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The Wonders of Arithmetic from Pierre Simon de Fermat. Youri Veniaminovich Kraskov
Читать онлайн.Название The Wonders of Arithmetic from Pierre Simon de Fermat
Год выпуска 2021
isbn 978-5-532-98629-9
Автор произведения Youri Veniaminovich Kraskov
Жанр Прочая образовательная литература
Издательство ЛитРес: Самиздат
Thus, in a sequence of numbers from δ1 to δi, all of them must be S-numbers i.e. each of them will consist of a sum minimum of k k-angle numbers, while our supposed number N will consist minimum of k+1 k-angle numbers. From (1) and (2) it follows:
N− mi =Si (3)
Thus, if we subtract any smaller polygonal number mi from our supposed number N then according to our assumption, the result should be only an S-number. Of course, this condition looks simply unbelievable and it seems that we are already at target, but then how can one prove that this is impossible? …
If we gave an answer to this question here, then this famous Fermat's theorem would immediately turn into the most common school problem and interest to it would be lost. To prevent this from happening, we will stay on the fact that the proof is presented here only by 99% and the remaining 1% will be offered to those who will be interested in order to appreciate the true magnificence of this scientific achievement of Fermat, especially in comparison with the Cauchy’s GFT proof.42
Pic. 34. Title Page the Cauchy's Proof
of the Fermat's Golden Theorem
Pic. 35. One of 43 Pages the Cauchy's Proof
of the Fermat's Golden Theorem
3.4.3. Archimedes-Fermat Problem
The problem statement is as follows:
Let any non-square number be given, you need to find an infinite number of squares, which after multiplication by this number and increasing by unit, will make a square.
Fermat proposed finding solutions for the numbers 61, 109, 149, and 433 [36].
The English mathematician John Wallis managed to find a way to calculate the required numbers using the Euclidean method of decomposing an irrational number into an infinite common fraction. He published his decision under the name "Commercium epistolicum" see pic. 37-38.
Pic. 36. John Wallis
Pic. 37. Title Page of Wallis's Publication Commercium Epistolicum
Although Wallis did not give a complete proof the validity of this method, Fermat nevertheless admitted that he had coped with the task. Euler came very close to the solution when he showed that this fraction is cyclical, but he was not able to complete the proof and this task was finally solved by Lagrange. Later, this Fermat's task also was solved by Gauss in his own way, but for this purpose the extensive theory he created called “Arithmetic of deductions” was involved. And everything would be fine if the Lagrange's proof was not in the category of highest difficulty and the Gauss decision was not based on the most complicated theory. Fermat himself clearly could not follow such ways. About how he himself solved this problem, he reports in the letter-testament to Carcavy in August 1659 [36]: “I recognize that Mr. Frenicle gave various special solutions to this question as well as Mr. Wallis, but a common solution will be found using the method of descent applied skillfully and appropriately." However, this Fermat's solutions so remained as the secret behind seven seals!
Pic. 38. Page 64 Commercium Epistolicum
Demonstrating Wallis Method
We will try here slightly to open the veil over this mystery. To do this, we will look at a simple example of Wallis calculations and then compare it with how one could do these calculations using Fermat's method. So, we need to find the smallest numbers x and y that satisfy the equation Ax2 + 1 = y2. Let A = 29 then calculations by the Wallis method look as follows [32]:
From this sequence of calculations, a chain of suitable fractions is obtained by backward i.e. from a5 to a0 and looks like: 5/1; 11/2; 16/3; 27/5. As a result, we get 70/13. Then the minimum solution would be:
x1√29+у1=(13√29+70)2=1820√29+9801; x1=1820; y1=9820
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, Am2−1 = z2. 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 61m2 – z2 = 1. Then the chain of calculations is obtained as follows:
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
Facsimile of the edition with the Cauchy's GTF proof was published by Google under the title MEMIRES DE LA CLASSE DES SCIENCES MATHTÉMATIQUES ET PHYSIQUES DE L’INSTITUT DE France. ANNEES 1813, 1814, 1815: https://books.google.de/books?id=k2pFAAAAcAAJ&pg=PA177#v=onepage&q&f=false What we need is on page 177 under the title DEMONSTRATION DU THÉORÉME GÉNÉRAL DE FERMAT, SUR LES NOMBRES POLYGONES. Par M. A. L. CAUCHY. Lu à l’Académie, le 13 novembre 1815 (see Pics 34, 35). The general proof of Cauchy takes 43 (!!!) pages and this circumstance alone indicates that it does not fit into any textbook. Such creations are not something that students, but also academics are not be available because the first cannot understand anything about them and the second simply do not have the necessary time for this. Then it turns out that such proofs are hardly possible to check how convincing they are i.e. are they any proofs in general? But if Cauchy applied the descent method recommended by Fermat, then the proof would become so convincing that no checks would be required. A very simple conclusion follows from this: