Алексей Виноградов – Fermat's Last Theorem (Conditions and decisions) (страница 2)

At the same time, it becomes unclear: either the validity of Taniyama’s conjecture is being proven using Fermat’s unproven theorem, or Fermat’s theorem is being proven using the unproven Taniyama-Shimura-Weil conjecture. The absence of a relationship between exponents of powers n > 2 and powers n = 1 and 2 raises doubts; the distribution of the conditions of Fermat’s theorem over the XOY plane and, in particular, to negative integers is not shown.

The conclusion about the truth of Fermat's equation is made on incorrect postulates.

First, it is assumed that there is some solution to Fermat's equation in positive integers.

Secondly, this solution is arbitrarily inserted into an algebraic form of a known form (a plane curve of degree 3) under the assumption that the elliptic curves thus obtained exist (the second unconfirmed assumption).

Thirdly, since other methods prove that the particular curve constructed is not modular, it means that it does not exist. This leads to the conclusion: there is no integer solution to Fermat’s equation and, therefore, the theorem is true.

There is a discrepancy between Fermat's equations and algebraic (elliptic) curves of the 3rd order. Frey proposed that the supposed elliptic curve representing the hypothetical integer solution of Fermat's equation could not be modular.

Wiles decided that every semistable elliptic curve defined over the field of rational numbers is modular. This led to the conclusion about the impossibility of integer solutions of Fermat’s equation and, consequently, about the validity of Fermat’s statement. He refers to the work of I. Elleguarche, in which he found a way to associate a hypothetical third-order curve with Fermat's equation. The author conducts his reasoning in terms of projective geometry. The curve was a priori perceived as elliptical. However, the segments of the Euclidean line, when adding points on it, are taken on a non-linear scale, and the curve is not actually elliptic.

An appeal in this case to derive the Fermat equation seems illegitimate. Despite the fact that it satisfies some criteria for the class of elliptic curves, it does not satisfy the main criterion of being an equation of the 3rd degree in a linear coordinate system.

To date, there is not a single scientific objection to the recognition of the fallacy of Wiles's proof.

4. Fermat's Last Theorem and Soviet Science

It is believed that for some reason they did not try to solve Fermat’s Last Theorem in the USSR. This is not true. At one time, it became clear that in order to review a solution, a written review from a competent reviewer was required, and the review fell within the competence of the Order of Lenin Mathematical Institute. V.A. Steklov of the USSR Academy of Sciences (MIAN). And in the presence of such support (Doctor of Mathematical Sciences), it turned out that this problem does not relate to computational mathematics and in general the Steklov Mathematical Institute “has never advised anyone and does not advise anyone to deal with the Fermat problem,” and its “Scientific Council decided not to consider any more materials devoted to this problem ". We are not looking at the problem, and the problem is gone!

From the author's archive

True, in Soviet times they at least answered, but in the new Russia they no longer do this. It is not surprising that in the end Russian science has reached the modern level, for which it is ready to blame anyone but itself.

From the author's archive

However, the problem of proving Fermat's Last Theorem was considered not only by academicians Postnikov and Fomenko, but also by less titled people. Let us note that, in the opinion of «Izvestia ANSSSR», most, not all, of the proofs of Fermat’s Last Theorem that came to them turned out to be incorrect. And a minority, clearly more than one, of the decisions were correct. However, Soviet science, being deeply secondary to the West, did not allow itself to get ahead of it in anything.

From the author's archive

Known published proofs do not match the dimensions of Pierre Fermat's proof, and in most cases they use mathematical apparatus developed after 1680. We will not argue how correct these constructions turned out to be (we will present them further).

5. Fermat's Last Theorem - proof options

Some decisions were published in the republics of the USSR. Although the mathematical journals of the Russian Academy of Sciences, «Algebra and Analysis», for example, do not consider elementary proofs of Fermat’s theorem, other departments of the Academy of Sciences were not prohibited from publishing. As far as is known, other decisions were published in the USSR.

It is impossible to determine how correct they are, because inaccuracies may have been made during publication.

1. First proof 1993-2012

Bobrov A.V. in 1993-2012 wrote a series of proofs of Fermat’s theorem. Option No. 1 was published in the journal of the Russian Academy of Sciences “Questions in the history of natural science and technology” No. 3, 1993.

Fermat's Last Theorem - two short proofs

Fermat's last theorem, sometimes called the Great Theorem, is formulated as follows: in the equality aⁿ + bⁿ = cⁿ, the numbers aⁿ, bⁿ, cⁿ and n cannot simultaneously be positive integers if n > 2.

