Fermat's Last Theorem (Conditions and decisions)

Fermat's Last Theorem

(Conditions and decisions)

The book of the outstanding scientist A. G. Vinogradov is devoted to the problem of solving some indeterminate equations. It is known that at present the search for an elementary solution to Fermat's Last Theorem has been unsuccessful and cannot be considered completed. This work answers this question. The book provides possible evidence made by researcher

Pierre de Fermat

The French lawyer and mathematician Pierre de Fermat or Petri de Fermat (1601–1665), with his work had a great influence on the further development of mathematics.

Pierre Fermat is one of the founders of number theory. He took this important step in his works on the largest and smallest quantities, which opened a series of studies by Fermat, which is one of the largest links in the history of the development of not only higher analysis in general, but also the analysis of infinitesimals in particular.

Fermat made the following statement: if a number a is not divisible by a prime number p, then there is an exponent k such that a-1 is divisible by p, and k is a divisor of p-1. This statement is called Fermat's little theorem. It is fundamental in all elementary number theory.

Pierre Fermat, according to the rules accepted in mathematics of the 19th-21st centuries, found tangents to algebraic curves.

In mathematical analysis, Fermat's lemma or the necessary criterion for an extremum is used: at extremum points, the derivative of a function is equal to zero.

Fermat developed a method for systematically finding all divisors of a number and formulated a theorem on the possibility of representing an arbitrary number by a sum of no more than four squares.

In the field of the infinitesimal method, he systematically studied the process of differentiation, gave a general law for differentiation of powers, and applied this law to the differentiation of fractional powers. In the preparation of modern methods of differential calculus, his creation of the rule for finding extrema was of great importance.

P. Fermat formulated the general law of differentiation of fractional powers and extended the formula for integrating powers to the cases of fractional and negative exponents. Pierre Fermat developed the foundations of probability theory. In Fermat's works, both basic processes of the infinitesimal method were systematically developed, but he ignored the connection between the operations of differentiation and integration. This connection was made later.

Fermat was the first to come up with the idea of coordinates and created analytical geometry, introducing an infinitesimal quantity. He solved the problem of squaring any curve, and on this basis solved a number of problems on finding the centers of gravity. In the work “Introduction to the Theory of Plane and Spatial Places,” he was the first to classify curves depending on the order of their equation, establishing that a first-order equation defines a straight line, and a second-order equation defines a conic section. Developing these ideas, he applied analytical geometry to space.

In the field of physics, Fermat is associated with the establishment of the basic principle of geometric optics, by virtue of which light in an inhomogeneous medium chooses the path that takes the least time (Fermat believed that the speed of light is infinite and formulated the principle more vaguely). With this thesis begins the history of the main law of physics – the principle of least action.

The first collected works of P. Fermat, “Various Works,” was published in 1679.

The problem of dividing a square into the sum of two squares

P. Fermat often wrote down not proofs, but only brief instructions about the method he used.

Page from the book “Arithmetic” by Diophantus 1670

In one of Diophantus's books, the problem of dividing a square into the sum of two squares was considered, and he meant the squares of positive numbers. In the margins of the book, Fermat wrote:

«Cubum      autem      in      duos      cubos,      aut      quadrato-quadratum in duos      quadrato-quadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duas ejusdem nominis fas est dividere; cujus rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet».

«"But the cube into two cubes, or a square-square into two square-squares, and in general it is right to divide any power infinitely beyond the square into two of the same name; Of course I discovered a wonderful demonstration of this matter. The smallness of the margin would not take him».

There was no general proof in Fermat's papers. He published a proof only for the case n = 4. A proof for n = 3 was given in 1768, for n = 5 in 1823, for n = 7 in 1837, for n = 67 in 1851, for powers up to 2521 in 1955, for degrees up to 4002 in 1966, for degrees up to 100000 in 1977.

And finally, at the end of the 20th century, through enormous efforts and calculations, the theorem was allegedly proven. Andrew John Wiles discovered a technical method that, with the help of Richard Taylor, allowed him to complete the proof in 1994. Wiles's proof, containing 129 pages, was published in the «Annals of Mathematics» in 1995.

Without touching on the fallacy of this solution, it can be noted that it cannot be the one sought by Pierre Fermat in the 17th century. The margins of the book were not wide enough for his entry, but still, it could fit on wider margins or a separate sheet, and not on 129 pages. The width of the book margin is 5 cm, the longest length is 63 cm (edge). Thus, the proof occupied 315 – 450 sq. cm of handwritten font, less than an A 4 page (623.7 sq. cm), while some of the characters (integral, etc.) were replaced by Latin words.

