Question:

<p><strong>Step 1:</strong> Let the nth roots of unity be $1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}$. These are roots of $z^n - 1 = 0$, or equivalently, roots of the polynomial equation: $$z^n = 1$$</p><p><strong>Step 2:</strong> We can factor this as: $$z^n - 1 = (z-1)(z - \alpha_1)(z - \alpha_2)\cdots(z - \alpha_{n-1})$$</p><p><strong>Step 3:</strong> Dividing both sides by $(z-1)$, we get the polynomial whose roots are $\alpha_1, \alpha_2, \ldots, \alpha_{n-1}$: $$\frac{z^n - 1}{z - 1} = z^{n-1} + z^{n-2} + z^{n-3} + \cdots + z + 1$$</p><p><strong>Step 4:</strong> By Vieta's formulas, for the polynomial $z^{n-1} + z^{n-2} + \cdots + z + 1$, the sum of products of roots taken two at a time equals the coefficient of $z^{n-3}$ with appropriate sign. This coefficient is 1.</p><p><strong>Step 5:</strong> More explicitly, from the polynomial $P(z) = z^{n-1} + z^{n-2} + \cdots + z + 1$, if we denote the elementary symmetric polynomial $e_2 = \sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j$, then by Vieta's formulas: $$e_2 = \frac{\text{coefficient of } z^{n-3}}{\text{leading coefficient}} = \frac{1}{1} = 1$$</p><p><strong>Step 6:</strong> Alternatively, we can verify using the identity: If $S_1 = \sum \alpha_i = -1$ (sum of all n-1 roots), and we use $(\sum \alpha_i)^2 = \sum \alpha_i^2 + 2\sum_{i<j} \alpha_i\alpha_j$. Since $\sum_{k=0}^{n-1} \alpha_k^2$ can be computed and $\sum_{i=1}^{n-1} \alpha_i^2 = -1$ (from the roots of unity), we get $(-1)^2 = -1 + 2e_2$, yielding $e_2 = 1$.</p><p><strong>∴ Answer:</strong> 1</p>