The Wonders of Arithmetic from Pierre Simon de Fermat

Youri Veniaminovich Kraskov, 2021

This book shows how the famous scientific problem called "Fermat Last theorem" (FLT) allows us to reveal the insolvency and incapacity of science, in which arithmetic for various historical reasons has lost the status of the primary basis of all knowledge. The unusual genre of the book was called "Scientific Blockbuster", what means a combination of an action-packed narrative in the style of fiction with individual fragments of purely scientific content. The original Russian text of this book is translated into English by its author Youri Kraskov.

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!!!⁴⁴ And as regards the “proof” of A. Wiles the conjecture of Taniyama — Shimura, he also himself admitted⁴⁵ 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,⁴⁶ 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) powers⁴⁷. 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 ⁴⁸. 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.⁴⁹

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.⁵⁰

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.

4.1.2. Diophantus' Task

The book entitled “Arithmetic” by Diophantus is very old however, probably it appeared not in III as it was thought until recently, but in the XIV or XV century. In those times when yet there were no print editions, it was a very impressive in volume manuscript consisting of 13 books, from which only six ones reached us. In today's printed form this is a small enough book with a volume of just over 300 pages [2, 27].

In France an original Greek version of this book was published in 1621 with a Latin translation and comments from the publisher, which was Bachet de Méziriac. This publication became the basis for Fermat's work on arithmetic. The contents of the book are 189 tasks and solutions are given for all them. Among them are both fairly simple and very difficult tasks. However, since they have been solved, a false impression is created that these tasks are not educational, but rather entertaining ones i.e., they are needed not to shape science, but to test for quick wit. In those times, it could not have been otherwise because even just literate people who could read and write were very rare.

However, from the point of view the scientific significance of the presented tasks and their solutions, the creation of such a book is not something that to the medieval Diophantus, but to all scientists in the entire visible history would be absolutely impossible. Moreover, even at least properly understanding the contents of the Euclid's"Elements"and the Diophantus'"Arithmetic"became an impossible task for our entire science. Then naturally, the question arises how did the authors of these books manage to do such creations? Of course, this question also arose in science, but instead of answering it only retains its proud silence. Well, then nothing prevents us from expressing our version here.

Apparently, there were somehow preserved and then restored written sources of knowledge from a highly developed civilization perished in earlier times. Only especially gifted people with extrasensory abilities allowing them to understand written sources regardless of the carrier and language, in which they were presented, could read and restore them. Euclid who was most likely a king, involved a whole team of such people, while Diophantus coped itself one and so the authorship of both appeared although in fact it was not the scientists who worked on the books, but only scribes and translators. But now we come back to the very task 8 from the second book of “Arithmetic” by Diophantus: Decompose a given square into the sum of two squares.

In the example of Diophantus, the number 16 is divided into the sum of two squares and his method gives one of the solutions 4²=20²/5²=16²/5²+12²/5² as well as countless other similar solutions⁵¹. However, this is not a solution to the task, but just a proof that any integer square can be made up of two squares any number of times either in integer or in fractional rational numbers. It follows that the practical value of the Diophantus method is paltry since from the point of view of arithmetic, the fractional squares are nonsense like, say, triangular rectangles or something like that. Obviously, this task should be solved only in integers, but Diophantus does not have such a solution and of course, Fermat seeks to solve this problem himself especially since at first, he sees it as not at all complicated.

So, let in the equation a²+b²=c² given the number c and you need to find the numbers a and b. The easiest way to find a solution is by decomposing the number c into prime factors: c=pp₁p₂…p_k; then

c²=p²p₁²p₂²…p_k²=p²(p₁p₂…p_k)²=p_i²N²

Now it becomes obvious that the number c² can be decomposed into a² + b² only if at least one of the numbers p_i² also decomposes into the sum of two squares.⁵² But this is a vicious circle because again you need to decompose square into a sum of two squares. However, the situation is already completely different because now you need to decompose a square of prime number and this circumstance becomes the basis for solving the task. If a solution is possible, then there must exist such prime numbers that decomposes into the sum of two squares and only in this case in accordance with the identity of the Pythagoreans, you can obtain:

p_i²=(x²+y²)²=(x²−y²)²+(2xy)²

i.e. the square of such a prime will also be the sum of two squares. From here appears the truly grandiose scientific discovery of Fermat:⁵³

All primes of type 4n+1 can be uniquely decomposed into the sum of two squares, i.e. the equation p=4n+1=x²+y² has a unique solution in integers. But all other primes of type 4n−1 cannot be decomposed in the same way.

