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Polignac: New Conjecture
ISSN: 2168-9679

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# Polignac: New Conjecture

Leichsenring IG*
Department of Mathematics, Karting International Aldeia da Serra, Aldeia da Serra, Barueri, São Paulo 06428-180, Brazil
*Corresponding Author: Leichsenring IG, Department of Mathematics, Karting International Aldeia da Serra, Aldeia Da Serra, Barueri, São Paulo 06428-180, Brazil, Tel: 55 11 99898-5544, Email: [email protected]

Received Date: Aug 03, 2018 / Accepted Date: Sep 12, 2018 / Published Date: Sep 20, 2018

#### Introduction

Initially we will summarize the proposal equivalent to Goldbach's conjecture, which can be examined [1].

All natural >1 can be represented by the mean of two primes p and q equidistant from a natural n, through an integer index k, such that [2]:

n 5(p 1 q) 4 2, being

p 5 n 2 k and

q 5 n 1 k.

There is symmetry involving n and both primes p and q with amplitude [3].

3 ••• n ••• 23n 23.

We will use these concepts as a foundation for the study that we will present about the twin primes, the pairs (g, h), with [4]

| h 2 g |5 2.

So we have, within our formulation, for a given kg [5]:

g 5(pg 1 qg) 4 2,

pg 5(g 2 kg),

qg 5 (g 1 kg);

And for a given kh:

h 5 (ph 1 qh) 4 2,

ph 5 (h 2 kh),

qh 5(h 1 kh).

For example, below are some symmetries with the pair (71, 73) (Table 1).

Pair (g, h) k p q
g 5 71
h 5 73
kg 5 12 pg 5 59 qg 5 83
kh 5 6 ph 5 67 qh 5 79
kg 5 18 pg 5 53 qg 5 89
kh 5 36 ph 5 37 qh 5 109
kg 5 30 Pg 5 41 qg 5 101
kh 5 30 ph 5 43 qh 5 103
kg 5 66 pg 5 5 qg 5 137
kh 5 66 ph 5 7 qh 5 139

Table 1: Symmetries pair of (71,73).

According to the conjecture, both symmetries exist -individually, of course and therefore the index k behaves randomly [6,7], as seen in the examples with

kg 5 12 and kh 5 6 or

kg 5 18 and kh 5 36,

Without connection between g and h.

However, we observed the possibility of finding in each pair chosen for testing, many cases where index k could be unique, as seen in two other examples [8,9], with

kg 5 kh 5 30 or

kg 5 kh 5 66,

And there is a link between g and h.

This condition -k 5 kg 5 kh -is the basis of this study and we are interested only when and if it can occur; in this situation [10].

For any pair (g, h) we can do

(g 2 k) 5p,

(g 1 k) 5q,

(h 2 k) 5 p 1 2,

(h 1 k) 5q 1 2.

Therefore we will have:

g 5 (p 1 q) 4 2 and

h 5 [ (p 12) 1 (q 1 2) ] 4 2.

Looking only for these solutions we had some success with several tests, which induced us to the theory that follows and to distinguish the twin primes in that kg ≠ kh of others in that kg≠ kh we adopt the following concept.

Identical twin primes

Are those in which at least one k, simultaneously, satisfies a pair (g, h).

Therefore, among the symmetries of the previous examples, only the following identities can be considered identical twin primes (Table 2).

Pair (g, h) k p q
g 5 71
h 5 73
30 pg 5 41 qg 5 101
ph 5 43 qh 5 103
66 pg 5 5 qg 5 137
ph 5 7 qh 5 139

Table 2: Symmetries of identical twin primes.

In iterative surveys with k, we were able to conduct symmetry in this way for identical twin primes of small magnitude, and we realized that it was possible to obtain them many times. In Table 3 we have the result of the first pairs.

Pairs k pg | ph qg | qh
3
5
impossible
5
7
impossible
11
13
6 5 17
7 19
17
19
12 5 29
7 31
29
31
12 17 41
19 43
41
43
30 11 71
13 73
59
61
42 17 101
19 103
71
73
30 41 101
43 103
101
103
90 11 191
13 193
107
109
90 17 197
19 199
137
139
132 5 269
7 271
149
151
42 107 191
109 193
179
181
168 11 347
13 349
191
193
90 101 281
103 283
197
199
impossible

Table 3: Symmetric survey of two identical twin primes.

