Текст книги "The Wonders of Arithmetic from Pierre Simon de Fermat"

Автор книги: Youri Kraskov

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3.4.3. Archimedes-Fermat Problem

The problem statement is as follows:

Let any non-square number be given, you need to find an infinite number of squares, which after multiplication by this number and increasing by unit, will make a square.

Fermat proposed finding solutions for the numbers 61, 109, 149, and 433 [36].

The English mathematician John Wallis managed to find a way to calculate the required numbers using the Euclidean method of decomposing an irrational number into an infinite common fraction. He published his decision under the name "Commercium epistolicum" see pic. 37-38.

Pic. 36. John Wallis

Pic. 37. Title Page of Wallis's Publication Commercium Epistolicum

Although Wallis did not give a complete proof the validity of this method, Fermat nevertheless admitted that he had coped with the task. Euler came very close to the solution when he showed that this fraction is cyclical, but he was not able to complete the proof and this task was finally solved by Lagrange. Later, this Fermat's task also was solved by Gauss in his own way, but for this purpose the extensive theory he created called “Arithmetic of deductions” was involved. And everything would be fine if the Lagrange's proof was not in the category of highest difficulty and the Gauss decision was not based on the most complicated theory. Fermat himself clearly could not follow such ways. About how he himself solved this problem, he reports in the letter-testament to Carcavy in August 1659 [36]: “I recognize that Mr. Frenicle gave various special solutions to this question as well as Mr. Wallis, but a common solution will be found using the method of descent applied skillfully and appropriately." However, this Fermat's solutions so remained as the secret behind seven seals!

Pic. 38. Page 64 Commercium Epistolicum

Demonstrating Wallis Method

We will try here slightly to open the veil over this mystery. To do this, we will look at a simple example of Wallis calculations and then compare it with how one could do these calculations using Fermat's method. So, we need to find the smallest numbers x and y that satisfy the equation Ax² + 1 = y². Let A = 29 then calculations by the Wallis method look as follows [32]:

From this sequence of calculations, a chain of suitable fractions is obtained by backward i.e. from a5 to a0 and looks like: 5/1; 11/2; 16/3; 27/5. As a result, we get 70/13. Then the minimum solution would be:

x₁√29+у₁=(13√29+70)²=1820√29+9801; x₁=1820; y₁=9820

Wallis was unable to prove that this method of computation gives solutions for any non-square number A. However, he guessed that the chain of computations ends where a6 will be computed by the same formula as a1. To understand the meaning of this chain of calculations, you need to study a very voluminous and extremely difficult theory [7, 14, 19, 23, 26, 32], which Fermat could not have developed at that time. Since no Fermat's manuscripts on arithmetic have survived, a natural question arises: how could he formulate such a difficult problem, about which there was very little information before him?

For today's science such a question is clearly beyond its capabilities since for it the pinnacle of achievements in solving Fermat's problems is any result even inflated to such incredible dimensions that we have today. However, it is difficult to imagine how much this our respected science will be dejected when from this book it learns that the problem was solved by Fermat not for great scientists, but … for schoolchildren!!! However, here we cannot afford to grieve science so much, so we only note that the example given in the textbooks is very unfortunate since it can be solved quite simply, namely: x = 2mz, where m2−1 = z². This last equation differs from the initial one only in sign and even by the method of ordinary tests without resorting to irrational numbers one can easily find the solution m = 13; z = 70; x = 2 x 13 x 70 = 1820; y = 9820.

Obviously, in textbooks it would be much more appropriate to demonstrate an example with the number 61 i.e. the smallest number proposed by Fermat himself. How he himself solved this problem is unknown to science, but we have already repeatedly demonstrated that it is not a problem for us to find out. We just need to look once more into the cache of the Toulousean senator and as soon as we succeeded, we quickly found the right example so that it could be compared with the Wallis method. In this example you can calculate x = 2mz, where m and z are solutions to the corresponding equation 61m² – z² = 1. Then the chain of calculations is obtained as follows:

61m²−z²=1

m=(8m₁±z₁)/3=(8×722+5639)/3=3805; z²=61×3805²−1=29718²

61m₁²−z₁²=3

m=(8m₁±z₁)/3=(8×722+5639)/3=3805; z₁²=61×722²−1=29718²

61m₂²−z₂²=9

m=(8m₁±z₁)/3=(8×722+5639)/3=3805; z₂²=61×137²−1=29718²

61m₃²−z₃²=27

m₃=(8m₄±z₄)/3=(8×5+38)/3=26; z₃²=61×26²−27=203²

61m₄²−z₄²=81

m₄=(8m₅±z₅)/3=(8×2−1)/3=5; z₄²=61×5²−81=38²

61m₅²−z₅²=243

m₅=2; z₅²=1

We will not reveal all nuances of this method, otherwise all interest to this problem would have been lost. We only note that in comparison with Wallis method where the descent method is not used, here it is present in an explicit form. This is expressed in the fact that if the numbers m and z satisfying the equation 61m²–z²=1 exist, then there must still exist numbers m₁11²–z₁²=3, as well as the numbers m₂1 and z₂1, from equation 61m₂²–z₂²=9, etc. up to the minimum values m₅4 and z₅4. The number 3 appearing in the descent is calculated as 64 – 61, that is, as the difference between 61 and the square closest to it. Calculations as well as in the Wallis method are carried out in the reverse order i.e. only after the minimum values of m₅ and z₅ have been calculated. As a result, we get:

m=3805; z=29718

x=2mz=2×3805×29718=226153980

y=√(61×2261539802+1)=1766319049

Of course, connoisseurs of the current theory will quickly notice in this example that the results of calculations obtained in it will exactly coincide with those that can be obtained by the Wallis' method. However, for this they will have to use the irrational number √61, and our example with Fermat's method showed that it is possible to do calculations exclusively in the framework of arithmetic i.e. only in natural numbers. There is no doubt also that experts without much effort will guess how to get the formulas shown in our example. However, it will not be easily for them to explain how to apply this Fermat's method in the general case because from our example it is not at all clear how it is possible to determine that the ultimate goal is to solve the equation 61m₅² – z₅² = 243 from which calculations should be performed with a countdown.

It would be simply excellent if today's science could explain Fermat's method in every detail, but even the ghostly hopes for this are not yet visible. It would be more realistic to expect that attempts will be made to refute this example as demonstration a method of solving the problem unknown to science. Nevertheless, science will have to reckon with the fact that this example is still the only one in history (!!!) confirmation of what Fermat said in his letter-testament. When this secret is fully revealed, then all skeptics will be put to shame and they will have no choice, but to recognize Fermat as greater than all the other greatest scientists because they were recognized as such mainly because they created theories so difficult for normal people to understand that they could only cause immense horror among students who now have to take the rap for such a science.4343
  Examples are in many videos from the Internet. However, these examples in no way detract from the merits of professors who know their job perfectly.


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https://www.youtube.com/watch?v=wFz8W2HsjfQ

https://www.youtube.com/watch?v=cUytn2SZ1n4

https://www.youtube.com/watch?v=ZhVNOgaBStY

In this sense, the following example of solving a problem using the descent method will be particularly interesting because it was proposed in a letter from Fermat to Mersenne at the end of 1636, i.e. the age of this task is almost four centuries! Euler's proof [8] was incorrect due to the use of «complex numbers» in it. However, even the revised version of André Weil in 1983 [17] is too complex for schooling.

3.4.4. Fermat’s Problem with Age 385 years

In the original version in 1636 this task was formulated as follows:

Find two square-squares, which sum is equal to a square-square,

or two cubes, which sum is a cube.

This formulation was used by Fermat's opponents as the fact that Fermat had no proof of the FLT and limited himself to only these two special cases. However, the very name «The Fermat’s Last Theorem» appeared only after the publication of «Arithmetic» by Diophantus with Fermat's remarks in 1670 i.e. five years after his death. So, there is no any reason to assert that Fermat announced the FLT in 1637.

The first case for the fourth power we have presented in detail in Appendix II. As for the case for the third power, Fermat's own proof method restored by us below, will not leave any chances to the solutions of this problem of Euler and Weil to remain in history of science, since from the point of view of the simplicity and elegance of the author's solution this problem, they will become just unnecessary.

