Fermat's Last Theorem. Elementary proof

Victor Sorokine · December 20, 2009

Why does C - B = a^n, or C - A = b^n? And shouldn't your first a, b, and c be A, B, and C?
=Uncool-

If the numbers C and B have no common factors and C-B divided not by n,

then the numbers C-B and P=(C^n-B^n)/(C-B) have no common factors.

Therefore if a' is a prime divider of the number C-B, then a'^n is a divider of the number C-B and is not a divider of the number P.

a is a designation of the product of all prime dividers a' of the number C-B,

b is a designation of the product of all prime dividers b' of the number C-A,

c is a designation of the product of all prime dividers c' of the number A+B.

Victor

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Strictly speaking, the essence of the represented proof of the FLT consists in all of the most primitive linear equation of 5°-11° with one unknown x and the interpretation of its solution. The lines of 1°-4° - these are well well-known facts, and the discrepancy of the formula of 11° can be shown by many methods. Here one more.

The number [(C-B)+(C-A)]=c[2r-c^(n-1)] in 11° it is divided only into c (since it is not divided by r), i.e.,

12°) [(C-B)+(C-A)]=c.

But with the aid of the reasonings of 1°-11° we can obtain analogous equalities, also, for the numbers A and B:

13°) [(A+B)-(C-B)]=b,

14°) [(A+B)-(C-A)]=a.

And if A+B-C=U, where, as is known, U is divided into abc, then, adding the equalities 13° and 14° and reading from the sum the equality 12°, we have:

15°) 2[(A+B)-(C-B)-(C-A)]=4(A+B-C)=4U=a+b-c.

Consequently, the number a+b-c is divided completely into 4abc, which is, obviously, impossible for natural a, b, c.

uncool · December 21, 2009

If the numbers C and B have no common factors and C-B divided not by n,
then the numbers C-B and P=(C^n-B^n)/(C-B) have no common factors.

Therefore if a' is a prime divider of the number C-B, then a'^n is a divider of the number C-B and is not a divider of the number P.

Ahhh. You probably should expand the proof that C - B and P have no common factors.

Proof:

(C^n - B^n)/(C - B) = sum from 1 to n of B^(n - i)C^(i - 1). If we look modulo C - B, the latter is equivalent to sum from 1 to n of B^n, or n B^(n - 1). Therefore, the greatest common factor of C - B and P is the same as the greatest common factor of C - B and nB^n. Then since n does not divide C - B and n is prime, n and C - B are relatively prime, so any common factor of C - B and nB^n is a common factor of C - B and B^n. This is then a common factor of B^n and C^n - which can only be 1.

=Uncool-

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Victor, you have an arithmetic error when going from step 10 to 11. You go from

(C-B)[T+(C-A)x]+(C-A)[T+(C-B)x]

to

[(C-B)+(C-A)](T+x)

when the correct outcome is

[(C-B)+(C-A)]T + 2(C-A)(C-B)x.

=Uncool-

Edited December 21, 2009 by uncool
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Victor Sorokine · December 21, 2009

You probably should expand the proof that C - B and P have no common factors.

C-B and P have no common factors, because the number P can be represented in the form:

P=T(C^2-B^2)+n(CB)^(n-1).

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Victor, you have an arithmetic error when going from step 10 to 11. You go from
(C-B)[T+(C-A)x]+(C-A)[T+(C-B)x]

to

[(C-B)+(C-A)](T+x)

when the correct outcome is

[(C-B)+(C-A)]T + 2(C-A)(C-B)x.

=Uncool-

Yes! Thank You very much!

Victor Sorokine · January 1, 2010

Fantastic fundamental contradiction of Fermat’s equality

It is completely obvious that for relatively prime natural A and B in the number

1°) D=a^n+b^n=(a+b)R occurs the inequality: a+b<R.

However, if D=c^n, then inequality sign changes by the opposite. The simple proof of this fact will be represented later.

==========

Prompt for the too impatient:

R<c^n-(c-1)^n<(c^n)^(0,5) (you will recall the formula of the expansion of the sum of degrees). Consequently, a+b>(c-1)^n>(c^n)^(0,5). From where (a+b)>R, that contradicts 1°.

uncool · January 2, 2010

Fantastic fundamental contradiction of Fermat’s equality

It is completely obvious that for relatively prime natural A and B in the number

1°) D=a^n+b^n=(a+b)R occurs the inequality: a+b<R.

However, if D=c^n, then inequality sign changes by the opposite. The simple proof of this fact will be represented later.

==========

Prompt for the too impatient:

R<c^n-(c-1)^n<(c^n)^(0,5) (you will recall the formula of the expansion of the sum of degrees). Consequently, a+b>(c-1)^n>(c^n)^(0,5). From where (a+b)>R, that contradicts 1°.

How are you getting that c^n - (c - 1)^n < sqrt(c^n)? Because that is pretty much never true.

=Uncool-

bascule · January 2, 2010

[math]\]\LaTeX\[[/math] ftw

Edited January 2, 2010 by Cap'n Refsmmat
fixed that for you

Victor Sorokine · January 4, 2010

How are you getting that c^n - (c - 1)^n < sqrt(c^n)? Because that is pretty much never true.
=Uncool-

Yes. Thank.

++++++++++

Last idea required refinement, the following lemma became result of which:

If for the integers a, b, c, p, q, r is the equality ap+bq-cr=abc>0 (c>a>b), then p+bq-cr>0.

Is there a simple proof of this lemma?

+ + +

If lemma is true, then Fermat's last theorem proves by one phrase.

Are actual, with relatively prime [math]A, B, C[/math]

1°) [math]A^n=C^n-B^n=(C-B)P=a^n*p^n[/math]

[or 1b°) [math]B^n=C^n-A^n=(C-A)Q=b^n*q^n] [/math], where, according to lemma,

2°) [math]p>a[/math] (or [math]q>b[/math]).

