Fermat's Last Theorem. Elementary proof

[Victor Sorokine]

Elementary proof of the FLT

 

The designations of the numbers, which participate in the proof, become clear from the following relationships after the elimination of the general cofactors

0°) - let us assume that this is possible, -

in the numbers A^n, B^n, C^n (for the beginning AB is’not divisible by n):

 

1°) A^n+B^n=C^n,

2°) A^n=C^n-B^n=(C-B)P=a^np^n=(ap)^n,

3°) B^n=C^n-A^n=(C-A)Q=b^nq^n=(bq)^n,

4°) C^n=A^n+B^n=(A+B)R,

5°) A+B-C=U=(A+B)-C=A+(B-C)=B+(A-C)=ap-a^n=bq-b^n=uabc*, where c* - greatest common divisor of the numbers C and A+B (it’s obviously that c*>1), from here

6°) C=A+B-uabc*.

 

7a°) It’s obviously that the numbers A^n+B^n, C^n-B^n, C^n-A^n, A^n-B^n, C^n+B^n, C^n+A^n are in pairs relatively prime.

7°) Obvious that the numbers A+B, C-B, C-A, A-B, C+B, C+A are also relatively prime.

 

Proof

 

Let us examine two numbers: C-A and C+A, or, taking into account 6°,

8°) C-A=B-uabc* and C+A=2A+B-uabc*.

 

And now it is easy to see that also the sum [2A+2B-2uabc*, or 2(A+B-uabc*)], and the difference [2a] of the numbers C-A and C+A ARE DIVIDED into c* (see 6° and 5°), that contradicts to 7°, 7a°, 0°.

 

If the number CB is not divisible by n, than the proof is perfectly analogy.

 

FLT is proven.

[Victor Sorokine]

Correction of the error

 

Fermat's last theorem

 

The designations of the numbers, which participate in the proof, become clear from the following relationships after the elimination of common factors in the numbers A, B, C (let us examine only the case, when AB not divided by simple n):

 

1°) A^n+B^n=C^n и

1a°) A^n=C^n-B^n=(C-B)P=(a^n)(p^n)=(ap)^n,

1b°) B^n=C^n-A^n=(C-A)Q=(b^n)(q^n)=(bq)^n,

1c°) C^n=A^n+B^n=(A+B)R.

2°) It is important that the numbers A-B, C, a, are relatively prime!

 

Classical proof of FLT

 

Let us examine the number D=A(C-B)-B(C-A).

 

From one side,

3°) D=A(C-B)-B(C-A)=(A-B)C.

 

But from the other side,

4°) D=A(C-B)-B(C-A)=Ab^n-Ba^n=apb^n-bqa^n=ab(pb^{n-1}-qa^{n-1}).

 

And comparing 3° from 4°, we see that the number (A-B)C is divided by ab, that contradicts to 2°.

 

FLT is proven.

 

 

23.09.2009

[uncool]

What are a, b, p, and q?

 

And why is (C - B) equal to b^n, and (C - A) equal to a^n?

=Uncool-

[Victor Sorokine]

What are a, b, p, and q?

 

And why is (C - B) equal to b^n, and (C - A) equal to a^n?

=Uncool-

 

My proof is wrong! BUT...

 

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

 

Fairy phenomenon

 

Foundations of the mathematics incorrect!

 

 

Example:

 

1) If: a=16,43; b=11,82; c=18,25; a+b-c=u=10, then a^3+b^3-c^3=0.

 

But if: a=(10+6,43); b=(10+1,82); c=(10+8,25); a+b-c=10, then a^3+b^3-c^3=8,06!

 

***

 

Calculation:

 

[1000+3*10*(41,34+3,31-68,06=-23,41)+(265,85+6,03-561,52=-289,64)=

 

=1000-702,3-289,64=8,06]

 

Thus: 0=8,06!!!

 

 

This phenomenon is true for ANY positives a, b, c! (There is the Proof.)

 

Еще один пример:

 

2) If: a=19,5013; b=11,1433; c=20,6446; u=10; then a^3+b^3-c^3=0.

 

But if: a=(10+9,5013); b=(10+1,1433); c=(20+6446; u=10); then a^3+b^3-c^3=-23,23.

 

Thus: 0=-23,23!!!

 

Calculation:

 

[-6000+3*(100*9,5013+100*1,1433-400*0,6446=806,62)+

 

-3*(10*90,269+10*1,3071-40*0,4155=899,141)+(857,727+1,4944-0,2678=-326,98) =

 

= -6000+2419,86+2697,423+859,489=-23,23]

 

Thanks!