Maximum field width of the book of Diophantus

Wiles–Taylor proof

At the end of the 20th century, through enormous efforts and calculations, Fermat's theorem was allegedly proven.

Andrew John Wiles discovered a technical method that, with the help of Richard Taylor, allowed him to complete the proof in 1994. Wiles's proof, containing 129 pages, was published in the «Annals of Mathematics» in 1995.

Meanwhile, it was not Fermat’s Last Theorem that was proven, but part of the Taniyama-Shimura-Weil theorem (the statement states that every elliptic curve over the field of algebraic numbers is automorphic), a frequent case of which some mathematicians considered Fermat’s Last Theorem. In the process of reasoning, Fermat's original statement was reformulated in terms of comparing a Diophantine equation of the p – degree with elliptic curves of the 3rd order. (That is, no one was going to prove the original theorem). This comparison forced the authors of the proof to announce that their method and reasoning lead to a final solution to Fermat's problem.

Fermat's Last Theorem

In 1999, Christophe Broglie, Brian Conrad, Fred Diamond, and Richard Taylor proved unstable cases of the Taniyama-Shimura-Weil theorem. Although they believe that they have proven the Taniyama-Shimura-Weil theorem, the scope of Wiles' method is limited. It only works for elliptic curves over rational numbers, while the Taniyama-Shimura-Weil conjecture covers elliptic curves over any algebraic number field. Therefore, it is assumed that there is a more general and more elegant proof of the modularity of elliptic curves.

Wiles considered the most general case of the Taniyama-Shimura-Weil theorem and proved this theorem for all semistable elliptic curves over the field of rational numbers in 1995, which he and his colleagues consider sufficient to prove Fermat's Last Theorem. In March 1996, Wiles and Robert Langlands received recognition from one of the institutions as the authors of the proof. Although neither of them completely proved the theorem, they were said to have made significant contributions that greatly facilitated further proof, which in 1996 the mathematical community believed they did not yet have.

Also, the proof of Fermat’s Last Theorem based on Taniyama’s theorem was attributed to 1993, 1995, 1998 or 2001. At the end of 2007, the scientific society “Göttingen Academy of Sciences” turned to some anonymous mathematical authorities and, having received from them good, awarded the prize for proving Fermat's Last Theorem to Wiles, and liquidated the bank account. The scientific community never learned the names of the mathematical authorities; out of modesty, they themselves did not tell us about themselves. As of 2024, no mathematician has ever held a published proof of Fermat's Last Theorem (based on Taniyama's theorem).

Wiles's work on the theorem was designed as a "proof without proof", that is, the proof is not direct and immediate. Wiles began working on Fermat's theorem after Ken Ribet showed the connection between semistable elliptic curves (a special case of the Taniyama-Shimura-Weil theorem) and Fermat's theorem. The main idea about such a connection belongs to the German mathematician Gerhard Frey. Fry suggested that the Taniyama-Shimura-Weil conjecture is a generalization of Fermat's Last Theorem, because any counterexample to Fermat's Last Theorem resulted in a non-modular elliptic curve.

Wiles in his proof proceeds from the fact that Fermat's theorem is a consequence of Taniyama's conjecture about modular elliptic formations. This conclusion was made on the basis of a limited number of points x, y, z from Fermat’s theorem, which allow the author to assert that these points characterize all combinations of x, y, z and n as being involved in modular elliptic curves.

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.

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!

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.

However, the problem of proving Fermat's Last Theorem was considered not only by academicians Postnikov and Fomenko, but also by less h2d 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.

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).

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.

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:

(c^n/2 + b^n/2) x (c^n/2 – b^n/2) = a₁ⁿ x a₂ⁿ = aⁿ (1)

Where a₁ⁿ = c^n/2 + b^n/2 , a₂ⁿ = c^n/2 – bⁿ/2 are real positive factors of the number aⁿ.

From (1) it follows: cⁿ = (aⁿ₁ + aⁿ₂)/2)², bⁿ = (aⁿ₁ – aⁿ₂)/2)² (2)

In accordance with the properties of the exponential function, for real positive numbers a₁ⁿ , a₂ⁿ and integer a ≠ 1, there are unique values of the exponents – <∞ q < p <+∞,, satisfying the equalities:

a ⁿ₁ = a ^p, a₂ⁿ = a ^q (3) , где a = a₁ a₂, n = p + q.

