Fermat’s Last Theorem: A one-operation proof

[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&#039; := (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&#039; := (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

 

[Victor Sorokine]

There is a question: is it THE END?

Dear ramsey2879, Hurkyl, robert Ihnot, learningphysics and moshek,
there is a question: is it THE END:

Start-situation:
Let a^n + b^n – c^n = 0 (1°)
u = a + b – c with k-zero-ending (k > 0);
r – maximum rank of c^n (1a°);

PROOF

1. Let’s multiply the equation (1°) by dn , where d = 1 + gn^r, in order to transform the digit
u_{r + k + 1} into 1 (2°). (The digits of u_(r + k) do not change.)
2. Now from
[a_(r + k) + a_{r + k + 1}]^n + [b_(r + k) + b_{r + k + 1}]^n + [c_(r + k) + c_{r + k + 1}]^n = 0 we have:
{[a_(r + k)] _(r + k + 1) + [b_(r + k)] _(r + k + 1) – [c_(r + k)] _(r + k + 1)} +
+ (n^(r + k + 2))(a_{r + k + 1} x 1 + b_{r + k + 1} x 1 – c_{r + k + 1} x 1) + (n^(r + k + 3))P = 0, where
{[a_(r + k)] _(r + k + 1) + [b_(r + k)] _(r + k + 1) – [c_(r + k)] _(r + k + 1)} = 0 (cf. 1a°) and
(a_{r + k + 1} x 1 + b_{r + k + 1} x 1 – c_{r + k + 1} x 1)_1 = u_{r + k + 1}.
From here u_{r + k + 1} = 0, that contradicts to (2°).
The proof is done.

Victor Sorokine

 

[Hurkyl]

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.

So? All you need to do is to (correctly) derive one contradiction in order to prove that one of your original assumptions must be false.

 

[moshek]

So? All you need to do is to (correctly) derive one contradiction in order to prove that one of your original assumptions must be false.

Do you think, that Victor prove for FLT is wrong ?

Moshe Klein

 

[moshek]

Dear ramsey2879, Hurkyl, robert Ihnot, learningphysics and moshek,
there is a question: is it THE END:

Start-situation:
Let a^n + b^n – c^n = 0 (1°)
u = a + b – c with k-zero-ending (k > 0);
r – maximum rank of c^n (1a°);

PROOF

1. Let’s multiply the equation (1°) by dn , where d = 1 + gn^r, in order to transform the digit
u_{r + k + 1} into 1 (2°). (The digits of u_(r + k) do not change.)
2. Now from
[a_(r + k) + a_{r + k + 1}]^n + [b_(r + k) + b_{r + k + 1}]^n + [c_(r + k) + c_{r + k + 1}]^n = 0 we have:
{[a_(r + k)] _(r + k + 1) + [b_(r + k)] _(r + k + 1) – [c_(r + k)] _(r + k + 1)} +
+ (n^(r + k + 2))(a_{r + k + 1} x 1 + b_{r + k + 1} x 1 – c_{r + k + 1} x 1) + (n^(r + k + 3))P = 0, where
{[a_(r + k)] _(r + k + 1) + [b_(r + k)] _(r + k + 1) – [c_(r + k)] _(r + k + 1)} = 0 (cf. 1a°) and
(a_{r + k + 1} x 1 + b_{r + k + 1} x 1 – c_{r + k + 1} x 1)_1 = u_{r + k + 1}.
From here u_{r + k + 1} = 0, that contradicts to (2°).
The proof is done.

Victor Sorokine

hi victor

well, even A.willes had a mistak in FLT until he fix it ( I hope)
Are all these calculation are in base n ?


Thank you
Moshe Klein

 

[moshek]

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.




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.

ramsey2879 hi

I am still study victor work on FLT , I know that it is Organic and very nice but I don't have a final opinion if it is really correct with no mistakes.

Does post 71 answer to your claim that statement 2 is palse ?

Thank you
Moshe Klein

 

[Victor Sorokine]

The same idea, BUT…

Dear Moshe,
thank you and excuse my mistake.
There is interesting correction.

It is obvious:

If a^n + b^n – c^n = 0 (1°), then there is such s, that
/(a + n^(–s))^n + (a + n^(–s))^n – (a + n^(–s))^n/ < n^(–sn) (2°), where s > r + 1 and r is maximum rank of the number c, and
/(an^s + 1)^n + (bn^s + 1)^n – (cn^s + 1)^n/ < 1 (3°).
From here: n^sn(a^n + b^n – c^n) = 0, nn^[s(n – 1)](a^(n – 1) + b^n(n – 1) – c^n(n – 1)) = 0
{or a^(n – 1) + b^(n – 1) – c^(n – 1) = 0}, … a + b – c = n^(–s) 0, that is impossible.
V.S.

