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 also been asked in math overflow since it was not answered by the time it was posted in math over flow.
Any help would be appreciated. Thanks in advance. 
 A: Let me first prove that the number of primes is infinite. This can be achieved using the identity
$$  x = \sum_{d \ge 1} \mu(d) \frac{x^d}{1-x^d} , $$
where $\mu$ denotes the Moebius function. 
If there are only finitely many prime numbers $p_1$, $\ldots$ , $p_n$, then no integer larger than $N = p_1 \cdots p_n$ can be squarefree, hence $\mu(d) = 0$ for all $d > N$ but $\mu(N) \ne 0$. But then the function on the right has a pole in $x = e^{2\pi i/N}$, whereas the polynomial on the left is entire. This is a contradiction. I am sure that there are more clever ways of obtaining such a contradiction.
The identity used for proving that there are infinitely many primes $q \equiv 3 \bmod 4$ is
$$   \sum_d x^d = \sum_m \mu(m) \frac{x^m}{1-x^{2m}}, $$
where the sum on the left is over all odd natural numbers $d$ not divisible by any prime $q \equiv 3 \bmod 4$, and the sum on the right is over all odd integers $m$ not divisible by any prime $p \equiv 1 \bmod 4$.
Assume that there are only finitely many primes $q \equiv 3 \bmod 4$. Then the sum on the right is finite, and the last nonzero term occurs when $m$ is equal to the product of all these primes. Setting $x = i$ we find $i^m = +i$ or $-i$ according as $m$ has an even or an odd number of prime factors, hence $i^m = \mu(m) \cdot i$ and $i^{2m} = (-1)^m = -1$. Thus
$$ \sum_m \mu(m) \frac{i^m}{1-i^{2m}} = \frac i2 \cdot M, $$
where $M$ is the number of nonzero terms on the right.
On the other hand, since the number of primes is infinite, there must be infinitely primes $p \equiv 1 \bmod 4$, hence the left hand side is unbounded as $x \to i$. This is a contradiction.
Again I am certain that there are more clever ways of exploiting these identities to prove the infinitude of primes.
