# Randomness can be a useful tool for solving problems.

Algebra

### Re: Randomness can be a useful tool for solving problems.

Dave wrote:Equation: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}$$.

That equation is quite beautiful!

And it may be quite important too!

Go Blue!

We recall two important and beautiful equations from the analysis of the complex Riemann Zeta Function:

1A. $$\ \ \ \sum_{k=2}^{\infty }\frac{cos((b_{n} \pm \triangle b_{n}) * log(k))}{k^{\frac{1}{2}}}= -1$$; (Real Part)

1B. $$\ \ \ \sum_{k=2}^{\infty }\frac{sin((b_{n} \pm \triangle b_{n}) * log(k))}{k^{\frac{1}{2}}}= 0$$. (Imaginary Part)

where $$b_{n+1} = b_{n} \pm \triangle b_{n}$$ for some nonzero real values, $$b_{n + 1}$$, $$b_{n}$$, and $$\triangle b_{n}$$.

Remark: $$b_{n}$$ may take positive or negative values.

Remark: $$z = \frac{1}{2} + b_{n} * i$$ is the $$n^{th}$$ complex nontrivial zero of the Riemann Zeta Function.

"In mathematics, Montgomery's pair correlation conjecture is a conjecture made by Hugh Montgomery (1973) that the pair correlation between pairs of zeros of the Riemann Zeta Function (normalized to have unit average spacing) is

$$1 - (\frac{sin(\pi u)}{\pi u})^{2} + \delta (u)$$."

Source link: https://en.wikipedia.org/wiki/Montgomery%27s_pair_correlation_conjecture.
Guest

### Re: Randomness can be a useful tool for solving problems.

Dave wrote:Equation 1: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}$$
where $$1 \le \lambda \le \frac{\sqrt{I}}{2}$$.

Remark: $$\sqrt{I + \lambda^{2}}$$ is a positive integer.

Remark: $$I = p * q$$ where p and q are consecutive distinct odd primes where p > q.

Remark: $$p - q = 2 \lambda$$.

Remark: $$b_{n}$$ may take positive or negative values.

Remark: $$z = \frac{1}{2} + b_{n} * i$$ is the $$n^{th}$$ complex nontrivial zero of the Riemann Zeta Function.

"In mathematics, Montgomery's pair correlation conjecture is a conjecture made by Hugh Montgomery (1973) that the pair correlation between pairs of zeros of the Riemann Zeta Function (normalized to have unit average spacing) is

$$1 - (\frac{sin(\pi u)}{\pi u})^{2} + \delta (u)$$."

Source link: https://en.wikipedia.org/wiki/Montgomery%27s_pair_correlation_conjecture.

From equation one (above) we extract the following equation two,

Equation 2: $$\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2 d(I, \lambda, k) }$$
where $$d(I, \lambda, k)$$ is a nonzero real-valued function and where $$1 \le \lambda \le \frac{\sqrt{I}}{2}$$.

We derived the following expression from equation two,

$$1 - (\frac{sin( \pi x(I, \lambda, k))}{ \pi x(I, \lambda, k)})^{2} + \triangle ( x(I, \lambda, k) )$$

where $$\triangle ( x(I, \lambda, k) )$$ is a real-valued function and where $$x(I, \lambda, k) = 2k \sqrt{I + \lambda^{2}}$$.

Hmm. What's next?

$$- (\frac{sin( \pi x(I, \lambda, k))}{ \pi x(I, \lambda, k)})^{2} + \triangle ( x(I, \lambda, k) ) \rightarrow 0$$ for all consecutive odd prime pairs, p and q, where p > q.

Remark: We have derived a pair correlation theorem for all consecutive odd primes, p and q.

Remark: Polignac's conjecture is true! (https://en.wikipedia.org/wiki/Polignac%27s_conjecture)

Remark: Professor H. L. Montgomery's pair correlation conjecture is true!
Guest

### Re: Randomness can be a useful tool for solving problems.

Dave wrote:Equation 1: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}$$
where $$1 \le \lambda \le \frac{\sqrt{I}}{2}$$.

