Fermat's Last Theorem. Elementary proof

[Victor Sorokine]

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-

 

You are completely right, and your counterexample is excellent! Enormous thanks!!!

I understood the reason for marriage. True theorem is formulated as follows:

 

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.

 

Difference as if is small, but is fundamental.

[uncool]

You are completely right, and your counterexample is excellent! Enormous thanks!!!

I understood the reason for marriage. True theorem is formulated as follows:

 

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.

 

Difference as if is small, but is fundamental.

The difference there is trivial, as you can take C' = -C, and you have the same theorem.

 

Also,

3^5 + 11^5 - 4^5 = 243 + 161051 - 1024 = 160270 has 2, 5, 11, 31, 47

=Uncool-

[Victor Sorokine]

The difference there is trivial, as you can take C' = -C, and you have the same theorem.

 

Also,

3^5 + 11^5 - 4^5 = 243 + 161051 - 1024 = 160270 has 2, 5, 11, 31, 47

=Uncool-

Thanks! This IS VERY interesting! I will in greater detail answer by evening.

[Victor Sorokine]

The difference there is trivial, as you can take C' = -C, and you have the same theorem.

 

Also,

3^5 + 11^5 - 4^5 = 243 + 161051 - 1024 = 160270 has 2, 5, 11, 31, 47

=Uncool-

I yet did not understand the reason for contradiction with dividers 31 and 47.

Searches continue.

There are hypotheses.

[Victor Sorokine]

Theorem about the dividers of exponential trinomial (precise formulation)

 

If natural A, B, C – relatively prime and the number [math]A^n+B^n-C^n[/math] is divided, but the number of [math]nABC[/math] is not divided into the prime number q, then the number [math]U=A+B-C[/math] is not divided into q.

[imatfaal]

Viktor - I love the fact that you are battling away against this problem - although I am certain it is a doomed task. I think you might get more challenges and thus learn more and be able to develop a more coherent argument if your posts were slightly less cryptic.

 

Your latest post seems to say talk about three numbers (A^N+B^n-C^n, n.A.B.C, and A+B-C) and whether they are divisors of a fourth number q. But then you say that q is a prime number! Prime numbers have no divisors except themselves and 1 - that's kinda the point. A^N+B^n-C^n will only divide q when A,B &C all equal 1, or when n=0, or when the expression equals q.

 

I think if you were more methodical and presented your argument in a more formal manner (perhaps with //comments to explain) then you would see errors and be able to hone your ideas.

[Victor Sorokine]

Your latest post seems to say talk about three numbers (A^N+B^n-C^n, n.A.B.C, and A+B-C) and whether they are divisors of a fourth number q.

 

On the contrary, prime number q is divider of three numbers: [math]A^n+B^n-C^n[/math], [math]A^q+B^q-C^q[/math], and [math]A+B-C[/math].

 

+++++++++++

 

Theorem about the dividers of the exponential trinomial

 

Theorem

 

If integers A, B, C are relatively prime and the number [math]d=A^n+B^n-C^n[/math] is divided, and the number of [math]nABC[/math] is not divided into the prime number q, then the number [math]U=A+B-C[/math] is not divided into q.

 

Proof by the method from the contrary (i.e. let us assume that the number U is divided into q)

 

1°). Let us examine the numbers A, B, C, D, U in the numeration system on the base q and will multiply the number D by this number [math]g^n[/math], that the number [math]gB[/math] finishes by digit 1 (see A1°).

 

2°). Let us write down the new values of the numbers A, B, C in the form: [math]A=qx+a[/math], [math]B=qy+1[/math], [math]C=qz+c[/math] (where a, b, c are last digits in the numbers A, B, C and, on the assumption, the number [math]u=a+1-c[/math] is divided into q) and let us substitute these values into the formula for D.

 

3°) Since the number D is divided into q, then, therefore, the number [math]d=a^n+1-c^n[/math] is divided into q.

 

4°) Since the number U is divided into q, then, therefore, the number [math]e=a^q+1-c^q[/math] is divided into q.

 

5°) Consequently, the number [math]d-e=(a^n-c^n)-(a^q-c^q)[/math] also is divided into q.

 

6°). However (according to the Theorem about the greatest common divisor of two exponential binomials see A6°) the greatest common divisor of the two numbers ([math]a^n-c^n[/math]) and ([math]a^q-c^q[/math]) is the number a-c, which, as can be seen from 2°, is not divided into q.

 

Thus Theorem is proven.

Edited October 29, 2010 by Victor Sorokine

[Victor Sorokine]

Viktor - I love the fact that you are battling away against this problem - although I am certain it is a doomed task. I think you might get more challenges and thus learn more and be able to develop a more coherent argument if your posts were slightly less cryptic.

 

As to say correctly in English:

Number 6 is divided into prime number 3?

Number 6 is divided by prime number 3?

Number 3 is a divider of number 6?

[imatfaal]

As to say correctly in English:

Number 6 is divided into prime number 3?

Number 6 is divided by prime number 3?

Number 3 is a divider of number 6?

