Infinetly many primes of form $4k+3$

Question

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?

You need to prove more. Note that $4p_1\cdots p_n+3$ is not necessarily prime, but it is divisible by a prime different from the $p_i$ and of the form $4k+3$. You need to show that part too. Not hard. — André Nicolas, Feb 26 '14 at 18:26

score 0 · Answer 1 · answered Feb 26 '14 at 20:27

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 $P+2$

Infinetly many primes of form $4k+3$

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