Proof of Sufficiency: Suppose equations (3) and (4) hold. To make the derivation clear, we use complex number operations of the form $x + y \sqrt{-3}$. From equation (3), we have
$$s = (\alpha + \sqrt{-3} \beta)(\alpha - \sqrt{-3} \beta),$$
thus
$$\begin{aligned} s^{3} &= (\alpha + \sqrt{-3} \beta)^{3}(\alpha - \sqrt{-3} \beta)^{3} \\ &= \left\{\left(\alpha^{3} - 9 \alpha \beta^{2}\right) + \sqrt{-3}\left(3 \alpha^{2} \beta - 3 \beta^{3}\right)\right\} \\ & \quad \times \left\{\left(\alpha^{3} - 9 \alpha \beta^{2}\right) - \sqrt{-3}\left(3 \alpha^{2} \beta - 3 \beta^{3}\right)\right\} \\ &= \left(\alpha^{3} - 9 \alpha \beta^{2}\right)^{2} + 3\left(3 \alpha^{2} \beta - 3 \beta^{3}\right)^{2}. \end{aligned}$$

From this and equation (4), we can deduce that $s^{3} = a^{2} + 3 b^{2}$. Given $(\alpha, 3 \beta) = 1$ (note: $2 \nmid s$, $2 \nmid \alpha \pm \beta$),
$$\begin{aligned} (a, b) &= \left(\alpha^{2} - 9 \beta^{2}, \alpha^{2} - \beta^{2}\right) = \left(8 \beta^{2}, \alpha^{2} - \beta^{2}\right) \\ &= \left(\beta^{2}, \alpha^{2} - \beta^{2}\right) = 1. \end{aligned}$$

This proves that equation (2) holds.

Proof of Necessity: Suppose equation (2) holds. In this case, $3 \nmid s$, so any prime factor $p$ of $s$ must satisfy
$$p > 3, \quad (p, ab) = 1.$$

This implies (why) that
$$\left(\frac{-3}{p}\right) = 1.$$

Let $\Omega(s)$ denote the number of prime factors of $s$ (counting multiplicities), and define $\Omega(1) = 0$. For example, $\Omega(6) = 2$, $\Omega(4) = 2$. We use induction on $\Omega(s)$ to prove the necessity. When $\Omega(s) = 0$, i.e., $s = 1$, we have $a = \pm 1$, $b = 0$. In this case, we can take $\alpha = \pm 1$, $\beta = 0$. Therefore, the necessity holds. Assume the necessity holds for $\Omega(s) = n (\geqslant 0)$. When $\Omega(s) = n + 1$, let $s = p t$, where $p$ is a prime and $\Omega(t) = n$. Since $p$ satisfies equation (6), by Theorem 4 in $\S 2$,
$$p = \alpha_{1}^{2} + 3 \beta_{1}^{2}$$
and clearly,
$$(\alpha_{1}, 3 \beta_{1}) = 1.$$

From the proof of sufficiency, we have
$$\begin{array}{c} p^{3} = c^{2} + 3 d^{2}, \quad (c, d) = 1 \\ c = \alpha_{1}^{3} - 9 \alpha_{1} \beta_{1}^{2}, \quad d = 3 \alpha_{1}^{2} \beta_{1} - 3 \beta_{1}^{3}. \end{array}$$

From this and equation (2), we get
$$\begin{aligned} p^{6} t^{3} &= p^{3} s^{3} = \left(c^{2} + 3 d^{2}\right)\left(a^{2} + 3 b^{2}\right) \\ &= \left\{\begin{array}{l} (c + \sqrt{-3} d)(a - \sqrt{-3} b)(c - \sqrt{-3} d)(a + \sqrt{-3} b) \\ (c + \sqrt{-3} d)(a + \sqrt{-3} b)(c - \sqrt{-3} d)(a - \sqrt{-3} b) \end{array}\right. \\ &= \left\{\begin{array}{l} (a c + 3 b d)^{2} + 3(a d - b c)^{2}, \\ (a c - 3 b d)^{2} + 3(a d + b c)^{2}. \end{array}\right. \end{aligned}$$

