Necessity If $q=2 p+1$ is a prime, then by the condition we know $q \equiv-1(\bmod 8)$, and thus by Theorem 3 we have $\left(\frac{2}{q}\right)=1$. From this and Theorem 1 (ii), we get
$$1 \equiv 2^{(q-1) / 2} \equiv 2^{p}(\bmod 2 p+1)$$

Sufficiency If equation (15) holds. Since $p$ is a prime, by Example 5 in Chapter 1, §3, we know that $p$ is the smallest positive integer $d$ satisfying
$$2^{d} \equiv 1(\bmod 2 p+1)$$

Furthermore, by Theorem 3 in Chapter 3, §3, we know $p \mid \varphi(2 p+1)$. Therefore, it must be that $\varphi(2 p+1)=p$ or $2 p$. Since $2 \mid \varphi(m)(m>2)$, we have $\varphi(2 p+1)=2 p$. This proves that $2 p+1$ is a prime (why).