Remark: $$\sqrt{I + \lambda^{2}}$$ is a positive integer.

Remark: $$I = p * q$$ where p and q are consecutive distinct odd primes where p > q.

Remark: $$p - q = 2 \lambda$$.

Remark: $$b_{n}$$ may take positive or negative values.

Remark: $$z = \frac{1}{2} + b_{n} * i$$ is the $$n^{th}$$ complex nontrivial zero of the Riemann Zeta Function.

"In mathematics, Montgomery's pair correlation conjecture is a conjecture made by Hugh Montgomery (1973) that the pair correlation between pairs of zeros of the Riemann Zeta Function (normalized to have unit average spacing) is

$$1 - (\frac{sin(\pi u)}{\pi u})^{2} + \delta (u)$$."

Source link: https://en.wikipedia.org/wiki/Montgomery%27s_pair_correlation_conjecture.

From equation one (above) we extract the following equation two,

Equation 2: $$\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2 d(I, \lambda, k) }$$
where $$d(I, \lambda, k)$$ is a nonzero real-valued function and where $$1 \le \lambda \le \frac{\sqrt{I}}{2}$$.

We derived the following expression from equation two,

$$1 - (\frac{sin( \pi x(I, \lambda, k))}{ \pi x(I, \lambda, k)})^{2} + \triangle ( x(I, \lambda, k) )$$

where $$\triangle ( x(I, \lambda, k) )$$ is a real-valued function and where $$x(I, \lambda, k) = 2k \sqrt{I + \lambda^{2}}$$.

Hmm. What's next?

$$- (\frac{sin( \pi x(I, \lambda, k))}{ \pi x(I, \lambda, k)})^{2} + \triangle ( x(I, \lambda, k) ) \rightarrow 0$$ for all consecutive odd prime pairs, p and q, where p > q.

Remark: We have derived a pair correlation theorem for all consecutive odd primes, p and q.

Remark: Polignac's conjecture is true! (https://en.wikipedia.org/wiki/Polignac%27s_conjecture)

Remark: Professor H. L. Montgomery's pair correlation conjecture is true!
Go Blue!
Attachments
Professor H. L. Montgomery's pair correlation conjecture is true!
Montgomery-Odlyzko's Law.png (9.31 KiB) Viewed 233 times
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### Re: Randomness can be a useful tool for solving problems.

Remark: $$1 \le \lambda < \frac{\sqrt{I}}{2}$$ where $$\lambda$$ is a positive integer.
Guest

### Re: Randomness can be a useful tool for solving problems.

Dave wrote:Remark: $$1 \le \lambda < \frac{\sqrt{I}}{2}$$ where $$\lambda$$ is a positive integer.

Hmm. We can do much better on that upper bound ($$\frac{\sqrt{I}}{2}$$) in accordance with Prime Number Theorem...

Remark: $$1 \le \lambda < \frac{log^{2}(I)}{2}$$ where $$\lambda$$ is a positive integer and where $$\lambda \rightarrow \frac{log^{2}(I)}{2}$$ as $$I \rightarrow \infty$$.
Guest

### Re: Randomness can be a useful tool for solving problems.

An Update:

Equation 1: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}$$
where $$1 \le \lambda < \frac{log^{2}(I)}{2}$$.

Remark: $$\sqrt{I + \lambda^{2}}$$ is a positive integer.

Remark: $$I = p * q$$ where p and q are consecutive distinct odd primes where p > q.

Remark: $$p - q = 2 \lambda$$.

Remark: $$b_{n}$$ may take positive or negative real values.

Remark: $$z = \frac{1}{2} + b_{n} * i$$ is the $$n^{th}$$ complex nontrivial zero of the Riemann Zeta Function.

"In mathematics, Montgomery's pair correlation conjecture is a conjecture made by Hugh Montgomery (1973) that the pair correlation between pairs of zeros of the Riemann Zeta Function (normalized to have unit average spacing) is

$$1 - (\frac{sin(\pi u)}{\pi u})^{2} + \delta (u)$$."

Source link: https://en.wikipedia.org/wiki/Montgomery%27s_pair_correlation_conjecture.