From (3) it follows a₁^p a₁^q = a₁^pa₂^p, a₂^p a₂^q = a₁^qa₂^q , or after reduction by numbers

a₁^p ≠ 0 , a₂^q ≠ 0 получим: a₁ ^q = a₂ ^p (4)

From (1), (2) and (3) it follows:

cⁿ – bⁿ = (a^p +a^q)/2)² – (a^p -a^q)/2)² = a^p+q =an (5)

or, taking into account equalities (3) and (4):

(a₁ ^p * a₁ ^q +a₁ ^q * a ₁^q)/2)² – (a₁ ^p * a₁ ^q – a₁ ^q * a ₁^q)/2)² = a₁ ^p * a₁ ^2q * a₂ ^q (6)

Let's take the common factor a₁^q out of brackets:

a₁ ^2q * (a₁ ^p +a₂ ^q )/2)² – a₁ ^2q * (a₁ ^p – a₂^q )/2)² = a₁ ^p*a₁ ^2q * a₂ ^q (7)

From (5) and (7) it follows that the numbers cⁿ , bⁿ and aⁿ contain a common factor a₁ ^2q = a₂^{2 p} ≠ 0, which contradicts the condition of their mutual simplicity, if a₁ ^2q = a₂^{2 p} ≠ 1. From a₁ ^2q = a₂^{2 p} = 1 it follows q = 0 , a₂ = 1, that is, n = p , a = a₁ *1, and equalities (5) and (7) take the form: cⁿ – bⁿ = (aⁿ +1 )/2)² – (aⁿ -1 )/2)² = aⁿ (8)

From (8) it follows that for odd a ≠ 1 the numbers cⁿ and bⁿ are also integers, and the identity always holds:

c^n/2 – b^n/2 = 1 (9)

that for simultaneously integer b, c and n is feasible only for n/2 ≤ 1, n = 1, 2, which is what needed to be proved.

The proof can be carried out in a slightly different way. All numbers of the equality aⁿ + bⁿ = cⁿ, where a ≠ 1, b and n are arbitrarily chosen natural numbers, c is a real positive number, through transformations (1)…(4) can be expressed as terms of identity (5).

Let us take the factor a^q ≠ 0 out of brackets and divide all terms of identity (5) by it:

(a^k +1)/2)² – (a^k – 1)/2)² = ak (10) where k = p – q.

In accordance with the properties of the exponential function, arbitrarily chosen natural numbers a, b and n, for example, from equality (5), correspond to a single value k satisfying the condition:

bⁿ = (a^k – 1)/2)² (11) then cⁿ = bⁿ + a^k = (a^k + 1)/2)², or cⁿ – bⁿ = a^k (12) where a, b and n are integers.

From (10), (11) and (12) it follows: : c^{n/ 2} – b ^n/2 = 1 (13) that is, the numbers b and c can be simultaneously integers only when n/2 ≤ 1, or n = 1, 2 . For n = 2, b and c are consecutive integers. Euclid also proved that any odd number can be expressed as the difference of the squares of two consecutive integers, which can be found using identity (10) for any integer k and odd a ≠ 1.

Note that equality (12) was obtained by dividing equality (5) by the factor a ^2q, while the number bⁿ in these equalities is the same, which ^2q = 1, q = 0 , p = k = n, and the identity (10) takes the form of identity (8).

Note also that identities (8) and (10) are valid not only for integer values of a. By substituting any rational fraction for a and setting k = n = 2, you can find all Pythagorean numbers.

The above transformations of the Fermat equality over the set of natural numbers show that with the help of a finite number of arithmetic operations it is always reduced to identity (13), which proves the theorem.

Proof of Fermat's theorem using elementary algebra methods

Later, Bobrov A.V. published another proof of Fermat’s Theorem.

Fermat's theorem states that the equality aⁿ + bⁿ = cⁿ for natural a, b, n can only hold for integers n ≤ 2.