 

[moshek]

Dear Victor,

I am really like your non conventual's way of thinking. for the first time I find real interest in FLT because of your organic attitude that can explain ( I hope so ) why Fermat wrote what he wrote about this problem ! if your prove is right [ I don't know that yet ] you should present it at the icm exactly 365 day's from today .

Please look at http://www.icm2006.org/

I hope to have my final opinion about your nice work at October
but since you correct it already few times in this thread ( and that's fine )can you sent here the most update version of your work.


Thank you
Moshe

 

[Victor Sorokine]

Error Of Principle Of My Critics

Counter-examples are done for such u where u'_{k + 1} = 0.
But in my published proof u'_{k + 1} =/ 0.(Not infrequently, to find the error in a counter-example it is more difficult then to solve the same problem.)
For 10 months 70 mathematicians (members of AMS) and
2000 amateurs could not find an error of principle in the demonstration.
I terminate the participation in the discussion about all following proofs.
So the discussion about the original proof continues.

***
A key of the proof:
For a and b:
if a + b = (a + b)_(k) = 0, then (a^n + b^n)_(k + 1) = 0,
if a + b = (a + b)_(k) =/ 0, then (a^n + b^n)_(k + 1) =/ 0 (cf. ADDENDUM).
For a, b, c:
if a + b – c = (a + b – c)_(k) = 0, then (a^n + b^n – c^n)_(k + 1) = 0 (cf. counter-examples),
if a + b – c = (a + b – c)_(k) =/ 0, then (a^n + b^n – c^n)_(k + 1) =/ 0 (cf. my proof).

Publications of the proof:
doc: Revista Foaie Matematică: www.fmatem.moldnet.md/1_(v_sor_05).htm
pdf: http://fox.ivlim.ru/docs/sorokine/flt.pdf

Correction in (9°) (cf. 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.
Victor

 

[Hurkyl]

Don't confuse "could not find an error" with "didn't bother looking for an error". Most people don't bother trying to fill in the gaps of an argument for which they have little to no expectation of validity.

The problem is that you don't actually write a proof: you write a bunch of equations, occasionally with a brief comment, and very few, if any, are self-evident.

 

[moshek]

Don't confuse "could not find an error" with "didn't bother looking for an error". Most people don't bother trying to fill in the gaps of an argument for which they have little to no expectation of validity.

The problem is that you don't actually write a proof: you write a bunch of equations, occasionally with a brief comment, and very few, if any, are self-evident.

Don't forget the story of Ramanujan who could not prove his 4000 amassing results. [ few of them are wrong ] Victor have a new interesting vision about FLT and he may need the help and the giddiness of the 'expert' to expand / check / correct / or maybe unfortunately cancel his attitude.

please read the strange situation Perelman paper on Poincaré conjecture : ( was taken from www.mathworld.com )

The Clay Mathematics Institute included the conjecture on its list of $1 million prize problems. In April 2002, M. J. Dunwoody produced a five-page paper that purports to prove the conjecture. However, Dunwoody's manuscript was quickly found to be fundamentally flawed (Weisstein 2002). A much more promising result has been reported by Perelman (2002, 2003; Robinson 2003). Perelman's work appears to establish a more general result known as the Thurston's geometrization conjecture, from which the Poincaré conjecture immediately follows (Weisstein 2003). Mathematicians familiar with Perelman's work describe it as well thought-out and expect that it will be difficult to locate any substantial mistakes (Robinson 2003, Collins 2004). In fact, Collins (2004) goes so far as to state, "everyone expects [that] Perelman's proof is correct."

And also from there about the strange situation Mihailescu paper on the Catalan problem:

..Finally, on April 18, 2002, Mihailescu sent a manuscript purporting to prove the entire conjecture to several mathematicians. The paper was apparently also accompanied by an expository paper by colleague Yuri Bilu that analyzes and summarizes Mihailescu's argument (van der Poorten 2002).

So I think that Victor ask for meangfull interaction with his work, I will try my best to have my final opinion on his work on FLT until October.

Moshe

 

[Victor Sorokine]

The simplest case for analysis

The simplest case for analysis: u_{k + 1} = u'_{k + 1} + u''_{k + 1} = 1

If u'_{k + 1} = 1, then u''_{k + 1} = 0, U'_{k + 2} =/ 0 (?), U''_{k + 2} = 0 and U' + U'' =/ 0;
if u''_{k + 1} = 1, then u'_{k + 1} = 0, U'_{k + 2} = 0 (?), U''_{k + 2} =/ 0 and U' + U'' =/ 0.

 

[Hurkyl]

Don't forget the story of Ramanujan who could not prove his 4000 amassing results.

I highly doubt that: as I told Victor, don't confuse "did not prove" with "cannot prove". Remember that many of his 4000 amazing results were in his private notebooks, not in papers he was trying to publish, so there wasn't any real reason to give a rigorous proof.

please read the strange situation Perelman paper on Poincaré conjecture : ( was taken from www.mathworld.com )

And mathematicians have not described Victor's paper as "well thought out", and have not stated they expect it to be correct, so I don't see your point. It sounds like Perelman actually tried to write a proof, rather than an outline he claims to work, as Victor has done.