From equation one (above) we extract the following equation two,

Equation 2: $$\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2 d(I, \lambda, k) }$$
where $$d(I, \lambda, k)$$ is a nonzero real-valued function and where $$1 \le \lambda < \frac{log^{2}(I)}{2}$$.

We derived the following function from equation two,

$$1 - (\frac{sin( \pi x(I, \lambda, k))}{ \pi x(I, \lambda, k)})^{2} + \triangle ( x(I, \lambda, k) )$$

where $$\triangle ( x(I, \lambda, k) )$$ is a real-valued function and where $$x(I, \lambda, k) = 2k \sqrt{I + \lambda^{2}}$$.

Hmm. What's next?

$$- (\frac{sin( \pi x(I, \lambda, k))}{ \pi x(I, \lambda, k)})^{2} + \triangle ( x(I, \lambda, k) ) \rightarrow 0$$ for all consecutive odd prime pairs, p and q, where p > q.

Remark: We have derived a pair correlation theorem for all consecutive odd primes, p and q.

Remark: Polignac's conjecture is true! (https://en.wikipedia.org/wiki/Polignac%27s_conjecture)

Remark: Professor H. L. Montgomery's pair correlation conjecture is true!

Go Blue!
Attachments
Professor H. L. Montgomery's correlation conjecture is true!
Montgomery-Odlyzko's Law.png (9.31 KiB) Viewed 219 times
Guest

### Re: Randomness can be a useful tool for solving problems.

"Remark: Polignac's conjecture is true! (https://en.wikipedia.org/wiki/Polignac%27s_conjecture)"

Hah! We are not convinced that Polignac's conjecture is true! There may be exceptional sets such that Poignac's conjecture is generally false!

Let's assume $$1 \le \lambda < \frac{log^{2}(pq)}{2}$$ is in accordance with the Prime Number Theorem when p > q are consecutive odd primes with $$p - q = 2 \lambda$$.

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =1$$?

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =2$$?

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =3$$?
...

What is the probability that Polignac's conjecture is false for any positive even integer, $$2 \lambda$$, where $$p - q = 2 \lambda$$?
Guest

### Re: Randomness can be a useful tool for solving problems.

Guest wrote:"Remark: Polignac's conjecture is true! (https://en.wikipedia.org/wiki/Polignac%27s_conjecture)"

Hah! We are not convinced that Polignac's conjecture is true! There may be exceptional sets such that Poignac's conjecture is generally false!

Let's assume $$1 \le \lambda < \frac{log^{2}(pq)}{2}$$ is in accordance with the Prime Number Theorem when p > q are consecutive odd primes with $$p - q = 2 \lambda$$.

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =1$$?

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =2$$?

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =3$$?
...

What is the probability that Polignac's conjecture is generally false for any positive even integer, $$2 \lambda$$?

Remark: The exponent,$$\frac{1}{2}$$, in the equation, $$\sqrt{pq + \lambda^{2}} = (pq + \lambda^{2})^{\frac{1}{2}}$$, is a big indicator of the truth of the Riemann Hypothesis!

Assumption: Polignac's conjecture is generally false for any positive even integer, $$2 \lambda$$.

To indicate that the Polignac's conjecture is false or $$p - q \ne 2 \lambda$$ for for all or almost all consecutive primes p > q, we define an exceptional and infinite set, E.

E = {(p, q) |$$p > q > 2 \lambda$$ are consecutive primes with $$p - q \ne 2 \lambda$$}.

The inequality, $$p - q \ne 2 \lambda$$, indicates $$n_{m } *q_{m } = p - 2 \lambda$$ where $$n_{m} > 1$$ is an odd integer associated with some odd prime, $$q_{m }$$.

Remark: $$n_{m} \ge q_{m}$$.

Remark: $$n_{m} = 1$$ indicates that Polignac's conjecture is true!

