Let's assume there are finite primes of the form \(4k+3\). Let \(p_0 = 3\) and let \(N\) be:
The set of primes \(\{ p_1, p_2, p_3, \ldots, p_r \}\) are all the finite primes of the form \(4k+3\) (other than 3). \(N\) itself cannot be prime because it's of the form \(4k+3\) and is greater than \(p_r\). Now let's think about the prime factors of \(N\). Since multiplying two numbers of the form \(4k+1\) gives another number of the form \(4k+1\), then all prime factors of \(N\) cannot be of the form \(4k+1\), at least one has to be of the form \(4k+3\).
Let \(p\) be a prime factor of \(N\) of the form \(4k+3\). If \(p=3\), then:
But 3 cannot divide 4 or any prime \(p_i\), so \(p\) cannot be 3. If instead \(p\) is one of \(\{ p_1, p_2, p_3, \ldots, p_r \}\), then \(p|4(p_1 p_2 p_3 \ldots p_r)\), but if \(p\) divides both \(N\) and \(4(p_1 p_2 p_3 \ldots p_r)\), then:
This is a contradiction, so \(p\) can't be one of \(\{ p_1, p_2, p_3, \ldots, p_r \}\), which mean's there exist primes (like \(p\)) of the form \(4k+3\) that don't belong to \(\{ p_1, p_2, p_3, \ldots, p_r \}\). This contradicts our assumption that there are finite primes of the form \(4k+3\).