Posthumous works - Chứng minh bài toán Ferma lớn

Chứng minh bài toán Ferma lớn

TABLE DES MATIERES

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

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

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

*

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

n n n

z y

i x i n n i

i n n

n

x x C x e e

1

)(

i n n

n

y y C y e e

1

)(

+

∈ Z

n

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

,y and z x

2 2

x x C x e e

x

1

)(

i n n

( ( II ) with y,e y ∈ Z+, y even , e y odd ;

Clearly, in the procedure ( I ) because then if there

i

i x i n i n

1 1

, or in the procedure ( II ) because

i

i y n i 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

i

i x i n

i

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

i

i y i n i

n

1 1

then the triplet obtained ( x, y, z = y+ )

Let ( x, y, z ) be a solution of ( 0 ) Then there are an even

number e x = z−x and an odd number e y =z− 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 )

1,y ,z x

1 1

n n

i

i x i n i n

1 1

1 1

have with In other words,

( ) is also a solution of the equation ( 0 )

n n n

z y

2,y ,z x

n n

i

i y i n i n

1 1

2 2

also a solution of the equation ( 0 )

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,

Definition1 Solution in the form (x,y,z) is called explicit solution

Definition 2 Solution in the form (e , x e y) is called implicit solution

Definition 3 Solution is called irreducible if its greatest common

divisor equals 1, i e gcd (x,y,z) = gcd (e , x e y) = 1

Definition 4 Solution is called k- proportional if its greatest

common divisor equals k, i e gcd (x,y,z) = gcd (e , x e y) = k

Definition 5 Solution is called generating if it is found by using

2

min =

= x

x e

e in ( I ) and by using e y=e ymin =1 in ( II ) Their

implicit forms are (e xmin,e y) and (e x,e ymin) respectively

Definition 6 Solution is called combining if its implicit form is

(e , x e y) where there are e x ≠e xmin and e y ≠e ymin

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

min min min,y ,z x

min

min, y

x e

e

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 )

min min, y

x e e

Furthermore : Z denotes the set of all positive integers; +

R denotes the set of all real numbers ( )

)!

(

!

i n i

In the procedure ( I ) with , for each odd x and each

even , there is always an

R e

x C y

n n

i

i x i n i

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

R e

y C

x

n n

i

i y i n i

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,

contradiction that Hence y’ = y

y

x e

e ,

n n

n

n

y x

y

n n n

n

y y x z

n n n n

x x y z e

(

n n n n

y x y

Property 2 - The values of each real triplet (x, y, z) of the

equation ( 0 ) is directly proportional to e xmin =2 in ( I ) and to

min R e

x C y

n n

i

i x i n i

on the real number line, say (X, Y, Z = X+ ) where

Z-X (3a) Evidently, X , Z and in general,

X

C

Y

n n

i

i x i n

(4a) But from (1a), there is

x k z k x z k e

k

e x = x xmin = x( − )= x − x (5a) From (3a) and (5a) we

have Z =k x z (6a) and X =k x x (7a) Noting e x = and

putting (7a) into (4a) we get Y =

min

x

x e k

n n

i

i x i n i n

x C x e k

1 1

In comparison (8a) with (2a) we get Y = k xy ∈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

x k e

e = 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 )

2 ) In the procedure ( II ) when using for each even y we

obtain a triplet of real numbers (x’, y’, z’= y’+ ) where =

i

i y i n i

min

'

(2b) Now when using e y =k y e ymin we will obtain a new triplet

on the real number line R, say (X’, Y’, Z’= Y+e y) , where

i

i y i n i

n Y e

C

1 1

min k z y k z k y

e

k

e y = y y = y − = y − y (5b) From (3b) and (5b) we

have Z’ = k yz’ (6b) and Y’ = k yy’ (7b) Noting e y =k y e ymin and

putting (7b) into (4b) we get X’ =

n n

i

i y i n i n

y C y e k

1 1

In comparison (8b) with (2b) we get X’ = k yx’ (9b)

