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Chứng minh bài toán Ferma lớn
N.V. TỪ-YÊN * ŒUVRES POSTHUMES Posthumous works * HÀ-NỘI , VIỆT-NAM MAI 2013 * Ecce homo Paroles de Pilate aux Juifs , Saint- Jean, XIX ,5. Homo sapiens – demens From Anthropology . TABLE DES MATIERES 1 - An elementary proof of the Fermat’s last theorem . …. 2 - Une preuve élémentaire de l’infinité des nombres premiers jumeaux (p,p+2) et des paires voisines de nombres premiers jumeaux (p,p+2; p+6,p+8) . …. 3 - Essai d’une preuve élémentaire de la conjecture de Goldbach . * Le détail du déroulement de ces recherches peut être trouvé dans mon autobiographie “ Méditation posthume ”, 2007 . à la mémoire de mes parents Hân-Đoàn à la mémoire de ma femme Thọ à ma fille Thu-Uyên et ma petite fille Hạnh-Duyên * PREFACE Voici le bilan des recherches non-professionnelles d’un amateur durant presque tout le temps libre de sa vie. Ce bilan a été récapitulé et écrit dans la 79-ième année de cet homo sapiens-demens pour le soumettre aux professionnels. Non-professionnel et amateur, sapiens et demens – c’est le trait général de l’auteur que le lecteur doive le méditer. Que ces recherches tombent aux mains d’un examinateur de haute responsabilité, qui n’est point allergique sur des preuves dites élémentaires, et surtout , sur des chercheurs au nom des amateurs. Que ces mémoires rencontrent les lecteurs curieux et méditatifs. Que j’aime exprimer ma profonde reconnaissance à cet examinateur et à ces lecteurs . * This is the researche’s recapitulation of an amateur in almost the spare time of his life. The review is recapituled and written at the age of 79 by this homo sapiens- demens with a view to submit to the professionals . Unprofessional and amateur, sapiens and demens this is the trait general of the author, on which the reader necessitate to meditate. Let these researches fall into the hand of one reviewer of high responsibility, who is not allergic to all elementary proofs for the difficult mathematic problems, and more particularly , to all researchers named unprofessional . Let these works are coming with the curious and meditative readers. Let me express my heartfelt gratitude to this reviewer and to these readers. * MÉMOIRE 1 * - soumis le 27 Juillet 1990 - mon père est décédé le 13 Octobre 1989 à mon âge de 54 . * * * 1 AN ELEMENTARY PROOF OF THE FERMAT’S LAST THEOREM N V. TU -YEN ( Vietnam ) ABSTRACT- To examine the structural properties of the solutions , if existent , of the Diophantine equation : ( 0 ) a pair of the two procedures ( I ) where x , ; x odd, even and nnn zyx =+ i x in n i i n nn x exCxex − = ∑ +=+ 1 )( + ∈ Zne x , x e ( II ) where y , ; y even, odd is proposed. i y in n i i n nn y eyCyey − = ∑ +=+ 1 )( + ∈ Zne y , y e The fives structural properties of these procedures reveal a secret that if the eq.( 0 ) has solution, then the existence of one consecutive solution as a first and smallest one is unavoidable . From this the Fermat’s last theorem is proved simply by the well- known fact in the number theory that when any consecutive solution does not exist for the equation ( 0 ) . 3≥n INTRODUCTION Although the Fermat’s last theorem had been proved in 1994 by the English mathematician A.Wiles, the question: “Is it possible to prove this theorem in an elementary way using only the mathematical knowledge in Fermat’s times ? ” posed during the whole last 371 years had been left unsolved. In honour of Fermat and of so many people, known and unknown after him, who had essayed to prove this famous theorem, I introduce here an elementary answer to this question. I . TWO PROCEDURES PROPOSED FOR THE PROOF 1 -Two procedures to find the triplet- solution (x, y, z) of eq.