61m2−z2=1
m=(8m1±z1)/3=(8×722+5639)/3=3805; z2=61×38052−1=297182
61m12−z12=3
m=(8m1±z1)/3=(8×722+5639)/3=3805; z12=61×7222−1=297182
61m22−z22=9
m=(8m1±z1)/3=(8×722+5639)/3=3805; z22=61×1372−1=297182
61m32−z32=27
m3=(8m4±z4)/3=(8×5+38)/3=26; z32=61×262−27=2032
61m42−z42=81
m4=(8m5±z5)/3=(8×2−1)/3=5; z42=61×52−81=382
61m52−z52=243
m5=2; z52=1
We will not reveal all nuances of this method, otherwise all interest to this problem would have been lost. We only note that in comparison with Wallis method where the descent method is not used, here it is present in an explicit form. This is expressed in the fact that if the numbers m and z satisfying the equation 61m2–z2=1 exist, then there must still exist numbers m1<m and z1<z satisfying the equation 61m12–z12=3, as well as the numbers m2<m1 and z2<z1, from equation 61m22–z22=9, etc. up to the minimum values m5<m4 and z5<z4. The number 3 appearing in the descent is calculated as 64 – 61, that is, as the difference between 61 and the square closest to it. Calculations as well as in the Wallis method are carried out in the reverse order i.e. only after the minimum values of m5 and z5 have been calculated. As a result, we get:
m=3805; z=29718
x=2mz=2×3805×29718=226153980
y=√(61×2261539802+1)=1766319049
Of course, connoisseurs of the current theory will quickly notice in this example that the results of calculations obtained in it will exactly coincide with those that can be obtained by the Wallis' method. However, for this they will have to use the irrational number √61, and our example with Fermat's method showed that it is possible to do calculations exclusively in the framework of arithmetic i.e. only in natural numbers. There is no doubt also that experts without much effort will guess how to get the formulas shown in our example. However, it will not be easily for them to explain how to apply this Fermat's method in the general case because from our example it is not at all clear how it is possible to determine that the ultimate goal is to solve the equation 61m52 – z52 = 243 from which calculations should be performed with a countdown.
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
https://www.youtube.com/watch?v=wFz8W2HsjfQ
https://www.youtube.com/watch?v=cUytn2SZ1n4
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.
3.4.4. Fermat’s Problem with Age 385 years
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
a3+b3 = c3 (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 c2, then equation (1) can be represented as
c3 = c2(x+y) = a3+b3 (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 = p3+q3= z3; x = p3; y = q3 (3)
Then equation (1) must have the form:
(z3)3 = (z2)3(p3+q3) (4)
Thus, we found that if there are numbers a, b, and c that satisfy equation (1), then there must be numbers p<a, q<b, and z<c that satisfy equations (3)
p3+q3= z3
If we now apply the same approach to solving this equation, that we applied to solving equation (1), we will get the same equation, only with smaller numbers. However, since it is impossible to infinitely reduce natural numbers, it follows that equation (1) has no solutions in integers.
At first glance, we have received a very simple and quite convincing proof of the Fermat problem by the descent method, which no one has been able to obtain in such a simple way for 385 years, and we can only be happy about it. However, such a conclusion would be too hasty, since this proof is actually incorrect and can be refuted in the most unexpected way.
