0; therefore k = 1.

(2°) u_{k+1} :: [=/ 0]
u_{k+1} = u_{1+1} = u_{2} = 5.

(3°, 3a°) u_{k+1} :: [= (u'_{k+1} + u''_{k+1})_1 = 5]
"(3°) We multiply the equality 1° by a number d_1^n (cf. §§2 and 2a in the Appendix) in order to transform the digit u_{k+1} into 5" (cf. the proof).
In this operation d_1^n = 1.

(4°) u' :: [= a_(k) + b_(k) – c_(k); !(a_(k) + b_(k) – c_(k))_ {k+1}! <= 1]
u' = a_(1) + b_(1) – c_(1) = a_1 + b_1 – c_1 = 7 + 5 – Z = 7 + 5 – Z = 0. !0_2! <= 1.

(5°) u'_{k+1} :: [!u'_{k+1}! <= 1]
u'_{k+1} = u'_2 = 0_2 = 0. !0_2! <= 1.

(7°) u'_{k+2} :: [= 0 always]
u'_3 = 0_3 = 0.

(4°) u'' :: [= u – u' = a – a_(k) + b – b_(k) – c + c_(k)]
u'' = u – u' = ...050 – 0 = ...050.

(6°) u''_{k+1} :: [= (u_{k+1} – u'_{k+1})_1 = (4, 5 or 6)]
u''_{k+1} = (u_{k+1} – u'_{k+1})_1 = (u_{2} – u'_{2})_1 = (5 – 0)_1 = 5.

etc

========

CONSIDERABLE SIMPLIFICATION OF THE PROOF.

There is interesting lemma (§3 – simple corollary from §2 and §2a in the ADDENDUM):
For numbers a (where a_1 =/ 0) and p_(k) there is such number d that (ad)_(k) = p_(k).

Considerable simplification of the proof:
After
"(3°) We multiply the equality 1° by a number (d_1)^n (cf. §§2 and 2a in the Appendix) in order to transform the digit u_{k+1} into 2 (instead 5 – cf. the proof)", the digit c_1 turns into new c_1. Then we multiply the new equality 1° by a number d^n (cf. §3) in order to transform the ending c_(k+2) into c_1 (i.e. c_2 = 0, … c_{k+2} = 0).
Now (if k > 1) the numbers
(a_(k) + b_(k) – c_(k))_{k+1} = (0 or 1),
(a*_(k) + b*_(k) – c*_(k))_{k+1} = (0 or 1),
(a_(k+1) + b_(k+1) – c_(k+1))_{k+2} = (0 or 1),
(a*_(k+1) + b*_(k+1) – c*_(k+1))_{k+2} = (0 or 1),
and the proof works for n = 3, 5, 7, 11, …

 

[ramsey2879]

About the difficult of the proof:
2. In the September I shall illustrate all operation with numerical calculations. But I am ready to explain in every detail any statement in my proof.

Victor's proof is all about looking at the digits of various numbers, and only the right-most few: it never looks at any of the digits to the left of the (k+3)-th digit.

So, there are two ways of looking at it...

(1) Anything Victor states ought to be true about p-adic values as well as integer values, because the proof doesn't care what's happening to the left of the (k+3)-th digit.

(2) A p-adic solution can be used to generate digits of a "problematic" set of integers a, b, and c, which do satisfy a^p + b^p = c^p, as long as you're only looking at the rightmost (k+3) digits.

...
a = 0CCCCCCC
b = 22454944 (sic b = 22456944)
c = 144444443

and we have u = B010500, so that k = 2, and also that the rightmost 5 digits of a^13 + b^13 = c^13 agree, so all the assumptions Victor actually uses in his proof ought to be satisfied, so he shouldn't be able to derive a contradiction.

The proof remains very difficult to follow. I would not care to verify each of statements 19-25 which is seem to yield a contradiction since I agree with the logic of Hurkyl's analysis. There is obviously an error in these lines, but since Victor does not give the details for each statement, it is very difficult to point out the error

 

[Victor Sorokine]

It's very interesting!

Quote:
Originally Posted by Hurkyl
Victor's proof is all about looking at the digits of various numbers, and only the right-most few: it never looks at any of the digits to the left of the (k+3)-th digit.

So, there are two ways of looking at it...