Therefore, as a result of E, we generate an infinite system of independent Diophantine equations with odd integer, $$n_{k } > 1$$, for appropriate odd primes, $$p_{k}$$ and $$q_{k}$$:

$$p_{1 } - 2 \lambda = n_{1 } *q_{1 }$$;

$$p_{2 } - 2 \lambda = n_{2 } *q_{2 }$$;

$$p_{3 } - 2 \lambda = n_{3 } *q_{3 }$$;

...

$$p_{\infty } - 2 \lambda = n_{\infty } *q_{\infty}$$;

Remark: $$p_{k} < p_{k+1}$$ are consecutive primes over E.

Remark: $$3 \le q_{k} \le \sqrt{ n_{k } *q_{k}} \le n_{k }$$.

Remark: $$\pi (x)$$ indicates the odd prime-counting function.

Remark: We assume the Riemann Hypothesis since it is true! Go Blue!

Remark: "The Riemann Hypothesis is equivalent to a much tighter bound on the error in the estimate for $$\pi (x)$$, and hence to a more regular distribution of prime numbers..." Source Link: https://en.wikipedia.org/wiki/Prime-counting_function#The_Riemann_hypothesis.

Remark: $$\pi ( \sqrt{p_{m } - 2})$$ is the maximum number of primes, $$q_{m }$$, that may divide $$p_{m } - 2$$.

The probability that Polignac's conjecture is false over E is Probability ($$p - 2 \lambda \ne q$$ over E ) or Prob( $$p - 2 \lambda \ne q$$ over E ).

Prob( $$p - 2 \lambda \ne q$$ over E ) = Prob ( $$n_{m} \ne 1 | p_{m } - 2 \lambda = n_{m } *q_{m }$$ ) * Prob ( $$p_{m } - 2 \lambda = n_{m } *q_{m }$$).

Remark: Prob ( $$p_{m } - 2 \lambda = n_{m } *q_{m }$$ ) = 1 since $$n_{m } \ne 1$$.

Therefore,

Prob( $$p - 2 \lambda \ne q$$ over E ) = Prob ( $$p_{m } - 2 \lambda = n_{m } *q_{m }$$)

= $$\prod_{m=1}^{\infty }\frac{\pi (\sqrt{p_{m} - 2 \lambda})}{\pi (\sqrt{p_{m} - 2 \lambda}) + 1} = 0$$.

Remark: We count $$n_{m} = 1$$, the exception that violates our assumption.

That result [ Prob( $$p - 2 \lambda \ne q$$ over E ) = 0 ] contradicts our assumption.

Remark: Polignac's conjecture is true!

Remark: The tighter error bound associated with the odd prime-counting function does not violate our final result.

Dave.

Go Blue!
Guest

### Re: Randomness can be a useful tool for solving problems.

Update: Prob( $$p - 2 \lambda \ne q$$ over E ) = Prob ( $$n_{m} \ne 1 | p_{m } - 2 \lambda = n_{m } *q_{m }$$ )

= $$\prod_{m=1}^{\infty }\frac{\pi (\sqrt{p_{m} - 2 \lambda})}{\pi (\sqrt{p_{m} - 2 \lambda}) + 1} = 0$$.
Guest

### Re: Randomness can be a useful tool for solving problems.

"To indicate that the Polignac's conjecture is false or $$p - q \ne 2 \lambda$$ for for all or almost all consecutive primes p > q, we define an exceptional and infinite set, E."

Remark: We should add that our exceptional and infinite set, E, must imply for any positive integer, $$2\lambda$$, Polignac's conjecture is true only a finite number of times.

Dave.

"Math is hard work!"
Guest

### Re: Randomness can be a useful tool for solving problems.

Remarks: We are attempting to develop tools and to advance theory that will aid us in solving integer factorization problems posted here and elsewhere. And we shall revisit old problems and other stuff (continued factorization algorithm, general number field sieve, etc.) posted here from time to time. Moreover, we seek to discover our mistakes and to correct them along the way. Hopefully, we can advance our work and meet our deadline (April 30, 2021, great Gauss's Birthday).

