On a very flawed proof of Polignac's Conjecture

[Guest]

There is a very flawed proof of Polignac's conjecture at the links below:

https://www.quora.com/What-great-conjectures-in-mathematics-combine-additive-theory-of-numbers-with-the-multiplicative-theory-of-numbers/answer/David-Cole-146;

https://www.researchgate.net/publication/300439567_Proof_of_Polignac_Conjecture.

Dave.

[Guest]

FYI: For the latest update of our proof of Polignac's conjecture, please refer to the link below.

'Randomness can be a useful tool for solving problems.'

https://www.math10.com/forum/viewtopic.php?f=1&t=8855&start=120.

Dave.

[Guest]

FYI: For the latest update of our proof of Polignac's conjecture, please refer to the link below.

'Randomness can be a useful tool for solving problems.'

https://www.math10.com/forum/viewtopic.php?f=1&t=8855&start=120.

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 [tex]n \ne 1[/tex] indicates k is nonprime. We can count the maximum number of primes that may divide k.

That value is either one or [tex]\pi (\sqrt{k}) \ge 1[/tex].

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

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

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!

[Guest]

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

Update: We can count the maximum number of primes (below [tex]\sqrt{k}[/tex]) that may divide k.

[Guest]

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

For positive real numbers, x and y > 1, [tex]x^{y} \rightarrow 0[/tex] as [tex]y \rightarrow \infty[/tex], if [tex]0 < x < 1[/tex].

[Guest]

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

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 [tex]2 \lambda_{0 }[/tex] over E.

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

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

[tex]\frac{Floor( \frac{log^{2}(p_{1 }p_{2})}{2} - 1) }{Floor( \frac{log^{2}(p_{1 }p_{2})}{2}) }[/tex].

However, over E, we generate the infinite product of similar values because of independence so that the chance [tex]2 \lambda_{0 }[/tex] 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]

Update:

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

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 [tex]2 \lambda_{0 }[/tex] over E.

Update:

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

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

[tex]\frac{Floor( \frac{log^{2}(max(p_{1 }, p_{2}))}{2} - 1) }{Floor( \frac{log^{2}(max(p_{1}, p_{2}))}{2}) }[/tex].

Remark: "Roughly" indicates too large.

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

Prob( [tex]p - q \ne 2 \lambda[/tex] over E )

= [tex]\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[/tex].
_______________________________________________________________________________________________________________________________________________
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!

"Math is hard work!"

[Guest]

Relevant Reference Link:

'LARGE GAPS BETWEEN CONSECUTIVE PRIME NUMBERS',

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