In the Fermat's letter-testament it was shows how this can be proven by the descent method. However, Fermat’s proof was not preserved and Euler who solved this problem had to use for this all his intellectual power for whole seven years.⁵⁴ Now already the solution to the Diophantine task seems obvious. If among the prime factors of number c there is not one related to the type 4n+1, then the number c² cannot be decomposed into the sum of two squares. And if there is at least one such number p_i, then through the Pythagoreans’ identity it can be obtain:

c²= N²p_i²= (Nx)²+(Ny)²

where x= u²−v²; y=2uv; a=N(u²−v²); b=N2uv

The solution is obtained, however it clearly does not satisfy Fermat because in order to calculate the number N you need to decompose the number c into prime factors, but this task at all times was considered as one of the most difficult of all problems in arithmetic.⁵⁵ Then you need to calculate the numbers x, y i.e. solve the problem of decomposing a prime of type 4n+1 into the sum of two squares. To solve this problem, Fermat worked almost until the end of his life.

It is quite natural that when there is a desire to simplify the solution of the Diophantine task, a new idea also arises of obtaining a general solution of the Pythagoras’ equation a² + b² = c² in a way different from using the identity of Pythagoreans. As it often happens, a new idea suddenly arises after experienced strong shocks. Apparently, this happened during the plague epidemic of 1652 when Fermat managed to survive only by some miracle, but it was after that when he quite clearly imagined how to solve the Pythagoras’ equation in a new way.

However, the method of the key formula for Fermat was not new, but when he deduced this formula and immediately received a new solution to the Pythagoras equation, he was so struck by this that he could not for a long time come to oneself. Indeed, before that to obtain one solution, two integers must be given in the Pythagoreans' identity, but with the new method, it may be obtained minimum three solutions with by only one given integer.

But the most surprising here is that the application of this new method does not depend on the power index and it can be used to solve equations with higher powers i.e. along with the equation a²+b²=c² can be solved in the same way also aⁿ+bⁿ=cⁿ with any powers n>2. To get the final result, it remained to overcome only some of the technical difficulties that Fermat successfully dealt with. And here such a way it appeared and became famous his remark to the task 8 of Book II Diophantus'"Arithmetic":

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

See Pic. 3 and the translation at the end of Pt. 1.

4.2. Fermat’s Proof

The reconstructed FLT proof presented here contains new discoveries unknown to today’s science. However, from this it does not follows that proof becomes difficult to understand. On the contrary, it is precisely these discoveries that make it possible to solve this problem most simply and easily. The phenomenon of the unprovable FLT itself would not have appeared at all if the French Academy of Sciences had been founded during the lifetime of P. Fermat. Then he would become an academician and published his scientific researches and among his theorems in all arithmetic textbooks there would be also such a most ordinary theorem:

For any given natural number n>2, there is not a single triplet of natural numbers a, b and c, satisfying the equation

aⁿ+bⁿ=cⁿ (1)

To prove this statement, suppose that a, b, c satisfying to (1) exist and then based on this, we can get the all without exception solutions to this equation in general form. To this aim we use the key formula method, in which one more equation is added to the initial equation so that it becomes possible to obtain solution (1) in a system of two equations. In our case the key formula is:

a+b=c+2m (2)

where m is a natural number.

To obtain formula (2) we note that a≠b since otherwise 2aⁿ=cⁿ what is obviously impossible. Consequently, a<b<c and we can state that (a^n-1+b^n-1)>c^n-1 whence (a+b)>c. Since in (1) cases with three odd a, b, c, as well as one odd and two even are impossible, the numbers a, b, c can be either all even or two odd and one even. Then from (a + b) > c follows formula (2) where the number 2m is even⁵⁶.