And now during the term-by-term multiplication of Fermat's equality by the sufficiently large number [math]d^{nn}[/math]

3°) the number [math]ad^n[/math] becomes more than the number [math]pd^{n-1}[/math], that contradicts 2°.

Edited January 4, 2010 by Victor Sorokine

Victor Sorokine · January 11, 2010

If lemma is true,..

There is reason to believe that last lemma is incorrect.

But to before end a study on FLT, I decided to rapidly examine the previous ideas, and in one of them (http://dxdy.ru/topic24331-15.html - Вт авг 11, 2009 16:28:24) I revealed the interesting passed moment:

If AB not divides by n then the number q-p must be divided by n(a-b);

however, in the field of real numbers (with prime n>2 and c>a>b>1) occurs the inequality n(a-b)>>q-p, which is magnificently confirmed by calculations on the computer:

if A=15,96, B=12, C=17,9596, n=3,

then the number q-p=0,722 are not divided by 3*(a-b)=1,659!!! And

if A=125, B=64, C=130,353, n=3,

then the number q-p=5,7096 are not divided by 3*(a-b)=6,9047!!!

And it remains to us only confirm this fact by the approximate analytical calculations.

[i will resemble: [math]A^n=C^n-B^n=(C-B)P=a^n*p^n; B^n=C^n-A^n=(C-A)Q=b^n*q^n[/math].]

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And it remains to us only confirm this fact by the approximate analytical calculations.

1. Hypothesis is incorrect with the large numbers.

2. However, I found very interesting tool. Here is the project of new proof.

Let us assume that for prime n>2 and integers A, B, C (C>A>B>0)

1°) [math]A^n+B^n=C^n[/math], where A+B-C=U>0.

Let us designate

by the letters A', B', C' remainder terms after division of the numbers A, B, C by U;

by letters A", B", C" the number A'-U, B'-U, C'-U;

by letters a, b, c the smallest (in terms of the absolute value) numbers in the pairs (A', A"), (B', B"), (C', C").

Thus,

2°) |a|<U/2, |b|<U/2, |c|<U/2 and

3°) A=a+dU, B=b+eU, C=c+fU (where d, e, f are integers).

As it follows from 2°, the number v=a+b-с=U-TU can have only three values: U, 0, -U.

Let us examine these cases.

The I. a+b-с=U. Then (see 3°) d+e-f=0, and in this case

4°) [math] (dU)^n+(eU)^n-(fU)^n<0[/math]. However, addition to the numbers dU, eU, fU the numbers a, b, с with any values of them in the range of 2° CANNOT CHANGE this inequality:

[math](dU+a)^n+(eU+b)^n-(fU+c)^n<0[/math].

The II. a+b-с=0. Then (see 3°) d+e-f=1, and in this case all the more

5°) [math](dU)^n+(eU)^n-(fU)^n<0[/math]. However, addition to the numbers dU, eU, fU the numbers a, b, с with any values of them in the range of 2° CANNOT CHANGE this inequality:

[math](dU+a)^n+(eU+b)^n-(fU+c)^n<0[/math].

At last,

The III. a+b-с=-U. Then (see 3°) d+e-f=2, and in this case all the more

6°) [math](dU)^n+(eU)^n-(fU)^n<0[/math]. However, addition to the numbers dU, eU, fU the numbers a, b, с with any values of them in the range of 2° CANNOT CHANGE this inequality:

[math](dU+a)^n+(eU+b)^n-(fU+c)^n<0[/math].

Thus, the equality of 1° in the integers is impossible.

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Impression, that the key to the proof of the FLT is found, did not change also after the second view. Now it is possible to approach the thorough calculation. However, now the process of proof is similar more greatly to the entertaining quiz: it suffices to know how to count only bis three. It is convenient to examine all numbers in the base U=1.

Thus, the case I: a+b-с=U.

Since in this case d+e-f=0 (see 3°) and (4°) [math](dU)^n+(eU)^n-(fU)^n<0[/math] (with minimum [math](1)^n+(1)^n-(2)^n<0[/math]), then it is necessary to maximize the number a and to minimize the number c. limiting values in this case: a=0.5, b=0.5, с=0.

But since the minimum values of the numbers d, e, f are in any event equal to 1 (this it follows from A+B-C=U>0 and C>A>B>0), and the numbers a, b, c are not equal to zero (and to 0.5) and, in addition to this, d+e=f, then it is necessary to introduce correctives into the values of a=0.5, b=0.5, с=0. But even with these values of the numbers a, b, c the maximum value of the number [math]H=(dU+a)^n+(eU+b)^n-(fU+c)^n[/math] takes the form: [math]H=(1+0.5)^n+(1+0.5)^n-(2)^n[/math], and now even with n=3 (worst case) the value of the number H (=6.8-8) is nevertheless negative! And an increase in the numbers d and f by the equal number does not can to convert the number H from the negative into the positive.

We pass now to the case OF THE II: a+b-с=0.

Since in this case d+e-f=1 (see 3°) and in this case (5°) [math](dU)^n+(eU)^n-(fU)^n>0[/math] [with the minimum values [math](1)^n+(1)^n-(1)^n>0] [/math], then necessary to minimize the number a and to maximize the number c. The limiting values of the numbers a, b, c in this case are such: a=0.5, b=0, с=0.5. But even with these values a, b, c and d=e=f=1 the minimum value of the number [math]H=(dU+a)^n+(eU+b)^n-(fU+c)^n[/math] takes the form: [math]H=(1+0.5)^n+(1)^n-(1+0.5)^n[/math], and even this value of the number H is positive. Those more it will be positive and with any other values of the numbers a, b, c, d, e, f. I.e., with a+b-с=0 the number H is always positive.

Finally, the case OF THE III: a+b-с=-U.