Dave.
Guest

### Re: Randomness can be a useful tool for solving problems.

Dave wrote:Remarks: We are attempting to develop tools and to advance theory that will aid us in solving integer factorization problems posted here and elsewhere. And we shall revisit old problems and other stuff (continued fraction factorization algorithm, general number field sieve, etc.) posted here from time to time. Moreover, we seek to discover our mistakes and to correct them along the way. Hopefully, we can advance our work and meet our deadline (April 30, 2021, great Gauss's Birthday).
Guest

### Re: Randomness can be a useful tool for solving problems.

Guest wrote:"Remark: Polignac's conjecture is true! (https://en.wikipedia.org/wiki/Polignac%27s_conjecture)"

Hah! We are not convinced that Polignac's conjecture is true! There may be exceptional sets such that Poignac's conjecture is generally false!

Let's assume $$1 \le \lambda < \frac{log^{2}(pq)}{2}$$ is in accordance with the Prime Number Theorem when p > q are consecutive odd primes with $$p - q = 2 \lambda$$.

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =1$$?

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =2$$?

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =3$$?
...

What is the probability that Polignac's conjecture is generally false for any positive even integer, $$2 \lambda$$?

An Update:

Remark: The exponent,$$\frac{1}{2}$$, in the equation, $$\sqrt{pq + \lambda^{2}} = (pq + \lambda^{2})^{\frac{1}{2}}$$, is a big indicator of the truth of the Riemann Hypothesis!

Assumption: Polignac's conjecture is generally false for any positive even integer, $$2 \lambda$$.

To indicate that the Polignac's conjecture is false or $$p - q \ne 2 \lambda$$ for for all or almost all consecutive primes p > q, we define an exceptional and infinite set, E.

E = {(p, q) |$$p > q > 2 \lambda$$ are consecutive primes with $$p - q \ne 2 \lambda$$}.

The inequality, $$p - q \ne 2 \lambda$$, indicates $$n_{m } *q_{m } = p - 2 \lambda$$ where $$n_{m} > 1$$ is an odd integer associated with some odd prime, $$q_{m }$$.

Remark: $$n_{m} \ge q_{m}$$.

Remark: $$n_{m} = 1$$ indicates that Polignac's conjecture is true!

Therefore, as a result of E, we generate an infinite system of independent Diophantine equations with odd integer, $$n_{k } > 1$$, for appropriate odd primes, $$p_{k}$$ and $$q_{k}$$:

$$p_{1 } - 2 \lambda = n_{1 } *q_{1 }$$;

$$p_{2 } - 2 \lambda = n_{2 } *q_{2 }$$;

$$p_{3 } - 2 \lambda = n_{3 } *q_{3 }$$;

...

$$p_{\infty } - 2 \lambda = n_{\infty } *q_{\infty}$$;

Remark: $$p_{k} < p_{k+1}$$ are consecutive primes over E.

Remark: $$3 \le q_{k} \le \sqrt{ n_{k } *q_{k}} \le n_{k }$$.

Remark: $$\pi (x)$$ indicates the odd prime-counting function.

Remark: We assume the Riemann Hypothesis since it is true! Go Blue!

Remark: "The Riemann Hypothesis is equivalent to a much tighter bound on the error in the estimate for $$\pi (x)$$, and hence to a more regular distribution of prime numbers..." Source Link: https://en.wikipedia.org/wiki/Prime-counting_function#The_Riemann_hypothesis.

Remark: $$\pi ( \sqrt{p_{m } - 2})$$ is the maximum number of primes, $$q_{m }$$, that may divide $$p_{m } - 2$$.

The probability that Polignac's conjecture is false over E is Probability ($$p - 2 \lambda \ne q$$ over E ) or Prob( $$p - 2 \lambda \ne q$$ over E ).

Prob( $$p - 2 \lambda \ne q$$ over E ) = Prob ( $$n_{m} \ne 1 \forall m \in \N | p_{m } - 2 \lambda = n_{m } *q_{m }$$ ) * Prob ( $$p_{m } - 2 \lambda = n_{m } *q_{m }$$).

Remark: Prob ( $$p_{m } - 2 \lambda = n_{m } *q_{m }$$ ) = 1 since we assume $$n_{m } \ne 1$$.

