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In August 2014, Polymath group showed that subject to the generalized Elliott–Halberstam conjecture, one can show the existence of infinitely many pairs of consecutive primes that differ by at most 6.

The list of primes less than 100 shows that the first prime after 89 is 97. Doesn't this counter-example disprove the "potential" proof of the idea that infinitely many pairs of consecutive primes can differ by only 6?

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  • Infinitely many pairs of consecutive primes may differ by $6$, but that doesn't mean there aren't also infinitely many that differ by $8$. – Robert Israel Mar 24 '17 at 06:13
  • "Infinitely many" doesn't mean all. – Galc127 Mar 24 '17 at 06:13
  • @RobertIsrael Doesn't that make that quoted sentence meaningless? I do not understand what the quote means given that the difference does not have to be limited to 6... – StopReadingThisUsername Mar 24 '17 at 06:16
  • @Galc127 Doesn't that make that quoted sentence meaningless? I do not understand what the quote means given that the difference does not have to be limited to 6... – StopReadingThisUsername Mar 24 '17 at 06:20
  • @StopReadingThisUsername, why meaningless? Is the fact that there are infinitely many primes $p$ such that $p\equiv 1\pmod 4$ is meaningless? It gives us some information about the distribution of primes. One open question is whether there are infinitely many pairs of primes that differ by 2. – Galc127 Mar 24 '17 at 06:30

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No, the idea is that there are infinitely many primes seperated by six, but it says nothing about where those primes live. For example, 131 and 137 satisfy this condition.

The idea is that no matter how far into the integers you go, you will always find another pair of primes seperated by exactly six,

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