And also from there about the strange situation Mihailescu paper on the Catalan problem:

And I have no idea what point you're trying to make with this one.

 

[Victor Sorokine]

Specially for MocheK

Specially for MocheK

(1°) Let (an^t)^n + (bn^t)^n – (cn^t)^n = 0, where t >> maximum rank R(c) = r.
From here
(2°) n^r > (an^t)^(n – 1) + (bn^t)^ (n – 1) – (cn^t)^ (n – 1) > 0,
and
(3°) (an^t + 1)^n + (bn^t + 1)^n – (cn^t + 1)^n > 0 and hence
(4°) (an^t + e)^n + (bn^t + 1)^n – (cn^t + 1)^n = 0, where e < 1.
From (4°) we have:
(5°) n^t(a^n + b^n – c^n) + n^[t(n – 1) + 1][a^(n – 1) + b^(n – 1) – c^(n – 1)e] + P = 0,
where P < n^[t(n – 2) + r] << n^[t(n – 1) + 1].
From (5°) we have: [a^(n – 1) + b^(n – 1)]_(t) = [c^(n – 1)e]_(t), or
(6°) a^(n – 1) + b^(n – 1) = c^(n – 1)e.
But it's easy to show that it is impossible!

Victor

 

[Victor Sorokine]

Condition at present and thanks

Condition at present

1. The last (from 1 to {k+s}) and first (from {r – k – s + 2}) digits of the number u do not show the contradiction.
2. Counter-examples are right.
3. My error: the digit u_{k+2} shows error.
4. The right solution: the digit u_{k + s + 1} shows error (k + s is maximum rank of the number u).

Deciding Lemma:
If U_(t + 1) = 0, then U*' _{t + 1} = U_{t + 1} (corollary from 21° - 25° in the Proof).

Idea of the new proof:
After transformation of the digit u_{k + s + 1} into 1 (with multiplication by 1 + n^(s+1)), we see that
U* _{ k + s + 1} = U*'' _{ k + s + 1} = 1 and hence U*'_{ k + s + 1} = –1.
Therefore up to multiplication U' _{ k + s + 1} = –1 and hence U _{ k + s + 1} = –1 =/ 0.

Let's come back to the discussion of the original proof in new version.
Dear Hurkyl, moshek, Robert Ihnot and Ramsay2879,
I would like to thank you all for your active participation and for your help.
Victor

 

[moshek]

Dear Victor,

Please call me Moshe. Since 1980 my investigation is about the Organic unity of mathematics, and it's hidden connection to physics ( Please read the end of Hilbert lecture at Paris 1900).

I would like now to start and studies carefully all the details your interesting Organic work about FLT at your web-site :

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

Did you correct there already your mistake?

Another very interesting attitude to FLT I found yesterday here:

http://noticingnumbers.net/FLTsummary.htm

Yours
Moshe

 

[moshek]

I highly doubt that: as I told Victor, don't confuse "did not prove" with "cannot prove". Remember that many of his 4000 amazing results were in his private notebooks, not in papers he was trying to publish, so there wasn't any real reason to give a rigorous proof.




And mathematicians have not described Victor's paper as "well thought out", and have not stated they expect it to be correct, so I don't see your point. It sounds like Perelman actually tried to write a proof, rather than an outline he claims to work, as Victor has done.




And I have no idea what point you're trying to make with this one.

Hurky:

Could it be, that mathematical true today come to a very exited point that it depend on a personal opinion of the mathematitioan ?

Moshe

 

[robert Ihnot]

According to the Russian paper Pravda, "Fermat's theorem is the unproved theorem indicating that the equation xn + yn = zn has no solution for x,y,z nonzero integers when n is greater than 2."

However, the matter has changed: "Doctor of Technical Sciences Alexander Ilyin will present his proof of Fermat's theorem at a meeting to be held at the Academy of Aviation and Aeronautics. His colleagues in Omsk believe Alexander's proof is flawless and simple." [/I][PLAIN]http://english.pravda.ru/science/19/94/377/16036_Fermat.html[/URL]

Remember you heard it here first!

 

[Hurkyl]

Could it be, that mathematical true today come to a very exited point that it depend on a personal opinion of the mathematitioan ?

Your question is unclear... anyways, I think the answer is "Mathematical facts do not depend on opinion"... in fact, the sciences are generally set up to reduce or eliminate subjective influences.

 

[vantheman]

According to the Russian paper Pravda, "Fermat's theorem is the unproved theorem indicating that the equation xn + yn = zn has no solution for x,y,z nonzero integers when n is greater than 2."