At first, we verify the effectiveness of the method for the case n = 2 or the Pythagoras’ equation a²+b²=c². Here the key formula (2) applies and you can get a solution to the system of equations (1), (2) if you substitute one into another. To simplify it, we will square both sides of (2) to make the numbers in (1) and (2) proportionate. Then (2) takes the form:

{a²+b²−c²}+2(c−b)(c−a)=4m² (3)

Substituting the Pythagoras’ equation in (3), we obtain:

A_iB_i=2m² (4)

where taking into account the formula (2):

A_i=c−b=a−2m; B_i=c−a=b−2m (5)

Now we decompose the number 2m² into prime factors to get all the A_iB_i options. For primes m there are always only three options: 1×2m²=2×m²=m×2m. In this case A₁=1; B₁=2m²; A₂=2; B₂=m²; A₃=m; B₃=2m. Since from (5) it follows a=A_i+2m; b=B_i+2m; and from (2) c=a+b−2m; then we end up with three solutions:

1. a₁=2m+1; b₁=2m(m+1); c₁=2m(m+1)+1

2. a₂=2(m+1); b₂=m(m+2); c₂= m(m+2)+2 (6)

3. a₃=3m b₃=4m; c₃=5m

Equations (6) are the solutions of the Pythagoras’ equation for any natural number m. If the number m is composite, then the number of solutions increases accordingly. In particular, if m consists of two prime factors, then the number of solutions increases to nine⁵⁷. Thus, we have a new way of calculating all without exception triples of Pythagoras’ numbers, while setting only one number m instead of two numbers that must be specified in the Pythagoreans identity. However, the usefulness of this method is not limited only to this since the same key formula (2) is also valid for obtaining a general solution of equations with higher powers.

Using the method to obtain solutions of (1) for the case n=2, it is also possible to obtain solutions for n>2 by performing the substitution (1) in (2) and exponentiating n both sides of (2). To do this, first we derive the following formula⁵⁸:

(x+y)ⁿ=zⁿ=zz^n-1=(x+y)z^n-1=xzz^n-2+yz^n-1=

x(x+y)z^n-2+yz^n-1=x²zz^n-3+y(z^n-1+xz^n-2)+…

(x±y)ⁿ=zⁿ=xⁿ±y(x^n-1+x^n-2z+x^n-3z²+…+xz^n-2+z^n-1) (7)

We will name the expression in brackets consisting of n summand a symmetric polynomial and we will present it in the form (x ++ z)_n as an abridged spelling. Now using formula (7), we will exponentiating n both sides of formula (2) as follows.

[a−(c−b)]ⁿ=aⁿ+{bⁿ−cⁿ+(cⁿ−bⁿ)}−(c−b)[a^n-1+a^n-22m+…

+ a(2m)^n-1+(2m)^n-1]=(2m)ⁿ

Now through identity

(cⁿ−bⁿ)=(c−b)(c^n-1+c^n-2b+…+cb^n-2+b^n-1) we obtain:

{aⁿ+bⁿ−cⁿ}+(c−b)[(c++b)_n−(a++2m)_n]=(2m)ⁿ (8)

Equation (8) is a formula (2) raised to the power n what can be seen after substituting c−b=a−2m in (8) and obtaining the identity⁵⁹:

{aⁿ+bⁿ−cⁿ}+(cⁿ−bⁿ)−[aⁿ−(2m)ⁿ]=(2m)ⁿ (9)

In this identity natural numbers a, b, c, n, m of course, may be any. The only question is whether there are such among them that {aⁿ+bⁿ−cⁿ} will be zero? However, the analogy with the solution of the Pythagoras’ equation ends on this since the substitution of (1) in (8) is not substantiated in any way. Indeed, by substituting (1) in (3), it is well known that the Pythagoras’ equation has as much as you like solutions in natural numbers, but for cases n>2 there is no single such fact. Therefore, the substitution of the non-existent equation (1) in (8) is not excluded, what should lead to contradictions. Nevertheless, such a substitution is easily feasible and as a result we obtain an equation very similar to (4), which gives solutions to the Pythagoras equation. Taking into account this circumstance, we yet substitute (1) in (8) as a test, but at the same time modify (8) so, that factor (c−a) take out of square brackets.⁶⁰

Then we obtain:

A_iB_iE_i=(2m)ⁿ (10)

where A_i = c−b=a−2m; B_i=c−a=b−2m; E_i — polynomial of power n−2.