Edited November 10, 2009 by Victor Sorokine

[the tree]

1) If: a=16,43; b=11,82; c=18,25; a+b-c=u=10,

What?

 

then a^3+b^3-c^3=0.

Even if the first statement, then still, what?

[Victor Sorokine]

What?

 

Even if the first statement, then still, what?

 

In the first case we find the value of the number a^3+b^3-c^3 WITHOUT the use of Newton's binomial, in the second case - with the use.

[uncool]

Um. a^3 + b^3 - c^3 is clearly not 0, but 8.20.

=Uncool-

[Victor Sorokine]

Um. a^3 + b^3 - c^3 is clearly not 0, but 8.20.

=Uncool-

 

1) If: a=16,43; b=11,82; c=18,25; a+b-c=u=10, then a^3+b^3-c^3=

=4435,19+1651,40-6086,6=0.

[uncool]

1) If: a=16,43; b=11,82; c=18,25; a+b-c=u=10, then a^3+b^3-c^3=

=4435,19+1651,40-6086,6=0.

 

18.25^3 = 6 078.390625

=Uncool-

[Dave]

Yeah. So Victor has previously been on here multiple times with the same 'proof', which eventually ended in a ban because he refused to listed to reason. So please report this if it gets out of hand.

[Victor Sorokine]

18.25^3 = 6 078.390625

=Uncool-

Yes!

 

Thank!

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Yeah. So Victor has previously been on here multiple times with the same 'proof', which eventually ended in a ban because he refused to listed to reason. So please report this if it gets out of hand.

 

More precise calculations give identical results.

 

Theme is closed.

 

Sorry...

[the tree]

I would put the 'error' down to a simple mistake, but basically the same false claim is made in the second example.

 

19.5013³+11.1433³-20.6446³=1.338181, not zero.

 

Although, there's no real indication of what this is meant to show.

[Victor Sorokine]

I would put the 'error' down to a simple mistake, but basically the same false claim is made in the second example.

 

19.5013³+11.1433³-20.6446³=1.338181, not zero.

 

Although, there's no real indication of what this is meant to show.

 

If A=a+pU, B=b+qU, C=c+rU (A+B-C=U), then (after the multiplication of Fermat’s equality by the sufficiently large number d^n) A^n+B^n>C^n!!! In all cases! Proof is transferred in the English.

 

Thanks

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++++++++

 

If A^n+B^n-C^n=0, then, after the substitution A=a+pU, B=b+qU, C=c+rU (where U=A+B-C), the number A^n+B^n-C^n has the form: U^n-VU^(n-1), where V is constant. And after the multiplication of Fermat’s equality by the sufficiently large number d^n the number A^n+B^n>C^n!!! In all cases! Proof is transferred in the English.

[the tree]

If A=a+pU, B=b+qU, C=c+rU (A+B-C=U), then (after the multiplication of Fermat’s equality by the sufficiently large number d^n) A^n+B^n>C^n!!! In all cases!

But there are infinitely many counter examples to that, it doesn't take much effort to set U to 0 and then find some a,b,c,n such that aⁿ+bⁿ<cⁿ.

[Victor Sorokine]

But there are infinitely many counter examples to that, it doesn't take much effort to set U to 0 and then find some a,b,c,n such that aⁿ+bⁿ<cⁿ.

 

If we will take multiplier d from the general solution of the linear diophantine equation c*(d+ut)-u(e+c*t)=1, where C* and u is the greatest and relatively prime dividers of the numbers C and U and C=c'c*:

d = d+u, d+2u, d+3u, …,

then C=[u(e+c*t)+1]c'=TU+c', where c' is CONSTANT!

So:

U^n>>>T*U^(n-1)+c'^n!

[the tree]

You'll have to make your notation clearer.

Define which are sets and which are entities, and try to avoid recursive definitions.

State your formulae and then their implications, if you give a worked example then use a calculator to avoid calculation errors.

[DJBruce]

It would also help if you could try to use LaTex as it is much easier to read. Especially when there are a lot of different equations being used.

[Victor Sorokine]

You'll have to make your notation clearer.

Define which are sets and which are entities, and try to avoid recursive definitions.

State your formulae and then their implications, if you give a worked example then use a calculator to avoid calculation errors.

***

Elementary proof of FLT

 

Let us assume that for natural A, B, C

1°) A^n+B^n=C^n [or E=A^n+B^n-C^n=0], where prime n>2, the number

2°) A+B-C=U, C>A>B>U>0 and, therefore,

3°) C-U>A-U>B-U>0;

4°) c' (>1) - the greatest common divisor of the numbers C and U; C= c'c*, U=uc'.

