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FIRST NUMBERS AND ATTEMPTS OF THE DEMONSTRATION OF THE GOLDBACH CONJUNCTURE AND MODERN CONJUNCTURE !!!

I-Itroduction In these lines that follow I try to make my modest contribution to the understanding of prime numbers. I have also tried to attack the mountain that seems insurmountable since 1742: I want to talk about the famous situation of Christian Goldbach. I am far from saying that I got to do it. I am a teacher of mathematics in a middle school in (Dakar, Senegal) and who did not graduate from higher education. For my part I do not address this subject with advanced mathematics because I have no advanced degree in the field but I approached in another way that thinks deserves to be taken into account because I am persuaded that even if my demonstration does not close the debate, it will advance it only remains one line towards its resolution. I confess that I consciously knew this conjecture in less than four months even though the prime numbers always fascinated me. When I was a student, what fascinated me most was that any natural whole number could be decomposed into a prime factor product. What guided me was to see if we could write these numbers by combining prime numbers and see if these scripts have a certain regularity compare to non-primes. Looking closely I noticed that all are combinations of 2 and 3.

II-NEW WRITING OF NATURAL AND COORDINATED ENTIRE NUMBERS.

Any natural whole number > 1 can be written using 2 or 3 or the combination of both. Examples: 2 = 2; 3 = 3; 4 = 2 + 2; 5 = 2 + 3; 6 = 2 + 2 + 2 = 3 + 3; 7 = 2 + 2 + 3; 8 = 2 + 3 + 3; 9 = 2 + 2 + 2 + 3; 10 = 2 + 2 + 3 + 3 If we fix a, the number of 2 and b, the number of 3 needed to write a natural integer> 1 such that n and m are very close, we obtain a great result (or powerful tool) of which formidable theorems and conjectures will naturally follow from this, and perhaps even the resolution of possible conjectures, including that of Goldbach.

Here is an example: in red you will recognize the prime numbers in the table below

If we have the natural numbers:

- HORIZONTAL way from 1 to 10 forms aisles.

An aisle n denoted An is the alley of natural numbers such as 10n-9≤An≤10n where n is a non-zero natural integer.

VERTICALLY and from top to bottom we get Valleys.

A valley Vi is the valley of integers whose unit is i with i a number in the base of the decimal number that is to say n∈ {1; 2; 3; 4; 5; 6; 7; 8; 9; 0}.

A number in valley i is denoted Pi, k = 10k + i with k∈N.

For example, V1 contains P1.0 = 1, P1, 1 = 11, P1, 2 = 21; P1.3 = 31;

V2 contains the numbers 2; 22; 32; 42; .........

V9 contains the numbers 9; 19; 29; 39; 49; ..........

NB: For the valley V0, Pi, k = 10k + i with k∈N *.

We thus distinguish n ranged distributed between 10 valleys of n numbers each to see TABLE OF DISTRIBUTION OF FIRST NUMBERS in joined parts.

COORDINATES OF NATURAL ENTIRE NUMBERS

It is possible to consider a special mark because unprecedented allowing to know the position of each whole number (even, odd, first, compound) through its coordinates.

Every nonzero natural integer has coordinates (An, Vi) with An aisle n and Vi valley i.

For example A1V1 = 1; A2V1 = 11;

Any nonzero natural integer can be written as AnVi = (n-1) x10 + i so A25V5 = 245.

A253V7 = 2527. A53643V9 = (53643-1) x10 + 9 = 536429

It is important to note that all prime numbers are distributed in four valleys except 2 and 5. That is why there are mainly four fully first valleys that we can call first valleys: that are V1, V3, V7, and V9.

The valleys V2 and V5 being the only ones respectively containing the prime numbers 2 (the first first pair) and 5 are called the 2-mono-first valley and the 5-mono-first valley (or mono-first valleys).

CONJUNCTURE 1: Every even number has the form 2n + 3m with m = 2k (k∈N) and | n-m | ∈ {0; 1; 2}

Proof : whatever the natural integer n, 2n is even. or m = 2k is also even where 3m = 3 (2k) = 2 (3k) is even. The sum of two even numbers being even so every even number is of the form 2n + 3m with m = 2k and | n-m | ∈ {0; 1; 2}.