Let's assume such numbers exist. Then the following conditions must be met:

The equality is valid for relatively prime numbers a, b and c that do not have common integer factors other than 1, i.e. two numbers are always odd.

There are numbers d₁= c - b and d₂ = c - a, or d₁ - d₂ = a - b, that is, for arbitrarily chosen natural numbers a > b there is an infinite set of rational, real or complex numbers d₁ and d₂ satisfying the above equality, if Arithmetic operations are feasible in this set. For integers c, the numbers d₁ and d₂ will also be integers.

Option 1

The equality cⁿ - bⁿ - aⁿ = 0 (1) by sequential division by numbers d₁ and d₂ is always transformed into two polynomials (equations) (n-1)th degree with respect to c:

(cⁿ - bⁿ)/ d₁ - aⁿ/ d₁ = c^n-1+ c^n-2 x b+…+c x bn-2 + b^n-1 - (aⁿ/ d₁) = 0 (2)

(cⁿ - aⁿ)/ d₂ - bⁿ/ d₂ = c^n-1+ c^n-2 x a+…+c x an-2 + a^n-1 - (bⁿ/ d₂) = 0 (3)

Equalities (2) and (3) were obtained by identical transformations of equality (1), i.e. must be executed for the same values of positive integers a, b and c. By definition, a necessary and sufficient condition for the identity of two polynomials over a certain number field (in our case, over a set of integers) is the equality of the coefficients of terms containing the same arguments in the same powers, that is, the following must be true:

a = b, a² = b², … a^n-2 = b^n-2, (aⁿ/ d₁) – b^n-1 = (bⁿ / d₂) – a^n-1 (4)

From (1) and (4) it follows that cⁿ = 2aⁿ, c =aⁿ√2, that is, the number c, as the common arithmetic root of equations (1), (2) and (3), cannot be rational for integer a, b. d₁ and d₂.

From the equality of free members it follows:

(aⁿ – (b^n-1 x c) + bⁿ ) / (c-b) = (bⁿ – (a^n-1 x c) + aⁿ ) / (c-a), or

(c x (c^n-1 - b^n-1) / (c-b) = (c x (c^n-1 - a^n-1) / (c-a), or

c^n-2 + с^n-3 x b +…+ с x b^n-3 +bn-2 = c^n-2 + с^n-3 x a +…+ с x a^n-3 +an-2 (5)

Subtracting the left side from the right side of equality (5), we obtain:

c^n-3 (a - b)+ c^n-4 x (a² – b² ) +... + c (a^n-3 - b^n-3) + (a^n-2 – b^n-2 ) = 0 (6)

or, if a ≠ b, reducing by a – b ≠ 0, we get:

c^n-3 + c^n-4 x (a + b ) +... + c (a^n-4 +... + b^n-4) +(a^n-3 +... + b^n-3 ) = 0 (7)

From equality (7) it follows that for a ≠ b the numbers a, b and c cannot be simultaneously positive.

The presented transformations allow us to draw the following conclusions:

- for polynomials (2) and (3) that are identical over the set of rational numbers for n > 2, the number c is the common arithmetic root of equations (1), (2) and (3), cannot be rational for positive integers a , b , d₁ and d₂;

- polynomials (2) and (3) for n > 2 and natural a and b are not identical over the set of rational numbers if the divisors d₁ and d₂ of equality (1) are irrational, which implies that the number c is irrational;

- the numbers a, b and c in equality (1) for n > 2 cannot be rational at the same time.

For n ≤ 2, the contradiction disappears, the coefficients of c^n-1 are equal to 1, and the equality of free terms after substituting the values d₁ and d₂ becomes an identity:

(a² – bc + b² ) / c – b =(b² – ac + a² ) / c – a (8)

If the right and left sides of equality (5) are denoted respectively by q₁ and q, where q₁ and q are positive integers, then polynomials (2) and (3) are transformed into quadratic equations with respect to c:

q x² – (b q - b ^n-1) x – (aⁿ + bⁿ ) = 0

q₁ x² – (a q₁ - a ^n-1) x – (aⁿ + bⁿ ) = 0 (9)

where the unknown c is denoted in the generally accepted way by x, that is, x = c

The same conclusions follow from the conditions of equivalence or analysis of the reasons for the non-equivalence of these equations.

Option 2

Let in the equality cⁿ – bⁿ = aⁿ the numbers a, b and c be relatively prime, a ≠ 1 - be an odd number. For any positive numbers, we can perform the operation of finding the arithmetic value of the square root, that is, we can write: