# 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 $S = \set {p_1, p_2, \ldots, p_k}$.

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 $4 n + 1$ 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:

$\displaystyle p \divides 4 \prod_{i \mathop = 1}^k p_i$

and so:

$\displaystyle p \divides 4 \prod_{i \mathop = 1}^k p_i - p$

So:

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

where:

$q = \dfrac {\displaystyle 4 \prod_{i \mathop = 1}^k p_i - p} p$

So applying the Euclidean Algorithm:

 $\displaystyle \gcd \set {N, p}$ $=$ $\displaystyle \gcd \set {p - 1, p}$ $\displaystyle$ $=$ $\displaystyle \gcd \set {p - 1, 1}$ $\displaystyle$ $=$ $\displaystyle 1$

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.

$\blacksquare$

## Also presented as

Some sources present this result as:

There are infinitely many prime numbers congruent to $3 \pmod 4$.