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With $2k \in \mathbb N$ greater or equal to the bound in Yitang Zhang's proof about prime number gaps (I put it that way since it's constantly decreasing).

As far as I know the proof states that there are infinitely many primes which differ by an even number lesser than the bound. Does that mean that there are only finitely many that differ by an even number greater or equal to the bound?

Edit:

The proof stated in the beginning: There are infinitely many primes that differ by a number less that 70 million. Now the (collective) proof states that there are infinitely many primes that differ by a number less than 246. Does that mean that the numbers between 246 and 70 million are excluded from being the distance of infinitely many prime pairs? (And also the ones above 70M ofc.)

Comment/answer by AndréNicolas:

They are not excluded. There is an old conjecture (Polignac's) that for every $k$ there are infinitely primes that differ by $2k$. That conjecture may very well be correct. It is now known that there is at least one $k$ such that there are infinitely primes that differ by $2k$, and that in fact there is such a $k$ with $k \leq 123$.

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  • $\begingroup$ I don't see how that would follow. Infinitely many even integers differ by 2, but it doesn't follow that only finitely many differ by more than 2. $\endgroup$
    – coffeemath
    Sep 19, 2015 at 23:58
  • $\begingroup$ @coffeemath So the twin prime number theorem is already proven? $\endgroup$
    – 355durch113
    Sep 19, 2015 at 23:59
  • $\begingroup$ No I didn't say that, only that it can't be decided just from Zhang's result. $\endgroup$
    – coffeemath
    Sep 19, 2015 at 23:59
  • $\begingroup$ No.There are arbitrarily large gaps. Consider for n>1 that n!+s for 1<s< n+1 is greater than s and divisible by s. We can also deduce the existence of arbitrarily large gaps by knowing that the asymptotic density of the primes is 0: If P(n) is the number of primes less than n then P(n)/n goes to 0 as n goes to infinity. $\endgroup$
    – DanielWainfleet
    Sep 20, 2015 at 0:00
  • $\begingroup$ @user254665 I know that. My question refers to the amount of primes that differ by a fixed number. $\endgroup$
    – 355durch113
    Sep 20, 2015 at 0:01

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You argue that because an infinite amount of numbers have some property, therefore a finite amount of numbers do not have that property.

That is not true; for example an infinite amount of integers are even, but there are also an infinite amount of odd integers.

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    $\begingroup$ no it doesn't mean that. it just means what it says $\endgroup$
    – DanielWainfleet
    Sep 20, 2015 at 0:08
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    $\begingroup$ They are not excluded. There is an old conjecture that for every $k$ there are infinitely primes that differ by $2k$. That conjecture may very well be correct. It is now known that there is at least one $k$ such that there are infinitely primes that differ by $2k$, and that in fact there is such a $k$ with $k\le 123$. $\endgroup$
    – André Nicolas
    Sep 20, 2015 at 0:11
  • $\begingroup$ (For those interested in that old conjecture, it is Polignac's conjecture $\endgroup$
    – Krijn
    Sep 20, 2015 at 0:15
  • $\begingroup$ You are welcome. The recent results are a major achievement, but for almost all $k$, the problem remains open. And we do not even know precisely the $k$ for which the problem has been settled! $\endgroup$
    – André Nicolas
    Sep 20, 2015 at 0:17
  • $\begingroup$ @AndréNicolas Is there anywhere to read up on the results? Does Zhangs breakthrough somehow say anything about more than one $k$ or something related? $\endgroup$
    – Krijn
    Sep 20, 2015 at 0:19

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