There are infinitely many primes of the form $4n+3$. Proof: Define $q$ by $q=2^2.3.5...p-1$. Then $q$ is of the form $4n+3$, and is not divisible by any of the primes up to $p$. It cannot be product of primes $4n+1$ only, since the product of two numbers of this form is of same form; and therefore it is divisible by a prime $4n+3$, greater than $p$.
I don't understand how "It cannot be product of primes $4n+1$ only, since the product of two numbers of this form is of same form"have conclusion "therefore it is divisible by prime greater than $p$" ? Also how $4n+1$ came out from thin air? Mine version of explanation will exclude $4n+1$ since you can literally say $4n+3$ is prime the nothing to prove else there are prime not included in that weird $q$ so using fundamental theorem of arithmetic there is one more prime number. Also how $q$ is derived? My may is you first have multiple of prime then you need to force this to be of form $4n+3$ by $4(d+q)+3$ where $q$ is $-1$ and $d$ is multiple of prime number finite amount I think it should be prime number of form $4n+3$ in $d$ which is missing from textbook definition. I hope I get absolute correct and detailed explanation. Thanks for attention!