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’

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 )

min

x

Property 3 - The equation ( 0 ) has no solution when either using

in the procedure ( I ) we have no any or using

in the procedure ( II ) we have no any

Proof - As evidenced by the argumentation in the proof of

property 2, we see that :

1 ) By the use of procedure ( I ) to find the solution ( x, y, z ) of

eq ( 0 ) with , because the two variables x and z are always

chosen as positive integers, the triplet obtained ( x, y, z ) will be a

solution of ( 0 ) if and only if the third variable calculated as

min

x

e

n n

i

i x i n

and > 1, a - proportional triplet (X,Y,Z)=

( ) will be obtained where X, Z

i

i x i n i

n

x C x e

k

1 1

will be positive integer if and only if the variable y in (2a)

calculated by using is a positive integer

Therefore if using in the procedure ( I ) we find no any

then there will be no any to find the

solutions of the equation ( 0 ) for the procedure ( II ) Thus,

without using the procedure ( II ) it can be concluded in this case

that the equation ( 0 ) has no solution

2 ) By the use of procedure ( II ) to find the solution ( x’, y’, z’)

of eq ( 0 ) with , because the two variables y’ and z’ are

always chosen as positive integers, the triplet obtained ( x’, y’, z’ )

will be a solution of ( 0 ) if and only if the third variable

i

i y i n i

n y e C

1 1

Now when using all other odd greater than , say

where is odd integer greater than 1, a -

proportional triplet (X’, Y’, Z’) = ( ') can be obtained

where Y’ and Z’ are positive integers and X’ =

n n

i

i y i n

be positive integer if and only if the variable x’ in (2b) calculated

by using e ymin is a positive integer

Therefore if using in the procedure ( II ) we find no any x’

Property 4 - If the equation ( 0 ) has some irreducible solution

then this solution will be found either by the procedure ( I ) at

Proof - From the property 2, the existence of some solution (x,

y, z) of ( 0 ) is uniquely depended on its existence either by the procedure ( I ) at or by the procedure ( II ) at Now at

the procedure ( I ) or at ( ) in the procedure ( II ) Hence after eliminating the common factors of proportionality

or we get always an irreducible solution (x, y, z) of the equation ( 0 )

z k y k x

k x , x , x

z k y k x

*5 Generating characteristic of the set { } found by using

in the procedure ( II ) and the set { } found by using

in the procedure ( I )

i x

About the last fifth property I would like to note that this

property does not play any role for the proof of the Fermat’s last theorem It is introduced here to understand the creation of the combining solutions, found by each pair of the Cartesian product

{ } x { } In other words for an elementary proof presented in

this paper only four properties given above are necessary and sufficient

x e

e min,,e

e

2 ) Each pair ( ) of the Cartesian product { } x { }

where i, j = 2, 3, , n , corresponds with one solution of the

equation ( 0 ), which is either irreducible if gcd ( ) = 1 or

k – proportional if gcd ( ) = k

j y i

x e

j y i

x e

e ,

j y i

x e

e ,

Proof - 1 ) The affirmation is evident by the fact that there are

always gcd (e xmin,e y j) = 1 for all j and gcd ( e x i,e ymin) = 1 for all i

2 ) Suppose that the implicit pair ( ) at some i,

j = 2,3, , n does not correspond with a solution, irreducible or

proportional , of the equation ( 0 ) Then when using the

But this leads to , which contradicts the given

As the same, when using the procedure ( II ) with

we have no any But this leads to , which contradicts the given Hence each combining pair ( ) at any i and j must be

corresponding with one triplet of positive integers ( ),

which satisfies the equality ( 0 )

j y i

x e

e ,

j j

j y z

For resuming, we present here a distribution’s picture of the

considered solutions in the form of a ( n x n ) matrix as follows :

Table 1 Representation of the solutions in the procedures (I), (II)

