# Randomness can be a useful tool for solving problems.

Algebra

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

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Dave wrote: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!

Oops! Our current proof of Polignac's conjecture is wrong!! The proof of Polignac's conjecture should be about the spacing between consecutive odd primes.

Example: Suppose we want to exclude $$2 \lambda_{0 }$$ over E.

We have $$p_{2 } - p_{1 } \ne 2 \lambda_{0 }$$ such that $$1 \le \lambda_{0 } < \frac{log^{2}(p_{1 }p_{2})}{2}$$.

The chance that $$2 \lambda_{0 }$$ is the wrong spacing between consecutive odd primes, $$p_{2 } > p_{1 }$$, is roughly

$$\frac{Floor( \frac{log^{2}(p_{1 }p_{2})}{2} - 1) }{Floor( \frac{log^{2}(p_{1 }p_{2})}{2}) }$$.

However, over E, we generate the infinite product of similar values because of independence so that the chance $$2 \lambda_{0 }$$ is the wrong spacing between consective odd primes over E equates to zero.

In short, Polignac's conjecture is still correct! But our previous reasoning was wrong! We hope we have right this time. We will review it later.

Dave.

Go Blue!
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Math works! But beware of flawed reasoning!
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### Re: Randomness can be a useful tool for solving problems.

Dave wrote:
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!

Oops! Our current proof of Polignac's conjecture is wrong!! The proof of Polignac's conjecture should be about the spacing between consecutive odd primes.

Example: Suppose we want to exclude $$2 \lambda_{0 }$$ over E.

We have $$p_{2 } - p_{1 } \ne 2 \lambda_{0 }$$ such that $$1 \le \lambda_{0 } < \frac{log^{2}(p_{1 }p_{2})}{2}$$.

The chance that $$2 \lambda_{0 }$$ is the wrong spacing between consecutive odd primes, $$p_{2 } > p_{1 }$$, is roughly

$$\frac{Floor( \frac{log^{2}(p_{1 }p_{2})}{2} - 1) }{Floor( \frac{log^{2}(p_{1 }p_{2})}{2}) }$$.

However, over E, we generate the infinite product of similar values because of independence so that the chance $$2 \lambda_{0 }$$ is the wrong spacing between consecutive odd primes over E equates to zero.

In short, Polignac's conjecture is still correct! But our previous reasoning was wrong! We hope we have it right this time. We will review it later.

Dave.

Go Blue!
Guest

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

Dave wrote:Update: Prob( $$p - 2 \lambda \ne q$$ over E ) = Prob ( $$n_{m} \ne 1 \forall m \in \N | p_{m } - 2 \lambda = n_{m } *q_{m }$$ )
= $$\prod_{m=1}^{\infty }\frac{\pi (\sqrt{p_{m} - 2 \lambda})}{\pi (\sqrt{p_{m} - 2 \lambda}) + 1} = 0$$.

Update: Prob( $$p - q \ne 2 \lambda$$ over E )

= $$\prod_{m=1}^{\infty }\frac{Floor( \frac{log^{2}(p_{m }p_{m+1})}{2} - 1) }{Floor( \frac{log^{2}(p_{m }p_{m+1})}{2}) } = 0$$.

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

Remarks: "Roughly" indicates less than ideal or too large or too small. For our latest proof of Polignac's conjecture, "roughly" indicates generally too large. But it is good enough for a probabilistic proof of Polignac's conjecture.

Dave.

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

Dave wrote:Update:

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

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

Remark: This problem is a big headache! Ouch!

Oops! Our current proof of Polignac's conjecture is wrong!! The proof of Polignac's conjecture should be about the spacing between consecutive odd primes.

Example: Suppose we want to exclude $$2 \lambda_{0 }$$ over E.

Update:

We have $$p_{2 } - p_{1 } \ne 2 \lambda_{0 }$$ such that $$1 \le \lambda_{0 } < \frac{log^{2}(max((p_{1 }, p_{2}))}{2}$$.
_____________________________________________________________________________________________________________________________________________
Update:

The chance that $$2 \lambda_{0 }$$ is the wrong spacing between consecutive odd primes, $$p_{2 } > p_{1 }$$, is roughly

$$\frac{Floor( \frac{log^{2}(max(p_{1 }, p_{2}))}{2} - 1) }{Floor( \frac{log^{2}(max(p_{1}, p_{2}))}{2}) }$$.

