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Does there exist a natural number $n$ such that all numbers

$2n+1, 2n+3, 2n+5, ... , 2n+2015$ be nonprime?

Can somebody help to solve this problem? I guess that it doesn't exist, but I have no idea how to start proving it.

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    $\begingroup$ To justify this duplicate: If you find 4031 consecutive prime numbers, the odd ones will satisfy this problem. $\endgroup$ – Matthew Daly Oct 13 '19 at 16:52
  • $\begingroup$ Sorry, ^ should be "4031 consecutive composite numbers", obvi. $\endgroup$ – Matthew Daly Oct 13 '19 at 17:14
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Let be $ 2n = 2017! + 2=k. $ Since for each $k \geq 2$ it is well known that $$ k! + 2,k! + 3 \cdots ,k! + k $$ are all composite, we have that $$ \begin{gathered} 2017! + 3 = \left( {2017! + 2} \right) + 1 = 2n + 1 \hfill \\ 2017! + 5 = \left( {2017! + 2} \right) + 3 = 2n + 3 \hfill \\ 2017! + 7 = \left( {2017! + 2} \right) + 5 = 2n + 5 \hfill \\ \vdots \hfill \\ 2017! + 2017 = \left( {2017! + 2} \right) + 2015 = 2n + 2015 \hfill \\ \end{gathered} $$ are all composite.

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