III APPLICATION OF THE PROPOSED PROCEDURES

TO FIND THE PYTHAGOREAN TRIPLETS

Before presenting a proof of the Fermat’s last theorem I would

like to illustrate an application of the two proposed procedures (I )

and ( II ) for finding all triplet – solutions of the Pythagorean case

n = 2, which are represented in form of a square matrix with

infinitely many rows and columns ( see Table 1 )

Table 2 Pythagorean triplets found by ( I ) and ( II )

e yj

xi e 12 2

3 2

5 2

7 2

9

2.1.1 3, 4, 5 15, 8, 17 35, 12, 37 63, 16, 65 99, 20,101

2.2.2 5, 12, 13 21, 20, 29 45, 28, 53 77,36,85 117, 44, 125 2.3.3 7, 24, 25 27, 36, 45 55, 48, 73 91, 60, 109 135, 72, 153

2.4.4 9, 40, 41 33, 56, 65 65, 72, 97 105, 88, 137 153,104,185

2.5.5 11, 60, 61 39, 80, 89 75, 100, 125 119,120,169 171,140,221

2.6.6 13,84,85 45,108,117 85,132,157 133,156,205 189,180,261

2.7.7 15,112,113 51,140,149 95,168,193 147,196,245 207,224,305

2.8.8 17,144,145 57,176,185 105,208,233 161,240,289 225,272,353

2.9.9 19,180,181 63,216,225 115,252,277 175,288,337 243,324,405

Let i , j be the i- th row and the j- th column respectively, then the Pythagorean triplets ( ) and the sets { } , { } are firstly represented here by the following formulas, which are easily verified by the identity :

j j j y z x , , e x i j y e 2 2 2 ij ij ij y z x + = x j =(2j−1)[2(i+ j)−1] y j =2i(2j+i−1) z j =(2j−1)[2(i+ j)−1]+2i2= 2

) 1 2 ( ) 1 2 ( 2i j+i− + j− {e x i} = { 2} and { } = { } with i , j = 1, 2, 3, ,

2i e y j (2j−1)2 ∞

From the Table 2 it is easy to verify all the five properties presented above In particular, the property 5 shows us the capacity for finding the existent solutions of the eq ( 0 ) by the generating sets { }e x i given by using e ymin =1 in the procedure (II) and { }e y j given by using e xmin =2 in the procedure ( I ), with considering each and each of them as an indivisible unit such as This important fact is evidenced from the irreducibility of all the pairs

i x

min min y

x and e

e

) , ( )

, (e xmin e y j and e x i e ymin

IV - PROOF OF THE FERMAT’S LAST THEOREM

1 - Lemma The Diophantine equation ( 0 ) for x,

y, z, n

n n n

z y

+

∈ Z has solution if and only if it has a consecutive

solution as the smallest one

Proof - First of all, remember some facts given above From

the definition 8, a consecutive solution must be either of the implicit form ( ) or of the explicit form ( x, x+1, x+2 ) Then from the use of the procedures ( I ) and ( II ); and the property 2 , this consecutive solution must be either the first and the smallest one found by the procedure ( I ) using , or the

min min, y

x e e

e

first and the smallest one found by the procedure ( II ) using Following the property 1, this solution is also the first and the smallest one found by both procedures ( I ) and ( II ) simultaneously

( 0 ) is empty This means that there does not exist any solution

or ( x, x+1, x+2 )

1 1

1,y ,z

2 - The Fermat’s last theorem The Diophantine equation

( 0 ) has no non- zero positive integer solution for x,

Proof - The lemma established the equivalence between the

statement of this theorem and the non- existence of the consecutive solution for the equation ( 0 ) But the non- existence

of the consecutive solution for all n > 2 of the equation ( 0 ) had been easily and beautifully proved long ago in the number theory

in form of an exercise, for example in [ 1 ] ( Chapter 2, section

18, exercise 5 ) For convenience let us copy out this proof

“ Exercise 5 - If n is a natural number greater than 2, then the equation : has no solution in natural number

n n

n

x x

x +( +1) =( +2)