Since in this case d+e-f=2 (see 3°) and in this case also (6°) [math](dU)^n+(eU)^n-(fU)^n>0[/math], then it is necessary to maximize on the module the numbers a and b and to minimize on the module the number c. The limiting values of the numbers a, b, c in this case are such: a=-0.5, b=0, с=0.5, and the numbers d, e, f are such: d=2, e=2, f=2. And the minimum value of the number [math]H=(dU+a)^n+(eU+b)^n-(fU+c)^n[/math] take the form: [math]H=(2-0.5)^n+(2)^n-(2-0.5)^n[/math], and even this value of the number H is positive. Those more it will be positive with any other values of the numbers a, b, c, d, e, f. I.e., with a+b-с=-U the number H is always positive.

Thus, with any of the value of the numbers a, b, c the number H is not equal to zero. The FLT is proven. It is possible to approach the thorough checking of the proof. It is interesting that for this proof it suffices to have only general idea about the degree.

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Proof of Fermat's last theorem

Let us assume that for the integers A, B, C (C>A>B>0) и n>2

1°) [math](H=) A^n+B^n-C^n=0[/math], where A+B-C=U>0. (From what it is evident that C>A>B>U.)

Let us accept for the base (for the unit of measurement) the number U: U=1.

Now A=d*U+x, B=e*U+y, C=f*U+z, or A=d*+x, B=e*+y, C=f*+z, where x, y, z are positive remainders from the division of the numbers A, B, C on U.

If x<0.5U, then let us designate the remainder x by letter a and the number d* let us designate by letter d.

If x>0.5U, then let us designate the number x-U by letter a and the number d*+U let us designate by letter d.

Analogously let us to make also with the numbers e*, f*, y, z. As a result this the numbers A, B, C are taken the form: [math]A=d+a, B=e+b, C=f+c[/math], where the positive or negative numbers a, b, c in absolute value less than 0.5.

It is evident from the equality [math]A+B-C=(dU+a)+(eU+b)-(fU+c)= [/math]

2°) [math]=(d+e+f)U+(a+b-c)=U[/math] that the number h=a+b-c is divided into U. But taking into account that in absolute value of the numbers a, b, c do not exceed number 0.5U, or 0.5, then

3°) the number [math]h=a+b-c[/math] can have only three values: U, 0, -U, or 1, 0, -1.

And now it is evident from the equality 2° that

4°) the number [math]g=d+e+f[/math] can have only these three values: 0, U, 2U, or 0, 1, 2.

Finally, for these three pairs of values h and g we can write down three inequalities:

The I. If h=1, then g=0 and with the complete obviousness

[math]max(H)=max(A^n+B^n-C^n)=max[(d+a)^n+(e+b^n-(f+c)^n]= [/math]

[math]=(1+0.5)^n+(1+0.5)^n-(2+0)^n<0[/math] (here d=1, e=1, f=2; a=0.5, b=0.5, c=0).

The II. If h=0, then g=1 and with the complete obviousness [math]min(H)=min[(d+a)^n+(e+b^n-(f+c)^n]=(1+0.5)^n+(1+0)^n-(1+0.5)^n>0[/math] (здесь d=1, e=1, f=1; a=0.5, b=0, c=0.5).

The III. If h=-1, then g=2 and with the complete obviousness [math]max(H)=[/math]

[math]=(2-0.5)^n+(2-0.5)^n-(2-0)^n<0[/math] (here d=2, e=2, f=2; a=0.5, b=0.5, c=0).

With any changes in the numbers a, b, c, d, e, f, but, of course, with the observance of the conditions of 1°, 3°, 4°, these inequalities can only BE STRENGTHENED.

Thus, the number H, or [math]A^n+B^n-C^n[/math], IS NOT EQUAL to ZERO

Fermat's last theorem is proven fully.

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P.S. Possibly, in the proof of the point of 5° it is necessary to use the simple lemma:

U<B<2U. This means that e*=1; consequently, with b>0 e=1, and with b<0 e=2.

Edited January 11, 2010 by Victor Sorokine
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Victor Sorokine · January 13, 2010

Proof of the Fermat’s Last Theorem (edited text)

Let us assume that for integers A, B, C (C>A>B>0) and n>2

1°) [math] (H=) A^n+B^n-C^n=0[/math], where [math]A+B-C=U>0[/math] [is obvious, C>A>B>U] and

2°) [math]U<B<1,5U[/math]. Actually, from [math] [b^n=] BB^(n-1)=(C-A)Q[/math] it follows: [math]BB^(n-1)<(C-A)[nB^(n-1)][/math]

(see formula for Q). From here [math]B<3(C-A), B<3(B-U), 3U<2B, B>1,5U[/math].

Let us accept for the base (for the unit of measurement) the number U: U=1.

Now A=d*U+a*, B=e*U+b*, C=f*U+c*, or A=d*+a*, B=e*+b*, C=f*+c*,

where a*, b*, c* are positive remainders after the division of the numbers A, B, C by U.

If a<0.5U, then the remainder a* let us designate by letter a, and the number d* let us designate by letter d.

If a*0.5U, then the number a*-U let us designate by letter a, and the number of d*+U let us designate by letter d.

Analogously let us make also with the numbers e*, f*, b*, c*. As a result this number A, B, C is taken the form: A=d+a, B=e+b, C=f+c, where the positive or negative numbers a, b, c in terms of the absolute value less than 0.5. (Equality - for example, a=0.5 – is’t impossibly for that reason, which in this case, according to the theory of the Fermat’s equality, number A would have common divisors with the numbers B and C even after their division by the greatest common divisor.)

From the equality U=A+B-C=(dU+a)+(eU+b)-(fU+c), or

3°) [math] (d+e+f)U+(a+b-c)=U[/math] follows that the number h=a+b-c is divided by U. But taking into account that the absolute value of the number a, b, c do not exceed number 0.5U, or 0.5,

4°) the number [math]h=a+b-c[/math] can have only three values: U, 0, -U, or 1, 0, -1.

And now it follows from the equality 3° that

5°) the number [math]g=d+e+f[/math] can have respectively these three values: 0, U, 2U, or 0, 1, 2.