Equation (10) is a ghost that can be seen clearly only on the assumption that the number {aⁿ+bⁿ−cⁿ} is reduced when (1) is substituted into (8). But if it is touched at least once, it immediately crumbles to dust. For example, if A_i×B_i×E_i=2m²×2^n-1m^n-2 then as one of the options could be such a system:

A_iB_i=2m²

E_i=2^n-1m^n-2

In this case, as we have already established above, it follows from A_iB_i=2m² that for any natural number m the solutions of equation (1) must be the Pythagoras’ numbers. However, for n>2 these numbers are clearly not suitable and there is no way to check any other case because in a given case (as with any other variant with the absence of solutions) another substitution will be definitely unlawful and the ghost equation (10), from which only solutions can be obtained, disappears.⁶¹ Since the precedent with an unsuccessful attempt to obtain solutions has already been created, there can be no doubt that also all other attempts to obtain solutions from (10) will be unsuccessful because at least in one case the condition {aⁿ+bⁿ−cⁿ}=0 is not fulfilled i.e. the equation (10) has been obtained by substituting a non-existing (1) in the key formula (2), and the Fermat’s Last Theorem is proven.⁶²

So, now we have a restored author's proof of the Fermat's most famous theorem. Here are interesting ideas, but at the same time there is nothing that could not be accessible to science for more than three hundred years. Also, from the point of view a difficulty in understanding its essence, it matches at least to the 8th grade of secondary school. Undoubtedly, the FLT is a very important component of number theory. However, there is no apparent reason that this task has become an unsolvable problem for centuries, even though millions of professional scientists and amateurs have taken part in the search for its solution. It remains now only to lament, that's how he is, this unholy!

After everything was completed so well with the restoration of the FLT proof, many will be disappointed because now the fairy tale is over, the theme is closed and nothing interesting is left here. But this was the case before, when in arithmetic there were only rebuses, but we know that this is not so, therefore for us the fairy tale not only has not ended, but even did not begin! The fact is that we have so far revealed the secret of only two of the Fermat’s six recordings, which we have restored at the beginning of our study. To make this possible, we made an action-packed historical travel, in which the LTF was an extra-class guide. This travel encouraged us to take advantage of our opportunities and look into these forbidden Fermat’s “heretical writings” to finally make a true science in image of the most fundamental discipline of arithmetic available to our intelligent civilization and allowing it to develop and flourish on this heavy-duty foundation like never before.

We can honestly confess that so far not everything that keep in Fermat's cache is accessible and understandable to us. Moreover, we cannot even determine where this place is. But also, to declare that everything that we tell here, is only ours, would be clearly unfair and dishonest because nobody would have believed us then. On the other hand, if everything was so simple, then it would be completely to no one interesting. The worst thing that could be done, is to reveal the entire contents of Fermat’s cache so that everyone will forget about it immediately after reading.

We will act otherwise. If something will be revealed, only to give an opportunity to learn about the even more innermost mysteries of science, which will not only make everyone smarter, but will indicate the best ways to solve vital problems. Using the example of solving the FLT problem, it will be quite easily to make sure this since with a such solution science receives so a reliable point of support that it can do whatever it wants with the integer power numbers. In particular, it may be easily calculated as much as you like of integer power numbers, which in sum or difference will give again an integer power number. The fact that only a computer can shovel such a work, is very ashamed for current science because this task is too simple even for children.

The most quick-witted of them will clearly prefer that adults ask them to explain something more difficult for example, FLT proof, which in their time was completely inaccessible to them. Children of course, will not fail to get naughty and will be important like high-class nobles when answer to stupid questions of adults and indicating to them that it would be nice for someone to learn something else. But it will be only little flowers. But after that, the amazement of adults will become simply indescribable when they find out that the children are addicted to peeping and copying everything that interests them directly from Fermat's cache! Indeed, at their age they still do not realize their capabilities and it seems to them that this is at all not a difficult task.

However, if they had not read interesting books about science, then such an idea would never have occurred to them. But when they find out that someone is doing this, they will find that they can do it just as well if even not better! Do you not believe? Well, everyone who wants to be convinced of this, will have this opportunity now. But one more small detail remains. Fermat in his"heretical writings"although he pointed out that he had to provide proofs of three simple theorems for children, which he specially prepared for them, but so far he did not have time for this nevertheless firmly promised that as soon as he has a time, then he will certainly and sure do it.

But apparently, he did not have enough time and so he did not manage to add the necessary recordings. Or perhaps he changed his mind because didn’t want to deprive children of joy on their own to learn to solve just such problems that adults can't afford. If the children can't cope, then who them will reproach for it. But if they manage it, then adults will not go anywhere and will bring to children many, many gifts!

4.3. Theorems About Magic Numbers

The above presented proof of FLT not only corresponds to Fermat's assessment as"truly amazing", but is also constructive since it allows us to calculate both the Pythagorean numbers and other special numbers in a new way what demonstrate the following theorems.