Solution -1 ) n is an odd number greater than 2 Putting

This proves that is

a divisor of the number 2, which by y = x+1 > 1, is impossible

n n

n

y y

y =( +1) −( −1)

22

2

2 1 1− 3 3 − − 2 2 =

y C y

C y

shows that whence y > 2n The second equality shows that y is a divisor of 2n This is a contradiction ”

02

2

2 1 1 − 3 3 − − 1 =

y C y

C y

C

022

2 1 2 3 4

1− − − − − − =

−

n y

C y

ny y

3 Corollary –

1 ) For positive integers x, e x and n, the n- th root

n n

is not an integer when n > 2

2 ) The Diophantine equation in three unknowns x, y and e x :

n i

i n i

( IV ) has no non- zero positive

integer solution when n ∈ Z and n > 2 +

Proof - 1) and 2) : Evidently, if +

i

i x i n i n

1

1

This contradicts the Fermat’s last theorem

n n n n

x x y z

e

( x, y and z ∈ Z+

3 ) : This assertion is directly resulted from both

assertions 1 ) and 2 ) above

* * *

Acknowlegment - I would like to express my deep gratitude

to all mathematicians - referees for their important contribution to

evaluate this proof and to bring this paper to the readers

To them I would like to express my heartfelt thankfulness

References

[ 1 ] W Sierpin’ski , Elementary theory of numbers , Polski

Academic Nauk, Warzawa , 1964

-

Author’s Address

N.V TU -YEN E- mail : thyen28@gmail.com

NHA E 1 , P 406 VAN CHUONG

PREMIERS JUMEAUX ( p, p+2 ; p+6, p+8 )

( Vietnam )

*

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

, the formulas describing the density- limits of the twin- primes and of the neighbouring pairs of twin- primes

)

( p

D∞

)2,

∞ p p

D

)8,6

;2,

;2,

∞ p p p p

had been used to prove the infinity of the twin- primes and the neighbouring pairs of twin- primes If the proof proposed here is an accepted one, it will

be reasonable to place both these proved here conjectures into the Arithmetic’s edifice already constructed as one Corollary of the Euclid’s theorem on the infinitude of the primes

*

premiers ( p ), deux cribles pour cribler les nombres premiers jumeaux ( p,p+2 )

et les paires voisines de nombres premiers jumeaux ( p,p+2 ; p+6,p+8 ) sont proposés De la structure de ces cribles et suivant la déduction connue de

des paires voisines de nombres premiers jumeaux

)

( p

D∞

)2,

∞ p p D

)8,6

;2,

∞ p p p p D

respectives sont établies en même forme comme celle de nombres premiers :

)1(3

égale 1 pour des nombres premiers, égale 2 pour des nombres premiers jumeaux et égale 4 pour des paires voisines de nombres premiers jumeaux Enfin ces deux dernières formules ont utilisées effectivement pour démontrer l’infinité des nombres premiers jumeaux et l’infinité des paires voisines de nombres premiers jumeaux Si la preuve proposée ici est acceptable, il serait raisonnable de placer ces deux anciennes conjectures comme un corollaire du théorème d’Euclide de l’infinitude des nombres premiers dans l’édifice d’ Arithmétique déjà construit

i

*

MOTS CLEFS - Crible d’Ératosthène ; Suite binaire ; Bande criblée ; Bande

criblante ; Bande résultante ; Nombre premier jumeau ; Paire de nombres premiers jumeaux ; Densité et Densité limite

