(a) Let $8p + 7$ be prime where $p \ge 1$. Prove that $2^{4p+3}- 1$ is composite.
(b) Let $q= 4n− 1$ and $p= 8n− 1$ both be prime for $n > 1$. Prove that $p \mid (2^q - 1)$.
(a) We consider two cases for $(4p+3)$. It is either composite or prime.
Case $(4p+3)$ is composite
Corollary (4.11) tells us that if $n$ is composite, so is $2^n-1$. If $(4p+3)$ is composite, so is $2^{4p+3}-1$.
Case $(4p+3)$ is prime
If $(4p+3)$ is prime, then $2(4p+3)+1=8p+7$, which we're given as prime. And so $(4p+3)$ is a Germain prime.
Proposition (4.22) says that if $q \ne 3$ is Germain prime, and $q \equiv -1 \pmod 4$ then $M_q=2^q-1$ is composite. We do have $4p+3 \equiv -1 \pmod 4$, and we have $4p+3 \ge 7 \ne 3$ , and so $2^{4p+3}-1$ is composite.
Both cases conclude $2^{4p+3}-1$ is composite.
(b) We have $q=4n-1$ is prime. We also have $p = 2q + 1 = 8n-1$ is prime. This means $q$ is a Germain prime.
Since $q \equiv -1 \pmod 4$, and $q > 4(1)-1 \ne 3$, by Proposition (4.22) we have $p \mid (2^q-1)$.