Fermat’s Last Theorem: A one-operation proof

[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

 

[ramsey2879]

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.

Why do you say this without proof. It makes no sense to me. For instance the sum of the digits 7+4-5 is =12 in base 3
7^3+4^3-5^3 = 1201 in base 3 1+2 is odd while 1+2+1 is even. Where is the impossibility?

P.S. Let's get back to your original proof in revised form. Do you have a clean copy?

 

[HallsofIvy]

Unfortunately the "one operation" involved here appears to be hand-waving!

 

[arildno]

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

Last time I looked, modelling reality with the aid of maths worked very nicely.

 

[Victor Sorokine]

For Moshe

Dear Victor, 31.8.2005
Please explain to me how do you transform the digit u_{k+1} into 1.

- If n is prime and digit g =/ 0 then for any digit d there exists such digit e that (ge)_1 = d (cf. §2 in the ADDENDUM for published proof).

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 ?

- Yes.

Thank you

Victor

 

[Victor Sorokine]

For ramsey2879

1. Why do you say this without proof. It makes no sense to me. For instance the sum of the digits 7+4-5 is =12 in base 3
7^3+4^3-5^3 = 1201 in base 3 1+2 is odd while (1°) 1+2+1 is even. Where is the impossibility?

P.S. Let's get back to your original proof in revised form.
(2°) Do you have a clean copy?

(1°) In your sum 1+2+1 the number 1, 2, 1 are not the numbers a, b, c.
In my last proof at each step a_{i} == a, b_{i} == b, c_{i} == c.

(2°) Not yet. I am sorry...

Victor

 

[moshek]

Unfortunately the "one operation" involved here appears to be hand-waving!

Please explain to me what is the real different
between hand waving to hand writing ?

It's a real question and not a joke !

Moshe

 

[Hurkyl]

Often times, when consciously skipping past significant details of an argument, a person will wave their hands around.

Thus, the term "hand waving" has been adopted to refer to an argument whose significant details have been omitted.

 

[moshek]

Often times, when consciously skipping past significant details of an argument, a person will wave their hands around.

Thus, the term "hand waving" has been adopted to refer to an argument whose significant details have been omitted.

So Fermat like Our Victor are doing hand-waving when they wrote that a^n+b^-c^n=0 have no solution ( n>2) even if they may have the same intuition.

 

[HallsofIvy]

I've always thought of "hand waving" as meaning skipping over the crucial parts of a proof. As in writing the crucial equation on the blackboard, waving your hand at it and saying "of course, this is obvious to any intelligent person"!

 

[Victor Sorokine]

Regular interesting idea

So Fermat like Our Victor are doing hand-waving when they wrote that a^n+b^-c^n=0 have no solution ( n>2) even if they may have the same intuition.

Regular interesting idea for my friends

(1°) Let a^n + b^n – c^n = 0, where for positive integers a, b, c
(2°) the number u = a + b – c > 0, where (abc)_1 =/ 0, u_(k) = 0, u_{k+1} * 0, k > 0.
(3°) Let's transform the numbers u_{k+1} and u_{k+2} into n – 1 (or "9").

(4°) Now U' = a_(k+1)^n + b_(k+1)^n – c_(k+1)^n > 0 and
(5°) U'' = a^n + b^n – c^n – U' = U – U' < 0; from here U'' = – U'.

Lemma. (6a°) If a = bn – c > 0, where c_1 =/ 0, then a_1 = n – c.
And (6b°) if d = en – c_1 > 0 and f = gn – c_1 < 0, then d_1 = n – f_1.

At last:
(7°) U'_{k+2} = (a_{k+2} + b_{k+2} – c_{k+2})_1 > 0 (cf. binominal theorem for 1°).
But some POSITIVE ending of the NEGATIVE number U'' has equal value:
(a_{k+2} + b_{k+2} – c_{k+2})_1. Therefore (cf. 6a°) U''_{k+2} =/ (– U') _{k+2}.
And therefore U_{k+2} =/ 0.

vs

 

[HallsofIvy]

Regular interesting idea for my friends

(1°) Let a^n + b^n – c^n = 0, where for positive integers a, b, c
(2°) the number u = a + b – c > 0, where (abc)_1 =/ 0, u_(k) = 0, u_{k+1} * 0, k > 0.
(3°) Let's transform the numbers u_{k+1} and u_{k+2} into n – 1 (or "9").

(4°) Now U' = a_(k+1)^n + b_(k+1)^n – c_(k+1)^n > 0 and
(5°) U'' = a^n + b^n – c^n – U' = U – U' < 0; from here U'' = – U'.