However, the matter has changed: "Doctor of Technical Sciences Alexander Ilyin will present his proof of Fermat's theorem at a meeting to be held at the Academy of Aviation and Aeronautics. His colleagues in Omsk believe Alexander's proof is flawless and simple." [/I][PLAIN]http://english.pravda.ru/science/19/94/377/16036_Fermat.html[/URL]

Remember you heard it here first!

When is this proof going to be presented?

 

[ramsey2879]

Dear Victor,

Please call me Moshe. Since 1980 my investigation is about the Organic unity of mathematics, and it's hidden connection to physics ( Please read the end of Hilbert lecture at Paris 1900).

I would like now to start and studies carefully all the details your interesting Organic work about FLT at your web-site :

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

Did you correct there already your mistake?

Yours
Moshe

No! There are still corrections to be made in the web page that Victor cites above. A few corrections follow:

Dear Hurkyl,
1. Your counter-example is right.
2. Congratulation!
3. Thank You very much!

Counter-examples are done for such u where u'_{k + 1} = 0.
But in my published proof u'_{k + 1} =/ 0.(Not infrequently, to find the error in a counter-example it is more difficult then to solve the same problem.)
For 10 months 70 mathematicians (members of AMS) and
2000 amateurs could not find an error of principle in the demonstration.
I terminate the participation in the discussion about all following proofs.
So the discussion about the original proof continues.

Unfortunately, the cited web page still does not account for the counter examples.
First, it even does not agree with Victors allegation that in his proof u'_{k + 1} =/ 0! On the contrary the cited proof contains the following: “(5°) u'k+1 = (–1, 0 or 1) – because – nk < a'(k) < nk, – nk < b'(k) < nk, – nk < c'(k) < nk
and the numbers a, b, c have different signs”
Thus I propose the following correction here. Specify the following additional conditions (1) a,b,c must all be greater than 0 with c greater than a or b and less than a+b. It is easy to show that this does not impose a meaningful limitation on the proof and it overcomes any counterexample presented so far.
Second the cited web page further does not agree with his current position as of 8/25/05. See below.

Condition at present

1. The last (from 1 to {k+s}) and first (from {r – k – s + 2}) digits of the number u do not show the contradiction.
2. Counter-examples are right.
3. My error: the digit u_{k+2} shows error.
4. The right solution: the digit u_{k + s + 1} shows error (k + s is maximum rank of the number u).

Deciding Lemma:
If U_(t + 1) = 0, then U*' _{t + 1} = U_{t + 1} (corollary from 21° - 25° in the Proof).

Idea of the new proof:
After transformation of the digit u_{k + s + 1} into 1 (with multiplication by 1 + n^(s+1)), we see that
U* _{ k + s + 1} = U*'' _{ k + s + 1} = 1 and hence U*'_{ k + s + 1} = –1.
Therefore up to multiplication U' _{ k + s + 1} = –1 and hence U _{ k + s + 1} = –1 =/ 0.

Let's come back to the discussion of the original proof in new version.

The web page says nothing about these new definitions of u, U’, U* etc with respect to k+s+1. It still contains the allegation that the error shows up at U'_k+2 contrary to Victor's current position.

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

...Should not "n - 2 with result 566...662" be --n - 1 with result 566...661--?

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

It appears that Victor's current position is that a,b and c should be multiplied first by d to give u = ZZZZZZ00000 where Z=n-1, the number of Z's is "s" and the number of 0's is k. Secondly, multiply the new a,b,c by (n-1) to give u=YZZZZZ10000 where Y=n-2 the number of Z's is s-1 and the number of zeros remains k. He has presented this position piecemeal and his posts are only in outline form and contain typographical errors that are significant to an understanding of the proof. We have given plenty of time to respond to Victor in a meaningful way. He owes it to us (and especially to you given your willingness to verify his proof) to explain his present position in a clear manner in a single post which has been checked for errors and which can be followed without undue effort.

 

[moshek]

Dear Victor,

You may be happy but still I am sorry that there is a mistke in the link which I share with you before on FLT : ttp://noticingnumbers.net/FLTsummary.htm[/URL]


I understand already from your web-site that your work on FLT is base on splitting the numbers a, b, c into pairs of sums, grouping them in two sums U' and U'' and multiplying the equation a^n + b^n – c^n = 0 by 11^n and then looking on 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) . when you look on it on base n we can see that is not equal to 0 .

This is very nice [B]organic[/B] looking and I really like it !

But please notice to Ramsey 2897 note :

[QUOTE]It appears that Victor's current position is that a,b and c should be multiplied first by d to give u = ZZZZZZ00000 where Z=n-1, the number of Z's is "s" and the number of 0's is k. Secondly, multiply the new a,b,c by (n-1) to give u=YZZZZZ10000 where Y=n-2 the number of Z's is s-1 and the number of zeros remains k. He has presented this position piecemeal and his posts are only in outline form and contain typographical errors that are significant to an understanding of the proof. We have given plenty of time to respond to Victor in a meaningful way. He owes it to us (and especially to you given your willingness to verify his proof) to explain his present position in a clear manner in a single post which has been checked for errors and which can be followed without undue effort.[/QUOTE]

So, I understand , that you did not correct yet in your web-site of FLT
the mistaks that you admit in this thread.