However, as we can see in the table, we have already started with two pairs where we cannot obtain simultaneous symmetry and, later, we stop at pair (197, 199), also in the same situation; that is, there are impossible cases if we require n>k.

At this point we will pause in our study of twin primes.

Let's revisit the original conjecture considering what would happen if we could expand the symmetry to negative values, that is, if we could make k>n possible.

Without restriction for k, one immediately observes symmetry with infinite amplitude.

Similarly, as in the initial conjecture, equalities are maintained:

n 5 (p 1 q) 4 2, being

p 5 n 2 k and

q 5 n 1 k;

Where are primes:

| p | and q.

Note that any integers can now be obtained, and that, in particular:

n 5 0 with any primes, for p 1q 5 0;

n 5 1 with any pairs of twin primes;

n < 0 it is a reflection of n>0.

The search iteration can be obtained as follows:

For n even:

k 5 1, 3, 5, ••• ∞.

For n odd:

k 5 2, 4, 6, ••• ∞.

But, let's return to our study, when we have identical twin primes.

The proposition assumes the bond between twin primes g and h, when and if

k 5 kg 5 kh.

And, except for the pair (3, 5), we have the iteration of k boils down to:

k 5 6, 12, 18, ••• ∞

Until simultaneously appear the primes:

| p | and q;

| p 1 2 | and q 1 2.

In summary, we have:

pg 5 (g 2 k),

qg 5 (g 1 k),

ph 5 (g 2 k 1 2) and

qh 5 (g 1 k 1 2).

Without restriction for k let's see those impossible identities of Table 4.

Pairs k pg | ph qg | qh
3
5
8 25 11
23 13
5
7
12 27 17
25 19
197
199
630 2433 827
2431 829

Table 4: Impossible identities without restriction of k.

Interesting; it is possible to obtain symmetry.

In addition, among the set of the first 1048576 odd primes we have:

3199 identities representing the identical twin primes (5, 7);

1669 identities for (197, 199).

Curiously, even with infinite amplitude, there is only one identity for (3, 5), with k 5 8, and it is an exercise for the reader to demonstrate the fact.

Hint: other twin primes are of the form (6m 2 1, 6m 1 1) for some natural m and therefore, g ≡ 2 (mod 3) and h ≡ 1 (mod 3).

For symmetry of pairs of identical twin primes it is necessary that, in general, more than one coincidence occurs for g and h -in isolation - and such that at some point, for identical k values, we find equidistant primes.

For 12484 first pairs of twins primes, with the same set of primes already mentioned, we found multiple identities intended, the lowest number being 1035 for (1302017, 1302019) and the highest value was 9468 for (180179, 180181).

To illustrate: among 2188 identities for (41, 43) we selected some cases (Table 5):

Pair k pg | ph qg | qh
41
43
30 111 71
213 73
18000 217959 18041
217957 18043
1008000 21007959 1008041
21007957 1008043
2070000 22069959 2070041
22069957 2070043
2163000 22162959 2163041
22162957 2163043
3894000 23893959 3894041
23893957 3894043
4092000 24091959 4092041
24091957 4092043
5010000 25009959 5010041
25009957 5010043

Table 5: Identities illustrate twin primes for (41, 43).

So it seems that being infinite amplitude, with infinite prime numbers, it is impossible to determine for each chosen pair how many representations result in identical twin primes, excluding, as already mentioned, the pair (3, 5) with a single identity.

However, one remaining question remains: can all twin primes be identified as identical? That is: are sets equivalents?

Then, reiterating, if

(g, h) are identical twin primes, we have:

g 5 (p 1 q) 4 2,

h 5 [ (p 1 2) 1 (q 1 2) ] 4 2

And as a consequence, are also twin primes the pairs:

(p, p 1 2) and

(q, q 1 2).

Therefore, under these conditions, each pair of identical twin primes leads to other twin primes, however not necessarily identical!

But by exploring the previous question:

❖ If we could ensure that all twin primes can also be identical

and

❖ If there were one last pair of identical twin primes (gu, hu).

