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Feb 24, 2019 at 13:25 review Close votes
Feb 26, 2019 at 15:47
Feb 23, 2019 at 18:57 comment added Greg Martin Also, every two numbers form an arithmetic progression of length $2$, so $k=2$ is a trivial case.
Feb 23, 2019 at 17:03 history edited user136205 CC BY-SA 4.0
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Feb 23, 2019 at 16:57 comment added GH from MO I suggest that the OP closes this question as obsolete.
Feb 23, 2019 at 16:52 history edited user136205 CC BY-SA 4.0
added 137 characters in body
Feb 23, 2019 at 16:51 comment added user136205 @Lucia Ah, of course. I see where I made an error.
Feb 23, 2019 at 16:35 comment added Lucia If you start with a prime $p$, then the common difference of a progression that begins with $p$ must be coprime to $p$. But then you cannot have more than a $p$ term progression before hitting a multiple of $p$.
Feb 23, 2019 at 16:33 history edited user136205 CC BY-SA 4.0
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Feb 23, 2019 at 16:20 review First posts
Feb 23, 2019 at 16:22
Feb 23, 2019 at 16:19 history asked user136205 CC BY-SA 4.0