For today's science such a question is clearly beyond its capabilities since for it the pinnacle of achievements in solving Fermat's problems is any result even inflated to such incredible dimensions that we have today. However, it is difficult to imagine how much this our respected science will be dejected when from this book it learns that the problem was solved by Fermat not for great scientists, but … for schoolchildren!!! However, here we cannot afford to grieve science so much, so we only note that the example given in the textbooks is very unfortunate since it can be solved quite simply, namely: x = 2mz, where m>2−1 = z^>2. This last equation differs from the initial one only in sign and even by the method of ordinary tests without resorting to irrational numbers one can easily find the solution m = 13; z = 70; x = 2 x 13 x 70 = 1820; y = 9820.

Obviously, in textbooks it would be much more appropriate to demonstrate an example with the number 61 i.e. the smallest number proposed by Fermat himself. How he himself solved this problem is unknown to science, but we have already repeatedly demonstrated that it is not a problem for us to find out. We just need to look once more into the cache of the Toulousean senator and as soon as we succeeded, we quickly found the right example so that it could be compared with the Wallis method. In this example you can calculate x = 2mz, where m and z are solutions to the corresponding equation 61m^>2 – z^>2 = 1. Then the chain of calculations is obtained as follows:

61m^>2−z^>2=1

m=(8m_>1±z_>1)/3=(8×722+5639)/3=3805; z^>2=61×3805^>2−1=29718^>2

61m_>1^>2−z_>1^>2=3

m=(8m_>1±z_>1)/3=(8×722+5639)/3=3805; z_>1^>2=61×722^>2−1=29718^>2

61m_>2^>2−z_>2^>2=9

m=(8m_>1±z_>1)/3=(8×722+5639)/3=3805; z_>2^>2=61×137^>2−1=29718^>2

61m_>3^>2−z_>3^>2=27

m_>3=(8m_>4±z_>4)/3=(8×5+38)/3=26; z_>3^>2=61×26^>2−27=203^>2

61m_>4^>2−z_>4^>2=81

m_>4=(8m_>5±z_>5)/3=(8×2−1)/3=5; z_>4^>2=61×5^>2−81=38^>2

61m_>5^>2−z_>5^>2=243

m_>5=2; z_>5^>2=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 61m^>2–z^>2=1 exist, then there must still exist numbers m_>1>1>1^>2–z_>1^>2=3, as well as the numbers m_>2>1 and z_>2>1, from equation 61m_>2^>2–z_>2^>2=9, etc. up to the minimum values m_>5>4 and z_>5>4. 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 m_>5 and z_>5 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 61m