The Wonders of Arithmetic from Pierre Simon de Fermat. Youri Veniaminovich Kraskov
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:
AiBi=2m2 (4)
where taking into account the formula (2):
Ai=c−b=a−2m; Bi=c−a=b−2m (5)
Now we decompose the number 2m2 into prime factors to get all the AiBi 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=Ai+2m; b=Bi+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=zzn-1=(x+y)zn-1=xzzn-2+yzn-1=
x(x+y)zn-2+yzn-1=x2zzn-3+y(zn-1+xzn-2)+…
(x±y)n=zn=xn±y(xn-1+xn-2z+xn-3z2+…+xzn-2+zn-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)[an-1+an-22m+…
+ a(2m)n-1+(2m)n-1]=(2m)n
Now through identity
(cn−bn)=(c−b)(cn-1+cn-2b+…+cbn-2+bn-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:
AiBiEi=(2m)n (10)
where Ai = c−b=a−2m; Bi=c−a=b−2m; Ei – 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 Ai×Bi×Ei=2m2×2n-1mn-2 then as one of the options could be such a system:
AiBi=2m2
Ei=2n-1mn-2
In this case, as we have already established above, it follows from AiBi=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
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
Formula (7) is called
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
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 = сn-1− an-1+ cn-2b− an-22m+ cn-3b2− an-3(2m)2+ … +bn-1 − (2m)n-1; сn-1 − an-1 = (с − a)(c++a)n-1; cn-2b − an-22m = 2m(cn-2 − an-2) + cn-2(b − 2m) = (c − a)[2m(c++a)n-2 + cn-2]; cn-3b2 −an-3(2m)2 = (2m)2(cn-3 − an-3) + cn-3(b2 − 4m2) = (c − a)[4m2(c++a)n-3 + cn-3(b +2m)]; bn-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: cxby − ax(2m)y=(2m)y(cx − ax) + cx[by − (2m)y] = (c − a)(c++a)x(2m)y + (b − 2m)(b++2m)ycx = (c − a)[(c++a)x(2m)y + (b++2m)ycx] 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 AiBiCiDi = (2m)n, where Ai = c – b = a − 2m; Bi = c – a = b − 2m; Ci = a + b = c + 2m; Di – polynomial of power n − 3 [30].
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 AiBi=2m2 may contradict Ei=2n-1mn-2, when equating Ei 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 Ei consists of the same numbers (a, b, c, m) as Ai, Bi where there can be only integers.
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!!!