OTHER

-Si | n-m | = 0 then n-m = 0 => n = m = 2k; * Let 2 (2k) + 3 (2k) = 10k be the set of even integers of unit 0; noted N0

-si | n-m | = 1: * Let n-m = 1 => n = 1 + m => n = 1 + 2k; N = 2n + 3m = 2 (1 + 2n) + 3 (2k) = 10k + 2 the set of even natural integers of unit 2: noted N2

* Let nm = -1 => n = m-1 => n = 2k-1, N = 2n + 3m = 2 (2k-1) +3 (2k) = 10k-2 (with k> 0) set of natural numbers of unity 8: noted N8

-si | n-m | = 2:

* Let nm = 2 => n = m + 2 => n = 2k + 2, N = 2n + 3m = 2 (2k + 2) +3 (2k) = 10k + 4 the set of natural even integers of unit 4: noted N4

* Let nm = -2 => n = m-2 => n = 2k-2, N = 2n + 3m = 2 (2k-2) + 3 (2k) = 10k-4 (with k> 0) all natural integers even of unit 6: noted N6

NB: If there are only one even natural integer left that does not belong to one of these sets (N0, N2, N4, N6 and N8) let me know so that I get back to work.

Application: Write the following number A = 235356887557899754368900854 in form an + bm with a = 2, b = 3 and m = 2k and | n-m | ∈ {0; 1; 2} The number A ∈ N4 so A = an + bm with a = 2 and

b = 3 or n = m + 2 so A = 2 (m + 2) + 3m = 5m + 4 => 5m = A-4 => m = (A-4) / 5 = (235356887557899754368900854-4) / 5 = 47071377511579950873780170 Thus n = m + 2 = 47071377511579950873780172 So A = 47071377511579950873780172 a + 470713775115799508737801870b with a = 2 and b = 3

"CONJONCTURE 2": Any odd number I differ from 1 is of the form 2n + 3m with m = 2k + 1 and | n-m | ∈ {0; 1; 2}

Proof: Since 2n is even regardless of the natural integer n and m = 2k + 1 is odd regardless of the natural integer k. So 2n + 3m is odd because it is the sum of an even number and an odd number. OTHER :

-If | n-m | = 0 so, n-m = 0 => n = m = 2k + 1

* Let 2 (2k + 1) +3 (2k + 1) = 4k + 2 + 6k + 3 = 10k + 5 the set of odd natural unit integers 5: noted I5 -If | n-m | = 1:

* Let n-m = 1 => n = m + 1 = 2k + 1 + 1 = 2k + 2; I = 2n + 3m = 2 (2k + 2) + 3 (2k + 1) = 4k + 4 + 6k + 3 = 10k + 7 the set of odd integers of unit 7: noted I7

* Let n-m = -1 => n = m-1 = 2k + 1-1 = 2k; I = 2n + 3m = 2 (2k) +3 (2k + 1) = 4k + 6k + 3 = 10k + 3 the set of odd integers of unit 3: noted I3

-If | n-m | = 2:

* Let nm = 2 => n = m + 2 = 2k + 1 + 2 = 2k + 3, I = 2n + 3m = 2 (2k + 3) +3 (2k + 1) = 4k + 6 + 6k + 3 = 10k + 9 the set of odd integers of unit 9: noted I9

* Let n-m = -2 => n = m - 2 = 2k + 1 - 2 = 2k - 1; I = 2n + 3m = 2 (2k-1) +3 (2k + 1) = 4k-2 + 6k + 3 = 10k + 1 the set of odd integers of unit 1: noted I1.

NB: If there is only one odd natural integer left that does not belong to one of these sets (I1, I3, I5, I7 and I9) let me know so that I can go back to work.

APPLICATION: Written as an + bm a = 2, b = 3 and | n-m | ∈ {0; 1; 2} the following number B = 2 457 864 167 880 023 B ∈ I3, therefore B = 10k + 3 => k = (B-3) / 10 = 245 786 416 788 002 m = 2k + 1 = 491,572,833,576,005; n = 491 572 833 576 004 B = 491 572 833 576 004 a + 491 572 833 576 005 b 2 457 864 167 880 023 = 491 572 833 576 004 a + 491 572 833 576 005 b

'' CONJECTURE 3 '': Any integer natural number differ from 1 can be written as 2n + 3m with n, m of natural numbers and | n-m | ∈ {0; 1; 2}.

PROOF: I refer you to conjunctures 1 and 2 and their proofs.

'' CONJECTURE 4 ': Any prime number noted Pn, m greater than 5 can be written as 2n + 3m with | n-m | ∈ {1; 2}, m = 2k + 1 and n differ by 3k (Not multiple of 3).