Now then, to prove that there are no two cubes whose sum is a cube, we use the simplest approach based on divisibility of numbers, what means that in the original equation

a³+b³ = c³ (1)

the numbers a, b, and c can be considered as coprime ones, i.e. they do not have common factors, but in general case this is not necessary, since if we prove that equation (1) cannot have solutions in any integers, including those with common factors, then we will prove that coprime numbers also cannot be solutions of the original equation. Then we assume that both sides of equation (1) in all cases must be divisible by the number c², then equation (1) can be represented as

c³ = c²(x+y) = a³+b³ (2)

In this case, it is easily to see that there is only one way to get solutions to equation (1) when the numbers c, x, y, and x+y are cubes, i.e.

с = x+y = p³+q³= z³; x = p³; y = q³ (3)

Then equation (1) must have the form:

(z³)³ = (z²)³(p³+q³) (4)

Thus, we found that if there are numbers a, b, and c that satisfy equation (1), then there must be numbers p

p³+q³= z³

If we now apply the same approach to solving this equation, that we applied to solving equation (1), we will get the same equation, only with smaller numbers. However, since it is impossible to infinitely reduce natural numbers, it follows that equation (1) has no solutions in integers.

At first glance, we have received a very simple and quite convincing proof of the Fermat problem by the descent method, which no one has been able to obtain in such a simple way for 385 years, and we can only be happy about it. However, such a conclusion would be too hasty, since this proof is actually incorrect and can be refuted in the most unexpected way.

However, this refutation is so surprising that we will not disclose it here, because it opens the way not only for the simplest proof of the FLT, but also automatically allows to reduce it to a very simple proof of the Beal conjecture. The disclosure the method of refuting this proof would cause a real commotion in the scientific world, therefore we will include this mystery among our riddles (see Appendix V Pt. 41).

So, we have demonstrated here solving to Fermat's problems (only by descent method!):

1) The proof of the Basic theorem of arithmetic.

2) The proof of the Fermat's theorem on the unique solving the

equation p³ = q² + 2.

3) A way to prove Fermat's Golden Theorem.

4) A Fermat's way to solve the Archimedes-Fermat equation

Ax² + 1 = y².

5) The proof method of impossibility a³+b³=c³ in integers, which

opens a way to simplest proofs of the FLT and Beal conjecture.

6) A Fermat's proof his grandiose discovery about primes in the

form 4n + 1 = a² + b² which we have presented in another style in

Appendix IV, story Year 1680.

Over the past 350 (!!!) years after the publication of these problems by Fermat, whole existing science could not even dream of such a result!

3.5. Parity Method

Before we embarking on the topic «Fermat's Last Theorem» we note that this problem was not solved by Fermat himself using the descent method, otherwise in his FLT formulation there would be no mention of a «truly amazing proof», which certainly related to other methods. Therefore, to the above examples of the application of the descent method we will add our presentation of two methods unknown to today's science. The most curious of these is the parity method.

3.5.1. Defining Parity as a Number

The Basic theorem of arithmetic implies a simple, but very effective idea of defining parity as a number, which is formulated as follows:

The parity of a given number is the quantity of divisions this number by two without a remainder until the result of the division becomes odd.

Let's introduce the parity symbol with angle brackets. Then the expression ‹x› = y will mean:

the parity of the number x is equal to y. For example, the expression "the parity of the number forty is equal to three" can be represented as: ‹40›= 3. From this definition of parity, it follows:

– parity of an odd number is zero.

– parity of zero is infinitely large.

– any natural number n can be represented as n = 2^w (2N – 1)

where N is the base of a natural number, w is its parity.

3.5.2. Parity Law

Based on the above definition the parity, it can be stated that equal numbers have equal parity. In relation to any equation this provision refers to its sides and is absolutely necessary in order for it to have solutions in integers. From here follows the parity law for equations:

Any equation can have solutions in integers if and only if the parities of both its sides are equal.

The mathematical expression for the parity law is W_L = W_R where W_L and W_R are the parities of the left and right sides of the equation respectively. A distinctive feature of the parity law is that the equality of numbers cannot be judged by the equality of their parity, but if their parities are not equal, then this certainly means the inequality of numbers.

3.5.3. Parity Calculation Rules

Parity of a sum or difference two numbers a and b

If ‹a› < ‹b› then ‹a ± b› = ‹a›.

It follows in particular that the sum or difference of an even and an odd number always gives a number with parity zero. If ‹a› = ‹b› = x then either ‹a + b› = x + 1 wherein ‹a – b› > x + 1

or ‹a – b› = x + 1 wherein ‹a + b› > x + 1

These formulas are due to the fact that

‹(a + b) + (a – b)› = ‹2a› = ‹a› + 1

It follows that the sum or difference of two even or two odd numbers gives an even number.

