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The Wonders of Arithmetic from Pierre Simon de Fermat
Год написания книги
2021
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Wallis was unable to prove that this method of computation gives solutions for any non-square number A. However, he guessed that the chain of computations ends where a6 will be computed by the same formula as a1. To understand the meaning of this chain of calculations, you need to study a very voluminous and extremely difficult theory [7, 14, 19, 23, 26, 32], which Fermat could not have developed at that time. Since no Fermat's manuscripts on arithmetic have survived, a natural question arises: how could he formulate such a difficult problem, about which there was very little information before him?
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<x, z<y, Am
?1 = z
. 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
– z
= 1. Then the chain of calculations is obtained as follows:
61m
?z
=1
m=(8m
±z
)/3=(8?722+5639)/3=3805; z
=61?3805
?1=29718
61m
?z
=3
m=(8m
±z
)/3=(8?722+5639)/3=3805; z
=61?722
?1=29718
61m
?z
=9
m=(8m
±z
)/3=(8?722+5639)/3=3805; z
=61?137
?1=29718
61m
?z
=27
m
=(8m
±z
)/3=(8?5+38)/3=26; z
=61?26
?27=203
61m
?z
=81
m
=(8m
±z
)/3=(8?2?1)/3=5; z
=61?5
?81=38
61m
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