 

 

 

Number 6 is divided into prime number 3? DOES NOT EQUAL Number 6 is divided by prime number 3?

 

 

3 is a DIVISOR of 6 MEANS 6 is DIVIDED BY 3 with no remainder.

These are the two terms you need ie 'x is a divisor of y' and similarly 'y is divided by x with no remainder'

 

Victor - where or what is A1°?

To be honest even your 1st point is hard to follow.

Let us examine the numbers A, B, C, D, U in the numeration system on the base q and will multiply the number D by this number , that the number finishes by digit 1 (see A1°).

A B C D U are integers in base q. // should we presume that g and n are also base q

D=A^n + B^n - C^n // I have presumed that lower case d is same as upper case D

// Have you shown somewhere else that this equation is satisfiable? It could be - but I am rubbish at guessing.

There exists a number of the form g^n such that if we multiply D.g^n that the number gB has a unit digit of 1

// multiply both sides of D=A^n + B^n - C^n by g^n

// D(g^n) = A^n(g^n) + B^n(g^n) - C^n(g^n) to get to gB you need to take the nth root - surely you cannot mean that!

// Where is the A1° I am sure that the forum has a link facility somewhere

[Victor Sorokine]

3 is a DIVISOR of 6 MEANS 6 is DIVIDED BY 3 with no remainder.

These are the two terms you need ie 'x is a divisor of y' and similarly 'y is divided by x with no remainder'

 

Victor - where or what is A1°?

 

Thank!

 

A - Appendix

A1°. See the multiplication tables of two numbers in the numeration system with the prime base.

A6°. Lemma (theorem about the greatest common divisor of exponential binomials).

 

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

 

Theorem about the common divisor of the exponential binomials

 

Lemma (theorem about the greatest common divisor of exponential binomials)

 

If with relatively prime integers a and b and relatively primes n (> 2) and q (> 2) both numbers [math]a^n-b^n[/math] and [math]a^q-b^q[/math] are divided into the prime number q, then the number [math]a-b[/math] is divided into q.

 

Proof by the method from the contrary (i.e. let us assume that the number a-b[/math] it is not divided into q, but both numbers [math]a^n-b^n[/math] and [math]a^q-b^q[/math] divided into q)

 

Let us examine two numbers D and E, which, obviously, are divided into q:

 

[math]D=a^{xn}-b^{xn}[/math] and [math]E=a^{yq}-b^{yq}[/math], where x and y are the solution of linear diophantine the equations [math]xn-yq=1[/math], 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]D_2=a^{e-1}-b^{e-1}[/math] also is divided into q.

 

It 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.

 

***

 

This lemma appears the key in the proof of the Theorem about the dividers of exponential trinomial. Thus far there is no evidence of the fact that the Lemma and Theorem (two fundamental theorems in the theory of the numbers) were proven earlier.

Edited November 1, 2010 by Victor Sorokine

[Victor Sorokine]

Proof of Fermat's last theorem

 

Basic properties (known from 17 centuries) of the Fermats equality:

 

1°) [math]A^n+B^n=C^n[/math] for the base case:

2°) prime n> 2,

3°) integers A, B, C are relatively prime;

4°) a, b, c are greatest common divisors respectively in the pairs of numbers (A, C-B ), (B, C-A), (C, A+B );

5°) p, q, r are second cofactors in the numbers A, B, C: [math]A=ap[/math], [math]B=bq[/math], [math]C=cr[/math], where

6°) the number in the pairs (a, p), (b, q), (c, r) relatively prime;

7°) last digits in the numbers p, q, r are 1;

8°) the number [math]U=A+B-C[/math] is divided into abc;

9°) if [math]ABC[/math] is not divided into n (case 1), then [math]C-B=a^n[/math], [math]C-A=b^n[/math], [math]A+B=c^n[/math];

10°) if - for example, C - is divided into n (case 2), then [math]C-B=a^n[/math], [math]C-A=b^n[/math], [math]A+B= (c^n): n[/math].

11°) is not difficult to see (taking into account 5°-8°) that the number U does not have dividers of the number pqr [that evidently from the equalities [math]U=ap-(C-B )=ap-a^n[/math], [math]U=bq-(C-A)=bq-b^n[/math], [math]U=(A+B )-cr=c^n-cr)[/math]].

 

Set of instruments:

12°) The Theorem about the dividers of exponential trinomial.

 

***

 

Proof of the FLT in the numeration system with the prime base n>2

 

According to the theorem about the dividers of exponential trinomial, the set of prime dividers (or prime bases) in the number [math]U=A+B-C[/math] does not exceed the limits of the set of prime dividers in the number [math]nABC[/math].

 

In this case, as can be seen from 4°-5°, numbers a, b, c are the dividers of the number U, in this case the number U is not divided into the number of abcd, where d is any divider of the number of abc. Furthermore, the prime dividers of the number [math]ABC[/math], which are not the dividers of the number abc, are not the dividers of the number U, therefore, are not dividers of the numbers of [math]nABC[/math]. All these dividers, as is known, finish to digit 1.