( 0 ). Assuming that there is a solution ( x, y, z ) of the eq. ( 0 ). Then we have : x, y, z and from this, . Noting that the n- dimensional cubes are always represented correspondingly by the 3- dimensional right prisms with squares as their bases and as their altitudes, we conclude the identical parity of x, y and z for all n . Now let x be odd, y be even then z and will be odd and will be even. For solving ( 0 ) we consider constantly its two of three unknowns, say x and z as positive intergers to find the third interger y, or y and z as positive intergers to find the third interger x of the triplet- solution. With this aim, two following procedures are proposed to nnn zyx =+ + ∈ Z ++ ∈−=∈−= ZyzeZxze yx , nnn zandyx , 222 , zandyx 222 , −−− nnn zandyx 2≥ y e x e 2 find the solution of eq. ( 0 ) : ( I ) with x odd , even ; ∑ = − +=+ n i i x ini n nn x exCxex 1 )( + ∈ Zex x , x e i y in n i i n nn y eyCyey − = ∑ +=+ 1 )( ( II ) with , y even , odd ; + ∈ Zey y , y e Clearly, in the procedure ( I ) because then if there is an y = + ∈ Zex x , + = − ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ ZexC n n i i x ini n 1 1 , or in the procedure ( II ) because then if there is an x = + ∈ Zey y , + = − ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ ZeyC n n i i y ni n 1 1 1 we have certainly a solution ( x, y, z ) of the equation ( 0 ). 2 - The use of procedures ( I ) and ( II ). Using ( I ) : For all odd x successively in their increasing order taking by turns each even beginning with , we calculate ( I ). If for some odd x and some even , we have y = x e 2 min = x e x e + = − ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ ZexC n n i i x ini n 1 1 then the triplet obtained ( x, y, z = x+ ) is one solution of ( 0 ). x e Using ( II ) : For all even y successively in their increasing order taking by turns each odd beginning with = 1 , we calculate ( II ). If for some even y and some odd , we have x = y e miny e y e + = − ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ ZeyC n n i i y ini n 1 1 then the triplet obtained ( x, y, z = y+ ) is one solution of ( 0 ). y e 3 - Solution equivalence between equations ( I ), ( II ) and the equation ( 0 ). Let ( x, y, z ) be a solution of ( 0 ). Then there are an even number and an odd number xze x −= yze y − = . Intuitively the solution equivalence of ( I ), ( II ) and ( 0 ) is evident from the fact that, by using ( I ) only if we have then ( I ) and ( 0 ) will become identity, and by using ( II ) only if we have then ( II ) and ( 0 ) will become identity. More rigorously the proof of this solution -equivalence can be represented as follows : Putting two integers : x , z = x+ of this solution into ( I ), we have with the same third integer y in the solution of ( 0 ). Now let ( ) be a solution of ( I ). From this and + ∈ Zy + ∈Zx x e ni x in n i i n yexC = − = ∑ 1 111 ,, zyx 111 xze x −= + = − ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∑ ZexCy n n i i x ini n 1 1 111 , i. e. we 3 have with . In other words, ( ) is also a solution of the equation ( 0 ). nnn zyx 111 =+ + ∈+= Zexz x111 111 ,, zyx Similarly, let ( ) be a solution of ( II ). From this and 222 ,, zyx 222 yze y −= + = − ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∑ ZeyCx n n i i y ini n 1 1 222 , i. e. we have with . In other words, ( ) is also a solution