Consider the equality C – B = A, (1) where A, B, C are coprime natural numbers, that is, numbers that do not have common integer factors other than 1. In this case, two numbers are always odd. Let –A = aⁿ – an odd number, a ≠ 1 and n – natural numbers. For any real positive number, we can perform the operation of finding the arithmetic value of the root, that is, equality (1) can be written as:

(√C + √B) (√C –√B) = A₁ A₂ = A, (2) where A₁ = √C+√B и A₂ =√C-√B -

real positive factors of a number In accordance with the properties of the exponential function, for any of the real positive numbers A₁ and A₂ there are unique values of the numbers 0< q< p < ∞ satisfying the equalities A₁ = a ^p A₂ = a ^q, (3)

From equalities (2) and (3) it follows:

√C = (a^p + a^q)/2, √B = (a^p – a^q)/2, A = a ^p+q = a ⁿ (4)

Since p > q, always takes place p-q=k,, or а^p= а^k × а^q, that is, the numbers A,B,C contain a common factor, which contradicts the condition of their mutual you just. This condition is fulfilled only when a^q = 1, that is, when q= 0, p = n.

Then equalities (4) take the form:

√C = (aⁿ + 1)/2, √B = (aⁿ – 1)/2, A = a ⁿ (5)

whence follows

√C - √B = 1 (6)

that is, for coprime A,B,C, the numbers c =√C, b = √B are always two consecutive integers. Euclid also proved that every odd number is expressed as the difference of the squares of two consecutive integers, that is, equality (1) for natural coprime a, b, c can only be expressed as the equality c² -b² = a². (7)

The validity of the above proof can be illustrated by the following example.

Let the numbers a,b,c in Fermat’s equality be coprime integers, and let n be even. Then the numbers c^n/2, b^n/2, their sum A₁ = a^p and difference A₂ = a^q are also integers, the exponent p>q.

Integers c^n/2 = ( a^k+q + a^q)/2, b^n/2 = (a^k+q – a^q)/2 are coprime if they do not contain common integer factors other than 1. This condition is satisfied only if the common integer factor is a^q = 1, that is, q = 0, p = n.

Then the difference c^n/2 - b^n/2 = 1, which for simultaneously integer b,c,n can only occur when n/2 ≤ 1, that is, when n = 1, 2.

Second proof 2001-2004

In 2001-2004, V. A. Gorbunov published a simple proof of Fermat’s last theorem at MSTU.

Proof of Fermat's Last Theorem

Fermat's theorem states that there are no integer solutions to the equation

xⁿ + yⁿ = zⁿ (1) for any natural number n > 2 (xyz ≠0).

Proof

By introducing the transformations u = x/z, v = y/z, (2) equation (1) is reduced to the form uⁿ + v ⁿ = 1. (3)

It is obvious that the variables u and v satisfy the inequalities 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1. (4)

Let us denote the sum uⁿ + vⁿ by S_n , and call the graph of the equation S_n = 1 the Fermat curve of the n-order. Figure 1 shows a square containing: a unit circle S₂ =1 and a Fermat curve S_n =1 (n ≥ 3). Due to the symmetry of equation (3) with respect to the variables u and v, in what follows we will consider only the upper part of the square with respect to the secondary diagonal (matrix terminology). For any natural number z (z > 1), we apply a uniform grid on the unit square with a rational step h = z/2, Figure 1. The larger z, the smaller h, that is, the denser the grid in the unit square. The statement of Fermat's theorem is equivalent to the fact that whatever the number z (and, consequently, h), Fermat curves do not pass through the nodal (rational) points in the unit square, which is what remains to be proved.

Fig.1. Unit circle S₂ =1 and Fermat curve S_n = 1 (n =3,4,…) in a unit square covered with a uniform grid with a step h =1/z.

It is known that there is an infinite number of rational points on the unit circle. If, when dividing a unit square for a natural number z₀, the rational point of the unit circle P₀ (x₀/z₀, y₀/z₀) turns out to be a nodal point on the mesh, then we will call this point Pythagorean, and the ray passing through it (coming from the origin) will be called Pythagorean beam ƛ₀.

If the Pythagorean number z₀ is prime, then at the top of the unit square on the unit circle there will be one Pythagorean point on the grid for z₀.

1. Any point in the unit square M₀ (x₀/z₁, y₀/z₂) with rational coordinates can become a node on the grid for a natural number z = z₁ * z₂, if NOD (z₁ * z₂) = 1= 1. Thus, in what follows about a rational point, we will assume that the denominators of the coordinates of the point are the same.

