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Show that it is not possible for $k>3$ prime numbers, each greater than $k$, to be in arithmetic progression with common difference less than or equal to $k+1$.

$p$>$k$ and $p,p+d,...,p+(k-1)d$ are k prime numbers.

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closed as off-topic by TMM, Théophile, iadvd, Greg Martin, user91500 Oct 4 '16 at 5:54

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    $\begingroup$ What are your own thoughts on the question? $\endgroup$ – Théophile Oct 3 '16 at 18:30
  • $\begingroup$ @amWhy I'm happy to delete post-edit, but I don't follow your comment. It is entirely possible for there to exist $2$ primes in arithmetic progression, each $>2$, with common difference $\le 3$. Those are indeed called twin primes. $\endgroup$ – Erick Wong Oct 3 '16 at 22:09
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    $\begingroup$ @ErickWong: It is often useful to violate the hypotheses, find a counterexample, and think about how the hypotheses rule it out. Your example of twin primes is a good one in this spirit. Another is $\{3,5,7\}$ $\endgroup$ – Ross Millikan Oct 4 '16 at 4:41
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When I see a problem like this I first try small $k$. The smallest $k$ allowed is $4$. In that case, one of the numbers in the progression will be even unless the difference is $2$, when one will be a multiple of $3$. For larger $k$ the difference must be even so all the numbers can be odd. Similarly, the difference must be a multiple of $3$ so all the numbers can have a remainder on division by $3$. Now the difference is a multiple of $6$, so $k \ge 5$. Keep on this way until the product of the primes gets so large that you can use Bertrand's postulate to show that two must have a common factor so the higher cannot be prime.

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