of the equation ( 0 ). nnn zyx 222 =+ + ∈+= Zeyz y222 222 ,, zyx 4 - Definitions connected with the procedures. Suppose that ( 0 ) or ( I ) and ( II ) have some solution (x,y,z). Then we have x, y, z, and n and, evidently, and . + ∈ Z + ∈−= Zxze x + ∈−= Zyze y Definition1. Solution in the form (x,y,z) is called explicit solution. Definition 2. Solution in the form ( ) is called implicit solution. yx ee , Definition 3. Solution is called irreducible if its greatest common divisor equals 1, i. e. gcd (x,y,z) = gcd ( ) = 1. yx ee , Definition 4. Solution is called k- proportional if its greatest common divisor equals k, i. e. gcd (x,y,z) = gcd ( ) = k. yx ee , Definition 5. Solution is called generating if it is found by using 2 min = = xx ee in ( I ) and by using 1 min = = yy ee in ( II ). Their implicit forms are ( ) and ( ) respectively. yx ee , min min , yx ee Definition 6. Solution is called combining if its implicit form is ( ) where there are yx ee , minxx ee ≠ and minyy ee ≠ . Definition 7. Solution is called the first or the smallest if its explicit form is ( ), and its implicit form is ( ). Here “ first ” is clear from the use of procedures ( I ) and ( II ) stated above, and “ smallest ” will be evidenced by the property 2 presented below. minminmin ,, zyx minmin , yx ee Definition 8. Solution ( x, y, z ) is called consecutive if its explicit form is (x, x+1, x+2) and its implicit form is ( ). This is evident from the procedures ( I ) and ( II ) that : x+2 = y+1 = z, i. e. there is ( x, y, z ) = ( x, x+1, x+2 ). minmin , yx ee Furthermore : + Z denotes the set of all positive integers; R denotes the set of all real numbers. ( ) )!(! ! ini n C n i i n − == is the i- th binomial coefficient. II - STRUCTURAL PROPERTIES OF THE TRIPLET NUMBERS ( x, y, z ) ON THE REAL NUMBER LINE R. In the procedure ( I ) with , for each odd x and each even , there is always an + ∈ Zex x , x e . 1 1 RexCy n n i i x ini n ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∑ = − In the 4 particular case when then the triplet numbers (x, y, z=x+ ) obtained is a solution of ( 0 ). Similarly in the procedure ( II ) with for each even y and each odd , there is always an + ∈ Zy x e + ∈Zey y , y e . 1 1 ReyCx n n i i y ini n ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∑ = − In the particular case when then the triplet numbers (x, y, z=y+ ) obtained is a solution of (0) + ∈Zx y e 1. Two ways to determine each solution. Property 1 - If the equation ( 0 ) has some solution then this solution can be found by two ways : Either by using the procedure ( I ) or by using the procedure ( II ). Proof - Suppose the equation ( 0 ) has a solution (x, y, z) in explicit form and ( ) in implicit form. We will prove that if this solution is found by the procedure ( I ) then it will be found by the procedure ( II ) and vice versa. In fact if by using ( I ) : we obtain a solution (x, y, z) but by using ( II ) we obtain an other one, say (x’, y, z), satisfying .Then this leads to a contradiction that Hence x’ = x. Similarly if by using ( II ) : + but by using( I )we obtain an other one, say (x, y’, z ), satisfying .Then this leads to a contradiction that . Hence y’ = y. yx ee , nnn x yxex +=+ )( n z= nnnn y zyxey =+=+ ')( .' nnnn yxyx +=+ y( nnnn y zxye =+=) nnnn x zyxex =+=+ ')( nnnn yxyx '+=+ 2. Proportionality of the triplet numbers (x, y, z) to the values of and . x e y e Property 2 - The values of each real triplet (x, y, z) of the equation ( 0 ) is directly proportional to 2 min = x e in ( I ) and to in ( II ). 