Parity of a sum or difference two power number aⁿ and bⁿ

If ‹a› < ‹b› then ‹aⁿ ± bⁿ› = ‹aⁿ›. If ‹a› = ‹b› = x then

only for even n:

‹aⁿ – bⁿ› = ‹a – b›+ ‹a + b›+ x(n – 2) + ‹n› – 1

‹aⁿ + bⁿ› = xn + 1

only for odd n:

‹aⁿ ± bⁿ› = ‹a ± b› + x(n – 1)

When natural numbers multiplying, their parities are added up

‹ab› = ‹a› + ‹b›

When natural numbers dividing, their parities are subtracted

‹a : b› = ‹a› – ‹b›

When raising number to the power, its parity is multiplied

‹a^b› = ‹a› × b

When extracting the root in number, its parity is divided

‹ ^b√a› = ‹a› : b

3.6. Key Formula Method

To solve equations with many unknowns in integers, an approach is often used when one more equation is added to the original equation and the solution to the original is sought in a system of two equations. We call this second equation the key formula. Until now due to its simplicity, this method did not stand out from other methods, however we will show here how effective it is and clearly deserves special attention. First of all, we note an important feature of the method, which is that:

Key formula cannot be other as derived from the original equation.

If this feature of the method is not taken into account i.e. add to the original equation some other one, then in this case, instead of solving the original equation we will get only a result indicating the compatibility of these two equations. In particular, we can obtain not all solutions of the original equation, but only those that are limited by the second equation.

In the case when the second equation is derived from the initial one, the result will be exhaustive i.e. either all solutions or insolvability in integers of the original equation. For example, we take equation z³ = x² + y². To find all its solutions we proceed from the assumption that a prerequisite (key formula) should be z = a² + b² since the right-hand side of the original equation cannot be obtained otherwise than the product of numbers which are the sum of two squares. This is based on the fact that:

The product of numbers being the sum of two squares in all

cases gives a number also consisting the sum of two squares.

The converse is also true: if it is given a composite number being the sum of two squares then it cannot have prime factors that are not the sum of two squares. This is easily to make sure from the identity

(a²+b²)(c²+d²)=(ac+bd)²+(ad−bc)²=(ac−bd)²+(ad+bc)²

Then from (a²+b²)(a²+b²)=(aa+bb)²+(ab−ba)²=(a²−b²)²+(ab+ba)² it follows that the square of a number consisting the sum of two squares, gives not two decompositions into the sum of two squares (as it should be in accordance with the identity), but only one, since (ab−ba)2= 0 what is not a natural number, otherwise any square number after adding to it zero could be formally considered the sum of two squares.

However, this is not the case since there are numbers that cannot be the sum of two squares.

As Pierre Fermat has established, such are all numbers containing at least one prime factor of type 4n − 1. Now from

a²−b²=c; ab+ba=2ab=d; (a²+b²)²=c²+d²

the final solution follows:

z³=(a²+b²)³=(a²+b²)(c²+d²)=x²+y²

where a, b are any natural numbers and all the rest are calculated as c=a²−b²; d=2ab; x=ac−bd; y=ad+bc (or x=ac+bd; y=ad−bc). Thus, we have established that the original equation z³=x²+y² has an infinite number of solutions in integers and for specific given numbers a, b – two solutions.

It is also clear from this example why one of the Fermat's theorems asserts that:

A prime number in the form 4n+1 and its square can be decomposed into two squares only in one way; its cube and biquadrate only in two; its quadrate-cube and cube-cube only in three etc. to infinity.

4. The Fermat’s Last Theorem

4.1. The Thorny Path to Truth

4.1.1. The FLT up to now remains unproven

The scientific world has been at first learned about the FLT after publication in 1670 of “Arithmetic” by Diophantus with Fermat’s remarks (see Pic. 3 and Pic. 96 from Appendix VI). And since then i.e. for three and a half centuries, science cannot cope with this task. Moreover, perhaps this is namely why the FLT became the object of unprecedented falsification in the history of mathematics. It is very easily to verify this since the main arguments of the FLT “proof” 1995 are well known and look as follows.