Could you do that for me and I will continue to study your intersting work ?

Your sincerely
Moshe



[QUOTE]

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 formal account of the history of the Theorem and the Bibliography are not included in the Russian version
An elementary proof of Fermat’s Last Theorem


VICTOR SOROKINE



TOOLS: [Square brackets are used for additional explanation.]

Notations used:

All the numbers are written using a base n with n being a prime number and n > 10.

[All the cases where n is not prime, except n = 2k (which can be reduced to the case n = 4),

are reduced to a case with a prime n using a simple substitution.

We don’t study the cases with n = 3, 5 and 7.]

ak – the digit at the place k from the end, in the number a (thus a1 is the last digit).

[Example for a = 1043 in the base 5: 1043 = 1x53 + 0x52 + 4x51 + 3x50; a1 = 3, a2 = 4, a3 = 0, a4 = 1.]

a(k) – is the k digits’ ending (it is a number) of the number a (a(1) = a1; 1043(3) = 043).

Everywhere in the text a1 ≠ 0.

[If all three numbers a, b and c end by a zero, we need to divide the equation 1° by nn.]

(ain)1 = ai and (ain - 1)1 = 1 (cf. Fermat’s Little Theorem for ai ≠ 0). (0.1°)

(n + 1)n = (10 + 1)n = 11n = …101 (cf. Newton Binomial for a prime n).

A simple corollary from the Newton Binomial and the Fermat’s Little Theorem for s ≠ 1 [a1 ≠ 0]:

If the digit as undergoes a rise or is reduced by d (0 < d < n),

then the digit ans+1 undergoes a rise or is reduced by d (or d + n, or d – n). (0.2°)

[Digits in negative numbers are « negative ».]



***



(1°) Let us assume that an + bn – cn = 0 .



Case 1: (bc)1 ≠ 0.



(2°) Let u = a + b – c, where u(k) = 0, uk+1 ≠ 0, k > 0 [we know that both u > 0 and k > 0 in 1°].

(3°) We multiply the equality 1° by a number d1n (cf. §§2 and 2a in the Appendix) in order to transform

the digit uk+1 into 5. After that operation the numbers’ labeling is not changed

and the equality still keeps its index (1°).

It is clear that also in the new equality (1°) u = a + b – c, u(k) = 0, uk+1 = 5.

(1*°) Then let a*n + b*n – c*n = 0, where the sign “*” designates numbers of the equation (1°) written in a canonical way, after the multiplication of the equation (1°) by 11n .



(4°) Let’s introduce following numbers, in that order: u, u' = a(k) + b(k) – c(k),

u'' = u – u' = (a – a(k)) + (b – b(k)) – (c – c(k)), v = (ak+2 + bk+2 – ck+2)1, u*' = a*(k) + b*(k) – c*(k),

u*'' = u* – u*' = (a* – a*(k)) + (b* – b*(k)) – (c* – c*(k)), 11u', 11u'', v* = (a*k+2 + b*k+2 – c*k+2)1.

Let’s then calculate the two last significant digits in these numbers:

(3a°) uk+1 = (u'k+1 + u''k+1)1 = 5;

(5°) u'k+1 = (–1, 0 or 1) – because – nk < a'(k) < nk, – nk < b'(k) < nk, – nk < c'(k) < nk

and the numbers a, b, c have different signs;

(6°) u''k+1 = (4, 5 or 6) (cf. 3a° and 5°) [it is important: 1 < u''k+1 < n – 1];

(7°) u'k+2 = 0 [always!] – because \u'\ < 2nk ;

(8°) u''k+2 = uk+2 [always!];

(9°) u''k+2 = [v + (ak+1 + bk+1 – ck+1)2]1, where (ak+1 + bk+1 – ck+1)2 = (–1, 0 or 1);

(10°) v = [uk+2 – (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)] =

= [uk+2 – (–1, 0 or 1)]1;

(11°) u*k+1 = uk+1 = 5 – because u*k+1 and uk+1 are the last significant digits in the numbers u* and u;

(12°) u*'k+1 = u'k+1 – because u*'k+1 and u'k+1 are the last significant digits in the numbers u*' and u';

(13°) u*''k+1 = (u*k+1 – u*'k+1)1 = (3 – u*'k+1)1 = (4, 5 or 6) [it is important: 1 < u*''k+1 < n – 1];

(14°) (11u')k+2 = (u'k+2 + u'k+1)1 (then, reducing the numbers into a canonical form –

the value u'k+1 «goes» into u*''k+2, because u*'k+2 = 0);

(14a°) it is important: the numbers (11u')(k+2) and u*'(k+2) differ from each other

only by k+2-th digits, more exactly: u*'k+2 = 0, but (11u')k+2 ≠ 0 generally;

(15°) (11u'')k+2 = (u''k+2 + u''k+1)1;

(16°) u*k+2 = (uk+2 + uk+1)1 = (u''k+2 + uk+1)1 = (u''k+2 + 5)1;

(16а°) Let’s note that: u*'k+2 = 0 (cf. 7°);

(17°) u*''k+2 = (u*k+2 +1, u*k+2 or u*k+2 – 1)1 = (cf. 9°) = (u''k+2 + 4, u''k+2 + 5 or u''k+2 + 6)1;

(18°) v* = [u*k+2 – (a*(k+1) + b*(k+1) – c*(k+1))k+2]1

[where u*k+2 = (uk+2 + uk+1)1 (cf. 16°), but (a*(k+1) + b*(k+1) – c*(k+1))k+2 =

= (–1, 0 or 1) – cf. 10°] = [(uk+2 + uk+1)1 – (–1, 0 or 1)]1.



(19°) Let’s introduce the numbers

U' = (ak+1)n + (bk+1)n – (ck+1)n, U'' = (an + bn – cn) – U', U = U' + U'',

U*' = (a*k+1)n + (b*k+1)n – (c*k+1)n, U*'' = (a*n + b*n – c*n) – U*', U* = U*' + U*'';

(19а°) Let’s note that: U'(k+1) = U*'(k+1) = 0.



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

Indeed, from 1° we can find the following:
U = an + bn – cn =

= (a(k+1) + nk+1ak+2 + nk+2Pa)n + (b(k+1) + nk+1bk+2 + nk+2Pb)n – (c(k+1) + nk+1ck+2 + nk+2Pc)n =

= (a(k+1)n + b(k+1)n – c(k+1)n) + nk+2(ak+2a(k+1)n - 1 + bk+2b(k+1)n - 1 – ck+2c(k+1)n - 1) + nk+3P =

= U' + U'' = 0, where

U' = a(k+1)n + b(k+1)n – c(k+1)n,

(20a°) U'' = nk+2(ak+2a(k+1)n -1 + bk+2b(k+1)n -1 – ck+2c(k+1)n -1) + nk+3P,

where (ak+2a(k+1)n -1 + bk+2b(k+1)n -1 – ck+2c(k+1)n -1)1 = (cf. 0.1°)=

(20b°) = (ak+2 + bk+2 – ck+2)1 = U''k+3 = v (cf. 4°).



(21°) Corollary: (U'k+3 + U''k+3)1 = (U*'k+3 + U*''k+3)1 = 0.

(22°) Let’s calculate the digit (11nU')k+3:

[as the numbers (11u')(k+2) and u*'(k+2) differ only by k+2 digits, with the difference being equal

to (11u')k+2), then this will also be the difference between the digits (11nU')k+3 and U*'k+3, which

means that the digit (11nU')k+3 will be greater than the digit U*'k+3 by (11u')k+2 (cf. 0.2°)]

(11nU')k+3 = U'k+3 = (U*'k+3 + (11u')k+2)1 = (U*'k+3 + u'k+1)1.

(23°) From there we come to: U*'k+3 = U' k+3 – u'k+1.

(24°) Let’s calculate the digit U*'' k+3 :

U*'' k+3 = v* = (uk+2 + uk+1)1 – (–1, 0 or 1) – cf. (18°);

(25°) Finally, let’s calculate the digit (U*'k+3 + U*''k+3)1:

(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 =

(cf. 23° and 24°) = (– u'k+1 + v* – v) = (cf. 18° and 10°) =

= (– u'k+1 + [uk+2 + uk+1 – (–1, 0 or 1)] – [uk+2 – (–1, 0 or 1)])1 =

= (– u'k+1 + uk+1 + (–2, –1, 0, 1, or 2))1 = (cf. 3a°) =

( u''k+1 + (–2, –1, 0, 1, or 2))1 = (cf. 6°) = (2, 3, 4, 5, 6, 7 or 8) ≠ 0,

which contradicts to 21° ; therefore the expression 1° is an inequality.



Case 2 [is proven in a similar way, however it is much more simple]: b (or c) = ntb',

where b1 = 0 and bt+1 = b'1 ≠ 0.

(26°) Let’s introduce number u = c – a > 0, where u(nt – 1) = 0, but unt ≠ 0 (cf. §1 in the Addendum).

(27°) After multiplying the equality 1° by a number d1n (in order to transform the digit unt into 5)

(cf. §§2 and 2a in the Addendum), we will keep the numbers’ notations.

(28°) Let’s: u' = a(nt – 1) – c(nt – 1), u'' = (a – a(nt – 1)) – (c – c(nt – 1)) (where, evidently, u''nt = (ant – cnt)1);

U' = a(nt)n + bn – c(nt)n (where U'(nt + 1) = 0 – cf. 1° and 26°), U'' = (an – a(nt)n) – (cn – c(nt)n),

U*' = a*(nt)n + b*n – c*(nt)n (where U*'(nt + 1) = 0), U*'' = (a*n – a*(nt)n) – (c*n – c*(nt)n),

v = ant+1 – cnt+1.