(1) Anything Victor states ought to be true about p-adic values as well as integer values, because the proof doesn't care what's happening to the left of the (k+3)-th digit.

(2) A p-adic solution can be used to generate digits of a "problematic" set of integers a, b, and c, which do satisfy a^p + b^p = c^p, as long as you're only looking at the rightmost (k+3) digits.

...
a = 0CCCCCCC
b = 22454944 (sic b = 22456944)
c = 144444443

and we have u = B010500, so that k = 2, and also that the rightmost 5 digits of a^13 + b^13 = c^13 agree, so all the assumptions Victor actually uses in his proof ought to be satisfied, so he shouldn't be able to derive a contradiction.

The proof remains very difficult to follow. I would not care to verify each of statements 19-25 which is seem to yield a contradiction since I agree with the logic of Hurkyl's analysis. There is obviously an error in these lines, but since Victor does not give the details for each statement, it is very difficult to point out the error

It's very interesting!
Hurkyl and ramsey2879, you are absolutely right, IF the number a, b, c are WHOLE:
if (a^13 + b^13)_(5) = c^13_(5), then (11^13 x a^13 + 11^13 x b^13)_(5) = (11^13 x c^13)_(5).
BUT in my proof (11^13 x U'^13)_(5) =/ – (11^13 x U''^13)_(5).
There fore (U'^13)_(5) =/ - (U''^13)_(5).
And therefore (U'^13)_(5) + (U''^13)_(5) = (a^13 + b^13)_(5) – c^13_(5) =/ 0.
And THEREFORE the number a, b, c are NOT WHOLE!
vs

 

[Hurkyl]

Well, we have a problem then, don't we?

In base 13, if I use the following whole numbers:

a = 0ccccccc
b = 22456944
c = 14444443

(so that a + b - c = 1b012500)

Then, I compute the following:

a^13 + b^13 = ...3b43350cb3
c^13 = ...7b43350cb3


So, I've exhibited whole numbers a, b, and c, such that the last (k+3) digits of a^13 + b^13 agree with the last (k+3) digits of c^13.

In fact, the last nine digits agree, not just the last 5 = (k+3) digits.

 

[Hurkyl]

P.S. why would you think that

(U'^13)_(5) + (U''^13)_(5) = (a^13 + b^13)_(5) – c^13_(5)

is an equality?

 

[Victor Sorokine]

Well, we have a problem then, don't we?

In base 13, if I use the following whole numbers:

a = 0ccccccc
b = 22456944
c = 14444443

(so that a + b - c = 1b012500)

Then, I compute the following:

a^13 + b^13 = ...3b43350cb3
c^13 = ...7b43350cb3


So, I've exhibited whole numbers a, b, and c, such that the last (k+3) digits of a^13 + b^13 agree with the last (k+3) digits of c^13.

In fact, the last nine digits agree, not just the last 5 = (k+3) digits.

Dear Hurkyl,
You ignore one condition from (20°): U'_(k+2) = 0, or ((a_(k+1))^n + (b_(k+1))^n – (c_(k+1))^n)_{k+2} = 0!
In your example (ccc^n + 944^n – 443^n)_(k+4) =/ 0 !
Victor

 

[ramsey2879]

(20°) U(k+2) = U'(k+2) = U''(k+2) = U*(k+2) = U*'(k+2) = U*''(k+2) = 0 [always!]

Dear Hurkyl,
You ignore one condition from (20°): U'_(k+2) = 0, or ((a_(k+1))^n + (b_(k+1))^n – (c_(k+1))^n)_{k+2} = 0!
In your example (ccc^n + 944^n – 443^n)_(k+4) =/ 0 !
Victor

From my perspective, that is not a condition or else your proof at best proves FLT only for limited cases. I rather think that (20°): U'_(k+2) = 0, or ((a_(k+1))^n + (b_(k+1))^n – (c_(k+1))^n)_{k+2} = 0 is merely an unsupported statement that is proven to be wrong by Hurkyl. In either case, however, there is no valid elementary proof of the FLT.

 

[Hurkyl]

Dear Hurkyl,
You ignore one condition from (20°)

That wasn't a condition: it is something you have claimed to prove.


