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Is there infinite number of prime pairs $(p_k,p_n)$ that satisfy equality $p_n=2p_k-3$

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2 Answers 2

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This is surely open. The usual conjecture is that, given a collection of linear equations in primes, if there is no "local" obstacle to it having infinitely many solutions, then it does. Here local obstacles would mean either (1) the size of the real solutions is bounded or (2) there is some prime number $q$ such that any solution has at least one of the $p_i$ divisible by $q$. Green and Tao term this the "generalized Hardy-Littlewood conjecture".

Since neither of these obstacles applies, one would conjecture that there are infinitely many prime pairs of this form. Green and Tao have recently proved that this is true in many cases but your example is not among the cases to which their results apply; in the terminology of their paper, it has "complexity $\infty$". I am fairly certain no progress has been made on the complexity $\infty$ cases.

Note also that your conjecture is extremely close to the conjecture that there are infinitely many Sophie Germain primes, which remains notoriously open.

I highly recommend reading the introduction to Green-Tao for an overview of what can and cannot be done in this area.

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    $\begingroup$ Surely you don't need to say "surely" in the first sentence. :) $\endgroup$
    – KConrad
    Sep 5, 2011 at 15:25
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    $\begingroup$ And don't call me "Surely" $\endgroup$
    – Igor Rivin
    Sep 6, 2011 at 9:48
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maybe there is,it is not proved yet. It is a particular case of a general conjecture concerning linear equations in primes :

Conjecture : Let $a$ and $b$ two integers such that $a>0$, the product $ab$ is even and $a$ and $b$ are coprime. Then the equation $p' =ap+b$ has infinitely many solutions in pairs of prime $(p,p')$.

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    $\begingroup$ This conjecture seems to be a special case of the general conjecture David Speyer mentioned; the general form of the conjecture goes by 'generalized Hardy--Littlewood conjecture.' $\endgroup$
    – user9072
    Sep 6, 2011 at 9:47
  • $\begingroup$ If $p_{k}=6k+1$, then $p_{n}=2*(6k+1)-3=6(2k)-1$. If $p_{k}=6k-1$, then $p_{n}=6(2k)-5=6(2k) + 1$. And I read these two arithmetic progressions contain an infinite number of primes. How is the question different from what I wrote? $\endgroup$
    – user25406
    Feb 14, 2019 at 20:05

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