And the following inequalities testify about truth of the FLT:

I. If h=1 and g=0, then [math](1+t+0,5)^n+(1+0,5)^n-(2+t+0)^n<max(H)=max(A^n+B^n-C^n)=[/math]

[math]=(1+0,5)^n+(1+0,5)^n-(2+0)^n<-1,24<0[/math] with d=1, e=1, f=2; a=0,5, b=0,5, c=0 and any t>0[/math].

II. If h=0 and g=1, then [math](1+t+0,25)^n+(1+0,25)^n-(1+t+0,5)^n>min(H)=(1+0,25)^n+[/math]

[math]+(1+0,25)^n-(1+0,5)^n>0,53>0[/math] with d=1, e=1, f=1; a=0,25, b=0,25, c=0,5 and any t>0[/math].

III. If h=-1 and g=2, than in this case either [math]B>1.5U[/math]

(in the [math]H=(2-0.5)^n+(2-0.5)^n-(2-0)^n)[/math], or [math]B<U[/math]

(in the [math]H=(3-0.5)^n+(1-0.5)^n-(2-0)^n)[/math], that contradicts to 2°.

Victor Sorokine · January 18, 2010

Corrections:

2°) [math]U<B<1.5U[/math]. Actually, [math]from [b^n=] BB^(n-1)=(C-A)Q[/math] it follows: [math]BB^(n-1)>[/math]

[math]>(C-A)[nB^(n-1)][/math] (see formula for Q). From here [math]B>3(C-A), B>3(B-U), 3U>2B, B<1.5U[/math].

.....

I. If h=1 and g=0, then [math](1+t+0,5)^n+(1+0.5)^n-(2+t+0)^n<max(H)=max(A^n+B^n-C^n)=[/math]

[math]=(1+0.5)^n+(1+0.5)^n-(2+0)^n<-1.24<0[/math] with d=1, e=1, f=2; a=0.5, b=0,5, c=0 and any t>0.

[With [math]n=2 max(H)=(1+0.5)^2+(1+0.5)^2-(2+0)^2=0.5>0[/math]!!! And equality H=0 is possible.]

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Proof of the Fermat’s Last Theorem (finish text)

Let us assume that for integers A, B, C (C>A>B>0) and n>2

1°) [math] (H=) A^n+B^n-C^n=0[/math], where [math]A+B-C=U>0[/math] [is obvious, C>A>B>U] and

2°) [math]U<B<1.5U[/math]. Actually, from [math] [b^n=] BB^{n-1}=(C-A)Q[/math] it follows:

[math]BB^{n-1}>

>(C-A)[nB^{n-1}][/math] (see formula for Q). From here

[math]B>3(C-A), B>3(B-U), 3U>2B, B<1.5U[/math].

The numbers A, B, C can be written down on the module U in two forms: with deficiency and with the surplus. For example, A=27 with the module u=12 can be written down thus: A=2U+3=3U-9. Of two versions of the numbers A, B, C let us leave those, in which the remainder (3 or 9) will be LESS:

2a°) A=dU+a, B=eU+b, C=fU+c, where 0<(a, b, c)<0.5U.

It is easy to show (see the Appendix) that in the equality of 1° the numbers a, b, c are equal neither 0 nor U/2.

From the equality U=A+B-C=(dU+a)+(eU+b)-(fU+c), or

3°) [math] (d+e+f)U+(a+b-c)=U[/math] follows that the number h=a+b-c is divided by U. But taking into account that the absolute value of the number a, b, c do not exceed number 0.5U,

4°) the number [math]h=a+b-c[/math] can have only three values: U, 0, -U.

And now it follows from the equality 3° that

5°) the number [math]g=d+e+f[/math] can have respectively these three values: 0, U, 2U.

And the following inequalities testify about truth of the FLT:

I. If h=U and g=0, then

[math] (U+t+0.5U)^n+(U+0.5U)^n-(2U+t+0)^n<max(H)=max(A^n+B^n-C^n)=[/math]

[math]=(U+0.5U)^n+(U+0.5U)^n-(2U+0)^n<-1.24U<0[/math] with d=1, e=1, f=2; a=0.5U, b=0.5U, c=0 and any t>0.

[With n=2 [math]max(H)=(U+0.5)^2+(U+0.5)^2-(2U+0)^2=0.5U>0[/math]!!! And equality H=0 is possible.]

II. If h=0 and g=1, then

[math] (U+t+0.25U)^n+(U+0.25U)^n-(U+t+0.5U)^n>min(H)=(U+0.25U)^n+[/math]

[math]+(U+0.25U)^n-(U+0.5U)^n>0.53U>0[/math] with d=1, e=1, f=1; a=0.25U, b=0.25U, c=0.5U and any t>0.

III. If h=-1 and g=2, than in this case either [math]B>1.5U[/math]

(in the [math]H=(2U-0.5U)^n+(2U-0.5U)^n-(2U-0)^n)[/math], or [math]B<U[/math]

(in the [math]H=(3U-0.5U)^n+(U-0.5U)^n-(2U-0)^n)[/math], that contradicts to 2°.

(Equality – for example, a=0.5 or a=0, – it’s impossibly for that reason, which in this case, according to the theory of the Fermat’s equality, number A would have common divisors with the numbers B and C even after their division by the greatest common divisor.)

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Corrections:
5°) the number [math]g=d+e+f[/math] can have respectively these three values: 0, U, 2U.

5°) the number [math]g=d+e-f[/math] can have respectively these three values: 0, U, 2U.

Edited January 16, 2010 by Victor Sorokine
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Victor Sorokine · January 20, 2010

4°) the number [math]h=a+b-c[/math] can have only three values: U, 0, -U.

Really,

-0.5=<a=<+0.5

-0.5=<b=<+0.5

-0.5=<c=<+0.5

+++++++++++

-1.5=<h=<+1.5, where h is WHOLE.

