** Advances in Pure Mathematics** Vol.4 No.5(2014), Article ID:45613,14 pages DOI:10.4236/apm.2014.45021

A Comparison of Sufficiency Condtions for the Goldbach and the Twin Primes Conjectures

C. J. Mozzochi

Princeton, NJ, USA

Email: cjm@ix.netcom.com

Copyright © 2014 by author and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

Received 6 February 2014; revised 6 March 2014; accepted 15 March 2014

ABSTRACT

It is generally known that under the generalized Riemann hypothesis one could establish the twin primes conjecture by the circle method, provided one could obtain the estimate for the integral of the representation function over the minor arcs. One of the new results here is that the assumption of can be removed. We compare this and other such sufficiency results with similar results for the Goldbach conjecture.

**Keywords:**Goldbach Conjecture, Twin Primes Conjecture, Circle Method, Generalized Riemann Hypothesis

1. Introduction

Let and for. Let rapidly. When and

let denote the closed interval, a so-called major arc.

It is easily shown, for any choice of, that all the are disjoint and contained in the closed interval.

For each let be those points in which are not in any closed neighborhood (major arc)

of radius about any rational number, where and.

For each let be those points in which are not in any closed neighborhood (major arc)

of radius about any rational number, where and.

Let denote the number of ways (even) can be represented as a sum of two primes.

Let denote the number of twin primes less than or equal to.

In [1] the following two theorems are established:

Theorem 1.1 Under the generalized Riemann hypothesis with and if

, then for all even.

Theorem 1.2 Let and if

, then for all even.

In [2] the following two theorems are established:

Theorem 1.3 Under the assumption that Siegel zeros do not exist with and if

then for all even.

Theorem 1.4 Let and. If

, then for all even.

The proof of Theorem 1.3 and, in particular, the proof of Theorem 1.4 is very complicated.

In Section 6 of [2] it is shown, by a very complicated argument, that a particularly natural approach for eliminating the condition in Theorem 1.2 does not work.

As we will now see, the situation with regard to the twin prime conjecture is significantly, very less complicated. The reason is primarily because we only need to consider the Ramanujan sums rather than the sums, which appear in all of the theorems above, related to the Goldbach conjecture.

In Section 2 we will establish Theorem 1.5 Let and. if as goes to infinity in some suitable sequence.

2. A Proof of Theorem 1.5

We decompose the above integral

It is immediate by the prime numbers theorem that

By definition

where

Lemma 2.1 Let and. Then

This is Theorem 58 in [3] .

Lemma 2.2 Under the hypothesis of Lemma 2.1 we have

Proof. This follows immediately from Lemma 2.1 and the trivial inequalities and and the fact that if and, then.

Hence it is immediate that

By the change of variable we have

However, by (2.1) we have

Let so that if then

Let with the condition of summation and.

It is easy to see that

For we have

Hence for and

so that for some fixed we have:

But and by definition

so that it follows immediately that

Now summing over all we have

since by Theorem 327 page 267 in [3]

Hence

where. Now let

Lemma 2.3

Proof. [4] page 211.

It is immediate by Lemma 2.3 that

What remains to be done is to show is bounded away from 0.

Let

Since and are all multiplicative functions of, is a multiplicative function of. Also, by means of the trivial estimate on, namely 2, and a direct application of Theorem 327 page 267 in [5] we have

so that by Theorem 2 [3] page 3 we have

Hence

3. A Primitive Formulation of the Circle Method

1) Part I

We assume (even).

For each let.

Let be the number of representations of as the sum of two primes, each of which is less than.

Clearly,

We decompose this integral

where

Clearly, by Theorem 55 in [3] and the last paragraph on page 63 in [3] we have

By direct application of the easily established Equation (151) in [3]

(3.0)

and the Equation (204) in [3]

(3.1)

We have for

(3.2)

By the trivial inequalities and and the fact that if and, then with and, we have for

(3.3)

By the change of variable, we have

so that

(3.4)

Clearly

Also, the number of terms on the right hand side of (3.4) is and each term is greater than and less than 1 so that

Hence by definition of and Abel's lemma we have

so that

(3.5)

Hence,

(3.6)

Let. Then

so that

so that we have

Remark. Unfortunately, the integral cannot be; since for almost all, the integral is asymptotically where is the usual singular series.

We assume. For each let.

Let be the number of twin primes, each of which is less than.

Clearly,

We decompose this integral

where

where

Immediately, from (3.3) we have

By the change of variable, we have

so that

Let

Let

(3.7)

Clearly,

Also, the number of terms on the right hand side of (3.7) is and each term is greater than and less than 1 so that

Hence by definition of and Abel’s lemma we have

so that

(3.8)

Hence

(3.9)

Let Then;

so that

so that we have as goes to infinity in some suitable sequence.

2) Part II

For each (even) let be a prime in Let be those points in

which are not in and not in any closed interval of radius about any rational number where

.

Let be the number of representations of as the sum of two primes, which are limited to those primes in the arithmetic progressions mod.

Clearly,

We decompose this integral

where

Conjecture 1.

We now estimate.

But consider

so that uniformly

Hence for uniformly

So by (100) and (101) for

But since and since, we have by the inequalities immediately below (3.2) for

But

Hence for

(3.10)

By a change of variable we have

However,

Let

Hence, if,

By (3.5) we have

and

But

Hence, if,

so that

where

But by (3.6) we have

so that

so that

Let

By Theorem 272 in [5]

so that, since we assume.

Hence

so that if Conjecture 1 is true.

Let be the number of twin primes, each of which is in one of the arithmetic progressions mod defined above.

Clearly,

We decompose the integral

Conjecture 2.

infinity in some suitable sequence.

From (3.10) we have for

By change of variable we have

However,

Let

Hence, if

Clearly,

So that by (3.8)

and.

Hence, if,

so that

where

But by (3.6)

so that

Hence

Hence

so that if Conjecture 2 is true, as goes to infinity in a sequence that satisfies the conjecture.

4. Some Heuristics

Theorem 4.1 If

then every sufficiently large even integer is the sum of two primes, where is an exceptional set, whose measure goes to 0 with.

Proof. This is established in [1] .

Theorem 4.2 Let be an arbitrary fixed integer. Then

uniformly for almost all.

Proof. This deep result is immediate by (5-2) in [6] .

There is no compelling reason to assume Theorem 4.2 is not true for.

It is worthwile to investigate if Carleson’s proof can be modified to establish Theorem 4.2 with replaced with and replaced with.

In [7] Tao presents a heuristic argument to establish that the major arc contribution in the circle method is

. He states that his argument can be made rigorous.

However, it follows from the proof of Theorem 1.5 that the major arc contribution is not in any sequence of.

But it is well known that so that the contribution of the minor arc in the circle method approach to the twin primes conjecture (Theorem 1.5) is, which makes plausible that the required estimate of might be true.

It is plausible that in Theorem 1.3

where the latter integral is that of Theorem 1.5, which makes plausible that the required estimate of might be true.

Those, who seriously attempt Conjecture 2 have the advantage that there is some degree of freedom in the choice of and in the choice of for each; and the estimate is required only as goes to infinity in some suitable sequence.

Acknowledgements

I thank R. C. Vaughan for the Remark in Part I, Section 3.

References

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