How to prove that set of prime number such $p\equiv 7 [12]$ is infinite? We have $n \ne 3k$ with $k\in \mathbb{N}$ and the integer $4n^2+3$. I think the first thing is to prove that this integer has a prime factor $p\equiv 7 [12]$.
I don't have idea to begin.
Thanks in advance !
 A: The proof below is a variant of Euclid's proof that there are infinitely many primes, or more precisely that given a finite collection of primes, there is a prime not in the collection.
Let $p_1,p_2,\dots, p_t$ be primes of the form $12k+7$. Let 
$$N=(2p_1p_2\cdots p_t)^2+3.$$
We will show that $N$ has a prime divisor $p$ of the form $12k+7$.  
Note that if $p$ divides $N$ then $-3$ is a quadratic residue of $p$. By quadratic reciprocity it follows that $p\equiv 1\pmod{3}$. (Here, depending on what has been proved in your course so far, there may be some detail to fill in.) 
So $p\equiv 1\pmod{12}$ or $p\equiv 7\pmod{12}$. But the prime divisors of $N$ cannot be all of the shape $12k+1$, else $N$ itself would be of the shape $12k+1$, and it isn't. 
We conclude that  $N$ has a prime divisor $p$ of the shape $12k+7$. It is clear that this $p$ cannot be equal to any of the $p_i$, and the result follows.
Appendix:  Non-QR proof that $N$ will have no prime factors with shape $6k-1$
Suppose that there exist whole numbers $P, k$ such that $q=6k-1$ is prime and
$P^2\equiv -3\bmod q$  Eq. 1
Then we also have
$P^2\equiv (P+q)^2\bmod q$
and either $P$ or $P+q$ will be odd.  Therefore for some whole number $m$
$(2m+1)^2\equiv -3\bmod q$
Expanding and collecting:
$4m^2+4m+4=4(m^2+m+1)\equiv 0\bmod q$
As $4$ is prime to $q$ we divide by that factor and continue:
$m^2+m+1\equiv 0\bmod q$
$(m^2+m+1)(m-1)=m^3-1\equiv 0\bmod q$
$m^3\equiv 1\bmod q$
$m^{6k}=(m^3)^{2k}\equiv 1\bmod q$  Eq. 2
Now Fermat's Little Theorem forces $m^{6k-2}\in\{0,1\}\bmod q$.  Clearly $m^{6k-2}\equiv 0\bmod q$ contradicts Eq. 2.  If instead $m^{6k-2}\equiv 1\bmod q$ then Eq. 2 forces $m\in\{-1,1\}\bmod q$.  The latter then implies
$P^2\equiv (2m+1)^2\in\{1,9\}\bmod q$
which contradicts Eq. 1.  Since all possibilities for $m$ lead to a contradiction the assumption of Eq. 1 cannot be true for any $P, k$.
A: Dirichlet's theorem on arithmetic progressions tells you that there is infinitely many primes $p$ such that 
$$p=a\pmod k$$
if $a\wedge k=1$.
And in your case you have $7\wedge 12=1$ so you're good.
