I want to prove there is an infinite number of prime of the form $12k+11$. I worked hard to make the first part of my solution but I don't know how to do with the second part. First : I suppose there is a finite number of prime of the form $12k+11$, let'S call them $p_1, p_2,..., p_r$.
Let $N=12(p_1p_2...p_r)-1$.
$N$ is not divible by any of the prime in the list $(p_1, p_2,...,p_r)$ because $n+1=12(p_1p_2...p_r)$ and two consecutives numbers can't both be divible by $p_1$. The same argument goes for $p_2, p_3,...,p_r$. Now, we know that $N$ is divisible by a prime that isn't in the list, or $N$ is a new prime of the form $12k+11$ that is not in the list. We can see that $N$ has a form of $12k+11$ because $N=12((p_1p_2...p_r)-1)+11$ and $N\equiv{-1}\equiv{11}\mod{12}$.
Next: Now I have to prove that $N$ is divisible by a prime of the form $12k+11$ but I don't know how. I found that all the prime numbers can be written with the form $12k+5$, $12k+7$ and $12k+11$ except $2,3$ and $13$.
Can you help me with that last part and tell me if the first part is correct?