Therefore,

Prob( $$p - 2 \lambda \ne q$$ over E ) = Prob ( $$n_{m} \ne 1 \forall m \in \N | p_{m } - 2 \lambda = n_{m } *q_{m }$$ )

= $$\prod_{m=1}^{\infty }\frac{\pi (\sqrt{p_{m} - 2 \lambda})}{\pi (\sqrt{p_{m} - 2 \lambda}) + 1} = 0$$.

Remark: We count $$n_{m} = 1$$, the exception that violates our assumption.

That result, Prob( $$p - 2 \lambda \ne q$$ over E ) = 0, contradicts our assumption.

Remark: Polignac's conjecture is true!

Remark: The tighter error bound associated with the odd prime-counting function does not violate our final result.

Remarks: Our proof of Polignac's conjecture is still a work in progress. And we expect more modifications/updates in the near future.

Dave.

Go Blue!
Guest

### Re: Randomness can be a useful tool for solving problems.

Remarks: In prime number theory, we have two endless processes, 'fission' (the generation of composite integers, or integers greater one and are not prime) and 'fusion' (the generation of primes) both in accordance with the Fundamental Theorem of Arithmetic and in accordance with the distribution of primes (the Prime Number Theorem).
Guest

### Re: Randomness can be a useful tool for solving problems.

Guest wrote:"Remark: Polignac's conjecture is true! (https://en.wikipedia.org/wiki/Polignac%27s_conjecture)"

Hah! We are not convinced that Polignac's conjecture is true! There may be exceptional sets such that Poignac's conjecture is generally false!

Let's assume $$1 \le \lambda < \frac{log^{2}(pq)}{2}$$ is in accordance with the Prime Number Theorem when p > q are consecutive odd primes with $$p - q = 2 \lambda$$.

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =1$$?

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =2$$?

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =3$$?
...

What is the probability that Polignac's conjecture is generally false for any positive even integer, $$2 \lambda$$?

An Update:

Remark: The exponent,$$\frac{1}{2}$$, in the equation, $$\sqrt{pq + \lambda^{2}} = (pq + \lambda^{2})^{\frac{1}{2}}$$, is a big indicator of the truth of the Riemann Hypothesis!

Assumption: Polignac's conjecture is generally false for any positive even integer, $$2 \lambda$$.

To indicate that the Polignac's conjecture is false or $$p - q \ne 2 \lambda$$ for for all or almost all consecutive odd primes p > q, we define an exceptional and infinite set, E.

E = {(p, q) |$$p > q > 2 \lambda$$ are consecutive odd primes with $$p - q \ne 2 \lambda$$}.

The inequality, $$p - q \ne 2 \lambda$$, indicates $$n_{m } *q_{m } = p - 2 \lambda$$ where $$n_{m} > 1$$ is an odd integer associated with some odd prime, $$q_{m }$$.

Remark: $$n_{m} \ge q_{m}$$.

Remark: $$n_{m} = 1$$ indicates that Polignac's conjecture is true!

Therefore, as a result of E, we generate a system of infinite independent Diophantine equations with odd integer, $$n_{k } > 1$$, for appropriate odd primes, $$p_{k}$$ and $$q_{k}$$:

$$p_{1 } - 2 \lambda = n_{1 } *q_{1 }$$;

$$p_{2 } - 2 \lambda = n_{2 } *q_{2 }$$;

$$p_{3 } - 2 \lambda = n_{3 } *q_{3 }$$;

...

$$p_{\infty } - 2 \lambda = n_{\infty } *q_{\infty}$$;

Remark: $$p_{k} < p_{k+1}$$ are consecutive odd primes over E.

Remark: $$3 \le q_{k} \le \sqrt{ n_{k } *q_{k}} \le n_{k }$$.

Remark: $$\pi (x)$$ indicates the odd prime-counting function.

Remark: We assume the Riemann Hypothesis since it is true! Go Blue!

Remark: "The Riemann Hypothesis is equivalent to a much tighter bound on the error in the estimate for $$\pi (x)$$, and hence to a more regular distribution of prime numbers..." Source Link: https://en.wikipedia.org/wiki/Prime-counting_function#The_Riemann_hypothesis.