1 min = y e Proof - Note that any even in ( I ) can be written by where and x e minxxx eke = + ∈ Zk x 2 min = x e , and any odd in ( II ) can be written by y e minyyy eke = where and is odd; + ∈ Zk y y k .1 min = y e 1 ) In the procedure ( I ) when using , for each odd x, we obtain a triplet of real numbers (x, y, z = x + , where x, z ; (1a) and in general minx e ) minx e + ∈Z xze x −= min . 1 1 min RexCy n n i i x ini n ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∑ = − (2a). Now when using minxxx eke = we will obtain a new triplet 5 on the real number line, say (X, Y, Z = X+ ) where Z-X (3a). Evidently, X , Z and in general, x e = x e + ∈Z ReXCY n n i i x ini n ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∑ = − 1 1 (4a). But from (1a), there is xkzkxzkeke xxxxxx − = −== )( min (5a). From (3a) and (5a) we have Z = (6a) and X = (7a). Noting zk x xk x = x e and putting (7a) into (4a) we get Y = minxx ek n n i i x ini nx exCk 1 1 min ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ = − R∈ (8a). In comparison (8a) with (2a) we get Y = y x k R ∈ (9a). Therefore if by procedure ( I ) using we get a triplet of real numbers (x,y,z) positioned on the number line R, then when using minx e minxxx eke = this triplet will be shifted forward and get a new position at (X, Y, Z)= (x, y, z) = ( ). This conclusion is evidently true also in the case where , i. e. when (x, y, z) is a solution of the equation ( 0 ). x k zkykxk xxx ,, + ∈Zy 2 ) In the procedure ( II ) when using for each even y we obtain a triplet of real numbers (x’, y’, z’= y’+ ) where = z’-y’ (1b), y’, z’ miny e miny e miny e + ∈ Z and, in general, x’ = ReyC n n i i y ini n ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ = − 1 1 min ' (2b). Now when using minyyy eke = we will obtain a new triplet on the real number line R, say (X’, Y’, Z’= Y+ ) , where y e =Z’-Y’ (3b), Y’,Z’ y e + ∈ Z and in general X’= n n i i y ini n eYC 1 1 ' ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ = − R ∈ (4b). But from (1b) there is '')''( min ykzkyzkeke yyyyyy − = −== (5b). From (3b) and (5b) we have Z’ = z’ (6b) and Y’ = y’ (7b). Noting y k y k minyyy eke = and putting (7b) into (4b) we get X’ = n n i i y ini ny eyCk 1 1 min ' ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∑ = − R∈ (8b). In comparison (8b) with (2b) we get X’ = x’ (9b). y k Therefore if by procedure ( II ) using we get a triplet of real numbers (x’, y’, z’) positioned on the number line R, then when using this triplet will be shifted forward and get a new position at (X’, Y’, Z’)= (x’,y’,z’) = ( ). This conclusion is evidently true also in the case where x’ miny e minyyy eke = y k ,'xk y ',' zkyk yy + ∈ Z , i. e. when (x’, y’, z’) is a solution of the equation ( 0 ). 3- Decisive role of and in the existence or the non- existence of solution of the equation ( 0 ). minx e miny e [...]... dans la paragraphe III 2 II - TERMINOLOGIE : NOTATIONS ET DÉFINITIONS 1 - L’ Alphabet français : Pour noter les nombres entiers et les indices 2 - N : Axe des nombres naturels : 0 1 2 … n–1 n R N : Son opposé : n n-1 … 2 1 0 R N N : Leur superposé 1 3 5 … 2n-3 2n-1 3 - N I : Axe des nombres impairs : N R : Son opposé I : 2n-1 2n-3 … N I N : Leur superposé 2 3 4 - P : Axe des nombres premiers... quand n pair et de (n+1)/2 quand n impair : N : 1 3 5 … 2n-5 2n-3 2n-1 R N : 2n-1 2n-3 2n-5 …… 5 3 1 D’où l’on voit bien que les paires de Goldbach ( p i , p j ) pour p i + p j = 2n , si elles existent , peuvent être cherchées en superposant