 

Thus, we have a equality:

[math]U=A+B-C=un^k=abcn^t[/math], in which the last significant digits of the numbers un^k and abcn^t are equal.

 

However, this equality is disrupted after the multiplication of Fermats equality by [math]2^n[/math].

Actually, during the multiplication by [math]2^n[/math] the last significant digits of the number [math]un^k[/math] is multiplied by 2, and the last significant digits of the number [math]abcn^t[/math] is multiplied

- by 8 in the case of 1 (if ABC is not divided into n),

- by 4 in the case of 2 (if ABC is divided into n),

since the last significant digits of each of the numbers a, b, c are multiplied by 2.

 

And we arrived at the contradiction.

Edited November 1, 2010 by Victor Sorokine

[Victor Sorokine]

Popular science communication about proof of the FLT

 

After the countless attempts to find contradiction between THE EXISTING elements of the Fermat’s equality in me gradually began to ripen idea about the fact that the contradiction should be looked between the elements, WHICH ARE ABSENT in the equality farm. This idea finally led to the study “nonexistent” prime dividers of the number [math]U=A+B-C[/math]. And actually, soon began to appear the signs of the fact that, besides the dividers of the numbers A, B, C yes degree itself n, no other dividers the number U contains!

 

Without being limited to itself Fermat’s equality, I began to search for this property in any trinomials and find his proof for all exponential trinomials [math]D=A^n+B^n-C^n[/math] with relatively prime A, B, C.

 

It is shorter, the number U in the trinomial D does not exceed the number [math]ABCn^k[/math], but in the Fermat’s equality is that less: [math]U=abcn^k[/math], where a, b, c is the greatest common divisors in the pairs of numbers (A, C-B ), (B, C-A), (C, A+B ).

 

However, most interesting proved to be very equality [math]A+B-C=abcn^k[/math], in which k>0 and last significant digits in the left and right sides of the equality, where they are obligated to be equal.

 

BUT!!! During the multiplication of the Fermat’s equality by [math]2^n[/math] last significant digits in the numbers A, B, C and U are multiplied by 2, while the last significant digits in the number [math]abcn^k[/math] is multiplied by 8 (if among the numbers A, B, C there is no multiple n) or, at the worst, by 4 (if among the numbers A, B, C one multiply n), which leads to THE INEQUALITY of the last significant digits in “the equality” [math]A+B-C=abcn^k[/math]! Let us note that no elliptical functions for the proof of this contradiction it was required!

 

I do not expect that the proof of the FLT will be in the very near future acknowledged by mathematical leaders (“This it cannot be, because this it cannot be ever!!! ”). But to the beginning mathematicians there is above how to think…

 

Number 6 is divided into prime number 3? DOES NOT EQUAL Number 6 is divided by prime number 3?

 

 

3 is a DIVISOR of 6 MEANS 6 is DIVIDED BY 3 with no remainder.

These are the two terms you need ie 'x is a divisor of y' and similarly 'y is divided by x with no remainder'

 

Victor - where or what is A1°?

I will regret, which interfere withs poor transfer clear understanding. I hope that this problem will be soon solved.

[imatfaal]

Victor to be brutal - at present no one would acknowledge your "proof" because it is impossible to truly follow. English is a dreadful language, it is irregular, informal, and difficult; but it does allow exact expression; unfortunately it is not easy to get to this level of proficiency. In order to improve your chance of getting someone to look hard at your work you must simplify.

 

1. For an attempt that relies on division you must learn the correct use of the verb "to divide". This will allow you to create ideas and express them in a way others can be certain to understand

 

2. Try and be more formal. Define ALL your variables at the beginning - if you have different bases then make it clear which is base10 and which are base_n. If you must use upper and lower of the same letter - then it is vital to be consistent. Its really confusing to mix latex with plain type.

 

3. Break your ideas down into smaller sentences. Include only one idea per sentence. Make it clear with each new idea the status of that idea (axiomatic, proved, to be proved, reliant on previous proof etc )

 

Matthew

[Victor Sorokine]

Estimation of the situation

 

I consider with the large confidence that I proved the following assertions:

 

1. Theorem about the greatest common divisor of exponential binomials,

2. Theorem about the dividers of exponential trinomial,

3. Formula for the Fermat’s equality: A+B-C=abcn^k,

4. FLT for the case: prime n> 4 and ABC not is divided by n.

 

The search for the perfect proofs of the remaining cases continues.

 

+++

 

For Matthew:

I will try to consider your observations.

The many thanks!

 

Linguistic proof-reading

 

Theorem about the dividers of the exponential trinomial

 

Theorem

 

If integers A, B, C are relatively prime and the number [math]d=A^n+B^n-C^n[/math] is divisible, and the number of [math]nABC[/math] is not divisible by a prime number q, then the number [math]U=A+B-C[/math] is not divisible by q.

