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I've been thinking about this question and trying to apply Dirichlet's theorem on primes in arithmetic progressions, but can't seem to get the details of the argument. Dirichlet says there are infinitely many primes $p$ with $p \equiv -1 \pmod{m}$. My general idea is to assume that $p_1, \dots, p_n$ have been constructed with $\gcd(p_1 + 1, \dots, p_n + 1) = km$, and then construct a new prime that has $p \equiv -1 \pmod m$ that reduces the gcd (or ideally, is just coprime to $k$). However, I can't see how to do this in the case that $k$ is not coprime to $m - 1$.

If it matters, I also need this result to be true even when excluding primes smaller than some given $N$, but I don't think this will have much consequence. Thanks so much for the help!

Daniel New
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  • If $,\color{#c00}{m=(p_1!+!1,\ldots,p_{n-1}!+!1)},$ then $,m = (\color{#c00}{p_1!+!1,\ldots,p_{n-1}!+!1},p_n!+!1) = (\color{#c00}m,p_n!+!1)$ $\iff m\mid p_n!+!1$ $\iff p_n\equiv -1\pmod{!m},,$ by the gcd associative law $\ \ $ – Bill Dubuque Feb 27 '24 at 14:46
  • Sorry, I don't see how this helps. That working is clear, but the statement we're proving doesn't seem very relevant considering it starts at the assumption that we already have such a set. I do agree if we have such a set we can keep adding more primes such that the gcd remains to be m, that's essentially what I can see from that proof. – Daniel New Feb 27 '24 at 15:30
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    No, take $m=5$. $2$ cannot be one of your primes since $5\not |3$. But then every prime is odd and therefore $2|p_i+1$. Thus $\text{gcd}(p_1+1,...,p_n+1)$ is even, a contradiction. – QC_QAOA Feb 27 '24 at 15:44
  • Ah, I see. Thanks for that example! Do you happen to know if this is true for $m = 6$ with some finite set of primes excluded? – Daniel New Feb 27 '24 at 15:48
  • @Dan It wasn't clear if you knew how to prove the inductive step. So it reduces to (dis)proving the base (which is trivial if $n=1$ is permitted: it holds $\iff m-1$ is prime). – Bill Dubuque Feb 27 '24 at 15:59
  • Well sure, just have $p_1=5$ then set $p_i$ as the next prime equal to $5$ mod $6$. Then by definition the gcd$(5+1,11+1,17+1,...,p_n+1)=6$ – QC_QAOA Feb 27 '24 at 16:00
  • Ah sure. I guess the question comes because I am interested in the question with some finite set of primes excluded (so we can't necessarily take p = 5). So it's not clear to me why we can always get a gcd of 6 after excluding many small primes. My idea would be that, for example, if we (again, after excluding some primes like 5 so that we don't have an easy answer) have $\gcd(p_1, p_2) = 30$, while we can't find some large prime congruent to 5 mod 30 clearly, we can do it for 11 mod 30. But in general I don't see why at least one of 5, 11, 17, etc. will be coprime to 6m for any m. – Daniel New Feb 27 '24 at 16:15
  • Ah, that makes a lot of sense! Thanks so much! – Daniel New Feb 28 '24 at 13:08

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