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Given a $K$ prime number and $l$ $is$ $even$ that satisfy $l<K$.

Is it possible that all the $l+1$ numbers : $K, K+l, K+2l, K+3l,...,K+l^2 $ are primes?

I assume no, but I am struggling to prove it.

***Except from the case $K=3$ and $l=2$

Thanks in advance!

Shaq
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  • I assume $K$ must be an odd prime? – PrincessEev Apr 06 '19 at 11:14
  • yes, otherwise any even $l$ will take $k+l$ to be composite number – Shaq Apr 06 '19 at 11:18
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    If $l$ is not a multiple of $3$ (hence, of $6$), then one of $K,K+l,K+2l$ is a multiple of $3$, hence, not prime (for $K>3$). So assume $l$ is a multiple of $6$, hence, at least $6$. Then if $l$ is not a multiple of $5$ (hence, of $30$), then one of $K,K+l,\dots,K+4l$ is a multiple of $5$, hence, not prime. So $l$ is a multiple of $30$. But by the same reasoning, now $l$ has to be a multiple of every prime under $30$. And so on. – Gerry Myerson Apr 06 '19 at 11:26
  • Thanks a lot! Very nice method :) – Shaq Apr 06 '19 at 11:36

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