Fermat's Last Theorem. Elementary proof

[Victor Sorokine]

FINAL

 

(You will excuse me for the poor quality transfer)

 

Proof of the FLT

with the aid of

 

Theorem about adjacent dividends:

 

If the number [math]g=a^n+b^n[/math] (where a and b are natural) is divided by d, then the nearest to g numbers, which are divided by d, are the numbers [math](a-d)^n+b^n[/math] and [math](a+d)^n+b^n[/math].

 

(Simple proof of theorem is based on the lemma:

“If the number [math]a^n+b^n[/math] is divided by prime d>1 and the numbers a, b and prime e, where [math]0<e<d[/math], are not divided by d, then the number [math](a+e)^n+b^n[/math] is not divided by d”, whose proof will be examined later.)

 

 

Strictly proof of the FLT

 

It follows from the equality

1°) [math]A^n+B^n=C^n[/math], where n>2 and [math]A, B, C (C>A>B)[/math] are natural,

that

 

2°) [math]C/2<C/2^{1/n}<A<C[/math] (with A=B we have a equality: [math]A2^{1/n}=C^n[/math]), [math]B>1[/math] and

3°) [math]A+B<AB[/math].

 

Let us examine the inequality

4°) [math]F=D^n-(A^n+B^n)>0[/math], where D=A+B and, consequently, the number F is divided into AB.

 

Let us find a number G adjacent with the number F and which is divided by AB.

According to the theorem about adjacent dividends this there will be the number [math]G=(D-AB)^n-(A^n+B^n)[/math], which, accordingly 3°, is negative, and therefore between A+B and A+B-AB the entire value of the number D, satisfying conditions G=0 and G is divided into AB (with n>2) is absent.

 

(With n=2 the negative number A+B-AB in terms of the absolute value it can prove to be less than the number A+B and the equality G=0 can be possible.)

 

On June 6, 2010

---

Merged post follows:

Consecutive posts merged

Thus, for n>2 I proved inequality

[math](A+B)^n-(A^n+B^n)>[0=]C^n-(A^n+B^n)[=0]>(A+B-AB)^n-(A^n+B^n)[/math], where all three numerical expressions are divided into AB.

 

From the other side, according to the theorem about adjacent the divisible, extreme number-expressions as dividends on AB they are ADJACENT, i.e., between them there are no other dividends by AB!

 

Here, strictly speaking, ALL proof of Fermat's last theorum. (Subsequently all attention it will be switched to the theorem about adjacent dividends.)

[Victor Sorokine]

Proof of FLT for case 2

(material for the deep reflections)

 

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

 

1°) [math]C^n-A^n-B^n=0[/math] is possible

and let the number C [either A or B] finishes on k zeros.

 

Then the numbers

[math]C^n[/math] and [math]A^n+B^n[/math] [[math]=(A+B)R[/math]] are divided by [math]n^{kn}[/math],

 

2°) R is divided by [math]n^1[/math] and not divided by [math]n^2[/math] (well-known fact), therefore,

 

3°) A+B is divided by [math]n^{kn-1}[/math],

 

4°) the number [math]A+B-C [=U][/math] is divided by [math]n^k[/math].

 

 

Proof of the FLT

 

Let us examine the number [math]U^n[/math], or [math]U^n+0[/math]

 

5°) [math]U^n+C^n-(A^n+B^n)=(U+C)T-(A^n+B^n)[/math], where

 

[math]U+C=A+B[/math] (see 4°) and, therefore, is divided by [math]n^{kn-1}[/math] (see 2°),

 

and each term in the polynomial T is divided by [math]n^{k(n-1)}[/math], since both U and C are is divided by [math]n^k[/math].

 

6°) As a result the numbers [math]U^n+C^n, A^n+B^n [=C^n], U^n[/math] are divided by [math]n^{kn-1+k(n-1)}[/math] [[math]=n^{2kn-k-1}[/math]].

 

7°) Consequently, the number [math]U^n[/math] and [math]C^n[/math] are equal on the module [math]n^{2kn-k-2}[/math].

 

8°) But then the number A+B finishes on [math]2kn-k-2 [>2kn-1][/math] zeros, that contradicts 3°.