Remark: "Roughly" indicates too large.

However, over E, we generate the infinite product of similar values because of independence so that the chance $$2 \lambda_{0 }$$ is the wrong spacing between consecutive odd primes over E equates to zero:

Prob( $$p - q \ne 2 \lambda$$ over E )

= $$\prod_{m=1}^{\infty }\frac{Floor( \frac{log^{2}(max(p_{m }, p_{m+1}))}{2} - 1) }{Floor( \frac{log^{2}(max(p_{m }, p_{m+1}))}{2}) } = 0$$.
_______________________________________________________________________________________________________________________________________________
In short, Polignac's conjecture is still correct! But our previous reasoning was wrong! We hope we have it right this time. We will review it later.

Remark: We apologized for the sloppy (flawed) math in previous posts.

Dave.

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

Remark: "Math is hard work!"
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### Re: Randomness can be a useful tool for solving problems.

Comments: We are done with Polignac's conjecture! And we hope our readers (mathematicians) make sure that any proof of Polignac's conjecture is correct.

Our experience with this problem has been frustrating and embarrassing.

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

'LARGE GAPS BETWEEN CONSECUTIVE PRIME NUMBERS',

https://www.math10.com/forum/viewtopic.php?f=63&t=8263.
Guest

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

Dave wrote:We are given the integer, I (a 1000-digit integer: please see attachment below), which is a product two unknown primes, p (a 500-digit integer) and q (a 500-digit integer).

What is $$\sqrt{I}$$ ?

What are p and q?

Hmm. Can we combine tools (continued fraction factorization, general number field sieve algorithm, etc.) to achieve integer factorization of I (see attachment below) in polynomial time?

Dave.
Attachments
What are the prime factors, p and q, of I?
I is a 1000-Digit Integer that is a product of two unknown primes, p and q, each with 500 digits..gif (134.73 KiB) Viewed 513 times
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### Re: Randomness can be a useful tool for solving problems.

Dave wrote:We are given the integer, I (a 1000-digit integer: please see attachment below), which is a product two unknown primes, p (a 500-digit integer) and q (a 500-digit integer).

What is $$\sqrt{I}$$ ?

What are p and q?

Hmm. Can we combine tools (continued fraction factorization, general number field sieve algorithm, etc.) to achieve integer factorization of I (see attachment below) in polynomial time?

Dave.

The Problem: We are given a large positive integer, I, which is a product of two unknown odd primes, p > q.

What are p and q?

The Solution Formulation:

We generate a system of two equations with three unknowns since $$p - q = 2 \lambda$$ for some unknown positive integer, $$2 \lambda$$:

1. $$p * q = I$$;

2. $$p - q = 2 \lambda$$ such that $$1 \le \lambda < \frac{log^{2} (pq)}{2}$$.

The Question: Can we achieve integer factorization of I in polynomial time?

The Answer: Yes! We can affirmatively achieve integer factorization of I in polynomial time.

Dave,

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

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

\lambda
Dave wrote:We are given the integer, I (a 1000-digit integer: please see attachment below), which is a product two unknown primes, p (a 500-digit integer) and q (a 500-digit integer).

What is $$\sqrt{I}$$ ?

What are p and q?

Hmm. Can we combine tools (continued fraction factorization, general number field sieve algorithm, etc.) to achieve integer factorization of I (see attachment below) in polynomial time?

Dave.

The Problem: We are given a large positive integer, I, which is a product of two unknown odd primes, p > q.

What are p and q?

The Solution Formulation:

We generate a system of two equations with three unknowns since $$p - q = 2 \lambda$$ for some unknown positive integer, $$2 \lambda$$:

1. $$p * q = I$$;

2. $$p - q = 2 \lambda$$ such that $$1 \le \lambda < \frac{log^{2} (pq)}{2}$$.