DEMONSTRATION

Let Pn, m be a prime number then Pn, m = 2n + 3m. if n is odd then n = 3k ± 1

* If | m-n | = 1, then n-m = 1 or n-m = -1

If n-m = 1 => m = n + 1 then P = 2n + 3 (n + 1) = 5n + 3

For n = 3k +1 then P = 5 (3k+1)+3 P3 = 15k + 8

For n = 3k - 1 then P'3 = 15k -2 ,

For k odd P3 and P'3 give the set of first integers greater than 5 of unit 3 and the corresponding non-multiple compounds of 3.

NB :P₃ P’₃=

P₃= 15k+8

K	1	3	5	7	9	11	13	15	17
15K+8	23	53	83	113	143	173	203	233	263
	19	21	23	25	27	29	31	33	35
	293	323	353	383	413	443	473	503	533
	37	39	41	43	45	47	49	51	53
	563	593	623	653	683	713	743	773	803
	55	57	59	61	63	65	67	69	71
	833	863	893	923	953	983	1013	1043	1073
	73	75	77	79	81	83	85	87	89
	1103	1133	1163	1193	1223	1253	1283	1313	1343
	91	93	95	97	99	101	103	105	107
	1373	1403	1433	1463	1493	1523	1553	1583	1613
	109	111	113	115	117	119	121	123	125
	1643	1673	1703	1733	1763	1793	1823	1853	1883
	127	129	131	133	135	137	139	141	143
	1913	1943	1973	2003	2033	2063	2093	2123	2153
	145	147	149	151	153	155	157	159	161
	2183	2213	2243	2273	2303	2333	2363	2393	2423

Distributions of prime numbers of the form P3 = 15k + 8

P’₃=15k-2

k	1	3	5	7	9	11	13	15	17
15k-2	13	43	73	103	133	163	193	223	253
	19	21	23	25	27	29	31	33	35
	283	313	343	373	403	433	463	493	523
	37	39	41	43	45	47	49	51	53
	553	583	613	643	673	703	733	763	793
	55	57	59	61	63	65	67	69	71
	823	853	883	913	943	973	1003	1033	1063
	73	75	77	79	81	83	85	87	89
	1093	1123	1153	1183	1213	1243	1273	1303	1333
	91	93	95	97	99	101	103	105	107
	1363	1393	1423	1453	1483	1513	1543	1573	1603
	109	111	113	115	117	119	121	123	125
	1633	1663	1693	1723	1753	1783	1813	1843	1873
	127	129	131	133	135	137	139	141	143
	1903	1933	1963	1993	2023	2053	2083	2113	2143
	145	147	149	151	153	155	157	159	161
	2173	2203	2233	2263	2293	2323	2353	2383	2413

Distributions of prime numbers of the form P'3 = 15k-2

Characterization :

The prime numbers' units 3 constitute a family and are divided into two groups:

-The set of first whose sum of the digits is a term of a first-term arithmetic sequence 4 and reason 3 (P'3 = 15k-2)

Examples 13 => 1 + 3 = 4; 43 => 4 + 3 = 7; 73 => 7 + 3 = 10; 103 => 1 + 0 + 3 = 4; 163 => 1 + 6 + 3 = 10; 193 => 1 + 9 + 3 = 13; 223 => 2 + 2 + 3 = 7 ... 2383 => 2 + 3 + 8 + 3 = 16.

-The set of first whose sum of numbers is a term of a first-term arithmetic sequence 5 and reason 3. (P3 = 15k + 8)

Examples

23 => 2 + 3 = 5; 53 => 5 + 3 = 8; 83 => 8 + 3 = 11; 113 => 1 + 1 + 3 = 5; 173 => 1 + 7 + 3 = 11; 233 => 2 + 3 + 3 = 8; 263 => 2 + 6 + 3 = 11; ... 2423 => 2 + 4 + 2 + 3 = 11

If n-m = - 1 => m = n-1 then P = 2n + 3 (n-1) = 5n-3

For n = 3k + 1then P = 5 (3k +1)3 noted P7 = 15k + 2

For n = 3k - 1 then P = 15k - 8 noted P’7=15k - 8 .

For k odd P7 and P'7 give the set of prime numbers of unit 7 and the corresponding non-multiples of 3.

