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 (an-1+bn-1)>cn-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:
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
