# Understanding Sylvester' s $1871$ paper of primes in arithmetic progression of the forms $4n+3$ and $6n+5$

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 $$4n+3.$$

This question has been asked in the following link Any help would be appreciated. Thanks in advance.

• 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. Jun 4 '20 at 18:00
• Sorry sir..the question was not answered by the time I posted it in math overflow..I will add the link..Thank you. Jun 4 '20 at 18:20
• Okay, no problem. I enjoyed thinking about it. Are you going through all of Sylvester's papers? Jun 4 '20 at 18:45
• no, not really. They seem to be difficult. I am going through the papers which he proved infinitude of primes. Jun 4 '20 at 21:40

## 1 Answer

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.

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