 

The case, when ABC not is divided by n, will be examined later.

 

(On June 22, 2010)

[Victor Sorokine]

On one incomplete proof of the FLT

 

There is a very simple and brief (5 lines) proof of the FLT, based on the following theorem for the exponential binomials:

 

Theorem

 

For the relatively prime numbers A and B and A+B not divided by prime n>2

each prime divider m of the number R in the identity

[math]A^n+B^n=(A+B)R[/math], where [math]A=a^{n^k}[/math] and [math]b=b^{n^k}[/math], but number a or/and b are not n-th degrees,

each prime divider m of the number R is represented in the form [math]m=dn^{k+1}+1[/math],

where d not is divided by n.

 

***

 

Unfortunately, i do not have complete proof of this theorem. With the aid of the formula of the solution linear diophantine equation I found the proof only of that fact that the number R does not contain the dividers of the form of [math]m=dn^k+1[/math].

It will be possible to prove that R does not contain the dividers of the form of [math]m=dn^{k+2}+1[/math] and so forth? But can, it will be possible to prove FLT without this part…

 

Only the base case is thus far until today proven:

The number R (either P or Q) contains at least one divider [math]m=dn+1[/math]. (This, in particular, means that now the number A+B-C is divided by [math]n^3[/math].)

[Victor Sorokine]

Proof of Fermat's last theorem for relatively prime A, B, C and prime n>2.

(Most splendid joke)

 

Proof is based on the simple consequence from the Little fermat's theorem:

In the numeration system on the base n the number R in the equality [math]A^n+B^n=(A+B)R[/math], where A, B and A+B are not multiple n, finishes with number 1.

 

Case 1: ABC not is multiple n

 

As is known, in this case in the Fermat’s equality [math]A^n+B^n-C^n=0[/math], or

 

1°) [math](C-B)P+(C-A)Q-(A+B)R=0[/math], the following formulas occur:

 

2°) [math]C-B=a^n, P= p^n; C-A=b^n, Q=q^n; A+B=c^n, R=r^n[/math].

 

3°) In this case the numbers P, Q, R finish with 01.

 

Proof of case 1

 

Let us examine the equality of 1° at the two-digit ends of the numbers (i.e. on the module [math]n^2[/math]).

 

After rejecting coefficients P, Q, R as being finished with 01, we obtain the equation

 

4°) [math](C-B)+(C-A)-(A+B)=0[/math], or [math]a^n+b^n-c^n=0[/math] (on the module of [math]n^2[/math]), which also can be represented in the form 1°:

 

1_1°) [math](C_1-B_1)P_1+(C_1-A_1)Q_1-(A_1+B_1)R_1=0[/math], where

 

2_1°) [math]C_1-B_1=a_1^n, P_1= p_1^n; C_1-A_1=b_1^n, Q_1=q_1^n; A_1+B_1=c_1^n, R_1=r_1^n[/math].

 

3_1°) in this case the numbers [math]P_1, Q_1, R_1[/math] finish with 01. And so on.

 

Obviously, if the numbers A, B, C contain one divider, which is finished to the number, different from 1, then the process of release from the cofactors, which are finished with 01, will be INFINITE. This means that the numbers A, B, C are infinite.

 

Case of 2, when A (either B or C) is multiple of degree n, proves with the aid of the same apparatus, but somewhat more complex.

[Victor Sorokine]

Because of the misunderstanding my post is published in another theme: http://www.scienceforums.net/topic/50825-puzzle-on-the-course-theory-of-the-numbers/ I please apology.

[Victor Sorokine]

Theorem* (and FLT) for “Dummies”

 

Theorem is true for composite r (that is sufficient for the simplest the proof of FLT).

 

This evidently at least based on the example of decimal system:

Numbers 7, 9 and 10 are relatively prime, and the coefficient d=5 (to prime divider of base).

And now the numbers of 7x5 and 9x5, equal to 35 and 45, finish TO one and the same digit 5. (Understandable that for the proof of FLT as r it is necessary to take, for example, rr.)

 

400 years were required, in order to formulate this theorem!!!

 

____________

 

 

Theorem*

 

If integers a, b, a+b and r are mutually-prime, then there is such d, relatively prime with r, that the ends of the numbers ad and bd are equal on the module r.