For composite numbers of the form z = k * p₁* p₂ *…*p_l,, where p₁ p₂ ,…, p_l are prime numbers of the form 4n+1, and k is a natural number whose factorization does not contain Pythagorean numbers, then in the upper part of the unit square there will be l² Pythagorean points on the unit circle for a grid with a step h = 1/ z.

Let us denote by ƛ the ray forming an angle ƛ with the Ou axis. In addition to the designation, we will give a numerical value to the parameter ƛ/ƛ = tgȹ. Then the coordinates of the points on this ray are related by the relation v = ƛ*u. Substituting instead of v into equation (3) ƛ*u, we obtain the coordinates of the point Ф_n lying at the intersection of the ray ƛ with the Fermat curve S_n =1:

u_n = 1/ ⁿ√(1+ ƛ ⁿ), v_n = ƛ / ⁿ√(1+ ƛ ⁿ) (5)

In what follows, we will be interested only in rays passing through nodal (rational) grid points in a unit square. On such rays the parameter ƛ takes rational values. We will distinguish between rays ƛ ₀ passing through rational points on the unit circle and rays ƛ^* passing through rational points in the unit square and not lying (points) on the unit circle.

Let us prove a lemma that will be useful to us later.

Lemma.

1). If the rational point Ф* ( x^*/z₀, y^*/z₀) lies on the ray ƛ ₀, then the length of the radius vector R^* of this point is a rational number, (x^*, y^* are natural numbers).

2). If the rational point Ф* ( x^*/z, y^*/z) lies on the ray ƛ_*, then the length of the radius vector R^* is an irrational number (x^*, y^*, z^* are natural numbers).

Let us prove the first part of the lemma.

Let P₀ (x₀/z₀, y₀/z₀) be a point with rational coordinates that belongs to the unit circle (P₀ ϵ S₂ =1). Then the coordinates of this point satisfy the equation x²₀ + y²₀ = z²₀, (6) where x₀, y₀, z₀ are natural numbers. Let the point Ф^* ( x^*/z₀, y^*/z₀) with rational coordinates lie on the ray ƛ ₀ passing through the point P₀, (Ф^* ϵ S₂ =1).

Then ƛ 0 = y₀/ x₀ = y*/ x* and the length of the radius vector of point Ф* will be equal to R* = √((x*/z₀)² + (y*/z₀)²) = x*/ z₀ √(1+ ƛ ₀²) = x*/ z₀ √(1+(y₀/x₀)²) = x*/ x₀ – rational number.

2. The proof of this formula is given in Appendix 2.

Let us prove the second part of the lemma.

Let the point Ф*(x*/z*, y*/z*) with rational coordinates lie on the ray ƛ ^* which intersects the unit circle at the irrational point P* (u* ,v*), u*= m/k*, v* = n/k*.. Since ƛ ^* = y*/x* is a rational number, then the numerators of the fractions u* , v* are rational numbers: ƛ ^* = n/m. Whence it follows that the denominator k* = √(m²+n²) is an irrational number. The length of the radius vector of point Ф* will be equal to

R* = √( (x*/z*)² + (y*/z*)²) = x*/z* √(1+(y*/x*)²) = x*/z* √(1+(n/m)²) = x*k*/z*m is an irrational number.

Thus, the length of the radius vector of a rational point on the ray ƛ₀ (passing through the rational point of the unit circle) is a rational number, and the length of the radius vector of a rational point lying on the ray ƛ^* (intersecting the unit circle at the irrational point) is the number irrational.Докажем вторую часть леммы.

Let x₀, y₀, z₀ be a Pythagorean triple of numbers with pairwise coprime numbers. Such Pythagorean triplets of numbers are called primitive. Then on the grid in the unit square for the natural number z₀ on the unit circle there is a Pythagorean point P₀ (u₀, v₀), where u₀ = x₀/ z₀, v₀ = y₀/ z₀. The ray ƛ₀ passing through the point P₀ does not contain other grid nodes for z₀. Let's prove it.

Let the nodal point Ф*(x*/z₀, y*/z₀) of the grid for z₀ lie on the ray ƛ ₀. Then y* = ƛ ₀ x* = (y₀/x₀) x* (7)

Since x₀ and y₀ are relatively prime numbers, then y* will be an integer if x* is a multiple of x₀, that is, x x* = m x₀ where m is a natural number, m > 1. Then y* = my₀ the coordinates of the point Ф* will be: u* = m x₀ / z₀, v* = m y₀ / z₀. It follows that the value of the radius vector of the point Ф* will be R* = m, where m is an integer. Since the largest value of the length of the radius vector of any point inside the unit square is less than √2, the point will be outside the unit square.

