Fermat's Theorem: Equality (for prime degree n>2)
1*) aⁿ+bⁿ-cⁿ=0 in positive integers a, b, c does not exist.

The notation and lemmas /Pour les preuves des lemmes, voir l'annexe
https://vixra.org/pdf/1707.0410v1.pdf /

a’, a’’; a’’’ - 1st, 2nd, 3rd digit from the end in the number a in the number system with a prime base n>2;
a_[2], a_[3], a_[4] - two-, three-, four-digit ending of the number a;
nn - n*n.

L1. If digit a' is not 0, then (a^n-1)’=1. (Fermat's little theorem.)

L1a. Therefore: (a^n-1)ⁿ_[2]=01, (a^n-1)ⁿⁿ_[3]=001.

L2 (key!). There is such a digit d that the second digit  (dⁿ)'' is not zero. [ Indeed, if all second digits are equal to zero, then the second digit of the sum of the number series dⁿ, where d = 1, 2, ... n-1, is not zero and is equal to (n-1)/2, which is incorrect. ]

L3. For k>1, the k-th digit in the number aⁿ does not depend on the k-th digit of the base a.
L3a. Consequence. If a’ is not equal to 0, then digits aⁿ_[2] and aⁿⁿ_[3] are functions of only a’ and do not depend on the digits of higher ranks.

2*) In Fermat's equality 1* two-digit endings of numbers a, b, c, not multiples of n, there are two-digit endings of degrees a’ⁿ, b’ⁿ, c’ⁿ.

Therefore, the number a (like b and c) can be represented as a=a’n+An², where A=(a-a_[2])/n², and the number aⁿ (and bⁿ and cⁿ) can be represented as

3*) aⁿ=(a’ⁿ+An²)ⁿ=a’ⁿⁿ+An³*a’^n(n-1)+An⁵*a’^n(n-2)+...,  (and similarly bⁿ=... и cⁿ=...),

where [(A’+B’-C’)/n³]_[2] = -[(aⁿ+bⁿ-cⁿ)/n³]_[2] and [insofar as (a^n-1)’=(b^n-1)’=(c^n-1)’=1]

a’^n(n-1)_[2]=b’^n(n-1)_[2]=c’^n(n-1)_[2]=01 . 

And now the equality 1 * can be written by five-digit endings in the form:

4*) (a’ⁿⁿ+b’ⁿⁿ-c’ⁿⁿ)_[5] + (a+b-c)_[2]n³ + Dn⁵ = 0.

L4. If in the equality 1* the number a ends, for example, with k zeros (k is always greater than 1!), then by multiplying the equality by some number gⁿⁿⁿ one can convert the ending of the number b (or c) of length kn+5 digits into 1.

And now the very PROOF of Fermat's theorem.

5*) Multiply equalities 1* and, accordingly, 4* by the number dⁿ  from L.2.

And we see that the two-digit ending of the number (a+b-c)_[2] multiplied by the single-digit number d, and the two-digit ending of the number [(a’ⁿⁿ+b’ⁿⁿ-c’ⁿⁿ)/n³]_[2] - EQUAL IN VALUE (but with the opposite sign) - multiplied by the two-digit ending of the number dⁿ with a non-zero second digit. And, therefore, the equivalent equality 4* turned into INEQUALITY.

The second case (for example, the number a ends in k zeros) is proved similarly and even somewhat easier.

After converting the (kn+5)-digit ending of b into 1, we obtain the equality of the three-digit ending of the significant part of the power aⁿ to the three-digit ending of the base of the number cⁿ without the unit (kn)-digit ending. And now, after multiplying Fermat's equality by dⁿ (out of 5 *), the two-digit ending of the number c will be multiplied by a single-digit d, and the two-digit ending of the number aⁿ with the EQUAL ending will be multiplied by the two-digit ending of the number dⁿ with an equal last digit (dⁿ)'  [ ...=d' ] but with positive d'', thus turning equality into an equivalent inequality.

This proves the truth of Fermat's great theorem for a prime degree.
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Victor Sorokine
03.09.2020. Mézos, France.
victor.sorokine2@gmail.com