deux axes des nombres premiers opposés en direction P et P R dans cet intervalle : pk … 2n-1 P : 1 pi PR: 2 2n-1 pj pm … 1 Partage d’un pain en parties fractionnaires... Author’s Address N.V TU -YEN NHA E 1 , P 406 VAN CHUONG DONG DA - HANOI- VIETNAM E- mail : thyen28@gmail.com MÉMOIRE 2 * - soumis en 2007 à mon âge de 72 * * * UNE PREUVE ÉLÉMENTAIRE DE L’INFINITÉ DES NOMBRES PREMIERS JUMEAUX ( p, p+2 ) ET DES PAIRES VOISINES DE NOMBRES PREMIERS JUMEAUX ( p, p+2 ; p+6, p+8 ) N V TU- YEN ( Vietnam ) * - In imitation of the Eratosthenes sieve for... 23 Novembre 2008 N V TU- YEN -Adresse postale Adresse E- mail thyen28@gmail.com N V TU- YEN Nha E-1, Phong 406 Q Van chuong , P Dong da Hanoi - Vietnam 12 UNE PROPOSITION COMPLÉ MENTAIRE de l’article “Une preuve élémentaire de l’infinité des nombres premiers jumeaux ( p , p + 2 ) et des paires voisines de nombres premiers jumeaux ( p , p + 2 ; p + 6 , p + 8 ) - m Dans cet article nous... la relation P c (2n) < P(2n) , c’est –à-dire dans ce cas : P c ( 2n ) = P ( 2n ) - e ; où e est un entier positif En résumé , par ces deux facteurs , on a : P c ( 2n ) = P ( 2n ) – ( m + e ) D’où la proposition est prouvée MÉMOIRE 3 * - écrit le 17 Mai 2013 - 70 ième anniversaire de la mort de ma mère - 24 ième anniversaire de la mort de mon père - 4 ième anniversaire de la mort de ma... (p), two sieves for sifting the twin- primes (p, p+2) and the neighbouring pairs of twin- primes (p,p+2 ; p+6,p+8) are proposed From the structure of these sieves, after the Legendre’s manner of deducing the density- limit of the primes D∞ ( p) , the formulas describing the density- limits of the twin- primes ABSTRACT D∞ ( p, p + 2) and of the neighbouring pairs of twin- primes D∞ ( p, p + 2; p + 6, p... le reste total après ce partage ? ” La réponse peut être trouvée après avoir raisonné comme suit : Le premier reste du pain : r 1 = ( 1-1 /3 ) = 2/3 Le deuxième reste du pain : r 2 = ( 1-1 /3 ) ( 1-1 /5 ) = 8/15 Le reste total du pain : r 3 = ( 1-1 /3 ) ( 1-1 /5 ) ( 1-1 /7 ) = 48/105 Autrement dit, pour ce partage il est raisonnable de diviser le pain en 3.5.7=105 morceaux égaux, alors le reste total en... non- existence of any solution, including the first smallest ( x1 1 , y1 1 , z1 1 ) that is consecutive one : ( e x min , e y min ) or ( x, x+1, x+2 ) 2 - The Fermat’s last theorem The Diophantine equation x n + y n = z n ( 0 ) has no non- zero positive integer solution for x, y, z when n ∈ Z + and n >2 Proof - The lemma established the equivalence between the statement of this theorem and the non-... nombres premiers à partir de 9 est présenté ci- après 2 - Schéma 1 : Crible 1 d’Ératosthène pour nombres premiers 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 1 1 0 1 1 1 1 1 1 -0 1 1 0 1 1 0 1 0 0 1 1 0 0 1 2 Remarques - 1) Car 37 < p 4 = 7 2 , il est nécessaire... dans un intervalle quelconque ( 0,2n ) de l’axe des nombres naturels N , est- ce-que l’utilisation d’une superposition de deux cribles d’Eratosthène mutuellement renversés droite-gauche E et E R peut retenir sur ce crible superposé , noté E E seulement des paires de Goldbach ( p i ,p j ) où p i + p j =2n ? R Ensuite , est-ce-que l’on peut considérer chaque intervalle partiel ( 0,2n ) enlevé d’un axe . N.V. TỪ-YÊN * ŒUVRES POSTHUMES Posthumous works * HÀ-NỘI , VIỆT-NAM MAI 2013 * Ecce homo Paroles. mathematic problems, and more particularly , to all researchers named unprofessional . Let these works are coming with the curious and meditative readers. Let me express my heartfelt gratitude