The calculations, in all ways analogical to the case 1, show that the digit at the nt+2-th place in the Fermat’s equality is not zero. The number b in all calculations (except the very last operation and also in paragraph 27°) can be ignored, because the digits bnnt+1 and bnnt+2 , after multiplying the equality 1° by 11n, won’t change (this is because 11n(3) = 101).



Therefore, for any prime number n > 10, the theorem is proven.



==================



ADDENDUM


§1. If the numbers a, b, c don’t have common factors and b1 = (c – a)1 = 0,

then, from the number R = (cn – an)/(c – a) =

= cn –1 + cn –2a + cn –3a2 + … c2an - 3 + can - 2 + an - 1 =

= (cn –1 + an –1) + ca(cn –3 + an –3) + … + c(n –1)/2a(n –1)/2 =

= (cn –1 – 2c(n –1)/2a(n –1)/2 + an –1 + 2c(n –1)/2a(n –1)/2) + ca(cn –3 – 2c(n –3)/2a(n –3)/2 + an –3 + 2c(n –3)/2a(n –3)/2) +

+ … + c(n –1)/2a(n –1)/2 = (c – a)2P + nc(n –1)/2a(n –1)/2 comes that:

c – a is divisible by n2, therefore R is divisible by n but is not by n2;

as R > n, then the number R has a prime factor r not equal to n;

c – a is not divisible by r;

if b = ntb', where b'1 ≠ 0, then the number c – a is divisible by ntn – 1 but is not by ntn.



§2. Lemma. All n digits (a1di)1, where di = 0, 1, … n – 1, are different.

Indeed, admitting that (a1d1*)1 = (a1d1**)1, we come to the following conclusion: ((d1* – d1**)a1)1 = 0.

And then d1* = d1**. Therefore, the sets of digits a1 (here along with a1 = 0) and d1 are the same.

[Example for a1 = 2: 0: 2x0 = 0; 1: 2x3 = 11; 2: 2x1 = 2; 3: 2x4 = 13; 4: 2x2 = 4.

If n is not prime, then the Lemma is not true: in base 10 both (2х2)1 = 4, and (2х7)1 = 4.]

§2a. Corollary. For any digit a1 ≠ 0 there is such a digit di, that (a1di)1 = 1.

[Example for a1 = 1, 2, 3, 4: 1x1 = 1; 2x3 = 11; 3x2 = 11; 4x4 = 31.]

[/QUOTE]

 

[moshek]

According to the Russian paper Pravda, "Fermat's theorem is the unproved theorem indicating that the equation xn + yn = zn has no solution for x,y,z nonzero integers when n is greater than 2."

However, the matter has changed: "Doctor of Technical Sciences Alexander Ilyin will present his proof of Fermat's theorem at a meeting to be held at the Academy of Aviation and Aeronautics. His colleagues in Omsk believe Alexander's proof is flawless and simple." [/I][PLAIN]http://english.pravda.ru/science/19/94/377/16036_Fermat.html[/URL]

Remember you heard it here first!

Hi robert,

Thank you for the interesting link !
can you find for us some more information
about this new FLT work.

Thank you
Moshe

 

[moshek]

Your question is unclear... anyways, I think the answer is "Mathematical facts do not depend on opinion"... in fact, the sciences are generally set up to reduce or eliminate subjective influences.

Hurky,

The thread which was open by Victor about cecking of his interesting new work on FLT is about mathematics and not at all about Science .


Why you have confuse between
mathematics and sciences ?

Moshe

 

[Chronos]

Hurky,

The thread which was open by Victor about cecking of his interesting new work on FLT is about mathematics and not at all about Science .


Why you have confuse between
mathematics and sciences ?

Moshe

I'm confused too. I routinely confuse mathematics with science. Would it be fair to say math defines the set of all possible realities, whereas science defines the subset of realities that do not conflict with observation? In other words, either description without the other is, at best, philosophy.

 

[HallsofIvy]

I'm confused too. I routinely confuse mathematics with science. Would it be fair to say math defines the set of all possible realities, whereas science defines the subset of realities that do not conflict with observation? In other words, either description without the other is, at best, philosophy.

I wouldn't put it in terms of possible "realities" (I'm not at all sure what a "possible" but "not real" reality would be!). Mathematics constructs "templates" that, to one degree or another, can be fitted to reality.

 

[arildno]

I'm confused too. I routinely confuse mathematics with science. Would it be fair to say math defines the set of all possible realities, whereas science defines the subset of realities that do not conflict with observation? In other words, either description without the other is, at best, philosophy.

I use to say, rather flippantly, "Reality is a special case of maths"..