In the interval [-1.5, 1.5] are only THREE integers.

Victor Sorokine · January 24, 2010

FLT. Unproven case II (h=a+b-c=0) with integer d>1:

[math]A^n+B^n=C^n[/math], or taking into account 2°-5°

[math] (dU+a)^n+(U+b)^n-(dU+c)^n=0[/math],

where integers 0<(a, b, c)<0.5U.

Very interesting task.

+++

First approach:

Among the general solution [math](x+tmy, y+tax)[/math] of the linear diophantine equation [math]a(x+tmy)-m(y+tax)=1[/math]

to find such a particular solution that, for example,

[math]b(x+tmy)-mz=2[/math].

And now after multiplication Fermat’s equalitie by [math](x+tmy)^n[/math] the numbers A, B, C in the base m finish to 1, 2, 3 and the number

[math]a^n+B^n-C^n[/math] finishe by [math]1^n+2^n-3^n[/math], but NOT by ZERO.

Merged post follows:

Consecutive posts merged

I have a Question:

For relatively prime a and m the general solution of the linear diophantine equation

ax-my=1

there will be, as is known, such: (x+tm, y+ta).

Question: Are there such t and z, that

b(x+tm)-mz=1,

where a, b, m are relatively prime?

Thanks

Victor Sorokine · January 28, 2010

Proof of the Fermat’s Last Theorem. Project for the reflection

Let us assume that for the relatively prime integers A, B, C (C>A>B>0) and prime n>2

1°) [math]A^n+B^n-C^n=0[/math], where, as can easily be seen, the number

[math]A^n+B^n, A^n-B^n, C^n-B^n, C^n+B^n, C^n-A^n, C^n+A^n[/math] are relatively prime.

For the proof is used simple lemma about infinity of the set of primee numbers m of the form m=2kn+1 (proof it descends).

Let us thus, take prime [math]m=2kn+1>2C^n[/math]. Then, according to fermat's small theorem, the number

[math]A^{2kn}-B^{2kn}, C^{2kn}-B^{2kn} n, C^{2kn}-A^{2kn}[/math] in the base m finish with 0, from what it follows that

one of the algebraic cofactors in each of these three numbers finish with 0 (i.e. they are multiple m). Let us assume that its are three following cofactors:

2°) [math]C^{kn}-B^{kn} [=(C^n-B^n)P], C^{kn}-A^{kn} [=(C^n-A^n)Q], A^{kn}-B^{kn} [=(A^n-B^n)R][/math].

I advance the following hypothesis:

If the numbers c-a and c-b relatively prime and are more than 1, then the number [math]p=(c^n-a^n)/(c-a)[/math] and [math]q=(c^n-b^n)/(c-b)[/math] also relatively prime.

If hypothesis is confirmed, then at least one of the numbers [math]C^n-B^n, C^n-A^n, A^n-B^n[/math] is multiple m, that is impossible, since [math]m>2C^n[/math].

The matter remained for “the small”…

Victor Sorokine · February 2, 2010

Today I left to the theorem, whose proof indicates the gap of the closure of logic of FLT and, therefore, the inaccuracy of theorem (if the same exists) about the absence of elementary proof of FLT. Here is the formulation of theorem.

Theorem

For the relatively prime numbers a, b, c, d, greater than 1, the numbers

[math]p=(a^n-b^n)/(a-b)[/math] and [math]q=(c^{n-1}-d^{n -1})/(c-d)[/math], where n is simple, sont relatively prime.

Theorem proves with the aid of the following lemma (and lemma itself it proves with the aid of the linear diophantine equations):

Lemma

For the relatively prime numbers a and b everything - with exception perhaps of only (namely the number n) - the prime dividers m of the number [math]p=(a^n-b^n)/(a-b)[/math] have the form of m=tn+1.

Merged post follows:

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Following lemma

“Numbers [math]A+B-C (=U> 0)[/math] and [math]A-B[/math] under the conditions of the FLT are relatively prime”.

(I will resemble: the numbers [math]A, B, C, C-B, C-A, A-B, C+B, C+A[/math] are relatively prime and [math]A^n+B^n=C^n[/math].)

I proved thus:

Let us assume that the first number contains divider d of the number A-B. Then both numbers in the following pairs are divided by d:

[math]2A-C [=A+B-C+(A-B)][/math] and [math]A-B[/math];

[math](2A)^n-C^n [=(2A-C)R][/math] and [math]2A^n-C^n [=A^n-B^n=(A-B)T][/math];

[math](2^n-2)A^n [=(2A)^n-C^n-(2A^n-C^n)][/math] and [math]2A^n-C^n[/math];

[math]2(2^{n-1}-1)[/math] and [math]2A^n-C^n[/math] [since [math]A^n[/math] and [math]A^n-B^n[/math] relatively prime].

It is evident from this that the common divisors of the numbers A+B-C and A-B can be located only among the dividers of number [math]2(2^{n-1}-1)[/math].

***

Strictly speaking, proof of the FLT in essence is already published: this is the Lemma given above.

If we investigate the passages from the pair to the pair (in these passages THE NEW prime dividers from If we investigate the passages from the pair to the pair (in these passages THE NEW prime dividers from the set of dividers of the number [math](2^{n-1}-1)[/math] CANNOT appear in principle!!!), then it is possible to note that ALL dividers of the number [math](2^{n -1}-1)[/math] - including n!!! - are dividers of the number A-B.

And now either pair of relatively prime numbers A+B (multiple n) and A B, or all three relatively prime numbers A-B, C+B, C+A are divided by n. In my opinion "arrived"!

I prepare the complete text of proof of the FLT.

Victor Sorokine · February 17, 2010

I prepare the complete text of proof of the FLT.

Very simple idea

Proof of the FLT (yet not acknowledged with mathematical association)

Let us assume that for relatively primes [math]A, B, C (C>A>B>1)[/math] the equality

1°) [math]A^n+B^n=C^n[/math] exists.