 

Proof by the method from the contrary (i.e. let us assume that the number U is divisible into q)

 

1°). Let us examine the numbers A, B, C, D, U in the base-q numeral system and let us multiply the number D by such a number [math]g^n[/math] that the number [math]gB[/math] finishes by digit 1 (see A1°).

 

2°). Let us write down the new values of the numbers A, B, C in the form: [math]A=qx+a[/math], [math]B=qy+1[/math], [math]C=qz+c[/math] (where a, b, c are the last digits in the numbers A, B, C and, on the assumption, the number [math]u=a+1-c[/math] is divisible by q) and let us substitute these values into the formula for D.

 

3°) Since the number D is divisible by q, then, therefore, the number [math]d=a^n+1-c^n[/math] is divisible by q.

 

4°) Since the number U is divisible by q, then, therefore, the number [math]e=a^q+1-c^q[/math] is divisible by q.

 

5°) Consequently, the number [math]d-e=(a^n-c^n)-(a^q-c^q)[/math] also is divisible by q.

 

6°). However (according to the Theorem about the greatest common divisor of two exponential binomials – see A6°) the greatest common divisor of the two numbers [math] (a^n-c^n)[/math] and [math](a^q-c^q)[/math] is the number [math]a-c[/math], which, as can be seen from 2°, is not divisible by q.

 

Thus the Theorem is proven.

 

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

 

Appendix

 

A1° - See the multiplication tables of two numbers written in a numeral system with a prime base.

A6° - Lemma (theorem about the greatest common divisor of exponential binomials).

 

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

 

Theorem about the common divisor of the exponential binomials

 

Lemma (theorem about the greatest common divisor of exponential binomials)

 

If, given two relatively prime integers a and b and relatively primes n (> 2) and q (> 2) both numbers [math]a^n-b^n[/math] and [math]a^q-b^q[/math] are divisible by a prime number q, then the number [math]a-b[/math] is divisible by q.

 

Proof by the method from the contrary (i.e. let us assume that the number [math]a-b[/math] it is not divisible by q, but both numbers [math]a^n-b^n[/math] and [math]a^q-b^q[/math] are divisible by q)

 

Let us examine two numbers D and E, which, obviously, are divisible by q:

 

[math]D=a^{xn}-b^{xn}[/math] and [math]E=a^{yq}-b^{yq}[/math], where x and y are the solution of the linear diophantine equation [math]xn-yq=1[/math], 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] we can see that the number [math]D_2=a^{e-1}-b^{e-1}[/math] is also divisible by q.

 

And so on:

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] we can see that the number [math]a^{e-3}-b^{e-3}=D_3[/math] is also divisible by q.

 

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

 

***

 

This lemma appears to be the key in the proof of the Theorem about the dividers of exponential trinomial. So far I can not find any evidence that the Lemma and the Theorem (two fundamental theorems in the theory of the numbers) were proven earlier.

[Victor Sorokine]

Proof of Fermat's last theorem (Final text)

 

Basic properties (known from 17 centuries) of the Fermat’s equality:

1°) [math]A^n+B^n=C^n[/math] for the base case:

2°) prime n>2 (in this proof [math]n>4[/math]),

3°) integers A, B, C are relatively prime;

4°) a, b, c are greatest common divisors respectively in the pairs of numbers (A, C-B ), (B, C-A), (C, A+B );

5°) p, q, r are second cofactors in the numbers A, B, C: [math]A=ap[/math], [math]B=bq[/math], [math]C=cr[/math], where

6°) the number in the pairs (a, p), (b, q), (c, r) relatively prime;

7°) the number [math]U=A+B-C[/math] is divisible into abc;

8°) if [math]ABC[/math] is not divisible into n (case 1), then [math]C-B=a^n[/math], [math]C-A=b^n[/math], [math]A+B=c^n[/math];

9°) if - for example, C - is divided into n (case 2), [math]then C-B=a^n[/math], [math]C-A=b^n[/math], [math]A+B= (c^n): n[/math].

10°) is not difficult to see (taking into account 5°-8°) that the number U does not have dividers of the number pqr [that evidently from the equalities [math]U=ap-(C-B )=ap-a^n[/math], [math]U=bq-(C-A)=bq-b^n[/math], [math]U=(A+B )-cr=c^n-cr)[/math]].

 

***

 

Proof of the FLT in the numeration system with the prime base n>2

 

According to the theorem about the dividers of exponential trinomial, the set of prime dividers (or prime bases) in the number [math]U=A+B-C[/math] does not exceed the limits of the set of prime dividers in the number [math]nABC[/math].

In this case, as can be seen from 4°-5°, numbers a, b, c are the dividers of the number U, in this case the number U is not divisible into the number of abcd, where d is any divider of the number of abc.

 

Furthermore, the prime dividers of the number [math]ABC[/math], which are not the dividers of the number nabc, are not the dividers of the number U, therefore, they are not dividers of the number [math]nABC[/math].

 

As a result under the conditions of the FLT we have a equality:

11°) [[math]U=[/math]] [math]A+B-C=abcn^t[/math].

