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 [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]?

An Update:

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

odd primes

p > q, we define an exceptional and

infinite set,

E.

E = {(p, q) |[tex]p > q > 2 \lambda[/tex] are consecutive

odd 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

a system of infinite 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

odd primes

belonging to 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 \forall m \in \N | 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 we assume [tex]n_{m } \ne 1[/tex].

Therefore,

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: 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.

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!