 

[Consequence. With relatively prime a, b, a+b and r, where the value r is undertaken from the equality:

1*) [math]a^n+b^n=(a+b)r^n=c^n[/math] or [math]a^n+b^n=(a+b)nr^n=c^n[/math], equality 1* is contradictory in the base r, since in the equality

2*) [math](ad)^n+(bd)^n=(cd)^n[/math] ([math]=Pr[/math]) right side is divided by r, but leftist is not divided.]

[Victor Sorokine]

Small correction

 

It is insufficient for the proof of the FLT one Theorem alone, since the case of prime r (as the case of prime c in the equality [math]A+B=c^n[/math]) does not prove FLT.

 

However, if we instead of r take the number C=cr, then on the basis Theorems we obtain the conclusion that the number C=cr is prime. Which is impossible.

 

Thank to ALL

[Victor Sorokine]

Fermat's last theorem. Elementary proof

 

Joke of the amateur

 

[The key of the proof: If A+B=1, where A and B are integers and non-negative numbers, then either A or B is equal to zero]

 

Proof of the FLT

 

If for mutually-prime integers A, B, C and prime n>2 there is an equality

 

1°) [math]A^n+B^n=C^n[/math], where [math]C^n=(A+B)R[/math] or (if C divided by n) [math]C^n=(A+B)Rn[/math], where, as is known,

 

2°) the number A+B and R are relatively prime,

 

then there is a positive solution (x, y) of linear diophantine equation

 

3°) [math] (A+B)x-Ry=1[/math].

 

It follows from this solution that in the numeration system on the base R the number

 

4°) A+B finishes by digit 1.

 

However, since the numbers A and B are wholes and non-negative, the either number A or the number B finishes by zero, i.e., it is divided by R, which contradicts to condition.

 

The case of n=4 proves analogously with the equation C^4-A^4=B^4.

 

Theorem is proven.

 

 

The case of n=4 proves analogously with the equation [math]C^4-A^4=B^4[/math].

 

Theorem is proven

Edited August 19, 2010 by Victor Sorokine

[Victor Sorokine]

In what joke?

 

Complete proof of Fermat's last theorem

 

Let us assume that for relatively prime natural [math] a[/math], [math]b[/math], [math]c[/math] ([math]c>a>b[/math]) and [math]n>2[/math] there is a equality:

 

1°) [math]a^n+b^n=c^n[/math], where, as is known, [math]a-b>1[/math].

 

Then the number

 

2°) [math]D=c^n+a^n=a^n+b^n+a^n=2a^n+b^n[/math], i.e. [math]2a^n+b^n[/math] is the sum of two degrees: [math]c^n+a^n[/math].

 

However, it is completely obvious that not be can, since the number [math]b^n[/math] is n-th degree with any b, while the number [math]2a^n[/math] is not n-th degree on no account.

 

Thus, the equality 1° has no solution in integers. FLT is proven.

 

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

 

To show the inapplicability of proof in the case n=2 is given to the readers.

I will indicate only that in this case the number a-b=1 and Fermat’s equality is reduced to the form: [math]2ab+1=c^2[/math].

Edited September 16, 2010 by Victor Sorokine

[Victor Sorokine]

Clean proof of Fermat's last theorem

 

“I found the truly fairytale proof of this remarkable theorem, but, unfortunately, place in the fields it is insufficient in order to give it here” /P.Ferma/.

 

For lack of time today I will describe only about method and essence of proof.

 

In the essence school proof consists of the following:

 

In the equality [math]a^n=c^n-b^n=(c-b)P=AP[/math] the remainder from the division of the number P into A, calculated BY TWO methods, gives two DIFFERENT values.

 

* * *

 

Utilized mathematical tools: formula for the polynomial P and the square of a difference of two numbers. And all.

 

Continuation follows.

[Victor Sorokine]

Passed proof

 

Earlier this version of proof it was rejected on the misunderstanding.

 

Idea of the proof: if (with relatively prime integers A, B and C) [math]A^n+B^n=C^n[/math], then the numbers A-B and R in the equality [math]A^n-B^n=(A-B)R[/math] are not relatively prime, that is impossible.