 

However, after term-by-term multiplication equality 1° by [math]2^{nn}[/math] equality 11° (with n>4) is disrupted, since the numbers [math]C-B[/math] [[math]=a^n[/math]], [math]C-A[/math] [[math]=b^n[/math]], [math]A+B[/math] [[math]=c^n[/math]] and, therefore, the left side of the number U are multiplied by [math]2^n[/math], but the numbers a, b, c are multiplied altogether only by 2. As a result the right side of the equality 11° is multiplied by 8.

But if, for example, the number C is divisible into n, then in the formula [math]c^n=(A+B ) [/math]n the number [math]A+B[/math] is multiplied by [math]2^n[/math]:

12°) [math]c’^n=[(A+B)2^n]n=[(A+B)n]2^n=[c^n]2^n=[2c]^n[/math], which again leads to the disturbance of the equality of 11°, and truth of the FLT is present.

 

(“Difficult” case itself: n=3 is degenerate and proves separately.)

[Victor Sorokine]

In the canonical case from the Fermat’s equality follows the second equality

[u=] [math]A+B-C=abcun^k[/math].

Farm

From the term-by-term multiplication of the Fermat’s equality by [math]2^{nn}[/math] it is evident that the left side of the second equality is multiplied by [math]2^n[/math], and in the right side the number abc is multiplied by [math]2^3[/math]. Consequently, the number u is multiplied by [math]2^{n-3}[/math].

 

Question: what algebraic structure does have the number u in this case?

 

If [math]u=d(abc)^{n-3}[/math], where d is a positive integral function, then

 

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

 

[math]d(A+B )(C-A)(C-B )(n^k)2^n[/math] – if ABC is divided by n, or

 

[math]d[(A+B ):n](C-A)(C-B )(n^k)2^n[/math] – if ABC is not divided by n (here k>1).

 

BUT [math]A+B-C<<(A+B )(C-A)(C-B )(n^k)[/math] {or [math]A+B-C<<[(A+B ):n](C-A)(C-B )(n^k)[/math]}!

[Victor Sorokine]

Bottleneck in the last proof of the FLT is the theorem about the dividers q of exponential trinomial. Let us attempt to introduce the additional constraints (undertaken from the Fermat’s equality) to the numbers A, B, C so that the theorem would prove to be true.

 

Thus, instead of

 

“If integers A, B, C is relatively prime and the number [math]D=A^n+B^n-C^n[/math] is divided, and the number of nABC is not divided into the prime number q, then the number U=A+B-C is not divided into q.”

 

let us take:

 

Theorem

 

If integers A, B, C - relatively prime, [math]A+B=c^n[/math] [or [math]A+B=c^n[/math]], [math]C-B=a^n[/math], [math]C-A=b^n[/math] and the number

 

[math]D=A^n+B^n-C^n[/math] is divided, and the number [math]nABC[/math] is not divided into the prime number q,

 

then the number [math]U=A+B-C[/math] is not divided into q.

 

There are interesting ideas for the proof of this theorem.

[Victor Sorokine]

«First swallow» and phenomenon of the FLT

 

Last super-brief proof of the FLT (consisting of the only operation of multiplication) is based on the lemma: in the number [math]U=A+B-C=abcu[/math] the cofactor u=1.

 

The truth of lemma is determined by the divisibility of a certain number D on u. However, if the Fermat’s equality is not used in the proof of this divisibility, then D is not divisible by u, but if it is used, then D is divisible by u! How to explain phenomenon?

 

***

 

Proof of the FLT in the prime base of n>2. Development of last idea.

 

I will resemble known from 17 centuries of property Fermat’s equality

 

1°) [math]A^n+B^n=C^n[/math] for the base (or given) case:

2°) prime n>2;

3°) integers A, B, C are relatively prime;

4°) a, b, c – greatest common divisors respectively in the pairs of numbers (A, C-B ), (B, C-A), (C, A+B );

5°) p, q, r – second cofactors in the numbers A, B, C: A=ap, B=bq, C=cr, where

6°) the number in the pairs (a, p), (b, q), (c, r) relatively prime;

7°) the number [math]U=A+B-C=abcu[/math], where

8°) the number u does not have dividers of the numbers A, B, C;

9°) if AB (or BC, AC) is not divisible by n, then [math]C-B=a^n[/math] and [math]C-A=b^n[/math]; in the equalities [math]A^n= (C-B )P[/math] and [math]B^n=(C-A)[/math] the number [math]P=p^n[/math] and [math]Q=q^n[/math].

 

***

 

At the base of the simplest proof of the FLT lies the following lemma:

 

Under the conditions of the FLT in the equality [u=] [math]A+B-C=abcu[/math] (7°) the number u=1.

 

Proof is achieved by the method from the contrary.

 

Taking into account 5° and 9°, let us write down the equality 7° in the form: [math]ap-a^n=abcu[/math], from where

 

10°) [math]p-a^{n-1}=bcu[/math], where the number [math]p-a^{n-1}[/math] is divisible by u.