Lemma. (6a°) If a = bn – c > 0, where c_1 =/ 0, then a_1 = n – c.
And (6b°) if d = en – c_1 > 0 and f = gn – c_1 < 0, then d_1 = n – f_1.

At last:
(7°) U'_{k+2} = (a_{k+2} + b_{k+2} – c_{k+2})_1 > 0 (cf. binominal theorem for 1°).
But some POSITIVE ending of the NEGATIVE number U'' has equal value:
(a_{k+2} + b_{k+2} – c_{k+2})_1. Therefore (cf. 6a°) U''_{k+2} =/ (– U') _{k+2}.
And therefore U_{k+2} =/ 0.

vs

Wonderful example! He has not said what (abc)_1 means, he has defined u but not u_k, and he has a lemma that says "If a = bn – c > 0, where c_1 =/ 0, then a_1 = n – c." without telling us what c_1 or a_1 mean!

Handwaving at its finest!

 

[Victor Sorokine]

For recent disputants of the forum

Wonderful example! He has not said what (abc)_1 means, he has defined u but not u_k, and he has a lemma that says "If a = bn – c > 0, where c_1 =/ 0, then a_1 = n – c." without telling us what c_1 or a_1 mean!

Handwaving at its finest!

For recent disputants of the forum:

a_k, or a_{k} – the digit at the place k from the end, in the number a (thus a_1 is the last digit);
a_(k) – is the k digits’ ending (it is a number) of the number a (a_(1) = a_1)
[cf. Revista Foaie Matematică: www.fmatem.moldnet.md/1_(v_sor_05).htm].

V.S.

 

[Victor Sorokine]

Shorter, simpler, more clearly, more complete

Shorter, simpler, more clearly, more complete

I don't more participate in the discussion about the previous versions of the proof.
My final choice is last (September) proof. Here is:

Lemma: In prime base n, if whole numbers a = pn + d > 0 (< 0) and b = qn + d < 0 (> 0), where whole d > 0, then a =/ – b (– a =/ b ) by any p and q.
Example in base 7: 50 + 3 =/ – (– 50 + 3), 50 + 3 =/ – (– 60 + 3)…

PROOF of FLT

Case 1: The last digit of the number abc is not equal to zero, or (abc)_1 =/ 0.

(1°) Let a^n + b^n – c^n = 0, where for positive integers a, b, c
(2°) the number u = a + b – c > 0, u_(k) = 0, u_{k+1} * 0, k > 0.
(3°) If the digit u_{k+1} = 1 then we multiply the equality 1° by a 2n.
Now u_{k+1} = 2 and the digit (a_{k+1} + b_{k+1} – c_{k+1})_1 = v =/ 0 since v = or 1 ether 2.

a^n = a_(k)^n + (n^(k+1))a_{k+1} + (n^(k+2))P_a, b^n = …, c^n = …, and:
a^n + b^n – c^n = [a_(k)^n + b_(k)^n – c_(k)^n] +
(n^(k+1))[a_{k+1} + b_{k+1} – c_{k+1}] + (n^(k+2))P, where
(4°) [a_(k)^n + b_(k)^n – c_(k)^n = U',
(5°) (n^(k+1))[a_{k+1} + b_{k+1} – c_{k+1}] + (n^(k+2))P = U'',
and U'_(k+1) = U''_(k+1) = 0, U'_{k+1} == U''_{k+1} == v > 0.

BUT the number U' is positive/negative and number U'' is positive/positive. Therefore (cf. Lemma) U' =/ –U''. And therefore U' + U'' = a^n + b^n – c^n =/ 0.

Case 2: (ac)_1 =/ 0, b_(t) = 0, b_{t+1} =/ 0, [or (ab)_1 =/ 0 and c_(t) = 0, c_{t+1} =/ 0]

In this case u = a + bn^(nt – 1) – c [or u = a + b – c n^(nt – 1)]. The proof is analogous.

The proof is done.


P.S. For recent disputants of the forum:

a_k, or a_{k} (only for the forums) – the digit at the place k from the end, in the number a (thus a_1 is the last digit);
a_(k) – is the k digits’ ending (it is a number) of the number a (a_(1) = a_1)
[cf. Revista Foaie Matematică: www.fmatem.moldnet.md/1_(v_sor_05).htm].

V.S.

