Exercise 2.1.38 (There are infinitely many primes of the form $4n+1$)

Proof. If there are only finitely many primes $p_1,\dots ,p_r$ of the form $4n+1$, let $P=p_1\cdots p_r$ be their product. Consider $N=4P^2+1$. Then $N>1$, so there exists some prime number $p$ such that $p$ divides $N$. Note that $p\ne 2$, since $N$ is odd.

Since $p\mid a^2+1$, where $a=2P$, Theorem 2.12 shows that $p\equiv 1(\mathit{mod}4)$. Therefore $p\in \{p_1,\dots ,p_r\}$, thus $p\mid P$. But $p\mid 4P^2+1$, therefore $p\mid 1$. This is a contradiction, which proves that there exist infinitely many primes of the form $4n+1$. □