 

Tthe proof of the FLT is based on the simple lemma, repeatedly proven earlier on the forums.

 

Lemma 1. with relatively prime integers a and b and a-b not multiple prime n>2 each prime divider r of the number R in the identity

[math]a^n-b^n=(a-b)R[/math] takes the form: [math]r=pn+1[/math].

 

 

Proof of Fermat's last theorem

 

Let us assume that with relatively prime integers A, B and C and (let us assume) AB not multiple prime n>2 there is a equality

 

1°) [math]A^n+B^n=C^n[/math], where, as is known,

 

2°) [math]C-B=a^n[/math]; [math]C-A=b^n[/math]; [math]A^n-B^n=(A-B)R[/math]; [math]A-B=a^n-b^n[/math]; the numbers [math]C, A-B[/math] and [math]R[/math] are relatively prime.

 

Let us take the prime divider r of the number R of the form [math]r=pn+1[/math]. Then two numbers, namely:

 

3°) [math]A^n-B^n[/math] [[math]=(A-B)R[/math]] and, according to fermat's little theorem, [math](C-B )^{pn}-(C-A)^{pn}[/math] [[math]=(A-B)T[/math]], and means the numbers R and T, they are divided into r.

 

However, as we will attempt to show below, the numbers R and T are relatively prime.

 

(Continuation it follows).

Edited September 30, 2010 by Victor Sorokine

[Victor Sorokine]

Passed proof

 

Earlier this version of proof it was rejected on the misunderstanding.

 

Idea of the proof: if (with relatively prime integers A, B and C) [math]A^n+B^n=C^n[/math], then the numbers A-B and R in the equality [math]A^n-B^n=(A-B)R[/math] are not relatively prime, that is impossible.

 

Tthe proof of the FLT is based on the simple lemma, repeatedly proven earlier on the forums.

 

Lemma 1. with relatively prime integers a and b and a-b not multiple prime n>2 each prime divider r of the number R in the identity

[math]a^n-b^n=(a-b)R[/math] takes the form: [math]r=pn+1[/math].

 

 

Proof of Fermat's last theorem

 

Let us assume that with relatively prime integers A, B and C and (let us assume) AB not multiple prime n>2 there is a equality

 

1°) [math]A^n+B^n=C^n[/math], where, as is known,

 

2°) [math]C-B=a^n[/math]; [math]C-A=b^n[/math]; [math]A^n-B^n=(A-B)R[/math]; [math]A-B=a^n-b^n[/math]; the numbers [math]C, A-B[/math] and [math]R[/math] are relatively prime.

 

Let us take the prime divider r of the number R of the form [math]r=pn+1[/math]. Then two numbers, namely:

 

3°) [math]A^n-B^n[/math] [[math]=(A-B)R[/math]] and, according to fermat's little theorem, [math](C-B )^{pn}-(C-A)^{pn}[/math] [[math]=(A-B)T[/math]], and means the numbers R and T, they are divided into r.

 

However, as we will attempt to show below, the numbers R and T are relatively prime.

 

(Continuation it follows).

 

Good-bye, Fermat!

 

Classical proof of Fermat's last theorem

Actually school proof of the FLT is finally found!

(Simplest logical substantiations are omitted.)

 

Let us take Fermat's equality in the form:

 

1°) [math]A^n=(C-B)P=a^n*p^n[/math] [and [math]B^n=(C-A)Q=b^n*q^n[/math], [math]C^n=(A+B)R[/math]], where

 

2°) prime n is more than 2, the numbers A, B, C relatively prime, AB not is divided by n [the 1st case],

 

3°) r - prime divider of the number R of the form [math]r=tn+1[/math] (which, as is known, exists and, moreover, the numbers A+B and A-B is not divided by r).

 

Then, according to Fermat's little theorem, they are divided by r the numbers:

 

[math](C-B )^{tn}-1[/math], therefore, and

 

4°) [math]B^{tn}-1[/math] and

 

[math]p^{tn}-1[/math], or [math]P^t-1[/math], therefore, and

 

5°) [math]B^{t(n-1)}-1[/math], therefore, and

 

[math][b^{tn}-1]-[b^{t(n-1)}-1][/math], or [math]B^{t(n-1)}*(B-1)[/math], therefore, and

 

6°) B-1, и, согласно рассуждениям, аналогичным в п.п. 4-5°,

7°) A-1, следовательно, и

8°) (A-1)-(B-1), или A-B.