 

[ramsey2879]

Shorter, simpler, more clearly, more complete

I don't more participate in the discussion about the previous versions of the proof.
My final choice is last (September) proof. Here is:

Lemma: In prime base n, if whole numbers a = pn + d > 0 (< 0) and b = qn + d < 0 (> 0), where whole d > 0, then a =/ – b (– a =/ b ) by any p and q.
Example in base 7: 50 + 3 =/ – (– 50 + 3), 50 + 3 =/ – (– 60 + 3)…

PROOF of FLT

Case 1: The last digit of the number abc is not equal to zero, or (abc)_1 =/ 0.

In other words neither a,b,or c in the formula a^n+b^n=c^n is divisible by n. In this proof the numbers are in base n, therefore, any number not divisible by n can not end in 0. x_a or x_{a} means the digit of the number a positions from the right. x_1 = the last digit of the number x

(1°) Let a^n + b^n – c^n = 0, where for positive integers a, b, c
(2°) the number u = a + b – c > 0, u_(k) = 0, u_{k+1} * 0, k > 0.

"u_{k+1} * 0" makes no sense but from past writings, it is clear that Victor is saying that u mod n^k = 0, while u mod n^(k+1) =/ 0 for some k> 0.

(3°) If the digit u_{k+1} = 1 then we multiply the equality 1° by a 2n.

Victor is "forgetting" again to write this proof in a manner such that those unfamiliar to the thread can understand it. If u_{k+1} =/ 1 then we can multiply a,b and c by some d such that u_{k+1} = 1, and yet the new a,b and c still satisfy the original assumptions. Thus there is no limitation on the proof to assume that u_{k+1} = 1. By the same reasoning if a,b and c were multiplied by 2d then u_{k+1} would equal 2. But Victor is "wrong" to suggest that we should multipled a,b and c by 2n, since then u_{k+1} would equal 0, not 2. In this fashion, Victor obtains new a,b and c such that u_{k+1} = 2

Now u_{k+1} = 2 and the digit (a_{k+1} + b_{k+1} – c_{k+1})_1 = v =/ 0 since v = or 1 ether 2.

It is a fact that the sum of the digits a_{k+1} + b_{k+1} minus the digit c_{k+1} must equal 1 or 2 modulus n.

a^n = a_(k)^n + (n^(k+1))a_{k+1} + (n^(k+2))P_a, b^n = …, c^n = …, and:
a^n + b^n – c^n = [a_(k)^n + b_(k)^n – c_(k)^n] +
(n^(k+1))[a_{k+1} + b_{k+1} – c_{k+1}] + (n^(k+2))P, where
(4°) [a_(k)^n + b_(k)^n – c_(k)^n = U',

For k=1, n^(k+1)=100 and n^(k+2)=1000. Let a=11, b=12 and c = 10 such that a+b-c = 20. K = 1 then 1^3+2^3-0^3 = U' = 100 and a^n+b^n-c^n = 20000 = U''+U' = U''+100. However, 20000 is not equal to 100 + 100*(1+1-1)+1000*P. Thus this part of Victor's proof is clearly incorrect.

(5°) (n^(k+1))[a_{k+1} + b_{k+1} – c_{k+1}] + (n^(k+2))P = U'',
and U'_(k+1) = U''_(k+1) = 0, U'_{k+1} == U''_{k+1} == v > 0.

Although U'_2 = U''_2= 0 it is not correct to state that they should for any reason equal 1 or 2. Victor simply has not establish any contradiction.

BUT the number U' is positive/negative and number U'' is positive/positive. Therefore (cf. Lemma) U' =/ –U''. And therefore U' + U'' = a^n + b^n – c^n =/ 0.

U' must equal - U'' for a^n+b^n-c^n =0 but Victor has not shown that U' = an+d and U'' = bn+d must hold, hence the Lemma simply does not apply.

Case 2: (ac)_1 =/ 0, b_(t) = 0, b_{t+1} =/ 0, [or (ab)_1 =/ 0 and c_(t) = 0, c_{t+1} =/ 0]
In this case u = a + bn^(nt – 1) – c [or u = a + b – c n^(nt – 1)]. The proof is analogous.

The proof is done.

As shown above the proof for case 1 fails in more than one way, thus the "analogous proof" whatever that is must also fail.

 

[moshek]

I've always thought of "hand waving" as meaning skipping over the crucial parts of a proof. As in writing the crucial equation on the blackboard, waving your hand at it and saying "of course, this is obvious to any intelligent person"!

HallsofIvy:

I feel that Victor improve now his hand waving about FLT. but still he need to write the missing points in his work.