It would mean that the last pair of identical twin primes would forward to the another pair of identical twin primes of greater magnitude, which would be an incongruity.

#### Introduction

Initially we will summarize the proposal equivalent to Goldbach's conjecture, which can be examined [1].

All natural >1 can be represented by the mean of two primes p and q equidistant from a natural n, through an integer index k, such that [2]:

n 5(p 1 q) 4 2, being

p 5 n 2 k and

q 5 n 1 k.

There is symmetry involving n and both primes p and q with amplitude [3].

3 ••• n ••• 23n 23.

We will use these concepts as a foundation for the study that we will present about the twin primes, the pairs (g, h), with [4]

| h 2 g |5 2.

So we have, within our formulation, for a given kg [5]:

g 5(pg 1 qg) 4 2,

pg 5(g 2 kg),

qg 5 (g 1 kg);

And for a given kh:

h 5 (ph 1 qh) 4 2,

ph 5 (h 2 kh),

qh 5(h 1 kh).

For example, below are some symmetries with the pair (71, 73) (Table 1).

Pair (g, h) k p q
g 5 71
h 5 73
kg 5 12 pg 5 59 qg 5 83
kh 5 6 ph 5 67 qh 5 79
kg 5 18 pg 5 53 qg 5 89
kh 5 36 ph 5 37 qh 5 109
kg 5 30 Pg 5 41 qg 5 101
kh 5 30 ph 5 43 qh 5 103
kg 5 66 pg 5 5 qg 5 137
kh 5 66 ph 5 7 qh 5 139

Table 1: Symmetries pair of (71,73).

According to the conjecture, both symmetries exist -individually, of course and therefore the index k behaves randomly [6,7], as seen in the examples with

kg 5 12 and kh 5 6 or

kg 5 18 and kh 5 36,

Without connection between g and h.

However, we observed the possibility of finding in each pair chosen for testing, many cases where index k could be unique, as seen in two other examples [8,9], with

kg 5 kh 5 30 or

kg 5 kh 5 66,

And there is a link between g and h.

This condition -k 5 kg 5 kh -is the basis of this study and we are interested only when and if it can occur; in this situation [10].

For any pair (g, h) we can do

(g 2 k) 5p,

(g 1 k) 5q,

(h 2 k) 5 p 1 2,

(h 1 k) 5q 1 2.

Therefore we will have:

g 5 (p 1 q) 4 2 and

h 5 [ (p 12) 1 (q 1 2) ] 4 2.

Looking only for these solutions we had some success with several tests, which induced us to the theory that follows and to distinguish the twin primes in that kg ≠ kh of others in that kg≠ kh we adopt the following concept.

Identical twin primes

Are those in which at least one k, simultaneously, satisfies a pair (g, h).

Therefore, among the symmetries of the previous examples, only the following identities can be considered identical twin primes (Table 2).

Pair (g, h) k p q
g 5 71
h 5 73
30 pg 5 41 qg 5 101
ph 5 43 qh 5 103
66 pg 5 5 qg 5 137
ph 5 7 qh 5 139

Table 2: Symmetries of identical twin primes.

In iterative surveys with k, we were able to conduct symmetry in this way for identical twin primes of small magnitude, and we realized that it was possible to obtain them many times. In Table 3 we have the result of the first pairs.

Pairs k pg | ph qg | qh
3
5
impossible
5
7
impossible
11
13
6 5 17
7 19
17
19
12 5 29
7 31
29
31
12 17 41
19 43
41
43
30 11 71
13 73
59
61
42 17 101
19 103
71
73
30 41 101
43 103
101
103
90 11 191
13 193
107
109
90 17 197
19 199
137
139
132 5 269
7 271
149
151
42 107 191
109 193
179
181
168 11 347
13 349
191
193
90 101 281
103 283
197
199
impossible

Table 3: Symmetric survey of two identical twin primes.

However, as we can see in the table, we have already started with two pairs where we cannot obtain simultaneous symmetry and, later, we stop at pair (197, 199), also in the same situation; that is, there are impossible cases if we require n>k.

At this point we will pause in our study of twin primes.