 

6°) B-1, and, according to the reasonings, analogous into 4-5°,

 

7°) A-1, therefore, and

 

8°) (A-1)-(B-1), or A-B.

 

But, as it is known (and which it was many times shown by the author on the mathematical forums), the number A-B is not divided by prime r.

 

Cases, when the number AC (or BC) is not divided by n, prove analogously.

 

Thus, Fermat's last theorem is proven for all odd degrees.

 

 

On October 7 to 8, 2010

 

Good-bye, Fermat!

 

Classical proof of Fermat's last theorem

 

Actually school proof of the FLT is finally found!

(Simplest logical substantiations are omitted.)

 

Let us take Fermat's equality in the form:

 

1°) [math]A^n=(C-B)P=a^n*p^n[/math] [and [math]B^n=(C-A)Q=b^n*q^n[/math], [math]C^n=(A+B)R[/math]], where

 

2°) prime n is more than 2, the numbers A, B, C relatively prime, AB not is divided by n [the 1st case],

 

3°) r - prime divider of the number R of the form [math]r=tn+1[/math] (which, as is known, exists and, moreover, the numbers A+B and A-B is not divided by r).

 

Then, according to Fermat's little theorem, they are divided by r the numbers:

 

[math](C-B )^{tn}-1[/math], therefore, and

 

4°) [math]B^{tn}-1[/math] and

 

[math]p^{tn}-1[/math], or [math]P^t-1[/math], therefore, and

 

5°) [math]B^{t(n-1)}-1[/math], therefore, and

 

[math][b^{tn}-1]-[b^{t(n-1)}-1][/math], or [math]B^{t(n-1)}*(B-1)[/math], therefore, and

 

6°) B-1, и, согласно рассуждениям, аналогичным в п.п. 4-5°,

7°) A-1, следовательно, и

8°) (A-1)-(B-1), или A-B.

 

6°) B-1, and, according to the reasonings, analogous into 4-5°,

 

7°) A-1, therefore, and

 

8°) (A-1)-(B-1), or A-B.

 

But, as it is known (and which it was many times shown by the author on the mathematical forums), the number A-B is not divided by prime r.

 

Cases, when the number AC (or BC) is not divided by n, prove analogously.

 

Thus, Fermat's last theorem is proven for all odd degrees.

 

 

On October 7 to 8, 2010

 

++++++++++++++++++

 

Unfortunately (as always - small), in the fifth point is contained school error, and therefore as a whole proof is erroneous. It was impossible to remove error. Only useful result: all prime dividers of the form tn+1 of the numbers A, B, C in the base n finish by 01. Erroneous proof leaves for the archive.

Edited October 9, 2010 by Victor Sorokine

[Victor Sorokine]

One additional attempt with the same idea and with the same tools

 

Proof is divided off into three completely similar cases: 1) AB [2) AC and 3) BC] not divided by prime n>2.

 

Let us take Fermat’s equality [case 1] for relatively prime A, B, C in the form

 

1°) [math]A^n=(C-B )P=(C-B )p^n=a^n*p^n[/math], and also [math]B^n=(C-A)Q=(C-A)q^n=b^n*q^n[/math], [math]C^n=(A+B )R[/math].

 

Let us take the prime divider [math]m=tn+1[/math] of the number S from the equality

 

[math](C-B )+(C-A)=a^n+b^n=(a+b)S=2C-(A+B )[/math], which is not been the divider of the number A+B. (It is not difficult to show that this divider exists; if it is required, this will be shown later.)

 

Then, as follows from the fermat's theorem, the number

 

2°) [math]A^{tn}-B^{tn}[/math] [just as the number [math](C-B )^{tn}-(C-A)^{tn}[/math]] is divided by m.

 

However, quick calculation shows that the number [math]A^{tn}-B^{tn}[/math] is not divided by m.

 

It would be a good thing of this to be convinced.

[Victor Sorokine]

Unexpected turning in the study

 

New idea

 

Let us assume that for relatively prime A, B, C and prime n>2 there is a equality

 

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

 

Then two cases are possible:

Case 1: U=us, where the number ABCn is not divided by prime s.