*

1

PRÉLIMINAIRE

Dans cet article le crible d’Ératosthène sera concrétisé par un système de 3 composants : D’abord une bande criblée comportant une suite infinie de bits 1 qui représentent les nombres naturels que l’on devrait les faire trier, puis une infinité des bandes criblantes donnant pour chacune une suite binaire de période avec étant le i-ième nombre premier, ó les bits 0 codant les multiples de et les bits 1 codant ses non- multiples, et enfin, une bande résultante qui est la suite binaire restée de la bande criblée après le triage Cette approche est en fait une concrétisation des concepts comme fonction crible, fonction Ératosthène , rappelés dans les ouvrages d’Informatique, par exemple [2]

on a une suite infinie des bits 1, donnant une bande criblée pour trier tous les nombres premiers à partir de 9 Les bandes criblantes sont placées en lignes horizontales de l’ordre croissant de haut en bas avec le premier bit 0, ou le premier trou, situé à l’extrémité gauche de valeur , mais plus simplement pour écrire le schéma du crible, de valeur suivant

la propriété 1 ci- après :

la bande criblante

1 +

i

p

i

p

Preuve - Car tous les nombres considérés sont impairs, en

supposantp i = k2 +1 et p i+1 =2(k+m)+1, avec entiers positifs k

et m, la différence entre ces deux premiers bits 0, ou trous, est

2

nombres impairs consécutifs, c’est à dire à 2 unités de nombres

naturels, la propriété 1 est évidente ∗ ( Symbole ∗ utilisé dans

cet article pour noter la fin d’une preuve )

Utilisant les composants du crible et la propriété 1 décrits au-

dessus, un schéma du crible d’Ératosthène pour déterminer les

nombres premiers à partir de 9 est présenté ci- après

2 - Schéma 1 : Crible 1 d’Ératosthène pour nombres premiers

Remarques - 1) Car 37 < , il est nécessaire et suffisant

d’utiliser seulement 3 bandes criblantes dans le crible,

parmi lesquelles celle de

2 2

4 =7

p

3 2

1,p et p p

respectivement ) La dernière ligne est la bande résultante Sur la

bande résultante de cet exemple on trouve 8 bits 1 correspondant

8 nombres premiers notés à la 1

è è

2) Les points dans chaque bande criblante notent ses périodes p i

3) L’usage seul des bandes criblantes ‘ premières ’ pour trier les

nombres premiers (p), premiers jumeaux (p,p+2) et les paires

voisines de nombres premiers jumeaux (p,p+2 ; p+6,p+8)

eux-mêmes n’est pas un cercle vicieux car un système de n bandes

criblantes ‘premières’ , i = 1, 2, , n est capable toujours de

trouver un bon nombre d’autres nombres premiers plus grands

que , qui sont prêts à continuer sans cesse le criblage

I - Schémas du crible d’ Ératosthène utilisés

1 - Cas des nombres premiers jumeaux ( p,p+2 )

En utilisant la bande criblante p2 =3, ‘011011011 ’ pour la

bande criblée, qui se compose seulement des nombres impairs

3

jumeaux ‘011’, on voit bien qu’à partir de 9, il serait possible d’en trouver un nombre premier jumeau si tous les deux bits 1 d’une période 3, ‘011’, restaient conservés sans être annulés par des bandes criblantes du crible En codant chaque période 3,

Dans ce cas, parce que la longueur de chaque période de la bande criblée était diminuée 3 fois, toutes les bandes criblantes

à partir de

i

p p3 =5 devraient se rétrécir en une même échelle

1/3 d’après la propriété 2 suivante :

1 ) Propriété 2 - Pour cribler les nombres premiers jumeaux

(p,p+2), la i- ième bande criblante à partir de est une suite binaire périodique infinie de période bits binaires commencée par

i

p bits 1, puis continue par

)}

232

bits 1 Ici [ ]x est la partie entière de x

Preuve - L’action d’annulation des bits 1 causée par la bande

criblante sur la bande criblée à l’échelle 1 : 1 comme

‘011011011 ’ est représentée par une suite binaire périodique infinie de période bits, ó il y a périodes 3, ‘011’, sur la bande criblée et 3 périodes sur la bande criblante Dans chaque période commune de la bande criblante il y a 4 bits 0 pour la diviser en 3 périodes , parmi lesquels deux bits

