Theorem
There are infinitely many prime numbers of the form \(4m+3\)

Proof

Suppose there are only finitely many primes \(p_1,...,p_k\) of the form \(4n+3\). Consider the number

$$ \begin{align} u = 4p_1p_2....p_k - 1 \end{align} $$

\(u\) is not divisible by any prime \(p_1,...,p_k\) because if \(p_i\) divides \(u\) for any \(i=1,2,\ldots,k\), then observe that since \(p_i \mid 4p_1p_2....p_k\), then this implies that

$$ \begin{align} p_i \mid (u - 4p_1p_2....p_k) = - 1 \end{align} $$

But that’s impossible. So either \(u\) itself is a prime or \(u\) is a composite.
Case 1: \(u\) is a prime. Then \(u\) has the form

$$ \begin{align} u = 4p_1p_2\ldots p_k-1 = 4m - 1 = 4(m-1) + 3. \end{align} $$

So \(u\) is a new prime not in the list of the form \(4k+3\).

Case 2: \(u\) is composite. Then by the Fundamental Theorem of Arithmetic \(u\) can be expressed as a product of primes \(q_1q_2 \ldots q_j\) where \(j \geq 1\). But notice that we can re-write \(u\) to be

$$ \begin{align} u = 4m - 1 = 4(m-1)+3 \end{align} $$

\(u\) is odd since any \(q_i\) must either be of the form \(4k+1\) or \(4k+3\). However, it’s not possible to have all the factors be of the form \(4k+1\) since the product of these factors will also be of the form \(4k + 1\) (we can show this by induction). Then, we must have at least one factor of the form \(4k+3\) that is not in the original list of primes. This is a contradiction to the original assumption and so we have infinitely many primes of the form \(4k+3\). \(\ \blacksquare\).

References