Theorem
There are infinitely many prime numbers of the form \(4m+1\)

Note: The problem here is that we can’t use the same trick/argument we used last time because the product of primes of the form \(4m+3\) isn’t necessarily of the form \(4m+3\) so we need another argument. Here we use the fact that if a prime divides \(m^2 + 1\), then \(p = 2\) or \(p = 4n+1\) (TODO: Proof Link).

Proof

Suppose there are only finitely many primes \(p_1,...,p_k\) of the form \(4n+1\). Consider the number

$$ \begin{align} u = (2p_1p_2...p_k)^2 + 1 \end{align} $$
(Side note: Why this? well we want to use that special fact and also squaring \(p_1...p_k\) and adding 1 might give us something that is divisible by 2 so to make sure that doesn't happen, we will stick 2 inside the squared product)

Notice that \(u\) 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 \(p_i \mid (2p_1p_2...p_k)^2\), then this implies that

$$ \begin{align} p_i \mid (u - (2p_1p_2...p_k)^2) = 1 \end{align} $$

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

$$ \begin{align} u = (2p_1p_2...p_k)^2 + 1 = 4(p_1p_2...p_k)^2 + 1 = 4m + 1 \end{align} $$

Therefore, \(u\) is a new prime not in the list of the form \(4k+1\).

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 \(u\) has the form

$$ \begin{align} u = (2p_1p_2...p_k)^2 + 1 = m^2 + 1 \quad \text{ where } m=2p_1p_2...p_k \end{align} $$

We know that if a prime \(p\) divides a product of the from \(m^2 + 1\), then either \(p = 2\) or \(p = 4n+1\). But notice that \(u\) is odd since \(m\) is even and thus \(m^2 + 1\) must be odd. So none of the factors can be \(2\). Therefore, every factor must be of the form \(4n + 1\). This is a contradiction to the original assumption. Therefore, we have infinitely many primes of the form \(4m+1\). \(\ \blacksquare\).

References