Will you be kind and explain for me because of my bad English what the meaning is of: "Euclid alone has looked on beauty bare"

Thank you
Moshe

 

[moshek]

In other words neither a,b,or c in the formula a^n+b^n=c^n is divisible by n. In this proof the numbers are in base n, therefore, any number not divisible by n can not end in 0. x_a or x_{a} means the digit of the number a positions from the right. x_1 = the last digit of the number x

"u_{k+1} * 0" makes no sense but from past writings, it is clear that Victor is saying that u mod n^k = 0, while u mod n^(k+1) =/ 0 for some k> 0.

.

dear ramsey2879

thank you very much for doing the translation to victor work. he really miss many points and I hope he will complete them a.s.a possible. I have to admit that I am still very far to understand his prove (?) but I hope I can give my final opinion at October.

As shown above the proof for case 1 fails in more than one way, thus the "analogous proof" whatever that is must also fail

Let's hope for Victor and bless his great effort and kindness to share here his Organic vision [ the base n idea with a_{k+1} observing ] obout Fermat work.

Moshe

 

[Victor Sorokine]

For Moshe and ramsey2879

Dear Moshe and ramsey2879,
I am glad to hear you. Thanks.

One more precise definition:
The digit u_{k+1} = 2 does not work, since a_(k)^n + b_(k)^n – c_(k)^n < 0.
Correctly: u_{k+1} = n – 1 (or "9"), u_{k+2} > 1. For example u_{k+2} = 2; then v = a_{k+2} + b_{k+2} – c_{k+2} = (1 or 2) =/ 0.
Then U' = a_(k+1)^n + b_(k+1)^n – c_(k+1)^n > 0, where U'_{k+3} > 0,
and U'' < 0, where U''*_{k+3} = v > 0. Therefore U'_{k+3} = – v. But in the NEGATIVE number U'' = – qn + v U''_{k+3} = – v.
And therefore U_{k+3} = – 2v =/0!
The END?

Victor

P.S. Where are hurkyl and Robert Ihnot?

PPS. Numerical example in base 3: k = 2 and u_3 = 2.
In the worst case a_(3) = 211, b_(3) = 211, c_(3) = 222,
or in base 10: a_(3) = 22, b_(3) = 22, c_(3) = 28;
U' = 22^3 + 22^3 – 26^3 = 21296 – 17576 > 0! U" < 0, v = (1 or 2) > 0. And therefore U_{k+3} = (n – 2v or 2n – 2v) =/ 0.

 

[ramsey2879]

Dear Moshe and ramsey2879,
I am glad to hear you. Thanks.

One more precise definition:
The digit u_{k+1} = 2 does not work, since a_(k)^n + b_(k)^n – c_(k)^n < 0.
Correctly: u_{k+1} = n – 1 (or "9"), u_{k+2} > 1. For example u_{k+2} = 2; then v = a_{k+2} + b_{k+2} – c_{k+2} = (1 or 2) =/ 0.
Then U' = a_(k+1)^n + b_(k+1)^n – c_(k+1)^n > 0, where U'_{k+3} > 0,
and U'' < 0, where U''*_{k+3} = v > 0. Therefore U'_{k+3} = – v. But in the NEGATIVE number U'' = – qn + v U''_{k+3} = – v.
And therefore U_{k+3} = – 2v =/0!
The END?

Victor

P.S. Where are hurkyl and Robert Ihnot?

PPS. Numerical example in base 3: k = 2 and u_3 = 2.
In the worst case a_(3) = 211, b_(3) = 211, c_(3) = 222,
or in base 10: a_(3) = 22, b_(3) = 22, c_(3) = 28;
U' = 22^3 + 22^3 – 26^3 = 21296 – 17576 > 0! U" < 0, v = (1 or 2) > 0. And therefore U_{k+3} = (n – 2v or 2n – 2v) =/ 0.

This is a proof? Where is the logic to this?

 

[moshek]

This is a proof? Where is the logic to this?

Dear ramsey2879

Since you understand victor work 10 times better then me I want to ask you a very simple question . Can you see the picture that Victor have in his mind about FLT. can you help him to find it's right logic ?

 

[HallsofIvy]

Dear ramsey2879

Since you understand victor work 10 times better then me I want to ask you a very simple question . Can you see the picture that Victor have in his mind about FLT. can you help him to find it's right logic ?

Are you saying that you believe that Victor does not even understand the logic of his own proof? ramsey2879 was at least just asking him to clarify his logic!