Theorem 1. For any natural number n, it can be calculated as many

triples as you like from different natural numbers a, b, c such that

n = a² + b² — c². For example:

n=7=6²+14²–15²=28²+128²–131²=568²+5188²–5219²=

=178328²[phone]²–5300145931² etc.

n=34=11²+13²–16²=323²+3059²–3076²=

=247597²[phone]²–2043475820² etc.

The meaning of this theorem is that if there is an infinite number of Pythagoras triples forming the number zero in the form a²+b²−c²=0 then nothing prevents creating any other integer in the same way. It follows from the text of the theorem that numbers with such properties can be “calculated”, therefore it is very useful for educating children in school.

In this case, we will not act rashly and will not give here or anywhere else a proof of this theorem, but not at all because we want to keep it a secret. Moreover, we will recommend that for school books or other books (if of course, it will appear there) do not disclose the proof because otherwise its educational value will be lost and children who could show their abilities here will lose such an opportunity. On the other hand, if the above FLT proof would remain unknown, then Theorem 1 would be very difficult, but since now this is not so, even not very capable students will quickly figure out how to prove it and as soon as they do, they will easily fulfill the given above calculations.

Конец ознакомительного фрагмента.

Примечания

44

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.

45

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.

46

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.

47

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.

48

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.

49

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.

50

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.

51

The original solution to the Diophantus' task is as follows. “Let it be necessary to decompose the number 16 into two squares. Suppose that the 1st is x², then the 2nd will be 16-x². I make a square of a certain number x minus as many units as there are in site of 16; let it be 2x-4. Then this square itself is 4x²–16 x+16. It should be 16-x². Add the missing to both sides and subtract the similar ones from the similar ones. Then 5x² is equal to 16x and x will be equal to 16 fifths. One square is 256/25 and the other is 144/25; both folded give 400/25 or 16 and each will be a square” [2, 27].

52

If c²= p²N² and p² (as well as any other p_i² of prime factors c) does not decomposed into a sum of two squares i.e. p²=q²+r where r is not a square then c²=p²(q²+r)=(pq)²+p²r and here in all variants of numbers q and r it turns out that p²r also is not a square then the number c² also cannot be the sum of two squares.

53

This discovery was first stated in Fermat’s letter to Mersenne dated December 25, 1640. Here, in item 2-30 it is reported: “This number (a prime of type 4n+1) being the hypotenuse of one right triangle, its square will be the hypotenuse of two, cube — of three, biquadrate — of four etc. to infinity". This is an inattention that is amazing and completely unusual for Fermat, because the correct statement is given in the neighbor item 2-20. The same is repeated in Fermat’s remark on Bachet’s commentary to task 22 book III of Arithmetic by Diophantus. But here immediately after this obviously erroneous statement the correct one follows: “This a prime number and its square can be divided into two squares in only one way; its cube and biquadrate only two; its quadrate-cube and cube-cube only three, etc. to infinity"(see Pt. 3.6). In this letter Fermat apparently felt that something was wrong here, therefore he added the following phrase: “I am writing to you in such a hurry that I do not pay attention to the fact that there are errors and omit a lot of things, about which I tell you in detail another time”. This of course, is not that mistake, which could have serious consequences, but the fact is that this blunder has been published in the print media and Internet for the fourth century in a row! It turns out that the countless publications of Fermat's works no one had ever carefully read, otherwise one else his task would have appeared, which obviously would have no solution.

54

Euler's proof is not constructive i.e. it does not provide a method for calculating the two squares that make up a prime of type 4n+1 (see Appendix III). So far, this problem has only a Gauss' solution, but it was obtained in the framework of a very complicated system “Arithmetic of Deductions”. The solution Fermat reported is still unknown. However, see comment 172 in Appendix IV (Year 1680).

55

Methods of calculating prime numbers have been the subject of searches since ancient times. The most famous method was called the"Eratosthenes’ Sieve". Many other methods have also been developed, but they are not widely used. A fragment of Fermat’s letter with a description of the method he created, has been preserved the letter LVII 1643 [36]. In item 7 of the letter-testament he notes: “I confess that my invention to establish whether a given number is prime or not, is imperfect. But I have many ways and methods in order to reduce the number of divides and significantly reduce them facilitating usual work."See also Pt. 5.1 with comments 73-74.

56

Fermat discovered formula (2) after transforming the Pythagoras’ equation into an algebraic quadratic equation — see Appendix IV story Year 1652. However, an algebraic solution does not give an understanding the essence of the resulting formula. This method was first published in 2008 [30].

