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Suppose we have $n$ consecutive prime numbers $p_k$ with $n \in\mathbb{N}$ such that $p_k-p_{k-1}=m$ and $k=2..n$. Is it possible to find $m$ and $n$ in order to have a finite number of these consecutive primes? For example, if we put $m=6,n=3$, for the first $23$ primes, we get only the triplet $(47,53,59)$ satisfying this condition. Obviously, for example, for $n=3,m=2$ (twin primes), there are only three consecutive primes $(3,5,7)$. Is this the only case we can have a finite number of consecutive primes with $n\gt 2$? Thanks.

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  • $\begingroup$ You mean $m=2$ for the twin primes? Beyond that, I'm not sure what you are asking. Pick some prime $p$ not dividing $m$ and work $\pmod p$. That gives you a bound for the chain. $\endgroup$
    – lulu
    Commented Nov 17, 2022 at 12:30
  • $\begingroup$ There is another trio below 100. Then there is 251,257,263,269 $\endgroup$
    – Empy2
    Commented Nov 17, 2022 at 12:40
  • $\begingroup$ Are you sure you want the n primes to be consecutive? $\endgroup$ Commented Nov 17, 2022 at 12:42
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    $\begingroup$ The Green Tao theorem guarantees arbitary long arithmetic progressions consisting of only prime numbers, but this theorem is not proven for consecutive primes. $\endgroup$
    – Peter
    Commented Nov 17, 2022 at 13:18
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    $\begingroup$ Please edit your post for clarity. As it stands, I can't sort out what you are asking. $\endgroup$
    – lulu
    Commented Nov 17, 2022 at 13:51

1 Answer 1

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A RATHER LONG COMMENT

For consecutive primes larger than $5$ in an arithmetic progression, the common increment must be a multiple of $6$, as those primes have a form $6t\pm 1$. Any five consecutive positive integers of the form $6t-1$ or $6t+1$ (or $6st \pm 1$ where $5 \not \mid s$) will have one member that is divisible by $5$, limiting sequences of primes to at most $4$ in length.

You can address this by making $5$ a factor of the increment, looking for strings among integers of the form $30t \pm r$ where $r \in \{1,7,11,13\}$, but then strings of $7$ such numbers will contain at least one member divisible by $7$. You could move to numbers of the form $210t \pm r$, but you will appreciate that as the increment of the arithmetic sequence gets larger, the likelihood of finding strings of consecutive primes diminishes.

This does not prove or even suggest that indefinitely long strings of consecutive prime numbers in an arithmetic sequence cannot be found, only that one very quickly is forced to look among very large numbers to have any chance of finding them.

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