Let $p$ be an odd prime number. Prove that 2 is a quadratic residue modulo $p$ when $p \equiv 7 \quad(\bmod 8)$. Deduce that, if $q$ is a prime number strictly greater than 3 with $q \equiv 3$ $(\bmod 4)$ such that $2 q+1$ is also a prime number, then $2^{q}-1$ is necessarily composite. Why does the argument break down for $q=3$ ?