 

[Victor Sorokine]

Fantastic idea for my friends

Condition at present

Fantastic idea for my friends

Right contradiction: the number u is infinite

(1°) Let a^n + b^n – c^n = 0,
(2°) where for integers a, b, c the number u = a + b – c > 0, where (a_1b_1c_1)_1 =/ 0, u_(k) = 0, u_{k+1} * 0, k > 0.
(3°) Let's transform the digit u_{k+1} into 1.

(4°) Let's assume that the number u contains only one non-zero digit (u_{k+1}). Then:
(4a°) if ((a_(k) + b_(k) – c_(k))_{k+1} = 0, then a_{k+1} + a_{k+1} – a_{k+1} = 1,
U''_{k+2} = a_{k+1} + a_{k+1} – a_{k+1} = 1 and the number U' contains only one non-zero digit (U'_{k+2} = 1).
Or: u is even, but a^n + b^n – c^n is odd, that is impossible.
(4b°) if ((a_(k) + b_(k) – c_(k))_{k+1} = 1, then a_{k+1} + a_{k+1} – a_{k+1} = 0,
U''_{k+2} = 0 and U'_{k+2} = 1. Or: u is odd, but a^n + b^n – c^n even is, that is impossible.
Therefore there exists second non-zero digit in the number u: u_s.

(5°) Let's assume that the number u contains only one non-zero digit (u_{k+1}). Then:
(5a°) if ((a_(s) + b_(s) – c_(s))_{k+1} is odd, then u is even, but U''_{s+1} (and a^n + b^n – c^n) is odd, that is impossible.
(5b°) if ((a_(s) + b_(s) – c_(s))_{k+1} is even, then u is odd, but U''_{s+1} (and a^n + b^n – c^n) is even, that is impossible.
Therefore there exists third non-zero digit in the number u: u_r.

(6°) Let's assume…
AND SO AD INFINITUM!

Victor Sorokine

 

[Victor Sorokine]

Or:

Fantastic idea for my friends
Right contradiction: the number u is infinite

OR:
If the sum of the digits of the number a + b – c is odd/even then the sum of the digits of the number a^n + b^n – c^n is even/odd.
Therefore:
If u = a + b – c is odd/even then a^n + b^n – c^n is even/odd, that is impossible!

V.S.

 

[moshek]

Dear Victor, 31.8.2005

Please explain to me how do you transform the digit u_{k+1} into 1.
Do you really mean that this is the entire prove to FLT
when you assume that a^n + b^n – c^n = 0,
and you define the number u = a + b – c > 0,
and look on the numbers on the base of n ?

Thank you
Moshe

Right contradiction: the number u is infinite

(1°) Let a^n + b^n – c^n = 0,
(2°) where for integers a, b, c the number u = a + b – c > 0, where (a_1b_1c_1)_1 =/ 0, u_(k) = 0, u_{k+1} * 0, k > 0.
(3°) Let's transform the digit u_{k+1} into 1.

(4°) Let's assume that the number u contains only one non-zero digit (u_{k+1}). Then:
(4a°) if ((a_(k) + b_(k) – c_(k))_{k+1} = 0, then a_{k+1} + a_{k+1} – a_{k+1} = 1,
U''_{k+2} = a_{k+1} + a_{k+1} – a_{k+1} = 1 and the number U' contains only one non-zero digit (U'_{k+2} = 1).
Or: u is even, but a^n + b^n – c^n is odd, that is impossible.
(4b°) if ((a_(k) + b_(k) – c_(k))_{k+1} = 1, then a_{k+1} + a_{k+1} – a_{k+1} = 0,
U''_{k+2} = 0 and U'_{k+2} = 1. Or: u is odd, but a^n + b^n – c^n even is, that is impossible.
Therefore there exists second non-zero digit in the number u: u_s.

(5°) Let's assume that the number u contains only one non-zero digit (u_{k+1}). Then:
(5a°) if ((a_(s) + b_(s) – c_(s))_{k+1} is odd, then u is even, but U''_{s+1} (and a^n + b^n – c^n) is odd, that is impossible.
(5b°) if ((a_(s) + b_(s) – c_(s))_{k+1} is even, then u is odd, but U''_{s+1} (and a^n + b^n – c^n) is even, that is impossible.
Therefore there exists third non-zero digit in the number u: u_r.

(6°) Let's assume…
AND SO AD INFINITUM!

Right contradiction: the number u is infinite


OR:
If the sum of the digits of the number a + b – c is odd/even then the sum of the digits of the number a^n + b^n – c^n is even/odd.
Therefore:
If u = a + b – c is odd/even then a^n + b^n – c^n is even/odd, that is impossible!

V.S.

 

[moshek]

I use to say, rather flippantly, "Reality is a special case of maths"..

arilno

I like your definition very much
do you see now the big mistake of modeling the world
by using mathematics since the true is the opposite ?

Thank you
Moshe