Anyways:


In your PDF, at 19°, you make the definition $U' := (a_{k+1})^n + (b_{k+1})^n - (c_{k+1})^n$, but looking at your post here, I guess that's supposed to be $U' := (a_{(k+1)})^n + (b_{(k+1)})^n - (c_{(k+1)})^n$

Anyways, now that I've made that correction, I compute:

U' = ...4a1274cc0000
U'' = ...8607a58010000

So that the last (k+2) digits of U' are zero.

I repeat:

ccc^n + 944^n - 443^n = 10136c7a32838914191aa52239134a1274cc0000

So that (ccc^n + 944^n - 443^n)_(k+2) = 0.

(The last (k+4) digits do contain some nonzero digits, but why do you suddenly care about those?)

 

[Victor Sorokine]

Number a+b-c is INFINITE

... but why do you suddenly care about those?)

Dear Hurkyl,
1. Your counter-example is right.
2. Congratulation!
3. Thank You very much!
4. Present for You: impromptu-proof of FLT (for 24 hours):

in Fermat's equality 1° the number u is INFINITE!

Tools:
prime number n > 2; digit a_{t}, a_{1} = a_1 =/ 0, ending a_(t);
u = a + b – c, k > 1; u(t)' = a_(t) + b_(t) – c_(t); v(t) = a_t + b_t – c_t;
u_{t+1} + (u(t)'_{t+1} + v(t))_1;
U = a^n + b^n – c^n; (20°)–(21°); and
LEMMA:
For number a there are such d and s, that ad = n^s – 1 (corollary from Little Fermat's theorem). Or:
all digits u_t of the number u (where k < t < n^s) are equal n – 1 and u_{n^s} = 0.

Proof:
We multiply the equality 1° by a number d^n (cf. LEMMA) in order to transform the number u into 2 (or 3 – for n > 3).
Now it's easy to see that each U_{t} = 0, (a_(t)^n + b_(t)^n – t_(t)^n)_{t+1} = 1 and
for equality (a_(t)^n + b_(t)^n – t_(t)^n)_{t+1} + U''_{t+1})_1 = 0 it's necessary that
(a_t + b_t – c_t )_1 > 0 and u_t > 0 (t = k+1, k+2, … n^s-1).
And u_{n^s} =/0 !

 

[ramsey2879]

Dear Hurkyl,
1. Your counter-example is right.
2. Congratulation!
3. Thank You very much!
4. Present for You: impromptu-proof of FLT (for 24 hours):

in Fermat's equality 1° the number u is INFINITE!

Tools:
prime number n > 2; digit a_{t}, a_{1} = a_1 =/ 0, ending a_(t);
u = a + b – c, k > 1; u(t)' = a_(t) + b_(t) – c_(t); v(t) = a_t + b_t – c_t;
u_{t+1} + (u(t)'_{t+1} + v(t))_1;
U = a^n + b^n – c^n; (20°)–(21°); and
LEMMA:
For number a there are such d and s, that ad = n^s – 1 (corollary from Little Fermat's theorem). Or:
all digits u_t of the number u (where k < t < n^s) are equal n – 1 and u_{n^s} = 0.

Proof:
We multiply the equality 1° by a number d^n (cf. LEMMA) in order to transform the number u into 2 (or 3 – for n > 3).
Now it's easy to see that each U_{t} = 0, (a_(t)^n + b_(t)^n – t_(t)^n)_{t+1} = 1 and
for equality (a_(t)^n + b_(t)^n – t_(t)^n)_{t+1} + U''_{t+1})_1 = 0 it's necessary that
(a_t + b_t – c_t )_1 > 0 and u_t > 0 (t = k+1, k+2, … n^s-1).
And u_{n^s} =/0 !

Your proof is lacking too much detail. Also, your subscripts are inconsistent or omitted: what do you mean by "transform the number u into 2 (or 3 for n>3)". How can u be infinite and not finite? You also employ the same symbol "d" for two different numbers. You don't say so but it is assumed that if you multiply a by d to get ad = n^s-1 then you must multiply b and c by d also, how can the same d be used to transform u into 2 or 3? When you say that if all digits u_t (where k < t < n^s) are equal n – 1, don't you mean all digits where k<t<n+k? One clear error is that if all digits u_t (where k < t < n^s) are equal n – 1 then each u_{t} are not equal to 0 as claimed.