 

Let us write down the numbers A, B, C on the module U:

5°) A=a+pU, B=b+qU, C=c+rU (here a, b, c are positive number-digits, <U).

 

6°) if c is not equal of c', then let us multiply the equality of 1° by the sufficiently large number D= (d+ut) ^n, where the number (d+ut) is undertaken from the solution of the linear diophantine equation

7°) c*(d+ut)-u(e+c*t)=1. About the value of the number (d+ut)^n will be said below.

8°) But is important that the new value of the number C in 5° (WITH ANY (d+ut)^n!) will take the form: C=c'+rU, where c'<<U and it is constant by the change t. Possibly, 6°-8° are superfluous, nevertheless we will examine the infinite sequence of Ferman’s equalitie 1°, multiplied by D= (d+ut)^n (t=1,2,…).

 

It is easy to see that from 5° (taking into account 2° and 3°) escape two possibilities:

9°) a+b-c=0, and then pU+qU-rU=U, or

10°) a+b-c=U, and then pU+qU-rU=0.

11°) It is important that in both cases q=1 (otherwise, taking into account 2°, as the module one should take the number qU.)

 

Let us substitute 5° into 1°, let us discover Newton's binomials and let us group the similar terms:

12°) E=(a+pU)^n+(b+qU)^n-(c'+rU)^n=0, or

E=U^n(p^n+1-r^n)+nU^{n-1}(ap^{n-1}+b^{n-1}-c'r^{n-1})+…+

+ nU(a^{n-1}+b^{n-1}-c'^{n-1})+(a^n+b^n-c'^n)=0.

 

Let us show now that on the infinite set of multipliers D=(d+ut)^n (see 6°-8°) for the equality 1° the number E>0.

 

Case of 9° (taking into account of 8° and 11°): a+b-c' =0 and pU+U-ru =U, where p=r, U>u>c' >a>b>0.

 

Let us discover Newton's binomials and it will group the similar terms:

E=(pU+a)^n+(U+b)^n-(pU+c')^n=

=(pU)^n+U^n-(pU)^n+ n[(pU)^(n-1)a+U^(n-1)b-(pU)^(n-1)c']+ …

… +n[a^(n-1)(pU)+b^(n-1)U-c'^(n-1)pU]+(a^n+b^n-c'^n).

 

Let us take the now reduced value of the number E:

13°) E'=U^n- n(pU)^(n-1)c'- … -nc'^(n-1)pU-(c'^n-a^n-b^n), where 0

0<c'^n-a^n-b^n<nc'^(n-1)pU (taking into account that c'-a-b=0).

 

Let us decrease now and this value of E':

let us first, increase the last term of c'^n-a^n-b^n to the value of nc'^(n-1)pU,

in the second place, except the first, let us increase all binomial coefficients to maximum T (in the binomial of the n-th degree),

thirdly, the exponents with the bases pU and c' in all terms let us increase to n–1.

 

As a result the considerably reduced value of E' it will become equal

14°) E*=U^n-nT(pc')^(n-1)U^(n-1)=U^n-VU^(n-1), where the V is constant.

 

And having now taken in 6° the sufficiently large number (d+ut)^n as the coefficient of the equality of 1°, we can obtain the value of the number E*, therefore, and the numbers E' and E, greater than any given number. Which was required to prove.

 

Case of 10° (taking into account of 8° and 11°): a+b-c' =U and pU+U-rU=0, where p+1=r

 

This case is reduced to previous with the aid of the substitution, achieved only in the last term of expansion E into 12° (since the sum of all rest it takes the form V'U^(n -1), where V' is constant):

 

15°) a=U-a'', b=U-b'', c'=U-c'', where on the infinite set of the coefficients (d+ut)^n (see 6°-8°) for the equality of 1° the digit c'', consequently and the number c''^n, they are constants, and the number a''+b''-c''=0. And now (just as in the preceding case), having substantially decreased the number E into 12° –

after replacing positive terms, besides U^n and c''^n, with zeros,

and each negative item, except a''^n and b''^n, decreasing to -T(pc')^(n-1)U^(n-1),

and decreasing the number -(a''^n+b''^n-c''^n) to -nT'(c'')^(2n-2)U^(n-1), –

 

we obtain the negative number E*=U^n-nT*(pc')^(n-1)U^(n-1)-nT'(c'')^(2n-2)U^(n-1)= U^n-VU^(n-1), which with the sufficiently high coefficient D=(d+ut)^n (см. 6°-8°) for the equality 1° exceeds any given number (i.e. just as in the first case). The negative number E* after its multiplication by the positive D=(d+ut)^n becomes the number positive, which is impossible.