Case 2: s=1 and, therefore, [math]U=abcn^k[/math], where (according to Fermat’s little theorem) k>0 and numbers a, b, c are the greatest dividers respectively in the pairs of numbers (A and C-B ), (B and C-A), (C and A+B ). (Two of the numbers of C-B, C-A and A+B - for example, C-B and C-A - they are n-th degrees.)

 

Case 1. The impossibility of case 1 directly follows [in the numeration system on the base s with the sum of remainders a’, b’, c’ (with the surplus or with the deficiency) or [math]a’+b’-c’=s[/math]] from the simple lemma:

 

If [math]a+b-c=s[/math] and [math]0<(a, b, c)<s[/math], then the number [math]a^n+b^n-c^n[/math] is not divided by s. Here n and s are prime.

 

2. Case [math]A+B-C=U=abcn^k[/math], where k>0 (third case there does not exist) it is decomposed into two subcases:

2a. The number C is divided by n;

2b. The number C (and AB) C is not divided by n;

 

* * *

 

Continuation follows.

[Victor Sorokine]

On the second case ([math]A+B-C=U=abcn^k[/math])

 

If one of the numbers (for example, C) finishes on k of zeros (in the base n), then the sum of other numbers ([math]A+B[/math]) finishes on kn-1 zero. Therefore the coefficient [math]n^k[/math] can be included in the number C: C=cr, where the number c finishes on k of zeros. And now the number U can be recorded in the form:

2a°) [math]U=abc[/math].

 

But if the number ABC is not divided by n, then

2b°) [math]U=abcn^k[/math], where the numbers a, b, c are not divided by n and, as were many times shown earlier, k>1.

 

It is possible that case 2 can be proven so simply, as case 1.

Interesting method of proof consists of the passage in the field of real numbers. Thus, for n=3 computer gives value for U equally 3abc, while theoretically U=abc (if abc it is divided by 3) and U=9abc (if abc it is not divided by 3).

Edited October 15, 2010 by Victor Sorokine

[Victor Sorokine]

Oct 16, 2010

 

Universal centuries-old error

 

(Proof of the case of 2b°: [math]U=abcn^k[/math])

 

It is improbable, but with the relatively prime (and WHOLES!) A, B, C (and a, b, c) [with ABC (and abc) not divided by n] the number [math]A+B-C[/math] NOT CONVERTED into the form [math]abcn^k[/math], since the number [math]n^k[/math] is not the coefficient of the number abc, but its DIVIDER and the k-digit ENDING of the number [math]A+B-C[/math] (=U)! (And the absence of the integral solution of the Fermat’s equality is present. By the way, there is a second proof: in the numeration system on the prime base n>2 all three numbers A, B, C have at the end a digit 1.)

 

Example (in the decimal system):

 

The sum of the numbers a=123 and b=477 is equal 600 and, therefore, it is divided by 100, but it does not follow from this fact that a+b=(123+477)*100=60000!!!

 

Thus, we reduced Fermat's last theorem to the only case:

 

[math]A+B-C=abc[/math], where one of the numbers A, B, C (and a, b, c) has at the end k zeros.

 

There is reason to believe that in the field of real numbers with [math] (A, B, C)>1[/math] the number [math]a+B-C>abc[/math], but we will give another proof (block diagram):

 

It is easy to show that all significant parts of the numbers A, B, C have at the end a digit 1.

And then it is possible to show:

if the number U has at the end k zeros, then it has at the end and k+1 zeros.

The interesting proof of this fact will be given later.

[Victor Sorokine]

Resumes

 

In the recent days for me it was possible to find and to prove extremely important Lemma 1:

 

The number [math]U=A+B-C[/math] does not contain the prime dividers, different from the dividers of the number abc.

 

And now the case of the FLT, when the number ABC is not divided by n, proves exceptionally simply:

 

From the equality [math]A^n+B^n=C^n[/math] (with relatively prime A, B, C and prime n>2) follows the contradictory equality [math]A+B-C=abc[/math] (where [math]a^n=C-B[/math], [math]b^n=C-A[/math], [math]c^n=A+B[/math]), since, according to Little Fermat's theorem, left side are divided by n, but right is not divided.

