An Asymptotic Approach to Twin Primes in a Prime-Dependent Interval

1. Introduction

The distribution of prime numbers has been one of the most fundamental and challenging problems in analytic number theory. Understanding the behavior of prime numbers is of great importance not only for theoretical mathematics but also for modern cryptography and computer science [1,2].

In particular, the question of whether there are infinitely many prime pairs of the form $(p,p+2)$, known as twin primes, has remained unsolved for nearly two centuries [3,4]. The Twin Prime Conjecture, which asserts the existence of infinitely many twin prime pairs, is one of the most famous unproven hypotheses.

In this study, we asymptotically analyze specific prime intervals to prove the infinitude of twin primes and demonstrate that at least one twin prime must exist within these intervals.

Specifically, the interval $(p_k·p_{k+1},p_{k+1}^2)$ has been chosen for analysis due to previous studies on the subject [5].

2. Methods

Definition 1. (Prime Counting Function)

$\pi \left(x\right)$ = It is the number of prime numbers up to x.

The following formula is known for the distribution of prime numbers:

$$\pi \left(x\right)\sim {\displaystyle \frac{x}{logx}}$$

(1)

Definition 2. (Twin Prime Counting Function)

${\pi }_2\left(x\right)$ = It is the number of twin prime numbers up to x.

Hardy-Littlewood twin prime estimate:

$${\pi }_2\left(x\right)\sim 2C_2{\int }_2^x{\displaystyle \frac{dt}{{(logt)}^2}}$$

(2)

$C_2$ is a constant and is calculated as follows:

$$C_2=\prod _{p\ge 3}\left(1-{\displaystyle \frac{1}{{(p-1)}^2}}\right)$$

(3)

This is approximately:

$$C_2\approx 0.6601618$$

(4)

Definition 3. (Montgomery-Vaughan inequalities.)

Montgomery and Vaughan developed a powerful asymptotic relation that examines the distribution of prime numbers in specific intervals, particularly useful for predicting the average density of twin prime pairs.

Montgomery-Vaughan inequalities:

$$\sum _{n\le x}(\pi (n+2)-\pi \left(n\right))\approx {\displaystyle \frac{x}{{log}^{2}x}}$$

(5)

3. Theorems and Proofs

Lemma 1. There is at least one prime number in the interval $(p_k·p_{k+1},p_{k+1}^2)$.

Proof. According to the Prime Number Theorem:

$$\pi \left(x\right)\sim {\displaystyle \frac{x}{logx}}$$

(6)

The number of prime numbers in this interval is:

$$\pi \left(p_{k+1}^2\right)-\pi \left(p_k{p}_{k+1}\right)$$

(7)

Approximately:

$$\pi \left(p_{k+1}^2\right)\approx {\displaystyle \frac{p_{k+1}^2}{2log{p}_{k+1}}},\pi \left(p_k{p}_{k+1}\right)\approx {\displaystyle \frac{p_k{p}_{k+1}}{log{p}_k+log{p}_{k+1}}}$$

(8)

Hence:

$$\pi \left(p_{k+1}^2\right)-\pi \left(p_k{p}_{k+1}\right)\approx {\displaystyle \frac{p_{k+1}^2}{2log{p}_{k+1}}}-{\displaystyle \frac{p_k{p}_{k+1}}{log{p}_k+log{p}_{k+1}}}$$

(9)

By showing that this is $\ge 1$, we can prove that there is at least one prime:

$${\displaystyle \frac{p_{k+1}^2}{2log{p}_{k+1}}}\ge 1+{\displaystyle \frac{p_k{p}_{k+1}}{2log{p}_{k+1}}}$$

(10)

$${\displaystyle \frac{p_{k+1}^2-p_k{p}_{k+1}}{2log{p}_{k+1}}}\ge 1$$

(11)

In this case, we must examine the following expression to verify the inequality:

$$p_{k+1}^2-p_k{p}_{k+1}\ge 2log{p}_{k+1}$$

(12)

It is evident that this expression is true.

Lemma 2. There are at least two prime numbers in the interval $(p_k·p_{k+1},p_{k+1}^2)$.

Proof. Based on Lemma 1, we can obtain the following expression:

$${\displaystyle \frac{p_{k+1}^2}{2log{p}_{k+1}}}\ge 2+{\displaystyle \frac{p_k{p}_{k+1}}{2log{p}_{k+1}}}$$

(13)

Therefore:

$${\displaystyle \frac{p_{k+1}^2-p_k{p}_{k+1}}{2log{p}_{k+1}}}\ge 2$$

(14)

As a result:

$$p_{k+1}^2-p_k{p}_{k+1}\ge 4log{p}_{k+1}$$

(15)

It is also evident that this expression is true.

