Proof of the Polignac Conjecture using the Proof of GC

[Guest]

PC - Polignac Conjecture and GC - Goldbach Conjecture.

Reference links:

https://en.wikipedia.org/wiki/Polignac%27s_conjecture;

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

[ten2match]

Can every even whole number greater than 2 be written as the sum of two primes?

A prime is a whole number which is only divisible by 1 and itself. Let's try with a few examples:

4 = 2 + 2 and 2 is a prime, so the answer to the question is "yes" for the number 4.
6 = 3 + 3 and 3 is prime, so it's "yes" for 6 also.
8 = 3 + 5, 5 is a prime too, so it's another "yes".

the whole article is https://plus.maths.org/content/mathemat ... conjecture

[Guest]

A Grand Hypothesis:

"The repetition and the growth of prime gaps between consecutive prime numbers are essential for the efficient generation of all composites (all positive integers that are not prime) in accordance with the Fundamental Theorem of Arithmetic and in accordance with the Prime Number Theorem." -- David Cole.

A Grand Claim:

"As a result of our grand hypothesis, we claim the Polignac Conjecture is true!" -- David Cole.

Relevant Reference Link:

'What great conjectures in mathematics combine the additive theory of numbers with the multiplicative theory of numbers?',

...

An Update:

"Remark: Polignac's conjecture is true! (...)"

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 [tex]1 \le \lambda < \frac{log^{2}(pq)}{2}[/tex] is in accordance with the Prime Number Theorem when p > q are consecutive odd primes with [tex]p - q = 2 \lambda[/tex].

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

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

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

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

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

Assumption: Polignac's conjecture is generally false for any positive even integer, [tex]2 \lambda[/tex].

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

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

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

Remark: [tex]n_{m} \ge q_{m}[/tex].

Remark: [tex]n_{m} = 1[/tex] 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, [tex]n_{k } > 1[/tex], for appropriate odd primes, [tex]p_{k}[/tex] and [tex]q_{k}[/tex]:

[tex]p_{1 } - 2 \lambda = n_{1 } *q_{1 }[/tex];

[tex]p_{2 } - 2 \lambda = n_{2 } *q_{2 }[/tex];

[tex]p_{3 } - 2 \lambda = n_{3 } *q_{3 }[/tex];

...

[tex]p_{\infty } - 2 \lambda = n_{\infty } *q_{\infty}[/tex];

Remark: [tex]p_{k} < p_{k+1}[/tex] are consecutive primes over E.

Remark: [tex]3 \le q_{k} \le \sqrt{ n_{k } *q_{k}} \le n_{k }[/tex].

Remark: [tex]\pi (x)[/tex] 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 [tex]\pi (x)[/tex], and hence to a more regular distribution of prime numbers..." Source Link: https://en.wikipedia.org/wiki/Prime-counting_function#The_Riemann_hypothesis.

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

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

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

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

Therefore,

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

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

Remark: We count [tex]n_{m} = 1[/tex], the exception that violates our assumption.

That result [ Prob( [tex]p - 2 \lambda \ne q[/tex] 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!

Relevant Reference Link:

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

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

[Guest]

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

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

[Guest]

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

[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]

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

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

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.