57

For example, if m = p₁p₂ then in addition to the first three solutions there will be others: A₄=p₁; B₄=2p₁p₂²; A₅=p₂; B₅=2p₁²p₂; A₆=2p₁; B₆=p₁p₂²; A₇=2p₂; B₇=p₂p₁²; A₈=p₁²; B₈=2p₂²; A₉=p₂²; B₉=2p₁²

58

Formula (7) is called Fermat Binomial. It is curious that the same name appeared in 1984 in the novel"Sharper than the epee"by the Soviet science fiction writer Alexander Kazantsev. This formula is not an identity because in contrast to the identity of Newton Binomial in addition to summands, there is also a sum of them, but with the help of Fermat Binominal it is easy to derive many useful identities in particular, factorization of the sum and difference of two identical powers [30], see also Pt. 4.4.

59

In this case, identity (9) indicates that the same key formula is substituted into the transformed key formula (2) or that the equation (8) we obtained, is a key formula (2) in power n. But you can go the reverse way just give the identity (9) and then divide into factor the differences of powers and such a way you can obtain (8) without using the Fermat Binominal (7). But this way can be a trick to hide the understanding of the essence because when some identity falls from the sky, it may seem that there is nothing to object. However, if you memorize only this path, there is a risk of exposure in a misunderstanding of the essence because the question how to obtain this identity, may go unanswered.

60

Taking into account that c−a=b−2m the expression in square brackets of equation (8) can be transformed as follows: (c++b)_n − (a++2m)_n = с^n-1− a^n-1+ c^n-2b− a^n-22m+ c^n-3b²− a^n-3(2m)²+… +b^n-1 − (2m)^n-1; с^n-1 − a^n-1 = (с − a)(c++a)_n-1; c^n-2b − a^n-22m = 2m(c^n-2 − a^n-2) + c^n-2(b − 2m) = (c − a)[2m(c++a)_n-2 + c^n-2]; c^n-3b² −a^n-3(2m)² = (2m)²(c^n-3 − a^n-3) + c^n-3(b² − 4m²) = (c − a)[4m²(c++a)_n-3 + c^n-3(b +2m)]; b^n-1 − (2m)^n-1 = (b − 2m)(b++2m)_n-1 = (c − a)(b++2m) _n-1 All differences of numbers except the first and last, can be set in general form: c^xb^y − a^x(2m)^y=(2m)^y(c^x − a^x) + c^x[b^y − (2m)^y] = (c − a)(c++a)_x(2m)^y + (b − 2m)(b++2m)_yc^x = (c − a)[(c++a)_x(2m)^y + (b++2m)_yc^x] And from here it is already become clear how the number (c − a) is take out of brackets. Similarly, you can take out of brackets the factor a + b = c + 2m. But this is possible only for odd powers n. In this case, equation (10) will have the form A_iB_iC_iD_i = (2m)ⁿ, where A_i = c — b = a − 2m; B_i = c — a = b − 2m; C_i = a + b = c + 2m; D_i — polynomial of power n − 3 [30].

61

Equation (10) can exist only if (1) holds i.e. {aⁿ+bⁿ−cⁿ}=0 therefore, any option with no solutions leads to the disappearance of this ghost equation. And in particular, there is no “refutation” that it is wrong to seek a solution for any combination of factors, since A_iB_i=2m² may contradict E_i=2^n-1m^n-2, when equating E_i to an integer does not always give integer solutions because a polynomial of power n−2 (remaining after take out the factor c−a) in this case may not consist only of integers. However, this argument does not refute the conclusion made, but rather strengthens it with another contradiction because E_i consists of the same numbers (a, b, c, m) as A_i, B_i where there can be only integers.

62

In this proof, it was quite logical to indicate such a combination of factors in equation (10), from which the Pythagoras’ numbers follow. However, there are many other possibilities to get the same conclusion from this equation. For example, in [30] a whole ten different options are given and if desired, you can find even more. It is easy to show that Fermat's equation (1) is also impossible for fractional rational numbers since in this case, they can be led to a common denominator, which can then be reduced. Then we get the case of solving the Fermat equation in integers, but it has already been proven that this is impossible. In this proof of the FLT new discoveries are used, which are not known to current science: there are the key formula (2), a new way to solve the Pythagoras’ equation (4), (5), (6), and the Fermat Binomial formula (7)… yes of course, else also magic numbers from Pt. 4.3!!!