 

[Victor Sorokine]

Your proof is lacking too much detail.

For ramsey2879 and Hurkyl

Completion of the Impromptu-proof

Here is complete algorithm of the proof (in base 7):
0. b < a < c.
1. Transform the number u/u^k into u_{n^s} – 1 with result 666…666 (cf. Lemma 3);
2. Multiply u_{n^s} – 1 by n – 2 with result 566…662
and after multiply by u^k with result 566…661000…00 (s + 1 + k = t digits).

I. If c = c_(s+1+k), then 1° has no solution (2 examples):
6…^n + 5…^n >> 6…^n; 7…^n + 2…^n >> 2…^n.

II. c > c_(s+1+k). Let maximum number (or rank) of digits in a, b, c is equal to r. For r-th digits we have: a_{r} + b_{r} – c_{r} = 0!
Then 1° has no solution (2 examples ):
2…^n + 3…^n << 5…^n; 10…^n + 11…^n << 2…^n.

To be continued

VS - 2005.08.08

Thanks

 

[Victor Sorokine]

If the contradiction there is not in the last digits then it exists in the first...

If the contradiction there is not in the last digits then it exists in the first digits!

About the Proof-2

From (1°) it follows:
u (= a + b – c) > 0; (c – u)/u < 2/n.
Example for n = 5.
If c is given then max(a + b) reaches by a = c, from here 2a^5 = c^5, from here 1.1487f = c, from here a = 0.87055;
u = (2 x 0.87055 – 1)c = 0.7411; c = 1.349u = u + 0.349u.
Let r is maximum rank of number u. Then 0.349u = 0.349(u_(r – 1) + n^(r – 1)u_r) < 0.349(n^(r – 1) + n^(r – 1)u_r) =
= 0.349 n^(r – 1)(u_r + 1) = n^(r – 1)(0.349 (u_r + 1)) < n^(r – 1)(0.349n)). Therefore (c – u)_r < 0.349n = 0.349 x 5 = 1.75.
Always max(c – u)_r = 1.
And always (a – a_(r)) + (b – b_(r)) – (c – c_(r)) ==0 (2°).

Lemma
For any integer a (where a_1 =/ 0) there are such p and m that ap = n^m – 1.
Truth of the Lemma follows from Little Fermat’s theorem: any prime number g1 is a factor of the number n^(g1 – 1) – 1,
and n^(g1 х g2 х… gt – 1) – 1 divides by g1 х g2 х… gt.
After multiplication of 1° by p^n and hence of the number u by p we turn the number u/u^k into n^m – 1
which consists (except k-ending) only of digits n – 1.

Proof of FLT has 2 cases.

Case 1: (abc)_1 =/ 0

If s = r impossibility of solution of 1° is obvious: even in favorable case a^n + b^n >> c^n.
2 examples in base 7: 6…^n + 6…^n > 6…^n; 5…^n + 5…^n >> 4…^n.
If s = r + 1 then or c_r < n – 1 and inequality a^n + b^n >> c^n conserves,
either c_r = n – 1 and then c_{r + 1} = 1 and (a – a_(r)) + (b – b_(r)) – (a – a_(r)) =/ 0 – cf. (2°).

Case 2: b = b'n^k, (ac)_1 =/ 0 (or c = c'n^k, (ab)_1 =/ 0)
This case is proved analogously, but here u = a + b'n^(kn – 1) – c
(since the number c – a contains a factor n^(kn – 1) – cf. ADDENDUM to Proof-1).

So 1° has no positive solution.

The proof is done.

11 August 2005

 

[robert Ihnot]

Commenting on Hurkyl's early solution set a = 3333333
b = 7114C11 c = 4444444. Now since -1=...CCCCCCCC indefinitively in P-adic with P=13, -1/12 =...1111111, so that we easily arrive at -1/4=...3333333, and -1/3 =...44444444.

However a infinitive series for b, base 13, that I do not know, but using computer program Pari, I was able to find (-1/3)^13-(-1/4)^13 O(13^13),and take the 13th root, which gives:61217114C11, O(13^12), which gives the correct solution 0 to O(13^12), or if you like in p-adic: (-1/4)^13+(61217114C11)^13-(-1/3)^13 =5*13^13+13^14+O(13^15).