Remark: $$\pi ( \sqrt{p_{m } - 2})$$ is the maximum number of primes, $$q_{m }$$, that may divide $$p_{m } - 2$$.

The probability that Polignac's conjecture is false over E is Probability ($$p - 2 \lambda \ne q$$ over E ) or Prob( $$p - 2 \lambda \ne q$$ over E ).

Prob( $$p - 2 \lambda \ne q$$ over E ) = Prob ( $$n_{m} \ne 1 \forall m \in \N | p_{m } - 2 \lambda = n_{m } *q_{m }$$ ) * Prob ( $$p_{m } - 2 \lambda = n_{m } *q_{m }$$).

Remark: Prob ( $$p_{m } - 2 \lambda = n_{m } *q_{m }$$ ) = 1 since we assume $$n_{m } \ne 1$$.

Therefore,

Prob( $$p - 2 \lambda \ne q$$ over E ) = Prob ( $$n_{m} \ne 1 \forall m \in \N | p_{m } - 2 \lambda = n_{m } *q_{m }$$ )

= $$\prod_{m=1}^{\infty }\frac{\pi (\sqrt{p_{m} - 2 \lambda})}{\pi (\sqrt{p_{m} - 2 \lambda}) + 1} = 0$$.

Remark: We count $$n_{m} = 1$$, the exception that violates our assumption.

That result, Prob( $$p - 2 \lambda \ne q$$ over E ) = 0, contradicts our assumption.

Remark: Polignac's conjecture is true!

Remark: The tighter error bound associated with the odd prime-counting function does not violate our final result.

Remarks: Our proof of Polignac's conjecture is still a work in progress. And we expect more modifications/updates in the near future.

Dave.

Go Blue!
Guest

### Re: Randomness can be a useful tool for solving problems.

Guest wrote:"Remark: Polignac's conjecture is true! (https://en.wikipedia.org/wiki/Polignac%27s_conjecture)"

Hah! We are not convinced that Polignac's conjecture is true! There may be exceptional sets such that Poignac's conjecture is generally false!

Let's assume $$1 \le \lambda < \frac{log^{2}(pq)}{2}$$ is in accordance with the Prime Number Theorem when p > q are consecutive odd primes with $$p - q = 2 \lambda$$.

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =1$$?

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =2$$?

Are there infinitely many consecutive prime pairs, p and q, such that $$\sqrt{pq + \lambda^{2}}$$ is a positive integer when $$\lambda =3$$?
...

What is the probability that Polignac's conjecture is generally false for any positive even integer, $$2 \lambda$$?

An Update:

Remark: The exponent,$$\frac{1}{2}$$, in the equation, $$\sqrt{pq + \lambda^{2}} = (pq + \lambda^{2})^{\frac{1}{2}}$$, is a big indicator of the truth of the Riemann Hypothesis!

Assumption: Polignac's conjecture is generally false for any positive even integer, $$2 \lambda$$.

To indicate that the Polignac's conjecture is false or $$p - q \ne 2 \lambda$$ for for all or almost all consecutive odd primes p > q, we define an exceptional and infinite set, E.

E = {(p, q) |$$p > q > 2 \lambda$$ are consecutive odd primes with $$p - q \ne 2 \lambda$$}.

The inequality, $$p - q \ne 2 \lambda$$, indicates $$n_{m } *q_{m } = p - 2 \lambda$$ where $$n_{m} > 1$$ is an odd integer associated with some odd prime, $$q_{m }$$.

Remark: $$n_{m} \ge q_{m}$$.

Remark: $$n_{m} = 1$$ indicates that Polignac's conjecture is true!

Therefore, as a result of E, we generate a system of infinite independent Diophantine equations with odd integer, $$n_{k } > 1$$, for appropriate odd primes, $$p_{k}$$ and $$q_{k}$$:

$$p_{1 } - 2 \lambda = n_{1 } *q_{1 }$$;

$$p_{2 } - 2 \lambda = n_{2 } *q_{2 }$$;

$$p_{3 } - 2 \lambda = n_{3 } *q_{3 }$$;

...

$$p_{\infty } - 2 \lambda = n_{\infty } *q_{\infty}$$;

Remark: $$p_{k} < p_{k+1}$$ are consecutive odd primes belonging to E.

