The authentic proof of the Fermat's Last Theorem

In Memory of my MOTHER

All calculations are done with numbers in base n, a prime number greater than 2.

The notations that are used in the proofs:

A' – the first digit from the end of the number A;

A_[k] is the k-digit ending of the number A (i.e. A_[k] = A mod n^k);

nn=n*n=n².

Here the well known properties of Fermat’s equality for natural and coprime numbers A,B,C:

1°) Aⁿ=Cⁿ-Bⁿ [=(C-B)P] //and Bⁿ=Cⁿ-Aⁿ [=(C-A)Q], Cⁿ=Aⁿ+Bⁿ [=(A+B)R]//. From here

1a°) (C-B)P+(C-A)Q-(A+B)R=0, where we denote with the letters a, b, c the greatest common divisors, respectively, of the pairs of numbers (A, C-B), (B, C-A), (C, A+B).

Then,

2°) if (ABC)'≠0, then C-B=aⁿ, P=pⁿ, A=ap; C-A=bⁿ, Q=qⁿ, B=bq; A+B=cⁿ, R=rⁿ, C=cr;

2a°) but if, for example, B_[k]=0, but B_[k+1]≠0, then (C-A)_[kn-1]=0, where kn-1>k (what is important in 8-2°);

3°) the number U=A+B-C=un^k, where k>1.

From here we find that (A+B)-(C-B)-(C-A)=2U and (if k=2)

3a-1°) A_[2]=aⁿ_[2]=a'ⁿ_[2], B_[2]=bⁿ_[2]=a'ⁿ_[2], C_[2]=cⁿ_[2]=a'ⁿ_[2]; consequently (see 5°),

3b-1°) Aⁿ_[3]=a'ⁿⁿ_[3], Bⁿ_[3]=b'ⁿⁿ_[3] ; Cⁿ_[3]=c'ⁿⁿ_[3]; consequently (see 1°),

3c-1°) a'ⁿⁿ_[3]+b'ⁿⁿ_[3]-c'ⁿⁿ_[3]=(Aⁿ_[3]+Bⁿ_[3]-Cⁿ_[3])_[3]=0.

4°) The digit Aⁿ_(k+1) is uniquely determined by the ending of A_[k] (a simple consequence of the binomial theorem).

5°) Lemma. Every prime divisor of the factor R of the binomial

A^{n^k}+B^{n^k}=(A^{n^{k-1}}+B^{n^{k-1}})R, where k>1, the natural numbers A and B are mutually prime and the number A+B is not a multiple of a prime n>2, can be presented as: m=dn^k+1

(http://www.mathforum.ru/forum/read/1/20535/page/63/ and /65/ - in Russian).

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And now the proof itself FLT. It consists of an endless sequence of cycles in which the exponent k (3°), starting with the value 2, increases in 1.

Thus, we consider the equality 3c-1° on three-digit endings:

6°) a'ⁿⁿ_[3]=[c'ⁿⁿ_[3]-b'ⁿⁿ_[3]]_[3]=(c'ⁿⁿ_[3]-b'ⁿⁿ_[3])P_[3]]_[3], where:

a) according to Lemma 5°, each prime factor of the number P ends with 01, and

b) each prime cofactor of the number P is in degree n.

And consequently (see 4°) P_[3]=001. Similarly and Q_[3]=R_[3]=001.

And now from the equality 1a°, we have: [(C-B)+(C-A)-(A+B)]_[3]=0. Where

7-2°) the number U=A+B-C=un³, so NOW k=3, and we compose the source data for the next cycle (increasing k by 1):

3a-2°) A_[3]=aⁿⁿ_[3]=a'ⁿⁿ_[3], B_[3]=bⁿⁿ_[3]=a'ⁿⁿ_[3], C_[3]=cⁿⁿ_[3]=a'ⁿⁿ_[3]; consequently (see 4°),

3b-2°) Aⁿ_[4]=a'ⁿⁿⁿ_[4], Bⁿ_[4]=b'ⁿⁿⁿ_[4], Cⁿ_[4]=c'ⁿⁿⁿ_[4]; consequently (see 1°),

3c-2°) (Aⁿ_[4]+Bⁿ_[4]-Cⁿ_[4])_[4]=a'ⁿⁿⁿ_[4]+b'ⁿⁿⁿ_[4]-c'ⁿⁿⁿ_[4]=0.

[And if, for example, B_[2]=0, then (C-A)_[kn-1]=0 and from 1a° we find that

2B_[3]=0 и U_[3]=0.]

Then we repeat the reasoning of 6°-7° with obtaining k=4 and go to the next cycle. And so on to infinity. Finally, the end of the numbers A, B, C take the following form:

8°) A_[k+1]=a'^{n^k}_[k+1], B_[k+1]=b'^{n^k}_[k+1], C_[k+1]=c'^{n^k}_[k+1], where k tends to infinity,

that indicates the impossibility of the equality of 1° and of the truth of the FLT.

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Victor Sorokine. Mezos. 5.5.2017

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Control text in Word, see: http://rm.pp.net.ua/

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