 

Consequently, and the number [math]p^n-a^{n(n-1)}[/math], or [math]P-(C-B )^{n-1}[/math] also is divisible by u. I.e., the number

 

11°) [math][C^{n-1}+C^{n-2}B+…+ CB^{n-2}+B^{n-1}]-(C-B )^{n-1}[/math] is divisible by u.

 

It follows on the same base from [math]q-b^{n-1}=acu[/math] that also the number

 

12°) [math] [C^{n-1}+C^{n-2}A+…+ CA^{n-2}+A^{n-1}]-(C-A)^{n-1}[/math] is divisible by u.

 

Summarizing the numbers 11° and 12° and taking into account that [math]B^{n-1}-(C-A)^{n -1}[/math] and [math]A^{n-1}-(C-B )^{n-1}[/math] are divisible by A+B-C (consequently and by u), and also rejecting common for all terms not divisible by u factor C, we obtain that multiple u the number

 

13°) [math]D=(C^{n-2}+C^{n-3}B+…+B^{n-2})+(C^{n-2}+C^{n-3}A+…+A^{n-2})[/math].

 

However, with n=3 the number [math]D=C+B+C+A=2C+A+B=3C+(A+B-C)=3C+U[/math] is not divisible by u!

 

(Thus, we found spectacular proof of the FLT for n=3.)

 

A study of the number D continues.

[Victor Sorokine]

Mathematical entertainment

 

Fermat's last theorem. New approach

 

Obtained contradiction: each of three numbers A, B, C is divisible by n.

 

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

 

1°) [math]A^n+B^n-C^n=0[/math], where

 

2°) [math]A+B-C=U=un^k[/math], where u is not divisible by n and, as is known, k>1.

 

Let us examine the numbers

 

3°) [math]P=(C^{n-1}-B^{n-1})/(C-B )[/math], [math]Q=(C^{n-1}-A^{n-1})/(C-A)[/math], [math]R=(A^{n-1}-B^{n-1})/(A+B)[/math].

 

It is easy to calculate (calculation will be represented later), that under the conditions of the FLT the numbers

 

4°) [math]P+Q[/math], [math]P-R[/math] и [math]Q-R[/math] are divisible by n. From what it follows that the number

 

5°) [math]P+Q[/math] и [math]P-Q[/math], => and the numbers P, Q, R, => and the numbers

 

[math]C^{n-1}-A^{n-1}[/math], [math]C^{n-1}-B^{n-1}[/math], [math]A^{n-1}+B^{n-1}[/math], => and the numbers

 

[math]A^{n-1}+B^{n-1}[/math] and [math]A^{n-1}-B^{n-1}[/math] => and the numbers [math]A^{n-1}[/math], [math]B^{n-1}[/math], [math]C^{n-1}[/math],

 

=> and the numbers A, B, C are divisible by n, which contradicts condition.

 

Fermat's last theorem is proven.

Edited December 4, 2010 by Victor Sorokine

[Victor Sorokine]

Calculations

 

Numerator of number [math]P-R[/math] [denominator [math](A+B )(C-B )[/math]]:

 

[math]=(C^{n-1}-B^{n-1})(A+B )-(A^{n-1}-B^{n-1})(C-B )=[/math]

 

[math]=AC^{n-1}-AB^{n-1}+BC^{n-1}-B^n-CA^{n-1}+CB^{n-1}+BA^{n-1}-B^n=[/math]

 

[math]=(A+B)C^{n-1}-(C-B)A^{n-1}+(C-A)B^{n-1}-2B^n=[/math]

 

[math]=(C+U)C^{n-1}-(A-U)A^{n-1}+(B-U)B^{n-1}-2B^n=[/math]

 

[math]=C^n-A^n+B^n-2B^n+UC^{n-1}+UA^{n-1}-UB^{n-1}=[/math]

 

[math]=(C^n-A^n-B^n)+U(C^{n-1}+A^{n-1}-B^{n-1})=[/math]

 

[math]=U(C^{n-1}+A^{n-1}-B^{n-1})[/math].

-------------------------------------

 

Thus,

 

[math]P-R=[ U(C^{n-1}+A^{n-1}-B^{n-1})]:[(A+B )(C-B )][/math].

 

If [math]U=un^k[/math], [math]C=cn^k[/math], [math]A+B=c’n^{kn-1}[/math] then

 

exponent with the coefficient n in the numerator: [math]k+min[k(n-1), (kn-1)]=kn[/math],

 

exponent with the coefficient n in the denominator: [math]kn-1[/math],

 

exponent with the coefficient n in the number P-R: [math]kn-(kn-1)=1[/math], i.e., the number P-R is divisible by n.

 

But if denominator [math](A+B )(C-B )[/math] is not divisible by n, then numerator is divisible by [math]n^k[/math].

 

All other calculations are analogous.