Theorem 1.

$$\forall k,\mathrm{there}\mathrm{is}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{twin}\mathrm{prime}\mathrm{pair}\mathrm{in}\mathrm{the}\mathrm{interval}(p_k·p_{k+1},p_{k+1}^2).$$

(16)

Propositon 1. We obtain the same expression using the Hardy-Littlewood and Montgomery-Vaughan relations.

Proof. Using the Hardy-Littlewood formula, we obtain the following expression for the interval we have defined:

$${\pi }_2\left(p_{k+1}^2\right)-{\pi }_2(p_k·p_{k+1})\approx 2C_2{\int }_{p_k·p_{k+1}}^{p_{k+1}^2}{\displaystyle \frac{dt}{{(logt)}^2}}$$

(17)

Similarly, by using the Montgomery-Vaughan relations, we can obtain the same expression:

$$\sum _{n=p_k·p_{k+1}}^{p_{k+1}^2}(\pi (n+2)-\pi \left(n\right))\approx {\int }_{p_k·p_{k+1}}^{p_{k+1}^2}{\displaystyle \frac{dt}{{log}^{2}t}}$$

(18)

Proposition 2. The ratio of the error term to the integral converges to 0 for the obtained expression.

Proof. The Hardy-Littlewood formula, including the error term, is expressed as follows:

$${\pi }_2\left(x\right)=2C_2{\int }_2^x{\displaystyle \frac{dt}{{(logt)}^2}}+O\left({\displaystyle \frac{x}{{(logx)}^3}}\right)$$

(19)

The error term here is:

$$H_1\left(x\right)=O\left({\displaystyle \frac{x}{{log}^{3}x}}\right)$$

(20)

The Montgomery-Vaughan formula, including the error term, is expressed as follows:

$$\sum _{n=a}^b(\pi (n+2)-\pi \left(n\right))={\int }_a^b{\displaystyle \frac{dt}{{log}^{2}t}}+O\left({\displaystyle \frac{b}{{log}^{3}b}}\right)$$

(21)

The error term here is:

$$H_2\left(b\right)=O\left({\displaystyle \frac{b}{{log}^{3}b}}\right)$$

(22)

Let us recall our integral:

$$I_k={\int }_{p_k{p}_{k+1}}^{p_{k+1}^2}{\displaystyle \frac{dt}{{log}^{2}t}}$$

(23)

To calculate the error term, let’s compute the integral approximately:

$${\int }_a^b{\displaystyle \frac{dt}{{(logt)}^2}}\approx {\displaystyle \frac{b-a}{{(logb)}^2}}$$

(24)

$$I_k\approx {\displaystyle \frac{p_{k+1}^2-p_k{p}_{k+1}}{{(log{p}_{k+1}^2)}^2}}$$

(25)

Here,

$$log\left(p_{k+1}^2\right)=2log{p}_{k+1}$$

(26)

$${(log{p}_{k+1}^2)}^2={(2log{p}_{k+1})}^2=4{log}^2{p}_{k+1}$$

(27)

Therefore:

$$I_k\approx {\displaystyle \frac{p_{k+1}^2-p_k{p}_{k+1}}{4{log}^2{p}_{k+1}}}$$

(28)

To simplify further:

$$p_k\approx klogk$$

(29)

From here:

$$p_{k+1}-p_k=O\left(\sqrt{p_k}\right)$$

(30)

$$p_{k+1}^2-p_k{p}_{k+1}=p_{k+1}(p_{k+1}-p_k)$$

(31)

$$=p_{k+1}O\left(\sqrt{p_k}\right)$$

(32)

$$p_{k+1}^2-p_k{p}_{k+1}\approx O\left(p_{k+1}\sqrt{p_k}\right)$$

(33)

Therefore, our integral is:

$$I_k\approx {\displaystyle \frac{p_{k+1}\sqrt{p_k}}{4{log}^2{p}_{k+1}}}$$

(34)

The error term is:

$$H_k=O\left({\displaystyle \frac{p_{k+1}\sqrt{p_k}}{4{log}^3{p}_{k+1}}}\right)$$

(35)

