Читать книгу The Wonders of Arithmetic from Pierre Simon de Fermat - Youri Veniaminovich Kraskov - Страница 17

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!

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!!!