The following is the proof of infinitude of primes in arithmetic progression of the form $4n+3$ and $ 6n+5$ done by Sylvester in $1871$ in his paper "On the theorem that an arithmetical progression which contains more than one, contains infinite number of primes number." The screenshot is from the book /note "The collected mathematical papers Of James Joseph Sylvester".

I having difficulty in understanding the proof in the case $4n+3$ . I would be highly grateful if someone helps me in understanding the proof for the case $4n+3.$

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    $\begingroup$ I've just noticed this question was cross-posted to MathStackexchange: math.stackexchange.com/questions/3702088/…, where Franz Lemmermeyer has given a similar answer to mine. @mathisfun: please do not post the same question in multiple places without adding links; it is not respectful of the time of people on either site. $\endgroup$ Jun 4 '20 at 18:00
  • $\begingroup$ Sorry sir..the question was not answered by the time I posted it in math overflow..I will add the link..Thank you. $\endgroup$ Jun 4 '20 at 18:20
  • $\begingroup$ Okay, no problem. I enjoyed thinking about it. Are you going through all of Sylvester's papers? $\endgroup$ Jun 4 '20 at 18:45
  • $\begingroup$ no, not really. They seem to be difficult. I am going through the papers which he proved infinitude of primes. $\endgroup$ Jun 4 '20 at 21:40

I think the 'identical equation' Sylvester has in mind is

$$\sum_q \mu(q) \frac{x^q}{1-x^{2q}} = x+x^5+x^{13}+x^{17}+x^{25}+x^{29}+\cdots $$

where the left-hand sum is over all natural numbers $q$ divisible only by primes of the form $4s+3$ and the right hand side is the sum of all powers $x^r$ where $r$ is divisible only by primes of the form $4s+1$. (Sylvester specifies no repeated prime factors for $q$ on the left-hand side, but since I'm using $\mu$, any such summand is killed by $\mu(q) = 0$; note the first summand is for $q=1$.)

Proof. The left coefficient of $x^n$ in the left-hand side is $\sum_{q} \mu(q)$ where the sum is over all square-free $q$ divisible only by primes of the form $4s+3$ such that $n/q$ is odd. It is therefore zero for even $n$. If $n$ is odd let $n = Np_1\ldots p_t$ where $p_i \equiv 3$ mod $4$ for each $i$ and no such prime divides $N$. The sum is then

$$\sum_{q \mid p_1\ldots p_t} \mu(q) = \begin{cases} 1 & \text{if $t=0$} \\ 0 & \text{otherwise.} \end{cases} $$

Hence the coefficient of $x^n$ is $0$ unless $n$ is divisible only by primes of the form $4s+1$, in which case it is $1$. $\Box$

The rest of Sylvester's argument seems clear enough to me: if there are only finitely many primes of the form $4s+3$ then the left-hand side is a finite sum, and well-defined when $x=i$ since $i^{2q} = (-1)^q = -1$ as $q$ is odd, and so $1-x^{2q} = 2$. But then, by hypothesis (and infinitely many primes), there are infinitely many primes of the form $4s+1$, making the right-hand side infinite when $x=i$.

There is of course an easier argument, as Euclid take the product of finitely many primes of the form $4s+3$ including the prime $3$, multiply by $4$ and subtract $1$; the result is then divisible by another prime still of the form $4s+3$.

Since I had to make an edit anyway, I'll add that almost the same argument works for primes of the form $6s+1$ and $6s+5$; using the latter instead of primes of the form $4s+3$ to define the left-hand side, the right-hand side is the sum of all powers $x^r$ where $r$ is divisible only by the prime $3$ or primes of the form $6s+1$. But again one can show there are infinitely many primes of the form $6s+5$ by a variation on Euclid's argument.

One feature of interest to me is that Sylvester's argument uses Lambert series rather than the Dirichlet series ubiquitous in analytic number theory.

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    $\begingroup$ One needs to revise Euclid with more care. Gerhard "Three Can Be A Factor" Paseman, 2020.06.03. $\endgroup$ Jun 3 '20 at 13:59
  • $\begingroup$ Good point. I should have subtracted instead. $\endgroup$ Jun 3 '20 at 14:03

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