Infinitely many Primes in Arithmetic progression $10n+9$

Question

Can anyone provide How J. A. Serret proved infinitude of primes in the arithmetic progression $10n+9$? I know there are many general proofs available now. But I want this one. Any help would be appreciated. Thanks in advance.

Answer 1

Assume that there are only finitely many primes $p \equiv 9 \bmod 10$. Consider the number $n = 5N^2-1$, where $N = 2 \cdot 3 \cdots p$ is a product of primes containing these finitely many $p \equiv 9 \bmod 10$. If $q$ is a prime $q \mid n$, then $5N^2 \equiv 1 \bmod q$ and $q \equiv \pm 1 \bmod 5$ by quadratic reciprocity. Since $n \equiv -1 \bmod 5$, not all prime factors $q$ of $n$ can be $\equiv 1 \bmod 5$. Thus there is at least one prime $q \equiv -1 \bmod 5$ dividing $n$, and this $q$ is not among the finitely many primes $p \equiv 9 \bmod 10$.

This proof works because there are only two residue clases modulo $5$ containing squares.