If the FLT were wrong, then there would be exist an elliptical “Frey curve” (???): y²=x(x−aⁿ)(x+bⁿ) where aⁿ+bⁿ=cⁿ. But Kenneth Ribet has proven that such a curve cannot be modular. Therefore, it suffices to obtain a proof of the Taniyama – Shimura conjecture, that all elliptic curves must be modular, so that it simultaneously becomes a proof of the FLT. The proof was presented in 1995 by Andrew Wiles who became the first scientist that allegedly has proven the FLT.

However, it turns out that the “Frey curve" and together with it the works of Ribet and Wiles have with the FLT nothing to do at all!!!4444
  It must be admitted that the method of Frey's proof is basically the same as that of Fermat i.e. it is based on obtaining a solution to the equation aⁿ+bⁿ=cⁿ by combining it into a system with another equation – a key formula, and then solving this system. But if Fermat’s key formula a+b=c+2m is derived directly from the initial equation, while at Frey it is just taken from nothing and united to the Fermat equation aⁿ+bⁿ=cⁿ i.e. Frey's curve y²=x(x−aⁿ)(x+bⁿ) is a magical trick that allows to hide the essence of the problem and replace it with some kind of illusion. Even if Frey could prove the absence of integer solutions in his equation then this could in no way lead him to the proof of the FLT. But he did not succeed it also, therefore one “brilliant idea” gave birth to an “even more brilliant idea” about the contradiction of the “Frey curve” to the Taniyama – Simura conjecture. With this approach you can get incredibly great opportunities for manipulating and juggling the desired result, for example, you can «prove» that the equation a+b+c=d as well as the Fermat equation aⁿ+bⁿ=cⁿ in integers cannot be solved if take abc=d as a key formula. However, such «ideas» that obviously indicate the substitution the subject of the proof should not be considered at all, since magicians hope only for the difficulty of directly refuting their trick.


[Закрыть] And as regards the “proof” of A. Wiles the conjecture of Taniyama – Shimura, he also himself admitted4545
  Here is how E. Wiles himself comments on a mistake found in his “proof” in 1993: “Even explaining it to a mathematician would require the mathematician to spend two or three months studying that part of the manuscript in great detail”. See Nova Internet Publishing http://www.pbs.org/wgbh/nova/physics/andrew-wiles-fermat.html It turns out that this «proof» understood only by its author, while everyone else needs to learn and learn.


[Закрыть] that one needs much more to learn (naturally, from Wiles) in order to understand all of its nuances, setting forth on 130 pages (!!!) of scientific journal «Annals of Mathematics». Quite naturally that after the appearance of such exotic “proof”, scientists cannot come to their senses from such a mockery of science, the Internet is replete with all sorts of refutations,4646
  Such debunks are very detailed, but too redundant since the arguments of the main authors of the FLT “proof” by G. Frey and E. Wiles look so ridiculous that otherwise as by the hypnotic influence of the unholy it would impossible to explain why many years after 1995 for some reason none of the recognized pundits so have ever noticed that instead of FLT proof we have got a something completely different.


[Закрыть] and there is no doubt that any generally accepted proof of the FLT still does not exist.

The special significance of the FLT is that in essence, this is one of the simple cases to addition of power numbers when only the sum of two squares can be a square and for higher powers such addition is impossible. However, according to the Waring-Hilbert theorem, any natural number (including an integer power) can be the sum of the same (or equal to a given) powers4747
  Similarly, to the example from Pythagoras 3²+4²=5² Euler found a very simple and beautiful example of adding powers: 3³+4³+5³=6³. For other examples, see comment 22 in Pt. 2.


[Закрыть]. And this a much more complex and no less fundamental theorem was proven much earlier than the FLT.

We also note the fact that the FLT attracts special attention not at all because this task is simple in appearance, but very difficult to solve. There are also much simpler-looking tasks, which are not only not to be solved, but also even nobody really knows how to approach them 4848
  For example, the task of the infinity of the set of pairs of twin primes or the Goldbach task of representing any even natural number as the sum of two primes. And also, the solution to the coolest problem of arithmetic about an effective way to calculate prime numbers is still very far from perfect despite the tons of paper spent on research on this problem.


