Prove Corollary (4.21) (b).
Corollary (4.21)(b) is
Let $q$ and $p= 2q + 1$ both be primes. Note that $q$ is a Germain prime.
(b) If $q \equiv 1 \pmod 4$ then $p \mid (2^q + 1)$.
We'll make use of Proposition (4.19)(b)
Let $p= 2n + 1$ be prime. Then
(b) If $p \equiv \pm 3 \pmod 8$ then $p \mid (2^n + 1)$.
We start with Germain prime $q \equiv 1 \pmod 4$. This means, for some integer $k$,
$$ q = 1 + 4k $$
Then prime $p=2q +1$ becomes
$$ \begin{align} p &= 2(1 + 4k) + 1 \\ \\ & = 3 + 8k \end{align} $$
And so $p \equiv 3 \pmod 8$.
By Proposition (4.19)(b), setting $q=n$, this means $p \mid (2^q+1)$.
And so Corollary (4.21)(b) is proven.