 

Thus, Fermat's last theorum is completely proven.

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It would also help if you could try to use LaTex as it is much easier to read. Especially when there are a lot of different equations being used.

The many thanks! But

1) I do not speak in English,

2) if the sense appears...

 

Victor

[Cap'n Refsmmat]

What DJBruce means is could you please look at this:

 

[thread]4236[/thread]

 

Then you can make your equations look pretty:

 

[math]\iiint_D f(x,y,z) \, dx \, dy \, dz[/math]

[John Cuthber]

I agree that this could be a lot clearer but I can't help wondering something.

If you start by saying

"Elementary proof of FLT

 

Let us assume that for natural A, B, C

1°) A^n+B^n=C^n [or E=A^n+B^n-C^n=0], where prime n>2,

"

doesn't that mean you have assumed the truth of Fermat's last theorem?

[Bignose]

I agree that this could be a lot clearer but I can't help wondering something.

If you start by saying

"Elementary proof of FLT

 

Let us assume that for natural A, B, C

1°) A^n+B^n=C^n [or E=A^n+B^n-C^n=0], where prime n>2,

"

doesn't that mean you have assumed the truth of Fermat's last theorem?

 

Read it more closely, John, he actually is assuming the opposite of FLT. This is a classic technique of proof, a proof by contradiction. Specifically, you assume the opposite of what you want to prove is true -- then discover what logical consequences that assumption leads to. The consequences of allowing the assumption to be true usually end up breaking one of the other assumptions/axioms/initial conditions/etc. and therefore, the assumption must be false.

[Victor Sorokine]

Two days without the Internet

 

***

 

Proof of FLT by P.Ferma (hypothesis)

 

Continuation after 8°:

 

Let us multiply the equality 1° by this number D= (d+ut)^n, that in new equality

9°) U>c'^n.

 

If now

10°) a+b-c'=0, the last figures in the base U in the equality of 1° give the non-integral equality of a^n+b^n-c' ^n.

 

But if

11°) a+b-c' =U, then after the substitution a=U-a'' (or b=U-b''), where now a''+b-c'=0, last digits in the base U in the equality 1° give the non-integral equality

12°) (-a'')^n+b^n-c'^n=0 (or a^n+(-b'')^n-c'^n=0'), that with entire a'', b, c' it is impossible.

 

Great theorem is proven.

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Letter to the forum

 

Gentlemen,

The second proof of FLT consists, actually, of one operation of the multiplication of Fermat’s equality by the large number D, after which we immediately obtain conclusion about the discrepancy of equality. Try to understand my proof, and I will answer all your questions. I hope that the imperfect transfer will not become obstacle to the mutual understanding.

 

Victor Sorokin

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Answer to questions of Moshe Klein

 

1) What is wrong in this prove for the case n=2?

- Cf.4°:

4°) c' (>1) - the greatest common divisor of the numbers C and U (C= c'c*, U=uc').

If n=2, then always c' =1;

If prime n>2, then always c' >1, since A^n+B^n=(A+B)R, where the greatest common divisor of the numbers A+B and R always more then 1.

 

2) What is the improvement from your last version of FLT?

- Both last versions of proof are accurate, but the second version is considerably shorter and it is simpler.

 

3) Why you say by Ferma?

- Because 80% of proof this of FLT is a formula of the general solution of the linear diophantine equation 8°) c*(d+ut)-u(e+c*t)=1.

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1) What is wrong in this prove for the case n=2?

- Cf.4°:

4°) c' (>1) - the greatest common divisor of the numbers C and U (C= c'c*, U=uc').

If n=2, then always c' =1;

If prime n>2, then always c' >1, since A^n+B^n=(A+B)R, where the greatest common divisor of the numbers A+B and R always more then 1.

 

Corr.:

1) What is wrong in this prove for the case n=2?

- Cf.4°:

4°) c' (>1) - the greatest common divisor of the numbers C and U (C= c'c*, U=uc').

If n=2, then always c' =1;

If prime n>2, then always c' >1, since A^n+B^n=(A+B)R, where the greatest common divisor of the numbers A+B and U always more then 1.

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Analysis of proofs.

 

1. Both versions of proof are based on the well-known fact: the numbers C and U contain the dividers c*>1 and u>1 not being been dividers numbers A+B; the numbers c* and u are interprime (relatively prime). (With n=2 this condition is not satisfied. Hence the hypothesis: with relatively prime A, B, C the number C in the equality A^n+B^n=C^n is prime.)