Let the number AB not is divided by n.

2°) It’s known that if [math]C=C'n^{kn}[/math], where C' not is divided by n, then [math]A+B=C"n^{kn-1}[/math], where C" not is divided by n.

Let us compose the new equality

3°) [math]a^n+b^n=c^n[/math] with the condition:

4°) [math]c-a=C-A, c-b=C-B, a+b=n^n[/math] – if A+B of odd, and [math]a+b=(2n)^n[/math] – if A+B of even.

In the case of odd A+B, from these relationships we find:

[math] (c-a)+(c-b)+(a+b)=2c=(C-A)+(C-B)+n^n[/math];

[math] (c-a)-(c-b)+(a+b)=2b=(C-A)-(C-B)+n^n[/math];

[math]-(c-a)+(c-b)+(a+b)=2a=-(C-A)+(C-B)+n^n[/math],

where right sides are even and, therefore, the numbers a, b, c are WHOLES.

From the other side, accordingly 2°, solution [math] (a, b, c)[/math] of the equation 3° are not INTEGRAL.

In the case of even A+B the conclusion is analogous.

And we arrived to the contradiction.

Victor Sorokine · February 20, 2010

It is seemed that at the basis of my last proof the lemma must lie:

For any rational h and assigned entire (A+B), where A is not equal to B, there is such t, that

[math](A+t)^n+(B-t)^n=(A+B)R=h[/math].

And it is now necessary to show that with the given three integers (C-B), (C-A), (A+B) there are such rational P, Q, R (at least in the form common fractions), that

[math](C-B)P+(C-A)Q=(A+B)R[/math].

uncool · February 21, 2010

Very simple idea

Proof of the FLT (yet not acknowledged with mathematical association)

Let us assume that for relatively primes [math]A, B, C (C>A>B>1)[/math] the equality

1°) [math]A^n+B^n=C^n[/math] exists.

Let the number AB not is divided by n.

2°) It’s known that if [math]C=C'n^{kn}[/math], where C' not is divided by n, then [math]A+B=C"n^{kn-1}[/math], where C" not is divided by n.

Let us compose the new equality

3°) [math]a^n+b^n=c^n[/math] with the condition:

4°) [math]c-a=C-A, c-b=C-B, a+b=n^n[/math] – if A+B of odd, and [math]a+b=(2n)^n[/math] – if A+B of even.

You would have to prove that that is possible in the first place. The three equations in 4 are not independent.

Also, what is the relation between A, B, C, and a, b, c?

=Uncool-

Victor Sorokine · February 21, 2010

You would have to prove that that is possible in the first place. The three equations in 4 are not independent.

Also, what is the relation between A, B, C, and a, b, c?

=Uncool-

It is seemed that at the basis of my last proof the lemma must lie:

For any rational h and assigned entire (A+B), where A is not equal to B, there is such t, that

[math](A+t)^n+(B-t)^n=(A+B)R=h[/math].

And it is now necessary to show that with the given three integers (C-B), (C-A), (A+B) there are such rational P, Q, R (at least in the form common fractions), that

[math](C-B)P+(C-A)Q=n^nR[/math] [or [math] (C-B)P+(C-A)Q=nR[/math]],

where (A, B, C) is a solution of Fermat’s equation.

OR:

And it is now necessary to show that with the given three integers (c-b), (c-a), (a+b) there are such rational P, Q, R (at least in the form common fractions), that

[math](c-b)P+(c-a)Q=n^nR[/math] [or [math](c-b)P+(c-a)Q=nR[/math]],

where c-b=C-B, c-a=C-A and

(A, B, C) is a solution of Fermat’s equation.

Merged post follows:

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Interesting moment in light of the last idea – equality

[math](C-B)P-(C-A)Q-(A-B)T=0, or A^n=B^n+D^n[/math]

[from here [math]D^n=A^n-B^n=(A-B)S, U'=B+D-A[/math]],

where [math]A^n+B^n-C^n=0[/math] is Fermat’s equality, the numbers A, B, C are whole end the number D is fractional.

Edited February 21, 2010 by Victor Sorokine
Consecutive posts merged.

Victor Sorokine · February 23, 2010

Proof of the FLT (project)

Let us assume that for relatively prime integers [math]A, B, C[/math] and prime n>2

1°) [math]A^n+B^n=C^n[/math], from here

2°) [math]A^n-B^n=C^n-2B^n=D[/math].

3°) It’s known that the number [math]R[/math] in the equality [math]A^n-B^n=(A-B)R[/math] has the prime divider [math]m[/math] of the form [math]m=pn+1[/math].

Let us show, however, that the number [math]D=C^n-2B^n[/math] is not divided by m.

According to Fermat's little theorem, the number

4°) [math]E=C^{pn}-2^{pn}B^{pn}[/math] (as the number D) is divided by prime m.

However, the numbers D and E are relatively prime relative to m. Let us show this.

………………..

(Proof based on the formula of linear diophantine equation.)

Victor Sorokine · March 2, 2010

Clean proof of the FLT. A special case.

Let for relatively prime A, B, C (for the certainty [math]AB(A-B)[/math] is not divided by n) and prime [math]n>2[/math]

1°) [math]A^n+B^n=C^n[/math].

Are known the following properties of the Fermat’s equality:

2°) [math]C-B=a^n=p[/math]; [math]C-A=b^n=q[/math];

3°) [math]p^n-q^n=(p-q)R[/math], where each prime base m of the number R takes the form [math]m=kn+1[/math].

Let us prove the FLT for k not divided by n.

4°) According to Fermat's little theorem, the number [math]a^{kn}-b^{kn}[/math], or [math]p^k-q^k[/math], or [math] (p-q)T[/math] (just as the number [math]p^n-q^n=(p-q)R[/math]) contains divider m.

But, as it is known (and that easy to show with the aid of the linear diophantine equation [math]kx-ny=1[/math]), if p and q relatively prime, p-q is not divided by n, k and n are relatively prime, then the numbers T and R are relatively prime, that contradicts 4°.