0 à ses deux extrémités sont déjà cọncidés avec ceux de la bande criblée qui n’annulent aucun bit 1 des périodes 3, ‘011’, et seulement deux bits 0 à l’intérieur de chaque période jouent

le rơle des trous, dits ‘ trous actifs ’, pour faire annuler deux bits

1 situés à deux périodes 3, ‘011’ distinguées, c’est à dire ces deux trous actifs annulent deux paires possibles de nombres premiers jumeaux sur la bande criblée On voit bien qu’à l’échelle rétrécie 1/3, la période commune bits de la bande criblante devient bits, ó chacune de ses trois périodes composantes possède

4

i p p

p3 4

n i

i p p p

Période : 10101 ; Période p3 =5 p5 =11 : 11101110111 ; etc

2) Schéma 2 : Crible 2 d’Ératosthène pour les nombres premiers jumeaux (p, p+2)

la première ligne divisent 3 et sont codés par des bits 0, leur deux impairs consécutifs ó le deuxième est écrit à la ligne sont des non- multiples de

2 =

p et codés par des bits 1 On voit donc que tous les bandes considérées et leur écarts des bits binaires aux extrémités gauches dans ce crible sont rétrécies à l’échelle 1/3 Dans cet exemple sur la bande résultante on trouve 6 bits 1 correspondant 6 nombres premiers jumeaux qui sont les deux nombres impairs consécutifs après ceux notés à la 1ère ligne dans leurs mêmes colonnes respectivement Ce sont : 11,13 ; 17,19 ; 29,31 ; 41,43 ; 59,61 et 71,73

5

2) et 3) sont identiques à celles du Crible 1 précédent

2 - Cas des paires voisines de nombres premiers jumeaux

( p, p+2 ; p+6, p+8 )

1 ) Propriété 3 - Pour cribler les paires voisines de nombres

premiers jumeaux (p,p+2 ;p+6,p+8), la i- ième bande criblante

à partir de a une période de bits binaires avec 4 bits 0 actifs inclus, qui comporte :

Preuve - L’action d’annulation des bits 1 séparés sur la

bande criblée non- codée ‘011011011 ’ causée par la bande criblante est représentée par une suite binaire périodique infinie

de période bits ó il y a périodes 6 bits binaires

‘011011’ sur la bande criblée et 6 périodes bits binaires sur la bande criblante Dans chaque période commune il y a 7 bits 0 pour la diviser en 6 périodes ó deux bits 0 situés à ses deux extrémités et 1 bit 0 à son point médian sont toujours cọncidés avec ceux de la bande criblée qui n’annulent pas aucun bit 1 de la période 6 ‘011011’ Alors restent sur chaque période

de la bande criblante du crible, 4 bits 0 ‘actifs’ ou trous, pour faire annuler 4 bits 1 distinguées de la bande criblée, et

Évidemment à l’échelle 1/6, la longueur d’une période

bits de la bande criblante devient celle d’une période bits et chacune de ses 6 périodes composantes rétrécies possède

ó 3 bits 0 cọncidents notés par (0) ne seront comptés pas dans

les p i bits, d’ó une répartition de (p i −4) bits 1 dans ces 6

intervalles doive être réalisée En notant que cette période est un

palindrome ou une symétrie centrale et que le bit central (0) se

trouve à la médiane d’une période 6, ‘011011’, de la bande

criblée, tandis que deux autres bits (0) situés aux extrémités de

chacune période de la bande criblante sont toujours cọncidés

avec deux bits 0 de la bande criblée, nous déterminons les bits 1

restés pour ces 6 intervalles de ( II ) selon un ordre suivant :

pour le 2ème et le 5ème intervalle chacun *

D’après ces formules chaque période de la bande criblante

prend une suite binaire concrète pour chaque , par exemple :

2 ) Schéma 3 : Crible 3 d’Ératosthène pour les paires voisines

de nombres premiers jumeaux (p,p+2 ; p+6,p+8)