[Victor Sorokine]

Calculations

 

Numerator of number [math]P+Q[/math] [the denominator: [math](C-B )(C-A)[/math], or [math](A-U)(B-U)[/math]]:

 

[math](C^{n-1}-A^{n-1})(A-U)+(C^{n-1}-B^{n-1})(B-U)=[/math]

 

[math]=AC^{n-1}-A^n-UC^{n-1}+UA^{n-1}+BC^{n-1}-B^n-UC^{n-1}+UB^{n-1}=[/math]

 

[math]=(C+U)C^{n-1}-A^n-UC^{n-1}+UA^{n-1}-B^n-UC^{n-1}+UB^{n-1}=[/math]

 

[math]=C^n-A^n-B^n+U(A^{n-1}+B^{n-1}-C^{n-1})= U(A^{n-1}+B^{n-1}-C^{n-1})[/math].

 

I.e., [math]P+Q[/math] is divisible by n even if [math]C-B[/math] and [math]C-A[/math] are divisible by n.

 

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

 

Numerator of number [math]P-R[/math] [denominator: [math](A+B )(C-B )[/math] or [math](C+U)(A-U) [/math]]:

 

[math] (C^{n-1}-B^{n-1})(C+U)-(A^{n-1}-B^{n-1})(A-U)=[/math]

 

[math]=C^n-CB^{n-1}+UC^{n-1}-UB^{n-1}-A^n+AB^{n-1}+UA^{n-1}-UB^{n-1}=[/math]

 

[math]=C^n-A^n-(C-A)B^{n-1}+UC^{n-1}-UB^{n-1}+UA^{n-1}-UB^{n-1}=[/math]

 

[math]=C^n-A^n-(B-U)B^{n-1}+UC^{n-1}-UB^{n-1}+UA^{n-1}-UB^{n-1}=[/math]

 

[math]=C^n-A^n-B^n+UB^{n-1}+UC^{n-1}-UB^{n-1}+UA^{n-1}-UB^{n-1}=[/math]

 

[math]=U(C^{n-1}+A^{n-1}-B^{n-1})[/math].

 

Thus, [math]P-R=[u(C^{n-1}+A^{n-1}-B^{n-1})]:[(A+B )(C-B )][/math] .

 

If [math]U=un^k, C=cn^k, A+B=c’n^{kn-1}[/math].

 

Exponent with the coefficient n in the numerator is equal to: [math]k+min[k(n-1), (kn-1)]=kn[/math].

 

Exponent with the coefficient n in the denominator is equal to: [math]kn-1[/math].

 

Exponent with the coefficient n in the number [math]P-R[/math] is equal to:

 

[math]kn-(kn-1)=1[/math], i.e., the number [math]P-R[/math] is divisible by n.

 

But if denominator [math](A+B )(C-B )[/math] is not divisible by n, then numerator is divisible by n^k.

 

As a result the numbers [math]P+Q, P-R, Q-R[/math], consequently and P-Q, consequently and P, Q, R are divisible by n, and if ABC is divisible by n, then all three numbers A, B, C are divisible by n.

I.e. for the second case Fermat's last theorem is proven.

[Victor Sorokine]

Fermat's last theorem. Algebraic proof “into the forehead”

 

Proof of the FLT in the prime base [math]n>2[/math]. Completion of old idea.

 

Desired contradiction: the number [math]q-p[/math] [either p-r or q-r] is divisible AND is not divisible by the number [math]A-B[/math] [either A+C or B+C].

 

I will resemble known from 17 centuries of property Fermat’s equality

 

1°) [math]A^n+B^n=C^n[/math] for the base (or given) case:

2°) prime [math]n>2[/math];

3°) integers A, B, C are relatively prime;

4°) a, b, c – greatest common divisors respectively in the pairs of numbers (A, C-B ), (B, C-A), (C, A+B );

5°) p, q, r – second cofactors in the numbers A, B, C: [math]A=ap[/math], [math]B=bq[/math], [math]C=cr[/math], where

6°) the number in the pairs (a, p), (b, q), (c, r) relatively prime;

7°) the number [math]U=A+B-C=abcu[/math] and

if [math]AB[/math] [either BC or AC – proof in these cases are completely analogous] is not divisible by [math]n[/math], then

8°) [math]C-B=a^n[/math] and [math]C-A=b^n[/math] and in the equalities

9°) [math]A^n=(C-B )P[/math] and [math]B^n=(C-A)[/math] the cofactors [math]P=p^n[/math] and [math]Q=q^n[/math].

 

 

Proof of the FLT

 

10a°) [math]Q-P=q^n-p^n=(q-p)S[/math] and

10b°) [math]Q-P=(C^{n-1}+AC^{n-2}+[/math]…[math]CA^{n-2}+A^{n-1})-(C^{n-1}+BC^{n-2}+[/math]…[math]CB^{n-2}+B^{n-1})=[/math]

[math]=(A-B )D=(a^n-b^n)D=(a-b)TD[/math]. In this case the numbers [math]A-B[/math] and [math]D[/math] are coprime, without considering prime divider n.