***

Thus, for the completion of proof of the FLT it remains to show that the case, when k is divided by n, is also contradictory.

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With probability 99% under the conditions of theorem, the number k is divided by n. If this hypothesis is true (it is possible that it is proven), then the first case has no benefit.

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It is very probable that as the basis of its proof of the great theorem P.Ferma it placed precisely this theorem:

For mutually-prime numbers a and b each prime divider m (with exception n) of the number

[math]R=(a^{n^{t+1}}+b^{n^{t+1}})/(a^{n^t}+b^{n^t})[/math] has the form of [math]m=pn^{t+1}+1[/math].

For example: [math](2^9-1)/(2^3-1)=73=8[/math]* [math]9+1[/math].

It is interesting, is there a simple proof of this theorem?

Apparently, P.Ferma proved also the stronger theorem:

If the t-digit ends of the numbers A and B are the t-digit ends of some numbers a^n and b^n, then the t-digit end of EACH prime divider R in the equality [math]A^n+B^n=(A+B)R[/math] is equal to 1.

For example: [math]=(***9+***8)(9^2-72+8^2)=73=***8[/math]*[math]9+1[/math].

If this theorem is true, then FLT proves from the force into ten lines:

If the number A+B-C is divided by [math]n^t[/math], then it “automatically” is divided also into [math]n^{t+1} [/math]. And so to infinity.

But the very important: two the theorems indicated generate large NEW region in the theory of the numbers.

Victor Sorokine · March 13, 2010

Proof of Fermat's last theorem

Thus,

Let us assume that for relatively prime natural A, B, C (for the certainty A>B and AB(A-B) is not divided by n) and prime n>2

1°) [math]A^n+B^n=C^n[/math],

where, as is known,

2°) [math]A^n-B^n=(A-B)R[/math];

3°) the number C-B, C-A, A-B and R are relatively prime;

4°) [math]C-B=a^n, C-A=b^n[/math]; from where

5°) [math]a^n-b^n=A-B[/math];

6°) each prime divider m of the number R takes the form m=pn+1.

Proof of the FLT

Let us take any prime divider m of the number R. Then, according to Fermat's little theorem, the numbers

7°) [math] (C-B)^{pn}-(C-A)^{pn}[/math] and [math]a^{pn}-b^{pn}[/math], or [math] (C-B)^{pn-1}-(C-A)^{pn-1}[/math], are divided by m.

But so the numbers in the pairs (C-B, C-A) and (pn, pn-1) are relatively prime, then, according to the Lemma (its proof three times was brought on the forum of dxdy), the number m is the divider of the number A-B, that contradicts to 3°.

FLT is proven.

***

(The proof of the Lemma, based on the formula of the solution of linear diophantine equations, will be given after the elimination of all other questions, which arose in the proof of the FLT.)

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Proof of Fermat's last theorem (the completing text)

Let us assume that for relatively prime integers A, B, C (for the certainty A>B and AB(A-B) is not divided by n) and prime n>2

1°) [math]A^n+B^n=C^n[/math],

where, as is known,

2°) [math]A^n-B^n=(A-B)R[/math];

3°) the number C-B, C-A, C, A-B and R are relatively prime;

4°) [math]C-B=a^n, C-A=b^n[/math]; from where

5°) [math]a^n-b^n=A-B[/math];

6°) each prime divider m of the number R takes the form m=pn+1.

Proof of the FLT

Let us take any prime divider m of the number R. Then the numbers [math]A^n-B^n[/math] and therefore

7°) [math]A^{n(p-1)}-B^{n(p-1)} [/math] and, according to Fermat's little theorem,

8°) [math]a^{pn}-b^{pn}[/math], или [math](C-B)^{n(p-1)}-(C-A)^{n(p-1)}[/math]

9°) are divided by m.

But so the number C is relatively prime with the number (A-B)R, then, according to lemma (see the Appendix), the numbers 7° and 8° are relatively prime, that contradicts 9°.

The case, when CB(C+B) is not divided by n, proves analogously.

***

Appendix

Lemma

If the number C is relatively prime with the number [math]A^n-B^n=(A-B)R[/math], where n>2, then the numbers [math]A^n-B^n[/math] and [math](C-A)^n-(C-B)^n[/math] are relatively prime relative to the dividers of the number R.

Unfortunately, the author does not have available the proof of Lemma. However, it is possible that P.Fermat found its proof in Diophante's “Arithmetic”.

A study strictly of great theorem (without considering Lemma) ceases on this.

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Analogous proof, but WITHOUT the Lemma.

Let us assume that for relatively prime integers A, B, C (for the certainty A>B and AB is not divided by n) and prime n>2

1°) [math]A^n+B^n=C^n[/math],

where, as is known,

2°) the number C-B and C-A are relatively prime and [math]C-B=a^n, C-A=b^n[/math]; from where

3°) [math]a^n+b^n=(a+b)R[/math];

4°) the number R contains the prime divider m of the form [math]m=pn+1[/math] (author's proof uses a formula of the solution of linear diophantine equation).

Proof of the FLT

Let us take any prime divider m of the number R. Then the numbers [math]a^n+b^n[/math], or [math](C-B)+(C-A) [/math], and therefore

5°) [math]D=(C-B)^{n(p-1)}+(C-A)^{n(p-1)}[/math], where [math]n(p-1)[/math] is odd (!),

and, according to Fermat's little theorem,

6°) [math]E=(C-B)^{n(p-1)}-(C-A)^{n(p-1)}[/math],

7°) are divided by m.

But so the numbers (C-B) and (C-A) are relatively prime (without considering dividers 2), then the numbers D and E have no common factors, which contradicts 7°.

The case, when CB is not divided by n, proves analogously. FLT is proven.

On March 12, 2010

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

END

Edited March 11, 2010 by Victor Sorokine
Consecutive posts merged.