NB : P₇ P’₇=

P₇= 15k + 2

k	1	3	5	7	9	11	13	15	17	19	21	23
15k+2	17	47	77	107	137	167	197	227	257	287	317	347
	25	27	29	31	33	35	37	39	41	43	45	47
	377	407	437	467	497	527	557	587	617	647	677	707
	49	51	53	55	57	59	61	63	65	67	69	71
	737	767	797	827	857	887	917	947	977	1007	1037	1067

Distributions of prime numbers of the form P7 = 15k+2

P’₇= 15k – 8

k	1	3	5	7	9	11	13	15	17	19	21	23
15K-8	7	37	67	97	127	157	187	217	247	277	307	337
	25	27	29	31	33	35	37	39	41	43	45	47
	367	397	427	457	487	517	547	577	607	637	667	697
	49	51	53	55	57	59	61	63	65	67	69	71
	727	757	787	817	847	877	907	937	967	997	1027	1057

Distributions of prime numbers of the form P'7 = 15k-8

Characterization :

The prime numbers' units 7 constitute a family and are divided into two groups:

-The set of first whose sum of numbers is a term of an arithmetic sequence of first term 7 and reason 3 (P'7 = 15k-8). Example: see Distributions of prime numbers of the form P'7 = 15k - 8

-The set of first whose sum of numbers is a term of an arithmetic sequence of first term 8 and reason 3. (P7 = 15k + 2). Example: see Distributions of prime numbers of the form P7 = 15k + 2

* If | m-n | = 2, then n-m = 2 or n-m = -2 n-m = 2 => m = n-2 we have P = 2n + 3m = 2n + 3 (n-2)

= 5n-6 ,for n = 3k + 1 we have P = 5 (3k + 1) -6 = 15k-1 which we can name P9 = 15k-1 for k even for n = 3k-1 we have P = 5 (3k-1) -6 = 15k-11 which we can name P'9 = 15k-11 for k even.

For k even P9 and P'9 give the set of prime numbers of unit 9 and the corresponding non-multiples of 3.

NB : P₉ P’₉=

P’₉= 15k-11

K	2	4	6	8	10	12	14	16	18	20	22	24
15k-11	19	49	79	109	139	169	199	229	259	289	319	349
	26	28	30	32	34	36	38	40	42	44	46	48
	379	409	439	469	499	529	559	589	619	649	679	709
	50	52	54	56	58	60	62	64	66	68	70	72
	739	769	799	829	859	889	919	949	979	1009	1039	1069

Distributions of prime numbers of the form P9 = 15k - 11

P₉=15k-1

k	2	4	6	8	10	12	14	16	18	20	22	24
15K-1	29	59	89	119	149	179	209	239	269	299	329	359
	26	28	30	32	34	36	38	40	42	44	46	48
	389	419	449	479	509	539	569	599	629	659	689	719
	50	52	54	56	58	60	62	64	66	68	70	72
	749	779	809	839	869	899	929	959	989	1019	1049	1079

Distributions of prime numbers of the form P'9 = 15k - 1

Characterization :

The prime numbers' units 9 constitute a family and are divided into two groups:

-The set of first whose sum of digits is a term of a first-order arithmetic sequence 10 and reason 3

(P'9 = 15k-11). Example: see Distributions of prime numbers of the form P'9 = 15k - 11

-The set of first whose sum of numbers is a term of an arithmetic sequence of first term 11 and reason 3. (P9 = 15k-1). Example: see Distributions of prime numbers of the form P9 = 15k -1

If n-m = -2 => m = n + 2 we thus have P = 2n + 3 (n + 2) = 5n + 6 for n = 3k + 1 we have P = 5 (3k + 1) + 6 = 15k + 11 which we can name P1 = 15k + 11 for k even

for n = 3k - 1 we have P = 5 (3k-1) + 6 = 15k + 1 which we can name P'1 = 15k + 1 for k even

For k odd P1 and P'1 give the set of prime numbers of unit 1 and the corresponding non-multiples of 3. Exemple: see Distributions of prime numbers of the form P'1 = 15k + 1

NB :P₁ P’₁=

P₁=15k+11

K	0	2	4	6	8	10	12	14	16	18	20	22	24
15k+11	11	41	71	101	131	161	191	221	251	281	311	341	371
		26	28	30	32	34	36	38	40	42	44	46	48
		401	431	461	491	521	551	581	611	641	671	701	731
		50	52	54	56	58	60	62	64	66	68	70	72
		761	791	821	851	881	911	941	971	1001	1031	1061	1091

