Prove that there are infinitely many primes of the form $4k + 3$ (where $k$ is an integer).
Note that it is a special case of "Theorem 6 (Dirichlet). Let a and b be positive coprime integers. Then the sequence $b$, $b + a$, $b + 2a$, $b + 3a$, $b + 4a$, $b + 5a$, ....," contains infinitely many prime numbers
So far I got that suppose there are a finite number of primes
$p......p$ and if $4(p.....p)+3$ is prime it's a contradiction so the initial statement is proven?
We know there are infinite odd primes.
Suppose there are not. Then there is a finite set $A$ which contains all primes of the form $4k+3$. Let P be the product of all of those primes.
If $|A|$ is odd then $P+4$ is of the form $4k+3$ and is not divisible by any of the elements of $A$ but it must contain an odd number of prime factors of the form $4k+3$ Thus it is divisible by primes not in $A$ of the form $4k+3$. Thus $A$ is not finite.
If $|A|$ is even use the same argument with $P+2$