Let's revisit the original conjecture considering what would happen if we could expand the symmetry to negative values, that is, if we could make k>n possible.

Without restriction for k, one immediately observes symmetry with infinite amplitude.

Similarly, as in the initial conjecture, equalities are maintained:

n 5 (p 1 q) 4 2, being

p 5 n 2 k and

q 5 n 1 k;

Where are primes:

| p | and q.

Note that any integers can now be obtained, and that, in particular:

n 5 0 with any primes, for p 1q 5 0;

n 5 1 with any pairs of twin primes;

n < 0 it is a reflection of n>0.

The search iteration can be obtained as follows:

For n even:

k 5 1, 3, 5, ••• ∞.

For n odd:

k 5 2, 4, 6, ••• ∞.

But, let's return to our study, when we have identical twin primes.

The proposition assumes the bond between twin primes g and h, when and if

k 5 kg 5 kh.

And, except for the pair (3, 5), we have the iteration of k boils down to:

k 5 6, 12, 18, ••• ∞

Until simultaneously appear the primes:

| p | and q;

| p 1 2 | and q 1 2.

In summary, we have:

pg 5 (g 2 k),

qg 5 (g 1 k),

ph 5 (g 2 k 1 2) and

qh 5 (g 1 k 1 2).

Without restriction for k let's see those impossible identities of Table 4.

Pairs k pg | ph qg | qh
3
5
8 25 11
23 13
5
7
12 27 17
25 19
197
199
630 2433 827
2431 829

Table 4: Impossible identities without restriction of k.

Interesting; it is possible to obtain symmetry.

In addition, among the set of the first 1048576 odd primes we have:

3199 identities representing the identical twin primes (5, 7);

1669 identities for (197, 199).

Curiously, even with infinite amplitude, there is only one identity for (3, 5), with k 5 8, and it is an exercise for the reader to demonstrate the fact.

Hint: other twin primes are of the form (6m 2 1, 6m 1 1) for some natural m and therefore, g ≡ 2 (mod 3) and h ≡ 1 (mod 3).

For symmetry of pairs of identical twin primes it is necessary that, in general, more than one coincidence occurs for g and h -in isolation - and such that at some point, for identical k values, we find equidistant primes.

For 12484 first pairs of twins primes, with the same set of primes already mentioned, we found multiple identities intended, the lowest number being 1035 for (1302017, 1302019) and the highest value was 9468 for (180179, 180181).

To illustrate: among 2188 identities for (41, 43) we selected some cases (Table 5):

Pair k pg | ph qg | qh
41
43
30 111 71
213 73
18000 217959 18041
217957 18043
1008000 21007959 1008041
21007957 1008043
2070000 22069959 2070041
22069957 2070043
2163000 22162959 2163041
22162957 2163043
3894000 23893959 3894041
23893957 3894043
4092000 24091959 4092041
24091957 4092043
5010000 25009959 5010041
25009957 5010043

Table 5: Identities illustrate twin primes for (41, 43).

So it seems that being infinite amplitude, with infinite prime numbers, it is impossible to determine for each chosen pair how many representations result in identical twin primes, excluding, as already mentioned, the pair (3, 5) with a single identity.

However, one remaining question remains: can all twin primes be identified as identical? That is: are sets equivalents?

Then, reiterating, if

(g, h) are identical twin primes, we have:

g 5 (p 1 q) 4 2,

h 5 [ (p 1 2) 1 (q 1 2) ] 4 2

And as a consequence, are also twin primes the pairs:

(p, p 1 2) and

(q, q 1 2).

Therefore, under these conditions, each pair of identical twin primes leads to other twin primes, however not necessarily identical!

But by exploring the previous question:

❖ If we could ensure that all twin primes can also be identical

and

❖ If there were one last pair of identical twin primes (gu, hu).

It would mean that the last pair of identical twin primes would forward to the another pair of identical twin primes of greater magnitude, which would be an incongruity.

#### Conclusion

If it were so, forcibly, the twin primes numbers would be infinite.

#### References

Citation: Leichsenring IG (2018) Polignac: New Conjecture. J Appl Computat Math 7: 415. DOI: 10.4172/2168-9679.1000415

Copyright: © 2018 Leichsenring IG. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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