Distributions of prime numbers of the form P'1 = 15k + 11

P’₁=15k+1

k	0	2	4	6	8	10	12	14	16	18	20	22	24
15K+1	1	31	61	91	121	151	181	211	241	271	301	331	361
		26	28	30	32	34	36	38	40	42	44	46	48
		391	421	451	481	511	541	571	601	631	661	691	721
		50	52	54	56	58	60	62	64	66	68	70	72
		751	781	811	841	871	901	931	961	991	1021	1051	1081

Distributions of prime numbers of the form P'1 = 15k + 1

Characterization :

The prime numbers' units 1 constitute a family and are divided into two groups:

-The set of first whose sum of numbers is a term of a first-term arithmetic sequence 2 and reason 3 (P1 = 15k + 11). Example: see Distributions of prime numbers of the form P1 = 15k + 11

-The set of first whose sum of numbers is a term of a first-term arithmetic sequence 4 and reason 3. (P1 = 15k + 1). Example: see Distributions of prime numbers of the form P9 = 15k + 1

III-ATTEMPT TO DEMONSTRATE THE CONJECTURE OF Goldbach:

PROOF OF THE CONJECTURE OF Goldbach

"Any integer number greater than 3 can be written as the sum of two prime numbers. "

Let Pn, m and Pn ', m' be two prime numbers, then we have Pn, m = 2n + 3m and Pn ', m' = 2n '+ 3m'. n and n ' being integers not all nil at a time and not multiples of 3 so 2n and 2n' are even and not multiples of 3. m 'and m are odd so 3m and 3m' are all odd. Let P be a natural integer such that

P = Pn, m + Pn ', m'

= (2n + 3m) + (2n '+ 3m')

= 2n + 2n '+ 3m + 3m'

= 2 (n + n ') + 3 (m + m') (1)

Whatever the natural numbers n and n '2 (n + n') is even. (2)

Whatever m is odd and m 'odd then m + m' is even, that is m + m '= 2k hence 3 (m + m') = 3 (2k) = 2 (3k). So 3 (m + m ') is even. (3)

According to (1), (2) and (3) then P = 2 (n + n ') + 3 (m + m') = even number + even number = even number. If we put N = n + n 'and M = m + m' we have: P = 2N + 3M with M = 2k (m + m 'sum of two odd numbers) consequence of CONJONCTURE 2 (which has value of a theorem if the proof bring is size). So any even number greater than or equal to 4 is the sum of two prime numbers !!!

PROOF OF MODERN CONJUNCTURE

Let us apply the same reasoning with the modern conjuncture which stipulates that: "If every even number greater than 2 can be written as the sum of three first, one is necessarily two. Based on Goldbach's previous business demo P = 2N + 3M with M = 2k Suppose a natural integer W sum of P and 2 since P results from a sum of two prime numbers then W = P + 2 is the sum of Three prime numbers one of which is 2.

Let's check if any prime number can be written as W. W = P + 2

= 2N + 3M + 2

= 2N + 2 + 3M

W = 2 (N + 1) + 3M, with M = 2k. and | (N + 1) - M | ∈ {0; 1; 2}

-If | (N + 1) - M | = 0 then N + 1-M = 0 => N = M-1 = 2k-1 W = 2 (2k-1) +3 (2k) = 10k-2: N8 the set of unitary natural numbers 8.

-If | (N + 1) - M | = 1:

* let (N + 1) -M = 1 => N = M + 1-1 = M = 2k in this case W = 2 (2k) +3 (2k) = 10k: N0 the set of natural numbers even 0 unit.

* ie (N + 1) -M = -1 => N = M-2 = 2k-2 in this case W = 2 (2k-2) +3 (2k) = 10k-4: N6 the set of integers natural unit peers 6.

- If | (N + 1) - M | = 2:

* Let (N + 1) -M = 2 => N = M + 1 = 2k + 1 in this case W = 2 (2k + 1) +3 (2k) = 10k + 2: N2 the set of natural numbers unit 2 peers;

* Let (N + 1) -M = -2 => N = M-3 = 2k-3 in this case W = 2 (2k-3) +3 (2k) W = 10k-6: N4 the set of unitary natural numbers of unit 4;

From the foregoing it may be said that if every even number greater than 2 can be written as the sum of three prime, one of them is necessarily 2!!!

OUMAR .LY / AMATEUR SEARCHER who has not, no higher degree in mathematics but is passionate about prime numbers; Thank you for reading my reflection

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