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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.

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    $\begingroup$ 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. $\endgroup$ – river May 29 '20 at 9:02
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    $\begingroup$ 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). $\endgroup$ – Wojowu May 29 '20 at 9:03
  • $\begingroup$ Thanks.Where from did you download the paper? @wojowu. $\endgroup$ – math is fun May 29 '20 at 9:23
  • $\begingroup$ Nowhere from. I dug out the title and then googled it. $\endgroup$ – Wojowu May 29 '20 at 9:26
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    $\begingroup$ $$p \mid 5N^2 - 1 \iff 5N^2 \equiv 1 \pmod{p} \iff (5N)^2 \equiv 5 \pmod{p}$$ $\endgroup$ – Daniel Fischer May 29 '20 at 20:06
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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.

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  • $\begingroup$ 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? $\endgroup$ – math is fun May 30 '20 at 22:08
  • $\begingroup$ @mathisfun The two are equivalent since any prime above $2$ is odd. $\endgroup$ – Wojowu May 30 '20 at 22:51
  • $\begingroup$ 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} $. $\endgroup$ – math is fun May 31 '20 at 8:22

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