So, if the Pythagorean triple of numbers ( x₀ , y₀ , z₀ ) is primitive, then on the Pythagorean ray ƛ₀, except for the Pythagorean point P₀ on the unit circle, there are no other nodal points of the grid. Then on the grid in the unit square for a natural number z₀, all Fermat curves intersect the ray ƛ₀ not at nodal points and, therefore, according to equation (1), there will be no integer solutions at these points. Then the transformed equation (3) will have irrational solutions for all n >2:

u_n = 1 /ⁿ√(1+ ƛ ₀ⁿ ) = x₀ / ⁿ√( x₀ⁿ +y₀ⁿ ), v_n = ƛ ₀ /ⁿ√(1+ ƛ ₀ⁿ ) = y₀ / ⁿ√( x₀ⁿ +y₀ⁿ )

Note that according to formula (8), the coordinates of the points of the Fermat curves on the ray ƛ ₀ are formed from the coordinates of the Pythagorean point P₀ (x₀ / z₀, y₀ / z₀) and, therefore, do not depend on other partitions of the unit square (obviously, the coordinates of the points do not depend at all on the partition of the unit square).

A comment.

On the one hand, the lemma states that on the Pythagorean ray ƛ ₀ the length of the radius vector of the point Ф* with rational coordinates R* is a rational number. On the other hand, on a grid for z₀ on which the point P₀ on the unit circle is a nodal point (that is, the unit square for a given partition is integerly decomposed over the cells of the partition into the sum of two squares), there are no other nodal points on the ray ƛ ₀ and we draw the conclusion (which has yet to be formally proven) that all Fermat curves intersect the Pythagorean ray ƛ ₀ at irrational points. Let's give an example.

Example 1.

z₀ = 5, P₀ (3/5 , 4/5), ƛ ₀ = 4/3. With such a partition (partition step h =1/5), the unit square can be decomposed into the sum of two squares with sides 3h and 4h. It is obvious that there is an infinite number of rational points on the ray, for example Ф* (1/2, 2/3). On a grid in a unit square for z = 6, this point will be a nodal point and lie on the ray ƛ₀, Ф* (3/6, 4/6). The radius vector of this point is a rational number, R* =5/6. If we combine these two grids, that is, cover a unit square with a grid z₁ = z₀ , z = 30, then both points P₀ (18/30, 24/30) and Ф* (20/30, 15/30) will be nodal points on this grid and are on the same ray ƛ₀. The partitioning step of this grid is h = 1/30 and point P₀ (18/30, 24/30) still splits the unit square into the sum of two squares with sides 18h and 24h.

In other words, let us make the assumption that some rational point Ф*_n lies at the intersection of the Fermat curve S_n = 1 with the Pythagorean ray ƛ₀, then we can always select a grid in the unit square in which both P₀ and Ф*_n will be nodal at the same time.

Let's return to the proof of the theorem. Since the numerators of the coordinates of the point Ф*_n (u_n ,v_n), (formulas (8)), are natural numbers, then solutions (8) will be irrational if we prove that ⁿ√ ( xⁿ₀+yⁿ₀) for n > 2 is an irrational number . Let's prove it.

Let for some natural number n, (n > 2 ), ⁿ√ ( xⁿ₀+yⁿ₀) = z¹₀ /k – a rational number, where z¹₀ and k are coprime natural numbers and the point Ф*_n (kx₀ / z¹0 , ky₀ / z¹₀ ) lies at the intersection of the Fermat curve S_n = 1 with the Pythagorean ray ƛ₀. It is obvious that z¹₀ /k < z₀, since the length of the radius vector of the point Ф_n is greater than one, (Fig. 1)

R_n = √( (kx₀ / z¹₀ )² + (ky₀ / z¹₀)²) = kz₀ / z¹₀ > 1 (9)

From the assumption that the point Ф _n with rational coordinates lies on the Fermat curve S_n = 1, it follows that equation (1) must have an integer solution (kx₀)ⁿ + (ky₀)ⁿ = (z¹₀)_n (10)

From the last equation it follows that z¹₀ must be divisible by k. But, by assumption, z¹₀ and k are relatively prime numbers. Therefore, ⁿ√( (xⁿ₀ + yⁿ₀ ) cannot be a fractional rational number.

