It would be simply excellent if today's science could explain Fermat's method in every detail, but even the ghostly hopes for this are not yet visible. It would be more realistic to expect that attempts will be made to refute this example as demonstration a method of solving the problem unknown to science. Nevertheless, science will have to reckon with the fact that this example is still the only one in history (!!!) confirmation of what Fermat said in his letter-testament. When this secret is fully revealed, then all skeptics will be put to shame and they will have no choice, but to recognize Fermat as greater than all the other greatest scientists because they were recognized as such mainly because they created theories so difficult for normal people to understand that they could only cause immense horror among students who now have to take the rap for such a science.43

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https://www.youtube.com/watch?v=ZhVNOgaBStY

In this sense, the following example of solving a problem using the descent method will be particularly interesting because it was proposed in a letter from Fermat to Mersenne at the end of 1636, i.e. the age of this task is almost four centuries! Euler's proof [8] was incorrect due to the use of "complex numbers" in it. However, even the revised version of André Weil in 1983 [17] is too complex for schooling.

In the original version in 1636 this task was formulated as follows:

Find two square-squares, which sum is equal to a square-square,

or two cubes, which sum is a cube.

This formulation was used by Fermat's opponents as the fact that Fermat had no proof of the FLT and limited himself to only these two special cases. However, the very name "The Fermat’s Last Theorem" appeared only after the publication of "Arithmetic" by Diophantus with Fermat's remarks in 1670 i.e. five years after his death. So, there is no any reason to assert that Fermat announced the FLT in 1637.

The first case for the fourth power we have presented in detail in Appendix II. As for the case for the third power, Fermat's own proof method restored by us below, will not leave any chances to the solutions of this problem of Euler and Weil to remain in history of science, since from the point of view of the simplicity and elegance of the author's solution this problem, they will become just unnecessary.

Now then, to prove that there are no two cubes whose sum is a cube, we use the simplest approach based on divisibility of numbers, what means that in the original equation

a^>3+b^>3 = c^>3 (1)

the numbers a, b, and c can be considered as coprime ones, i.e. they do not have common factors, but in general case this is not necessary, since if we prove that equation (1) cannot have solutions in any integers, including those with common factors, then we will prove that coprime numbers also cannot be solutions of the original equation. Then we assume that both sides of equation (1) in all cases must be divisible by the number c^>2, then equation (1) can be represented as

c^>3 = c^>2(x+y) = a^>3+b^>3 (2)

In this case, it is easily to see that there is only one way to get solutions to equation (1) when the numbers c, x, y, and x+y are cubes, i.e.

с = x+y = p^>3+q^>3= z^>3; x = p^>3; y = q^>3 (3)

Then equation (1) must have the form:

(z^>3)^>3 = (z^>2)^>3(p^>3+q^>3) (4)

Thus, we found that if there are numbers a, b, and c that satisfy equation (1), then there must be numbers p