Victor Sorokine · March 15, 2010

Theorem about transformation

If in the prime base n the two-digit end of the number a (a is not divided by n) is a two-digit end of a certain number [math]a'^n[/math], then there is a number A^n with the end a, i.e., [math]a=A^n-Pn^k[/math], where [math]n^k>a[/math].

(Simple, but voluminous, the proof of Theorem is based on the fundamental theorem: In the numeration system in the prime base n>2 for any single-digit number a (a is not divided by n) and the given number d there is an only number g, that the last digit of the number ag is d.

I do not have a possibility to refer to sources, but entire apparatus for the conversion of numbers was repeatedly presented by me on the sites of “mmonline” and “dxdy”. This entire the simple apparatus, repeatedly checked by participants in the forums, I will present again, but after the completion of proof and of his acknowledgement by the true under the assumption to faithfulness to my theory of numeration in the base with the prime base.)

Corollary from Theorem about transformation

If in the prime base n the two-digit end of the number a (a is not divided by n) is a two-digit end of a certain number [math]A^n[/math], then there is this number [math]a^{n^t}[/math], that [math]a'^{n^t}=a+Pn^k[/math], where k is as desired great.

(Proof is based on the simple lemma:

Last k digits of the number a do not influence value (k+1)-th digit of the number a^n.)

Proof of the FLT

Let us represent the number a, b, c (thhey are not divided by n) in the equality

1°) [math]a^n+b^n=c^n[/math], according to corollary about the Theorem about transformation:

2°) [math]a=a'^{n^t}-Pn^k, b=b'^{n^t}-Qn^k, c=c'^{n^t}-Rn^k[/math], where k is as much as desired great.

Then as much as desired the long ends of the numbers [math]a^{n-1}, b^{n-1}, c^{n-1}[/math] (in the equality of 1°) finish to 1 and, therefore, the number of [math]u=a+b-c[/math] finishes to as much as desired number of zeros (if abc not multiply n and if one of the numbers a, b, c is not divided by n).

It is understandable, the text of even damp and on the course of events it will be refined

Victor Sorokine · March 18, 2010

Block diagram of the last (dated March 16) project of the proof

1. On the last (different from zero) number of each prime divider of the number a we determine (with the aid of the Fermat's little theorem) the last digit of the number [math]a^(n-1)[/math] in the prime base n>2; this be 1.

2. Since in the equality [math]a^n+b^n=(a+b)R[/math] (and two analogous) the number R (not divided by n) is [math]r^n,[/math] then the two-digit end of the number R is 1 (or 01).

3. From this it follows (on the basis one of the S-theorems) that the two-digit ends of the numbers (not divided by n) [math]a^{n-1}, b^{n-1}, c^{n-1}[/math] are 1.

4. From this it follows (on the basis one of the S- theorems) that the two-digit ends of each of the simple dividers of the numbers (not divided by n) R and two others (not divided by n) is 1.

5. Since in the equality [math]a^n+b^n=(a+b)R[/math] (and two analogous) the number R (not divided by n) is r^n, then the three-digit end of the number R is 1 (or 001).

Thus further - to infinity.

Victor Sorokine · April 7, 2010

There is no proof. A study continues.

For the amateurs of mathematics, managing general algebraic theory of FLT.

Theorem 5

If in the prime base n>2 2-digit end of the s-digit number A is the 2-digit end of the n-th degree, not multiple n, then the number A has the equivalent form:

[math]A=a^n-Pn^w[/math], where w are more than the given number v and the w-digit end of the number a^n is equal to A.

Theorem 5a (generalized)

If in the prime base n>2 t-digit end of the s-digit number A is the t-digit end of [math]n^{t-1}[/math]-th degree, not multiple n, then the number A has the equivalent form:

[math]A=a^{t-1}-Pn^w[/math], where v are more than the given number and the w-digit end of the number [math]a^{t-1}[/math] is equal to A.

The validity of these impressive theorems follows from the simplest fact: the k-th digit (from the end) in the number a and (k+1)-th digit in the number a^n (k>1) one-to-one are determined the values of each other (that it follows from the binomial of Newton and of fermat's litlle theorem in the prime base n>2). These two theorems make it possible to propose simple proof of FLT.

Actually, from the equality

1°) [math]A^n+B^n=C^n[/math] (A, B, C are coprime and prime n>2) it follows that the 2-digit ends of the numbers A, B, C are the 2-digit ends of n-th degrees, not of multiple n (with ABC not multiple n), and therefore with [math]n^{w+1}>C^n[/math] have the equivalent form:

2°) [math]A=a^{n^2}-Pn^w, B=b^{n^2}-Qn^w, C=c^{n^2}-Rn^w[/math].

Let us substitute these values into 1° and, after the disclosure of Newton's binomials, on the [math]n^w[/math]-digit ends of the all numbers in the equality 1° we obtain the second equality:

3°) [math]a^{n^2}+b^{n^2}=c^{n^2}[/math].

Question: how many numbers in the numbers a, b, c?

Victor Sorokine · June 2, 2010

Half-Joke

For relatively prime [math]A, B, C[/math] (where [math]C>A>B>1, AB[/math] is not divided by prime [math]m[/math]) and [math]m=n-1>1[/math] let us examine the equality

1°) [math]A^m+B^m-C^m=0[/math], or

2°) [math]A^m=C^m-B^m=(C-B)P, B^m=C^m-A^m=(C-A)Q[/math], from where

3°) [math]C^n-A^n-B^n= (P+Q)(C-B)(C-A)=D^n[/math].

To the proof of the FLT

The number D is either 1) whole or 2) fractional.

In the first case, as can easily be seen (taking into account that the number [math]C-B, C-A, P+Q[/math] are relatively prime),

4°) [math]C-B=x^{mn}, C-A=y^{mn}, P+Q=z^n[/math], and provability of the FLT is very real.

Impression is created, that the case 2 also is demonstrated.

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