To determine whether the error term can be neglected in the integral calculation, we need to compute the ratio of the error term to the integral:

$${\displaystyle \frac{H_k}{I_k}}={\displaystyle \frac{O\left(\frac{p_{k+1}\sqrt{p_k}}{4{log}^3{p}_{k+1}}\right)}{\frac{p_{k+1}\sqrt{p_k}}{4{log}^2{p}_{k+1}}}}$$

(36)

$${\displaystyle \frac{H_k}{I_k}}=O\left({\displaystyle \frac{1}{log{p}_{k+1}}}\right)$$

(37)

Since we are doing asymptotic analysis, we need to look at the limit at infinity:

$$\underset{p_{k+1}\to \infty }{lim}{\displaystyle \frac{H_k}{I_k}}=0$$

(38)

As a result, the error term can be neglected.

Proposition 3. The value of the integral is greater than or equal to 1.

Proof. At this stage, we need to calculate the integral in more detail.

$$I_k={\int }_{p_k·p_{k+1}}^{p_{k+1}^2}{\displaystyle \frac{dt}{{log}^{2}t}}$$

(39)

According to prime number theory:

$$p_k\approx klogk$$

(40)

$$p_k{p}_{k+1}\approx (klogk)·((k+1)log(k+1))$$

(41)

$$p_{k+1}^2\approx {(k+1)}^2{log}^2(k+1)$$

(42)

$$log\left(p_k{p}_{k+1}\right)\approx logk+log(k+1)+loglogk+loglog(k+1)$$

(43)

$$log\left(p_{k+1}^2\right)\approx 2(log(k+1)+loglog(k+1))$$

(44)

We can express the integral as follows:

$$I_k\approx {\int }_{log\left(p_k{p}_{k+1}\right)}^{log\left(p_{k+1}^2\right)}{\displaystyle \frac{e^{u}du}{u^2}}$$

(45)

$$\int e^u{u}^{-2}du\approx {\displaystyle \frac{e^u}{u^2}}$$

(46)

Hence:

$$I_k\approx \left[{\displaystyle \frac{e^u}{u^2}}\right]{log\left(p_{k}pk+1\right)}^{log\left(p_{k+1}^2\right)}$$

(47)

Here, the upper limit will be as follows:

$${\displaystyle \frac{p_{k+1}^2}{{(2log(k+1))}^2}}$$

(48)

The lower limit will be as follows:

$${\displaystyle \frac{p_k{p}_{k+1}}{{(logklog(k+1))}^2}}$$

(49)

Therefore, the integral becomes:

$$I_k\approx {\displaystyle \frac{p_{k+1}^2}{{(2log(k+1))}^2}}-{\displaystyle \frac{p_k{p}_{k+1}}{{(logklog(k+1))}^2}}$$

(50)

The proposition is true if the following condition is satisfied:

$${\displaystyle \frac{p_{k+1}^2}{{(2log{p}_{k+1})}^2}}\ge 1+{\displaystyle \frac{p_k{p}_{k+1}}{{(log{p}_k+log{p}_{k+1})}^2}}$$

(51)

For this reason, the following expression must be incorrect:

$${\displaystyle \frac{p_{k+1}^2}{{(2log{p}_{k+1})}^2}}<1$$

(52)

Again, according to prime number theory:

$$p_{k+1}\approx p_k+O\left(\sqrt{p_k}\right)$$

(53)

$$log{p}_k\approx log{p}_{k+1}$$

(54)

$$log{p}_k+log{p}_{k+1}\approx 2log{p}_{k+1}$$

(55)

Thence:

$${\displaystyle \frac{p_k{p}_{k+1}}{{(2log{p}_{k+1})}^2}}$$

(56)

$$={\displaystyle \frac{p_k{p}_{k+1}}{4{(log{p}_{k+1})}^2}}$$

(57)

The following expression can be formed:

$$p_k{p}_{k+1}<p_{k+1}^2$$

(58)

From here:

$${\displaystyle \frac{p_k{p}_{k+1}}{4{(log{p}_{k+1})}^2}}<{\displaystyle \frac{p_{k+1}^2}{4{(log{p}_{k+1})}^2}}$$

(59)

Hence:

$${\displaystyle \frac{p_{k+1}^2}{4{(log{p}_{k+1})}^2}}\ge 1+{\displaystyle \frac{p_k{p}_{k+1}}{4{(log{p}_{k+1})}^2}}$$

(60)

Then:

$${\displaystyle \frac{p_{k+1}^2-p_k{p}_{k+1}}{4{(log{p}_{k+1})}^2}}\ge 1$$

(61)

For this reason:

$$p_{k+1}^2-p_k{p}_{k+1}\ge 4{(log{p}_{k+1})}^2$$

(62)

It is evident that this expression is true.

