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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)

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