Infinitely many primes ≡ 1 (mod 4)

Construction. For $n\ge2$ let $N=(n!)^2+1$ and let $q$ be any prime divisor of $N$.

$q>n$. If $q\le n$, then $q\mid n!$, so $N\equiv1\pmod q$, contradicting $q\mid N$.

$q$ is odd. For $n\ge2$, $n!$ is even, so $N$ is odd.

The order of $x=n!$ modulo $q$. We have $x^2\equiv-1\not\equiv1\pmod q$ ($q$ odd), and $x^4\equiv1$. So the order of $x$ is exactly $4$. By Fermat's little theorem $x^{q-1}\equiv1$, hence $4\mid q-1$, i.e. $q\equiv1\pmod4$.

Conclusion. For every $n\ge2$ there is a prime $q\equiv1\pmod4$ with $q>n$. So there are infinitely many.

Check. $n=3$: $37=36+1$, and $37\equiv1\pmod4$. $n=4$: $577$ is prime.

Lesson. If $x^2\equiv-1\pmod q$, then $x$ has order $4$ and $4\mid q-1$.