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Are there (in)finitely many primes $p$ such that $1+kp$ is a prime for some positive integer $k$ ?

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    $\begingroup$ Aren't all integers $p$ (not only primes) with this property by Dirichlet ? $\endgroup$ Nov 28, 2019 at 12:20
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    $\begingroup$ I believe those voting to close have misread the question. Fix k, and ask if the map (1+kx) sends infinitely many primes to (possibly other) primes. This is a reasonable research question. Gerhard "Is Voting To Reopen This" Paseman, 2019.11.28. $\endgroup$ Nov 28, 2019 at 18:17
  • $\begingroup$ @Gerhard, are we misreading the question, or did OP miswrite it? $\endgroup$ Nov 28, 2019 at 21:35
  • $\begingroup$ @Gerry, most likely both. There are cases where the closers err on the side of caution regarding a non optimally worded question and do not consider the possibility that a better reading is so close by. In many cases, I see the missed opportunity: sometimes I react, sometimes I comment, and in rare cases I edit. This is the first case where I object. Gerhard "And Also Where I Vote" Paseman, 2019.11.28. $\endgroup$ Nov 28, 2019 at 22:40
  • $\begingroup$ Note Dmitri Krachun's response. He (I'm guessing at the pronoun) addresses the potential ambiguity and considers the two immediate readings. His response can invite the poster to clarify and rewrite, and in a collegial fashion. I prefer his response to those of the closers: it encourages community. For this, I give him one of my few upvotes. Gerhard "Shouldn't We All Encourage Community?" Paseman, 2019.11.28. $\endgroup$ Nov 28, 2019 at 22:45

2 Answers 2

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Note that for fixed even $k$ this is open. If you mean $p$ such that $1+kp$ is prime for $\textbf{some}$ $k$, then this is an elementary case of Dirichlet's theorem: For any prime $p>2$ take any positive integer $n>2$ divisible by $p$, then any prime divisor of $(n^p-1)/(n-1)$ is congruent to $1$ modulo $p$.

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For al $p$ there are infinitely many primes of the form $kp+1$, since it is a particular case of Dirichlet’s Theorem.

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