Here is my problem :
I would like to prove that there is an integer $n_0$ such that for all even integers $n \equiv k \pmod 5> n_0$ with $k \in [\![0, 4 ]\!]$, there exist a prime integer $p \not\equiv k \pmod 5$ (and $p \ne n-1$ and $p \ne 5$) such that : $p \not\mid n$.
Yet I don't know how to prove this result.
Here is what I've been thinking at so far :
we kow by Dirichlet that there is infinitely many primes of the form : $5a + k$, $k \in [\![0, 4 ]\!]$.
$n \# > n$ for all integers $n > 3$, and if we want an integer $n$ which doesn't respect the above condition we must have : $2 \cdot \displaystyle\prod w_i \mid n$, (the $w_i$ are the prime integers such that for all $i$ we have : $w_i \ne 5$ and $w_i \ne n-1$ and $w_i \not\equiv k \pmod 5$) so maybe I can prove that : $2 \cdot \displaystyle\prod w_i > n$ ? Yet, if I want to prove this result I must know the number of integers of the form : $5a+k$...
Any ideas ?