Prove that there exist infintely many primes $p$ with $p\equiv2$ or $-2 \pmod{5}$ I want to show that there exist infintely many primes $p$ with $p\equiv2$ or $-2 \pmod{5}$. 
I already know a few proofs for primes with given modulo, but I don't know how I would use that here. Besides, I don't want to use Dirichlet's theorem on arithmetic progressions.
So I started assuming that there are only finite number of primes with $p\equiv2$ or $-2 \pmod{5}$, $p_1,...,p_n$. In many proofs one would look at $N := 5\cdot p_1...p_n-2$ or ($+2$?).
I would appreciate any hints.
 A: It is simple to prove in a elementary fashion that there are infinite primes of the form $5k+1$, for instance by studying the prime factors of $\Phi_5(n)$ and building a sequence $\{\Phi_5(n_k)\}_{k\geq 1}$ of mutually coprime integers. The existence of infinite primes of the form $5k-1$ can be proved in a similar way, by studying the prime divisors of $\Psi(n)=5n^4-10n^2+1$. Now we may use an argument similar to the one outlined at page 12 of my notes. Both
$$ L(\chi_1,s)=\sum_{n\geq 0}\left(\frac{1}{(5n+1)^s}-\frac{1}{(5n+2)^s}-\frac{1}{(5n+3)^s}+\frac{1}{(5n+4)^s}\right) $$
$$ L(\chi_2,s)=\sum_{n\geq 0}\left(\frac{1}{(5n+1)^s}+\frac{i}{(5n+2)^s}-\frac{i}{(5n+3)^s}-\frac{1}{(5n+4)^s}\right) $$
are convergent for any $s$ such that $\text{Re}(s)>0$ and they have an associated Euler product. Since $L(\chi_1,1)=\frac{2}{\sqrt{5}}\log\varphi\neq 0$ we have infinite primes of the form $5k\pm 2$. Primes of the form $5k+2$ (or $5k-2$) have to be infinite as well, since the opposite assumption would contradict the boundedness of the imaginary part of $L(\chi_2,1)=\frac{\pi\sqrt{4+2i}}{5^{5/4}}$.