The Question: Can we achieve integer factorization of I in polynomial time?

The Answer: Yes! We can affirmatively achieve integer factorization of I in polynomial time.

Dave,

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

Go Blue!

"Simple seeks simplest (best) solution."

Moreover,

1a. $$q = -\lambda + \sqrt{I + \lambda^{2}}$$ where $$1 \le \lambda < \frac{log^{2} (pq)}{2}$$;

1b. $$p = \frac{I}{q}$$.

For our particular problem, we have $$1 \le \lambda < 2,650,949$$.

Dave.

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

Oops! $$(log(p * q))^{2}$$ is the wrong formula! The $$(log(p * q))^{2}$$ works for the largest gap between consecutive primes less than p * q.

Correction! $$2 \le 2 \lambda < \frac{\sqrt{I}}{2}$$ is probably closer to the truth.

Is integer factorization achievable in polynomial time? WE DO NOT KNOW!

Simple Dave
Guest

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

Dave wrote:Oops! $$(log(p * q))^{2}$$ is the wrong formula! The $$(log(p * q))^{2}$$ works for the largest gap between consecutive primes less than p * q.

Correction! $$2 \le 2 \lambda < \frac{\sqrt{I}}{2}$$ is probably closer to the truth.

Is integer factorization achievable in polynomial time? WE DO NOT KNOW!

Simple Dave

"We will know!" -- David Hilbert, a great mathematician.
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### Re: Randomness can be a useful tool for solving problems.

Equation 0: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}$$

where $$\lambda$$ is a positive integer such that $$1 \le \lambda < \frac{\sqrt{I}}{4}$$.

Given the integral value, $$I = p * q$$, what is $$\lambda$$?

Can the Newton's Method solve equation zero? It is worth a try.

Dave.

'Newton's Method',

https://en.wikipedia.org/wiki/Newton%27s_method.
Guest

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

Dave wrote:Equation 0: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}$$

where $$\lambda$$ is a positive integer such that $$1 \le \lambda < \frac{\sqrt{I}}{2}$$. (Update)

Given the integral value, $$I = p * q$$, what is $$\lambda$$?

Can the Newton's Method solve equation zero? It is worth a try.

'Newton's Method',

https://en.wikipedia.org/wiki/Newton%27s_method.
Guest

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

FYI: Newton's Method may not work, but it is worth a try. We must beware of the global minimum, 0, versus many local minima in the open interval, (0, 1), oscillations, and the failure analysis associated with Newton's Method.

However, Newton's Method could be a valuable complementary tool for other tools (continued fraction factorization algorithm, general number field sieve algorithm, etc.) in regards to solving our problem.

'Failure Analysis associated with Newton's Method',

https://en.wikipedia.org/wiki/Newton%27s_method#Failure_analysis;

'Integer Factorization in PT (Polynomial Time)',

https://www.math10.com/forum/viewtopic.php?f=63&t=8985&p=17162#p17162.

Dave.
Guest

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

Dave wrote:Equation 0: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}$$

where $$\lambda$$ is a positive integer such that $$1 \le \lambda \le \frac{p - 3}{2}$$. (Update)

Given the integral value, $$I = p * q$$ for odd primes, p > q, what is $$\lambda$$?

Can the Newton's Method solve equation zero? It is worth a try.

'Newton's Method',

https://en.wikipedia.org/wiki/Newton%27s_method.
Guest

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

Update!

Dave wrote:Equation 0: $$\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = 0$$ if $$\sqrt{I + \lambda^{2}}$$ is a positive integer.

...

Given the integral value, $$I = p * q$$ for odd primes, p > q, what is $$\lambda$$?

Can the Newton's Method solve equation zero? It is worth a try.

'Newton's Method',

https://en.wikipedia.org/wiki/Newton%27s_method.
Guest

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

Our analysis is wrong!

Dave.

Where did our analysis begin to go astray?
Guest

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

What is the value of $$\sum_{k=1}^{\infty }\frac{cos(2 \pi k \sqrt{I + \lambda^{2}})}{k}$$?
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