This can all be transformed into the base 10 by observing the geometric series 1+p+p^2+++=[(P^n)-1]/(p-1), and thus for Order 13^12, we have:

{(13^12-1)/4}^13+839483858773^13-{(13^12-1)/3)^13 =5*13^13+13^14+O(13^15), or that 13^12 divides the expression, giving 12 right hand zeros in base 13.

So we have results as whole numbers and the suggestion this process can be extended, if we seek an even larger solution. (Note, Hurkyl has previously gone into a method for a larger solution.)

 

[Victor Sorokine]

FLT-2005. Theses

The proof contains 3 tools only:
1) Scale of notation in base prime;
2) Lemma: for number a (there last digit a_1 =/ 0) there are such d and s, that ad = n^s – 1 (corollary from Little Fermat's theorem). After multiplication u (= a + b – c) by p in new Fermat's equation the number u has the form (in base 7): 666…666000…000 with r digits equal to n – 1 or "9" (with k zeros in the end; r > k + 2).
3) The number h = (c – u)/u > 0.
From c > a and c > b we have: 2c > a + b, c > a + b – c, c – u > 0, h > 0.

THE PROOF of FLT

1. Let a^n + b^n – c^n = 0 (1°).

2. For any rank i, where k < i < r, it fulfils the equality: a_i + b_i – c_i = 0, and for rank r this equality has a form:
(a_r + b_r – c_r)_1 = n – 1 = «9».

Case 1. c_r = n – 1 (= "9") and c_t = 0, where t > r; then a_r = b_r = c_r = n – 1 (= "9") and
even in the best case { if a = [(n^k)n^(r + 1 – k) – 1)], b = [(n^k)n^(r + 1 – k) – 1)], c = n^(r + 1) – 1} a^n + b^n > c^n:
[(n^k)n^(r + 1 – k) – 1)]^n + [(n^k)n^(r + 1 – k) – 1)]^n > (n^(r + 1) – 1)^n (2°).

Case 2. c_r =/ n – 1 (= "9") and therefore there is c_t =/ 0, where t > r.
Then even in the best case {if a = n^(r + 1) – 1, b = 1, c = n^(r + 1) + 1} a^n + b^n < c^n:
(n^(r + 1) – 1)^n + 1^n < (n^(r + 1))^n (3°).

Therefore the equation (1°) has no integer solution.

Next materials: detailing and numerical examples.

 

[ramsey2879]

The proof contains 3 tools only:
1) Scale of notation in base prime;
2) Lemma: for number a (there last digit a_1 =/ 0) there are such d and s, that ad = n^s – 1 (corollary from Little Fermat's theorem). After multiplication u (= a + b – c) by p in new Fermat's equation the number u has the form (in base 7): 666…666000…000 with r digits equal to n – 1 or "9" (with k zeros in the end; r > k + 2).
3) The number h = (c – u)/u > 0.
From c > a and c > b we have: 2c > a + b, c > a + b – c, c – u > 0, h > 0.

THE PROOF of FLT

1. Let a^n + b^n – c^n = 0 (1°).

2. For any rank i, where k < i < r, it fulfils the equality: a_i + b_i – c_i = 0, and for rank r this equality has a form:
(a_r + b_r – c_r)_1 = n – 1 = «9».

{snip}
Next materials: detailing and numerical examples.

I don't think you have a proof of statement 2 above, i.e. that there are no carries in the addition of a and b or borrows in the substraction of c from a+b

 

[robert Ihnot]

Victor Sorokine
1 (corollary from Little Fermat's theorem). After multiplication u (= a + b – c) by p in new Fermat's equation the number u has the form (in base 7): 666…666000…000 with r digits equal to n – 1 or "9" (with k zeros in the end; r > k + 2).

And judging from ramsey2879, I conjucture that this example may have a bearing on the problem:

$$35^2+612^2=613^2$$

X+Y-Z=35+612-613 = 34, which in base 2 is: 100010, and multiplying by 2 gives us 1000100, which after a series of 0 gives us p-1=1, and then a new series of 0.

