WakeUp wrote: Does there exist a finite sequence of integers $c_1,c_2,\ldots ,c_n$ such that all the numbers $a+c_1,a+c_2,\ldots ,a+c_n$ are primes for more than one but not infinitely many different integers $a$? Yes. Choose for example $n=1$ and $(c_1,c_2,\ldots ,c_n)=(1)$

pco wrote: WakeUp wrote: Does there exist a finite sequence of integers $c_1,c_2,\ldots ,c_n$ such that all the numbers $a+c_1,a+c_2,\ldots ,a+c_n$ are primes for more than one but not infinitely many different integers $a$? Yes. Choose for example $n=1$ and $(c_1,c_2,\ldots ,c_n)=(1)$ I don't understand pco, are you saying there exists only finitely many $a$ such that $a+1$ is a prime?

WakeUp wrote: pco wrote: WakeUp wrote: Does there exist a finite sequence of integers $c_1,c_2,\ldots ,c_n$ such that all the numbers $a+c_1,a+c_2,\ldots ,a+c_n$ are primes for more than one but not infinitely many different integers $a$? Yes. Choose for example $n=1$ and $(c_1,c_2,\ldots ,c_n)=(1)$ I don't understand pco, are you saying there exists only finitely many $a$ such that $a+1$ is a prime? No, I'm sorry. I understood "not necessarily infinitely many". But the problem is clear : "not infinitely many". So my answer was stupid Sorry again.

Yes. Choose $n=5$ and $(c_1,c_2,c_3,c_4,c_5)=(3,5,29,101,107)$. For $a=0$ and $a=2$, the numbers $(3,5,29,101,107)$ and $(5,7,31,103,109)$ are all primes. Since $(c_1,\ldots,c_5)$ are pairwise distinct in modulo 5, one of $a+c_1,\ldots,a+c_5$ is divisible by 5, and hence must equal 5, so there are finitely many such $a$.

pco wrote: No, I'm sorry. I understood "not necessarily infinitely many". But the problem is clear : "not infinitely many". So my answer was stupid Sorry again. Don't worry, we all make mistakes