 

I assume that the case of the FLT, when the number ABC is divided by n, proves with the aid of the hypothesis:

 

Hypothesis:

If one of the relatively prime numbers A, B, C (in the Fermat’s equality) is divided by the prime n (>2), then the number U=A+B-C contains the prime divider, which is not been the divider of the number abc, which contradicts lemma 1.

 

Continuation follows.

[Victor Sorokine]

Repetition of that passed

 

At the basis of proof of the FLT lies simple lemma 1 (according to all signs, previously not known):

If [math]a^n+b^n-c^n=abcd[/math], where prime n>2 and the numbers a, b, c are relatively prime, the number d=1.

 

At the basis of proof the simple lemma lies:

If [math]a+b-c[/math] is equal to prime e (e not equally to n>2), the number [math]a^n+b^n-c^n[/math] is not divided by e.

 

With the aid of the lemma 1 proof of the FLT in the case of 1 (ABC not divided by prime n>2) it is exceptionally simple:

 

The equality [math]A^n+B^n=C^n[/math] is contradictory, since the contradictorily second equality [math]A+B-C=abc[/math], where the left side is divided by n, but right is not divided.

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

 

Proof of the FLT in the case 2 (for example, C is divided by prime n>2, but AB not is divided)

 

As is known, in this case the number

1°) [math]2(A+B-C)=(A+B )-(C-B )-(C-A)=[(c^n):n]-a^n-b^n=2abc[/math].

 

From the other side, according to lemma 1, the number

 

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

since the dividers of the number ab are not the dividers of the number [math]c^n-a^n-b^n[/math].

 

Сравнивая числа [math][(c^n):n]-a^n-b^n=2abc[/math] и [math]c^n-a^n-b^n=2c[/math], что увеличение левой части равенства приводит к уменьшению правой части равенства. И мы имеем противоречие.

 

Таким образом, оба случая ВТФ доказаны.

 

Comparing the equalities [math][(c^n):n]-a^n-b^n=2abc[/math] and [math]c^n-a^n-b^n=2c[/math], we see that an increase in the left side of the equality leads to the decrease of the right side of the equality. And we have a contradiction.

 

Thus, both cases of the FLT are proven.

[Victor Sorokine]

Simplification of the proof

 

Base lemma

 

At the basis of elementary proof of the FLT lies the following simple lemma:

 

Lemma

 

If the number a, b, c relatively prime and the number U=a+b-c is divided by prime n>2, the number [math]a^n+b^n-c^n[/math] is not divided by prime q, different from the dividers of the number abcn.

 

Proof of Fermat's last theorem

 

Possibly Fermat's equality contradicts to lemma.

[Victor Sorokine]

The third it is not given!

 

According to the basic law of the Logic, the given above lemma

 

A) or is true,

B ) or is not true.

 

In the case A) the number [math]a^n+b^n-c^n[/math] (=0!) is not divided completely by an integer q. Therefore, in this case FLT is true.

 

In the case B ) the finite number U ([math]=a+b-c[/math]) is divided completely by ALL integers q. Therefore, FLT is true in this case.

 

But THE THIRD IT IS NOT GIVEN!!!

[Victor Sorokine]

Post is rejected because of the inaccuracy.

Edited October 23, 2010 by Victor Sorokine

[Victor Sorokine]

Returning to the prime dividers q, different from the dividers of the number abcn.

 

0°. Earlier was given simple and brief proof of the FLT (for n>2) for the case [math]A+B-C=abc[/math], where a, b, c - greatest common divisors respectively in the pairs of numbers (A, C-B ), (B, C-A), (C, A+B ), where A, B, C - relatively prime numbers of the Fermat’s equality.

 

Now let us prove (in the theses) the equality [math]A+B-C=abc[/math], i.e., let us show that if [math]U=A+B-C[/math] is divided into prime n and prime q (which is not been the divider of the number [math]ABC[/math]), then the number [math]A^n+B^n-C^n[/math] is not divided into q. (Obvious that the number [math]U=A+B-C[/math] in theFermat’s equality does not have prime dividers from their set in the number [math]ABC[/math], which are not been the dividers of the number abc.)

