# Infinite Primes in Arithmetic progression $10n+9$

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.

• The book "Dickson - History of the Theory of Numbers, Volume I" cites that as appearing in the french journal "Jour. de Math, 17, 1852, 186-9". I wasn't able to find the original article but this might be a starting point. – river May 29 '20 at 9:02
• You can find the proof here. He consider numbers of the form $5N^2-1$ where $N$ is a product of primes, stating its prime factors are of the form $10n+1$ or $10n+9$ (not sure how he justifies this, I don't have time to translate the French right now). – Wojowu May 29 '20 at 9:03
• Thanks.Where from did you download the paper? @wojowu. – math is fun May 29 '20 at 9:23
• Nowhere from. I dug out the title and then googled it. – Wojowu May 29 '20 at 9:26
• $$p \mid 5N^2 - 1 \iff 5N^2 \equiv 1 \pmod{p} \iff (5N)^2 \equiv 5 \pmod{p}$$ – Daniel Fischer May 29 '20 at 20:06

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.
• Please look at my last comment above. I got $q \equiv \pm1 \pmod{10}$ first but not $q \equiv \pm1 \pmod{5}$ . Am I doing mistake? – math is fun May 30 '20 at 22:08
• @mathisfun The two are equivalent since any prime above $2$ is odd. – Wojowu May 30 '20 at 22:51
• Yeah..I know that. I applied Gauss' lemma and I got $q \equiv \pm1 \pmod{10}$ first. So I just curious how you guys got $q \equiv \pm1 \pmod{5}$. – math is fun May 31 '20 at 8:22