Let now ⁿ√(xⁿ₀+ yⁿ₀)be a natural number and z¹₀ –z₀. Then on the grid in the unit square for the natural number z¹₀, the point Ф_n (xⁿ₀/ z¹₀, y₀/ z¹₀) is a nodal point and lies on the Fermat curve S_n = 1, according to the proposal.

Obviously, for Pythagorean triples of numbers (x₀, y₀, z₀), for which z₀ = y₀ +1 (for example, (3,4,5), (5,12,13), etc.), ⁿ√(xⁿ₀+ yⁿ₀) cannot be an integer, since in this case the inequalities must be satisfied.

y₀ < ⁿ√(xⁿ₀+ yⁿ₀) < y₀ +1 (11)

In other cases, when z₀ > y₀ +1, for example (20, 21, 29) from the assumption that the point Ф_n (x₀/ z¹₀, y₀/ z¹₀) lies on the Fermat curve S_n = 1, it follows that this point will be a nodal point on grid in the unit square for a natural number z¹₀ that is not a multiple of z₀ , (z₀ > z¹₀ and z₀ is a prime number). According to formula (8), with n =2, the solution to equation (3) must be rational, and equation (1) must be integer. But on the grid in the unit square for z¹₀, the point P0 will not be a nodal point and, therefore, with n = 2 at this point, equation (1) will not have an integer solution.

Therefore ⁿ√(xⁿ₀+ yⁿ₀) is an irrational number.

Proving that ⁿ√(xⁿ₀+ yⁿ₀) is irrational is not necessary. This follows from the previously proven statement that on the Pythagorean ray ƛ₀ on the grid in the unit square for z₀, except for the point P₀ on the unit circle, there are no other nodal points. This is equivalent to the statement that for a right triangle with legs x₀ and y₀, the sum of legs to any other power n > 2 will not be a rational number (or rather an integer).

From the irrationality of ⁿ√(xⁿ₀+ yⁿ₀) for n > 2 it follows that the length of the radius vector of the point Ф_n lying at the intersection of the Fermat curve S_n = 1 with the Pythagorean ray ƛ₀ is also an irrational number.

R_n = √(u²_n+ v_n²) = √( (x₀/√(x₀ⁿ+ y₀ⁿ))² + (y₀/√(x₀ⁿ+ y₀ⁿ))²) = z₀/ⁿ√(x₀ⁿ+ y₀ⁿ) (12)

It is obvious that there are infinitely many rational points on the ray ƛ₀ and each of them, with an appropriate choice of the partitioning step of the unit square h + 1/z, can become a node on the grid for a natural number z. According to the lemma, the length of the radius vector of the point Ф* on the ray ƛ₀ is a rational number. Consequently, for any partitions of the unit square, on the Pythagorean ray ƛ₀ the inequality R* ≠ R_n holds, (13) where R_n is the length of the radius vector of the point Ф_n of the Fermat curve S_n = 1, and R* is the length of the radius vector of any rational point on the beam ƛ₀.

The coordinates of the point Ф_n on the Pythagorean ray ƛ ₀ can be written as

Ф_n ( x₀/ⁿ√(x₀ⁿ+ y₀ⁿ), y₀/ⁿ√(x₀ⁿ+ y₀ⁿ)) or Ф_n ( x₀ h_n), ( y₀ h_n), where h_n = R_n/ z₀ = 1/ ⁿ√(x₀ⁿ+ y₀ⁿ) (14)

The coordinates of the Pythagorean point P₀ can be written as P₀(x₀ h₀, y₀ h₀), where h₀ = 1/ z₀ is the step of partitioning the unit square. (15)

Obviously, h_n > h₀,, since R_n > 1, (h_n = h₀ R_n)

If we construct a square with a side equal to the length of the radius vector of the point Ф_n and cover this square with a uniform mesh with a step of h_n = R_n/z₀, then the point Ф_n in this square will be a nodal (Pythagorean) point. At this point, as well as for the Pythagorean point P₀ ϵ S₂ = 1, the Pythagorean theorem is fulfilled integer over cells (Fig. 2).

It remains for us to consider the rays ƛ^* passing through the nodal (rational) points of various partitions of the unit square, which are not Pythagorean rays for any partitions. These rays intersect the unit circle at irrational points. For example, ƛ