 

2. Positive integer relatively prime numbers c* and u generate the linear diophantine equation c*(d+ut)-u^n(e+c*t)=1.

 

3. After the multiplication of Fermat’s equality on D=(d+ut)^n with sufficiently large parameter t we obtain equivalent Fermat’s equality A''^n+B''^n-C''^n=0, where A''=a''+p''U'', B=b''+q''U'', C''=c''+r''U'' (где A''+B''-C''=U''), in which on the base U'' the number c'' is equal to the number c' from the previous equality, since C(d+ut)=c'c*(d+ut)=c'[c*(d+ut)]=c'[c*(d+ut)]=c'[u^n(e+c*t)+1]=c'. I.e. on the infinite set of the numbers D=(d+ut)^n (t=1, 2, …) the number c' is CONSTANT.

 

4. And now we see that both last versions of proof of the FLT ВТФ are accurate:

in the first, A''^n+B''^n-C''^n>0;

in the secondly, a''^n+b''^n-c'^n=0 (with a''+b''-c'=0) or a''^n-b''^n-c'^n=0 (with a''-b''-c'=0).

 

Problem of Fermat in the unofficial science is exhausted.

Edited November 19, 2009 by Victor Sorokine
Consecutive posts merged.

[Victor Sorokine]

For me it was possible to lead Fermat’s equality for prime n to the very interesting form:

 

either:

 

A=pU+a, B=U+b, C=pU+c, where

A, B, C is natural relatively prime,

U=A+B-C and multiple by n^2,

a+b-c=0,

0<(a, b, c)<U;

among them there are no equal,

among the numbers a, b, c two are not deliberately multiple n,

 

or:

 

A=pU+e, B=U+d-e, C=pU+d, where

U>d>e>0:

(pU+e)^n+(U+d-e)^n=(pU+d)^n.

 

The equality a+b-c=U, as it is very easy to show, under the conditions FLT is impossible.

Possibly, so easily it will be possible to prove the impossibility of the equality of a+b-c=0.

[Victor Sorokine]

Last nail

 

Proof of Fermat's last theorem for the prime n>2

 

Let us assume that for the relatively prime natural a, b, c there is a equality:

1°) A^n+B^n=C^n.

 

Let for the certainty the number A and B be not multiple n. In this case, as is well known, there are the following relationships and the ideas:

2°) A^n=(C-B)P, B^n=(C-A)Q, C^n=(A+B)R, A^n-B^n=(A-B)T;

the number in the sets

3°) [A, B, C], [C-B, C-A, A+B, A-B], [P, Q, R, T], [R, A+B, if A+B not is multiple n] relatively prime, but if A+B is multiple n, then R is multiple n and not multiply n^2;

4°) C-B=a^n, C-A=b^n, A-B=a^n-b^n, A+B= either c^n (if A+B not is multiple n), or (c^n)/n (if A+B multiply n).

 

 

Proof of the FLT

 

We have from the equalities A^n-B^n=(A-B)T=(C-B)T-(C-A)T and A^n-B^n=(C-B)P-(C-A)Q:

5°) (C-B)T-(C-A)T=(C-B)P-(C-A)Q, from where

6°) (C-B)(P-T)=(C-A)(Q-T).

Since the number C-A and C-B have no common factors, then the number P-T is divided by C-A, and the number Q-T is divided into C-B, i.e., we can write down:

7°) P-T=(C-A)x, Q-T=(C-B)y. Let us substitute these values into 6°:

8°) (C-B)(C-A)x=(C-A)(C-B)y, from where x=y. And now we have from 7°:

9°) P=T+(C-A)x, Q=T+(C-B)x.

 

Let us substitute these values into the equality A^n+B^n=(C-B)P+(C-A)Q:

10°) A^n+B^n=(C-B)P+(C-A)Q=(C-B)[T+(C-A)x]+(C-A)[T+(C-B)x]=

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

11°) =[(C-B)+(C-A)](T+x)=C^n.

 

It is easy to see that the number

12°) D=(C-B)+(C-A)=2C-(A+B)=2cr-c^n=c[2r-c^(n-1)] contains only one divider - the number c, the number D is not divided into the number r, and the number C does not contain dividers beyond the limits of the dividers of the number cr.

However, D=2C-(A+B)>c, since for C>n

13°) D=(C-B)+(C-A)>[nC^(n-1)]^(1/n)+[nC^(n-1)]^(1/n)=2[nC^(n-1)]^(1/n)>c.

 

But if C is multiple n, then the relationship 13° those more is fulfilled, since the number T not is multiple n, but the number x is multiple n.

[uncool]

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-