[Закрыть]. The FLT especially differs from other tasks that attempts to find its solution lead to the rapid growth of new ideas, which become impulses for the development of science. However, there was so much heaped up on this path that even in very voluminous studies, all this cannot be systematized and combined.4949
  In particular, Edwards in his very voluminous book [6, 38], was not aware of the fact that Gauss solved the Fermat's task of decomposing a prime number type 4n + 1 into a sum of two squares. But it was this task that became a kind of bridge to the subsequent discovering the FLT. Fermat himself first reported it in a letter to Blaise Pascal on 09/25/1654 and this is one of the evidences that of all his scientific works, the FLT is really his last and greatest discovery.


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Great scholars did not attach much importance to building the foundations of science apparently considering such creativity to be a purely formal matter, but centuries-old failures with the FLT proof indicate that they underestimated the significance of such studies. Now when it became clear where such an effective scientific tool as the descent method could come from, as well as other tools based on understanding the essence of number, it becomes clear why Fermat was so clearly superior to other mathematicians in arithmetic, while his opponents have long been in complete bewilderment from this obvious fact.

Here we come to the fact that the main reason for failures in the search for FLT proof lies in the difference between approaches to solving tasks by Fermat and other scientists, as well as in the fact that even modern science has not reached the knowledge that already was used by Fermat in those far times. This situation needs to be corrected because otherwise the FLT so will continue to discredit whole science.

One of the main questions in the studies on the FLT was the question of what method did Fermat use to prove this theorem? Opinions were very different and most often it was assumed that this was the method of descent, but then Fermat himself hardly called it "truly amazing proof." He also could not apply the Kummer method, from which the best result was obtained in proving the FLT proof over the last 170 years. But perhaps he besides the descent method had also other ones? Yes indeed, this is also described in detail in treatise "A New Discovery in the Art of Analysis" by Jacques de Billy [36]. There, he sets out in detail Fermat's methods, which allow him to find as many solutions as necessary in systems of two, three, or more equations. But here his predecessors Diophantus, Bachet and Viet at best found only one solution. After demonstrating Fermat's methods for solving the double equalities Billy also points to the most important conclusion, which follows from this: This kind of actions serves not only to solve double equalities, but also for any other equations.

Now it remains only to find out how to use the system of two equations to prove the FLT? Obviously, mathematicians simply did not pay attention to such an explicit clue from Fermat or did not understand its meaning. But for us this is not a problem because we can look into the cache and delve into the "heretical writings"! Based on what we have already been able to recover from Fermat’s works, we can now begin to uncover this greatest mystery of science, indicating also an effective method that allows us to solve the problem of FLT proof.

How it wouldn't be surprising, the essence of this method is quite simple. In the case when there are as many equations as there are unknowns in them, such a system is solved by ordinary substitutions. But if there is only one equation with several unknowns, then it can be very difficult to establish whether it can even have any solutions in integers. In this case, the numbers supposed as solutions can be expressed in the form of another equation called the “Key Formula” and then the result can be obtained by solving a system of two equations. Similar techniques when some numbers are expressed through others, have always been used by mathematicians, but the essence of the key formula is in another, it forms exactly that number, which reflects the essence of the problem and this greatly simplifies the way to solving the original equation. In such approaches and methods, based on an understanding the essence of numbers, in fact also lies the main superiority of Fermat over other scientists.5050
  The main and fundamental difference between Fermat's methods and the ones of other scientists is that his methods are universal enough for a very wide range of problems and are not directly related to a specific task. As a rule, attempts to solve a problem begin with trial calculations and enumeration of all possible options and those who think faster get correspondingly more opportunities to solve it. Fermat has another approach. He makes trying only for the purpose of bringing them to some universal method suitable for the given task. And as soon as it him succeeds, the task is practically solved and the result is guaranteed even if there is still a very large amount of routine calculations ahead. See for example, comment 30 in Pt. 2.


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To make it possible to follow the path that Fermat once laid, you need to find the starting element from the chain of events leading to the appearance of the FLT, otherwise there will be very little chance of success because everything else is already studied far and wide. And if we ask the question exactly this way, we will suddenly find that this very initial element was still in sight from 1670, but since then no one has paid any attention to it at all. However, in fact, we are talking about the very problem under number 8 from the book II of Arithmetic by Diophantus, to which also Fermat’s remark was written, became later a famous scientific problem. Everyone thought that this simple-looking problem has no difficulties for science and only Fermat had another opinion and worked for many years to solve it. As a result, he not only obtained it, but in addition to this he secured to his name unfading world fame.