We now prove that among $a d - b c$ and $a d + b c$, exactly one is divisible by $p$. Using equations (2), (9), and $s = p t$,
$$\begin{aligned} (a d - b c)(a d + b c) &= \left(a^{2} + 3 b^{2}\right) d^{2} - \left(c^{2} + 3 d^{2}\right) b^{2} \\ &= s^{3} d^{2} - p^{3} b^{2} = p^{3}\left(t^{3} d^{2} - b^{2}\right). \end{aligned}$$

Thus, $p$ divides at least one of these numbers. However, if $p \mid (a d - b c, a d + b c)$, then $p \mid (a d, b c)$ (since $p > 3$). By equation (9), $p \nmid c d$, so $p \mid (a, b)$, which contradicts $(a, b) = 1$. This proves the desired result, and by equation (12), the number divisible by $p$ must be divisible by $p^{3}$. Assume $p^{3} \mid a d - b c$ (the case $p^{3} \mid a d + b c$ can be treated similarly). From equation (11),
$$\begin{array}{c} t^{3} = u^{2} + 3 v^{2} \\ u = (a c + 3 b d) / p^{3}, \quad v = (a d - b c) / p^{3}. \end{array}$$

We now prove that
$$(u, v) = 1.$$

From equations (9) and (14),
$$\begin{array}{c} a c + 3 b d = u\left(c^{2} + 3 d^{2}\right) \\ a d - b c = v\left(c^{2} + 3 d^{2}\right). \end{array}$$

This implies (eliminating $b$ and $a$),
$$a = u c + 3 v d, \quad b = u d - v c.$$

From these two equations and $(a, b) = 1$, we conclude that equation (15) holds.
Thus, we have $\Omega(t) = n$ and equations (13) and (15) hold, so by the induction hypothesis, there exist $\alpha_{2}, \beta_{2}$ such that
$$t = \alpha_{2}^{2} + 3 \beta_{2}^{2}, \quad (\alpha_{2}, 3 \beta_{2}) = 1$$
and
$$u = \alpha_{2}^{3} - 9 \alpha_{2} \beta_{2}^{2}, \quad v = 3 \alpha_{2}^{2} \beta_{2} - 3 \beta_{2}^{3}.$$

From equations (7) and (17), we get (similar to equation (11))
$$s = p t = \left(\alpha_{1}^{2} + 3 \beta_{1}^{2}\right)\left(\alpha_{2}^{2} + 3 \beta_{2}^{2}\right) = \alpha^{2} + 3 \beta^{2},$$
where
$$\alpha = \alpha_{1} \alpha_{2} + 3 \beta_{1} \beta_{2}, \quad \beta = \alpha_{2} \beta_{1} - \beta_{2} \alpha_{1}.$$

To prove that the necessity holds when $\Omega(s) = n + 1$, we need to show that for the chosen $\alpha, \beta$, equation (4) holds and $(\alpha, 3 \beta) = 1$. For clarity, we use complex number operations. Equation (16) can be written as
$$a + \sqrt{-3} b = (c + \sqrt{-3} d)(u - \sqrt{-3} v).$$

Equation (10) can be written as
$$c + \sqrt{-3} d = \left(\alpha_{1} + \sqrt{-3} \beta_{1}\right)^{3}.$$

Equation (18) can be written as
$$u - \sqrt{-3} v = \left(\alpha_{2} - \sqrt{-3} \beta_{2}\right)^{3}.$$

From these three equations and equation (20), we get
$$\begin{aligned} a + \sqrt{-3} b &= \left\{\left(\alpha_{1} \alpha_{2} + 3 \beta_{1} \beta_{2}\right) + \sqrt{-3}\left(\alpha_{2} \beta_{1} - \beta_{2} \alpha_{1}\right)\right\}^{3} \\ &= (\alpha + \sqrt{-3} \beta)^{3}. \end{aligned}$$

Comparing the real and imaginary parts of the above equation, we deduce that equation (4) holds. From equation (2), we have $(a, 3 b) = 1$, and from equation (4), we get $(\alpha, 3 \beta) = 1$. Proof complete.