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I have recently become interested in reading a little more on certain directions regarding primes in arithmetic progressions (AP). I would appreciate specific paper references (with the journal and the year, if possible) and suggested textbook/monograph materials that may explain any of the following questions I recently had.

Given an odd prime $p$, is there a $3$-AP in primes that it belongs to? Moreover, is there a $3$-AP whose first term is $p$? For instance, given $3$, the AP $3,5,7$ is a $3$-AP in primes, and for $5$, the AP $5,11,17$ is such an AP. I am trying to code for this right now on Python for small primes, to check it for primes until $1000$ at least, but I was wondering if there is a general proof or a counterexample. More generally, given an odd-prime $p$, is there an AP in primes of length $k$ (where $2 < k \leq p$), whose first term is $p$? If this is not true in general for every odd-prime, that is, $$\exists p_{0} \in \mathbb{P}_{\text{odd}}: \exists k_{0} \in \{3, \dotsc, p_{0}\}: \forall d \in 2\mathbb{N}: \{p_{0}, p_{0} + d, \dotsc, p_{0} + k_{0}d\} \cap (\mathbb{N} - \mathbb{P}) \neq \emptyset, $$ then one could ask whether there are infinitely many primes for which it is true (call the set of these primes $S_{\text{true}}$). An analogous question can be asked about ($\mathbb{P}_{\text{odd}} - S_{\text{true}}$). A trivial remark is that $3 \in S_{\text{true}}$, so, one at least knows that $S_{\text{true}} \neq \emptyset$. In fact, for $5$, the AP $5,11,17,23,29$ is an AP in primes, and so, a $3$-, $4$- and $5$-AP in primes exists, whose first term is $5$, so $5 \in S_{\text{true}}$. Therefore, if one can show that given a prime $p$, there is an AP in primes of length $p$, with the first term of the AP being $p$, then, one has shown that $S_{\text{true}} = \mathbb{P}$.

I am just a beginner, and so, this question might be too standard for MathOverflow (I don't know), so please excuse me if this is the case. Also, I hope my formulation of the question is non-trivial.

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  • $\begingroup$ Based on the rest of your question, the first thing you ask, whether every prime belongs to a 3-AP in primes, should really be whether every prime begins a 3-AP in primes, right? $\endgroup$ – LSpice May 23 at 2:45
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    $\begingroup$ @LSpice Absolutely agree. However, if one can show that $\exists p \in \mathbb{P}_{odd}:$ $(p$ $is$ $not$ $in$ $any$ $3-A.P.$ $in$ $primes)$, then, the next question about whether every prime begins a 3-A.P. in primes becomes moot, although, I think the rest of the question is probably still valid, with some modification. I have added a sentence there to reflect your comment. Thank you. $\endgroup$ – user137748 May 23 at 2:59
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The question, whether there are infinitely many three-term arithmetic progressions with first term $3$, is asking whether there are infinitely many primes $p$ such that $2p-3$ is also prime. In general, the question, given integers $a,b$ are there infinitely many primes $p$ such that $ap+b$ is prime, is wide open (except in the cases where the answer is trivially "no", such as when $\gcd(a,b)>1$, or when $a\equiv b\equiv1\bmod2$). It is expected that the answer is "yes", and it is worth consulting the literature on Dickson's conjecture.

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    $\begingroup$ It is very rare to have a result in prime number theory that produces an example of something without proving that there are infinitely many. The reason is because almost all (and as far as I know, ‘almost’ can be dropped) methods that produce primes actually produces an asymptotic lower bound (or asymptotic formula) for how many such primes there are up to some height. So in practice, the question you asked and the question Gerry Myerson answered are equivalent. $\endgroup$ – Stanley Yao Xiao May 23 at 12:36
  • $\begingroup$ My original question asks only for one A.P. in primes, whose first term is $p$ (and in particular, $3$), but a natural extension would be to ask what you have stated- if there are infinitely many such 3-A.P.s, given a prime $p$. Of course, the Dickson's conjecture is far more general than that. Also, that is a neat reformulation in the first sentence- one is, after all, only really checking 'prime-ness' for the third term, since every prime greater than $3$ forms a 2-A.P. with $3$. Thanks! $\endgroup$ – user137748 May 23 at 12:38

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