 

[Victor Sorokine]

Condition at present

The last and first digits do not show the contradiction. The Lemma is necessaries but it do not work with the number u. THEREFORE the contradiction exists in middle digits and the Lemma works with the number c:
1. We transform number u into c = n^t – 1, or с = 999…999 [or: c = n^s(n^t – 1) = 999…999000…000]. Here 9 = n – 1, 8 = n – 2; maximal rank R of c R(c) = t.
2. If a = 8… and b = 8… then 8…^n + 8…^n < 9…^n. Here R(a) = R(b) = R(c) = t.
Therefore a_t = c_t = 9.
3. If b_t = 1 then 9000…^n + 1000…^n > 9999…^n and a^n + b^n – c^n =/ 0.
4. If a_{t – 1} = 8 and 98…^n + b^n = 99…^n then R(b) > t – 1 (cf. binomial theorem for n prim) and therefore (cf. point 3) a^n + b^n – c^n =/ 0. Therefore a_{t – 1} = 9.
Etc. to a_1. THEREFORE a = c and b = 0 < 1 and a^n + b^n – c^n =/ 0 for (a, b, c) > 0.
=============
The cause of impossibility of the Fermat's equation is cleared from next equation:
(n^t + n^r)^n – b^n = (n^t)^n, where r < t and b is true number.
Example: c = 1010, a = 1000, 1010^n – b^n = 1000^n. From here 1010^n –1000^n = b^n or (1000 + 10)^n –1000^n = b^n or 1000^n + n10x1000^(n – 1) + d – 1000^n = b^n or
b^n = n10x1000^(n – 1) + d = n^(2 + 3n – 3) + d = n^(3n – 1) + d from here b > n^[(3n – 1)/n] or (in the worst case n = 3) b > n^3.
Conclusion: the increase of the digit c_r by 1 is impossible to compensate by subtract any (r-digit number)^n!

Now the proof with c = 999…999(000…000) is obvious.

==========
Dear Hurkyl, Robert Ihnot and Ramsay2879,
thank you very much for your help.
Excuse me for unliterary and short texts.

 

[moshek]

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.

Victor Sorokine hi

I notice here about your paper only yesterday.

Do you know about this work on FLT in : www.fermatproof.com ?

Are you sure that A.Willes really know about your nice work to solve FLT ?

In any case I plane to study your sort mostly claim for the prove.

Thank you
Moshe

 

[ramsey2879]

The proof contains 3 tools only:
1) Scale of notation in base prime;
2) Lemma: for number a (there last digit a_1 =/ 0) there are such d and s, that ad = n^s – 1 (corollary from Little Fermat's theorem). After multiplication u (= a + b – c) by p {sic"d"} in new Fermat's equation the number u has the form (in base 7): 666…666000…000 with r digits equal to n – 1 or "9" (with k zeros in the end; r > k + 2).

It is possible to choose a "d" where ad = n^s-1 but r < k+2. If this is the case, then, for instance if r=2 then d must be multiplied by 10101...01 where the number of zeros between the 1's = r-1 to get a new R. It thus is possible to chose a "d" such that R > is any multiple of r > k+2.

3) The number h = (c – u)/u > 0.
From c > a and c > b we have: 2c > a + b, c > a + b – c, c – u > 0, h > 0.

THE PROOF of FLT

1. Let a^n + b^n – c^n = 0 (1°).

2. For any rank i, where k < i < r, it fulfils the equality: a_i + b_i – c_i = 0, and for rank r this equality has a form:
(a_r + b_r – c_r)_1 = n – 1 = «9».

statement 2 is false
let a= 251, b = 326502, c=53 base 7
Then a+b-c=330000; thus d = 2020202 and k=4, r=8
A= 2020202*a=540404032, B=2020202*b= 666266660004, C=2*c=140404036

You state that for k<i<r [i.e. 4<i<8] that A_i + B_i + C_i =0, but this is false.
You should use variables consistently and watch your statements. Also please do not make us have to refer to your earlier proof to understand what you are talking about in the current proof.

 

[moshek]

statement 2 is false


You state that for k<i<r [i.e. 4<i<8] that A_i + B_i + C_i =0,
but this is false.

Hi ramsey2879

thank you for your complicate example
You probably mean A_i+B_i - C_i = 0
let's see Victor replay to your contra example

But Isn't Victor FLT contradiction (?) appear only at the base n
which is the power of the original equation a^n+b^n=c^n.



Moshe