Remark: $$3 \le q_{k} \le \sqrt{ n_{k } *q_{k}} \le n_{k }$$.

Remark: $$\pi (x)$$ indicates the odd prime-counting function.

Remark: We assume the Riemann Hypothesis since it is true! Go Blue!

Remark: "The Riemann Hypothesis is equivalent to a much tighter bound on the error in the estimate for $$\pi (x)$$, and hence to a more regular distribution of prime numbers..." Source Link: https://en.wikipedia.org/wiki/Prime-counting_function#The_Riemann_hypothesis.

Remark: $$\pi ( \sqrt{p_{m } - 2})$$ is the maximum number of primes, $$q_{m }$$, that may divide $$p_{m } - 2$$.

The probability that Polignac's conjecture is false over E is Probability ($$p - 2 \lambda \ne q$$ over E ) or Prob( $$p - 2 \lambda \ne q$$ over E ).

Prob( $$p - 2 \lambda \ne q$$ over E ) = Prob ( $$n_{m} \ne 1 \forall m \in \N | p_{m } - 2 \lambda = n_{m } *q_{m }$$ ) * Prob ( $$p_{m } - 2 \lambda = n_{m } *q_{m }$$).

Remark: Prob ( $$p_{m } - 2 \lambda = n_{m } *q_{m }$$ ) = 1 since we assume $$n_{m } \ne 1$$.

Therefore,

Prob( $$p - 2 \lambda \ne q$$ over E ) = Prob ( $$n_{m} \ne 1 \forall m \in \N | p_{m } - 2 \lambda = n_{m } *q_{m }$$ )

= $$\prod_{m=1}^{\infty }\frac{\pi (\sqrt{p_{m} - 2 \lambda})}{\pi (\sqrt{p_{m} - 2 \lambda}) + 1} = 0$$.

Remark: We count $$n_{m} = 1$$, the exception that violates our assumption.

That result, Prob( $$p - 2 \lambda \ne q$$ over E ) = 0, contradicts our assumption.

Remark: Polignac's conjecture is true!

Remark: The tighter error bound associated with the odd prime-counting function does not violate our final result.

Remarks: Our proof of Polignac's conjecture is still a work in progress. And we expect more modifications/updates in the near future.

Dave.

Go Blue!
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### Re: Randomness can be a useful tool for solving problems.

A key idea behind our latest proof of Polignac's conjecture: Given a positive odd integer, k > 1, then k is either nonprime or prime.

If k = n * p, then n = 1 indicates k is prime (assuming p is prime). The value $$n \ne 1$$ indicates k is nonprime. We can count the maximum number of primes that may divide k.

That value is either one or $$\pi (\sqrt{k}) \ge 1$$.

Remark: Let's assume $$\pi (x)$$ is the exact odd prime-counting function.

Therefore, the chance that k is nonprime is roughly $$\frac{\pi (\sqrt{k})}{\pi (\sqrt{k}) + 1}$$.

Remarks: Roughly indicates an imperfect result. However, that result is good enough for a probabilistic proof of Polignac's conjecture.

Dave,

https://www.researchgate.net/profile/David_Cole29.

Go Blue!
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### Re: Randomness can be a useful tool for solving problems.

"We can count the maximum number of primes that may divide k."

Update: We can count the maximum number of primes (below $$\sqrt{k}$$) that may divide k.
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### Re: Randomness can be a useful tool for solving problems.

A key idea behind our latest proof of Polignac's conjecture:

For positive real numbers, x and y > 1, $$x^{y} \rightarrow 0$$ as $$y \rightarrow \infty$$, if $$0 < x < 1$$.
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### Re: Randomness can be a useful tool for solving problems.

Remark 1: $$p_{k+1 } - p_{k } \ne 2\lambda$$ over E for some $$\lambda$$ such that $$1 \le \lambda < \frac{log^{2}(p_{k+1 }p_{k})}{2}$$.

Remark: We must redefine our current exceptional set, E, to comply with remark one.

Remark: This problem is a big headache! Ouch!
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