 

1°. First of all, with the aid of the multiplication of the number U by the appropriate number g we convert the last digit of the number B, recorded in the numeration system on the base q, into 1. (possibility of this conversion it was repeatedly shown by the author on the mathematical forums.)

 

2°. If the sum of the remainders [math]A'+1-C'[/math] from the division of the new values of the numbers A, B, C on q (or last digits in the numbers A, B, C in the numeration system on the base q) is equal to 0, then let us replace positive number [math]A'[/math] by the negative number [math]q-A'[/math]:

[math]1-a-c=q[/math], where a=[math]q-A'[/math], of [math]c=C'[/math],..

 

3°. …in this case the numbers A, B, C can be recorded in the form: [math]A=qx-a[/math], [math]B=qy+1[/math], [math]C=qz-c[/math].

 

4°. Let us substitute the values of the numbers A, B, C from 3 into the Fermat’s equality, from where, after the disclosure of Newton's binomials, we find that the number

[math]d=1-a^n-c^n[/math] is divided into q.

 

5°. Known also that if [math]1-a-c=q[/math] (q is prime), then the number [math]e=1-a^q-c^q[/math] is divided into q.

 

6°. But then the difference in [math]d-e=(a^q-c^q)-(a^n-c^n)[/math] also is divided into q.

 

7°. However, it is known (which was repeatedly shown the author on the forums), that the greatest common divisor of the numbers [math](a^q-c^q)[/math] and [math](a^n-c^n)[/math] is the number [math]a-c[/math], which, as can be seen from 2, into q is not divided.

 

Thus, taking into account 0°, the solution of Fermat’s equation in the integers does not exist.

FLT is proven.

 

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

 

The author is ready to answer any questions, which are concerned this proof.

[Victor Sorokine]

Materials to the proof of the formula [math]A+B-C=abc[/math].

 

T1. Lemma (fundamental theorem, main tool for proof of the FLT).

 

With the relatively prime numbers a and b and relatively prime n (>2) and q (>2) the greatest common divisor of the numbers [math]a^n-b^n[/math] and [math]a^q-b^q[/math] is [math]a-b[/math].

 

Proof.

 

Let us examine the numbers [math]D=a^{xn}-b^{xn}[/math] and [math]E=a^{yq}-b^{yq}[/math], where [math]xn-yq=1[/math] (see linear they are diophantine the equation), or [math]D=a^d-b^d[/math] and [math]E=a^e-b^e[/math], where [math]d-e=1[/math].

 

From the multiplication [math]E(a+b )=a^{e+1}-b^{e+1}+ab(a^{e-1}-b^{e-1})=D+abD_2[/math] is evident that the number [math]a^{e-1}-b^{e-1}=D_2[/math] also is divided into q.

 

Is further analogous: From the multiplication [math] (E_2)(a+b )=a^{e-1}-b^{e-1}+ab(a^{e-3}-b^{e-3})=D_2+abD_3[/math] is evident that the number [math]a^{e-3}-b^{e-3}=D_3[/math] also is divided into q.

 

And so on to the number [math](a^2-b^2)(a+b )=a^3-b^3+ab(a-b )[/math], where the number a-b is divided into q, QED.

[Victor Sorokine]

Actually, I proved the following theorem:

 

The number [math]A^n+B^n+C^n[/math], where A+B+C is divided into prime n, and A, B, C - relatively prime, and A+B+C and ABC are not equal to zero, it does not contain prime bases beyond the limits of their set in the number ABC.

[uncool]

Actually, I proved the following theorem:

 

The number [math]A^n+B^n+C^n[/math], where A+B+C is divided into prime n, and A, B, C - relatively prime, and A+B+C and ABC are not equal to zero, it does not contain prime bases beyond the limits of their set in the number ABC.

Precisely where?

 

2^5 + 3^5 + 5^5 = 32 + 243 + 3125 = 275 + 3125 = 3400

 

A, B, and C are relatively prime, A + B + C = 10 =/= 0, ABC = 30 =/= 0, 17 divides ABC.

 

Similarly:

3^5 + 11^5 + 11^5 = 243 + 161051 + 161051 = 322345 is divisible by 5, 23, and 2803.

=Uncool-