 

In addition to this, from the equalities [math]A=C-B+U[/math] and [math]B=C-A+U[/math], or [math]ap=a^n+abcu[/math] and [math]bq=b^n+abcu[/math], we find:

 

11°) [math]p=a^{n-1}+bcu[/math] and [math]q=b^{n-1}+acu[/math], from where

 

12°) [math]q-p=(b^{n-1}-a^{n-1})+(b-a)cu[/math], i.e., the number [math]q-p[/math] is divisible by [math]a-b[/math].

 

Thus, we have: [math]Q-P[/math] [[math]=(q-p)S[/math]] is divisible by [math]A-B[/math] [[math]=(a-b)T[/math]], where [math]q-p=(a-b)g=ag-bg[/math].

 

The following theorem about the exponential binomials is true:

 

If integers a, b, q, p are coprime and [math]q-p=(a-b)g[/math], where g is whole, then the numbers T and S in the equalities [math]a^n-b^n=(a-b)T[/math] and [math]q^n-p^n=(q-p)S[/math] are coprime, without considering prime divider n.

The author's proof of Theorem will be examined later.

 

If theorem is true, then all prime dividers of the number [math]T[/math] are the dividers of the number [math]q-p[/math], i.e., the number [math]q-p[/math] is divisible by [math]A-B[/math]. However, simple calculation shows inequality (by the way, they are ideally confirmed by computer calculations):

 

13°) [math]q-p<A-B[/math] and, therefore, the number [math]q-p[/math] is not divisible by [math]A-B[/math].

 

And we arrived at the contradiction.

Edited December 25, 2010 by Victor Sorokine

[Victor Sorokine]

Fermat's last theorem. Anti-proof

 

Desired contradiction:

the numbers [math]A^n-B^n[/math], or [math]C^n-2B^n[/math], and [math]C^{pn}-B^{pn}[/math] (where prime [math]m=pn+1[/math] is the divider of the number [math]A^n-B^n[/math]) are AND are not coprime.

 

I will resemble the simplest properties of the Fermat’s equality

 

1°) [math]A^n+B^n=C^n[/math] for the base (or given) case:

 

2°) prime [math]n>2[/math];

 

3°) integers A, B, C are relatively prime (coprime);

 

4°) [math]AB[/math] [either BC or AC – proof in these cases is completely analogous] is not divisible by n,

 

5°) [math]m[/math] [[math]=pn+1[/math]] – a prime divider of the number R in the equality [math]A^n-B^n=(A-B )R[/math].

 

* * *

 

Proof of the FLT

 

We find from 1° that the number [math]A^n-B^n[/math], or [math]C^n-2B^n[/math], is divisible by prime [math]m=pn+1[/math].

 

But, according to Little Fermat's theorem, the number [math]C^{pn}-B^{pn}[/math] also is divisible by m.

 

And after change in the designations – [math]C^n=c[/math], [math]B^n=b[/math] – we have a pair of numbers, which are divisible by m:

 

6°) [math]d=c-2b[/math] and [math]e=c^p-b^p[/math] (where, as can easily be seen, the number b and c-b are coprime),

 

which, as it will be shown separately in the Theorem about the exponential binomial, are coprime, that contradicts to 6°.

 

Thus, if the theorem about the exponential binomial is true, FLT is proven.

 

* * *

 

Theorem about the exponential binomials. New version:

 

If integers a, b, a-b and d are coprime and >1, then the numbers [math]a^n-(b+d)^n[/math] and [math]a^n-(b-d)^n[/math], where [math]n>2[/math], also coprime.

 

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

 

More interesting hypothesis. Theorem about the prime divider of the exponential binomial:

 

For coprime integers C and B and prime [math]n>2[/math] number [math]C^n-2B^n[/math] does not have dividers m of the form [math]m=pn+1[/math]. If theorem is confirmed, then proof of the FLT is reduced by third.

Edited December 28, 2010 by Victor Sorokine

[Victor Sorokine]

More interesting hypothesis. Theorem about the prime divider of the exponential binomial:

 

For coprime integers C and B and prime [math]n>2[/math] number [math]C^n-2B^n[/math] does not have dividers m of the form [math]m=pn+1[/math]. If theorem is confirmed, then proof of the FLT is reduced by third.

One of the possible continuations of the proof of the last version of the Theorem about the exponential binomial brings to a special case of little Fermat's theorem:

 

If the number d not multiply prime number [math]m=pn+1[/math], where prime [math]n>2[/math], then the number [math]d^p-1[/math] is not divisible by m.

 

The proof of this, as if not complex, theorems indicates the presence of the single line proofs of two theorems: Theorem about the exponential binomial and FLT.

[Victor Sorokine]

I assume that the work is completed.

 

Simple proof of the FLT (n>4) based on the lemma (see the Appendix):

in the number [u=] [math]A+B-C=abcun^k[/math] (6°) the cofactor u=1.

 

Actually, after the term-by-term multiplication of the hypothetical Fermfat’s equality by [math]2^{nn}[/math] the number A+B-C into 6° is multiplied by [math]2^n[/math] [that > [math]2^4[/math]], while the number [math]abcn^k[/math] is multiplied by [math]2^3[/math]. And the contradiction of the formula of 6° presently. Thus FLT is proven.