Conclude there are infinitely primes with given residue class Things I know (already proved): 1. There are infinitely many primes $p$, such that $p \equiv 1 \pmod{3}$ 2. There are infinitely many primes $p$, such that $p \equiv -1 \pmod{3}$ 3. There are infinitely many primes $p$, such that $p \equiv 1 \pmod{4}$ 4. There are infinitely many primes $p$, such that $p \equiv -1 \pmod{4}$ 5. There are infinitely many primes $p$, such that $p \equiv \pm 2 \pmod{5}$ Is it possible to use this knowledge (maybe together with the Chinese Remainer Theorem) to conclude, that there are infinitely many primes $p$, such that $p \equiv -1 \pmod{5}$?

Yes, it is sort of possible except that it requires one additional piece of theory (not as strong as Dirichlet's theorem), namely quadratic reciprocity.

Given any finite set of primes $p_1,p_2,\ldots, p_k$, each of the form $5k-1$, we can form the value

$$N = 5(2p_1p_2\cdots p_k)^2 - 1.$$

It is clear that $N$ is divisible only by odd primes, none of which are equal to $5$ or any of the $p_i$. Furthermore, any prime $q$ which divides $N$ must have $5$ as a quadratic residue (since $(2p_1p_2\cdots p_k)^{-1}$ is a square root of $5$ modulo $q$), and by quadratic reciprocity this means $q \equiv \pm1 \pmod 5$.

Finally, not all of the primes dividing $N$ can be $1 \pmod 5$ since $N$ is itself positive and $-1 \pmod 5$. So there must be at least one $q \mid N$ which is $-1 \pmod 5$, and it is necessarily distinct from $p_1,\ldots, p_k$.