As a result:

$${\displaystyle \frac{p_{k+1}^2}{{(2log{p}_{k+1})}^2}}\ge 1+{\displaystyle \frac{p_k{p}_{k+1}}{{(log{p}_k+log{p}_{k+1})}^2}}$$

(63)

This situation can also be demonstrated using a limit. The limit expression will be as follows:

$$\underset{p_k\to \infty }{lim}\left({\displaystyle \frac{p_{k+1}^2}{{(2log{p}_{k+1})}^2}}-\left(1+{\displaystyle \frac{p_k{p}_{k+1}}{{(log{p}_k+log{p}_{k+1})}^2}}\right)\right)$$

(64)

From here, it is easy to conclude:

$$\underset{p_k\to \infty }{lim}\left({\displaystyle \frac{p_{k+1}^2}{{(2log{p}_{k+1})}^2}}-\left(1+{\displaystyle \frac{p_k{p}_{k+1}}{{(log{p}_k+log{p}_{k+1})}^2}}\right)\right)\ge 0$$

(65)

Corollaries 1. Proposition 3 naturally shows that there must be at least one twin prime pair in the asymptotic interval. This completes the proof of Theorem 1.

4. Results

Firstly, we showed that there must be at least one prime number in the interval $(p_k{p}_{k+1},p_{k+1}^2)$.

We used the asymptotic formula of the prime counting function to obtain this result:

$$\pi \left(x\right)\sim {\displaystyle \frac{x}{logx}}$$

(66)

According to this relation, the number of prime numbers in the specified interval was calculated as follows:

$$\pi \left(p_{k+1}^2\right)-\pi \left(p_k{p}_{k+1}\right)\approx {\displaystyle \frac{p_{k+1}^2}{2log{p}_{k+1}}}-{\displaystyle \frac{p_k{p}_{k+1}}{log{p}_k+log{p}_{k+1}}}$$

(67)

From here, it was shown that this difference is greater than or equal to 1, meaning that there must be at least one prime number in this interval. Secondly, we demonstrated that there must be at least two primes in the specified interval.

Finally, we showed that one of the at least two prime numbers in the interval must necessarily be a twin prime.

At this stage, the Hardy-Littlewood twin prime conjecture was used in the following form:

$${\pi }_2\left(p_{k+1}^2\right)-{\pi }_2\left(p_k{p}_{k+1}\right)\approx 2C_2{\int }_{p_k{p}_{k+1}}^{p_{k+1}^2}{\displaystyle \frac{dt}{{(logt)}^2}}$$

(68)

The result of this integral is always greater than 1 for large values of k. This proves that there must be at least one twin prime pair in every specified interval.

One of the most important points in this study was the analysis of the error term.

By denoting the error term as $H_k$ and the integral value as $I_k$, we showed that the error term converges to zero proportionally:

$$\underset{p_{k+1}\to \infty }{lim}{\displaystyle \frac{H_k}{I_k}}=0$$

(69)

5. Discussion

This study presents a systematic approach to proving the infinitude of twin primes. Our work provides a fresh perspective by directly analyzing the existence of twin primes in specified prime number intervals.

By using the Hardy-Littlewood twin prime conjecture and Montgomery-Vaughan relations directly within an analytical framework, we obtained definitive results. Specifically, by driving the limit of the error term to zero, we controlled the error margin, thus providing a mathematically rigorous proof rather than a probabilistic or heuristic estimate.

The results obtained not only apply to twin primes but could also facilitate similar analyses for other prime pairs such as $(p,p+4)$, $(p,p+6)$, and others. Moreover, this could contribute to the development of new approaches to the Riemann hypothesis and the broader distribution of prime numbers.

The distribution of twin primes is significant not only in theoretical mathematics but also in cryptography and computer science. The RSA algorithm relies on the difficulty of factoring large prime numbers. A better understanding of twin prime distribution could allow for the more efficient selection of prime numbers. Understanding the density of large primes in certain intervals could optimize random prime generation methods for cryptographic security. Determining the exact distribution of twin primes could lead to the development of faster and more effective algorithms for discovering prime numbers.

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