Infinite Number of Primes of form 4n - 1

Theorem
There are infinitely many prime numbers of the form $4 n - 1$.

Proof
Aiming for a contradiction, suppose the contrary.

That is, suppose there is a finite number of prime numbers of the form $4 n - 1$.

Let there be $k$ of them: $p_1, p_2, \ldots, p_k$.

Let $S = \left\{{p_1, p_2, \ldots, p_k}\right\}$.

Let $N$ be constructed as:
 * $\displaystyle N = 4 \prod_{i \mathop = 1}^k p_i - 1$

If $N$ is a prime number, then it is of the form $4 n - 1$.

But then we have that $N \notin S$, which means that $S$ is not the complete set of prime numbers of the form $4 n - 1$.

Therefore $N$ must be composite.

Suppose all the prime factors of $N$ are of the form $4 n + 1$.

Then from Product of Integers of form 4n + 1 it follows that $N$ itself is of the form $4 n + 1$.

Therefore $N$ must have at least one prime factor $p$ of the form $4 n - 1$.

Suppose $p \in S$.

We have that:
 * $\displaystyle p \mathop \backslash 4 \prod_{i \mathop = 1}^k p_i$

and so:
 * $\displaystyle p \mathop \backslash 4 \prod_{i \mathop = 1}^k p_i - p$

So:
 * $N = q p + \left({p - 1}\right)$

where:
 * $\displaystyle q = \frac {4 \prod_{i \mathop = 1}^k p_i - p} p$

So applying the Euclidean Algorithm:

So if $p \in S$ then it cannot be a divisor of $N$

Therefore there must be a prime numbers of the form $4 n - 1$ which is not in $S$.

Therefore $S$ is not the complete set of prime numbers of the form $4 n - 1$.

Therefore the assumption that $S$ is finite was false.

Hence the result.