Let $n \ge 2$ be a fixed integer. [list=a] [*]Determine the largest positive integer $m$ (in terms of $n$) such that there exist complex numbers $r_1$, $\dots$, $r_n$, not all zero, for which \[ \prod_{k=1}^n (r_k+1) = \prod_{k=1}^n (r_k^2+1) = \dots = \prod_{k=1}^n (r_k^m+1) = 1. \] [*]For this value of $m$, find all possible values of \[ \prod\limits_{k=1}^n (r_k^{m+1}+1). \] [/list]

[i]Kaixin Wang[/i]
1. **Determine the largest positive integer \( m \) (in terms of \( n \)) such that there exist complex numbers \( r_1, \dots, r_n \), not all zero, for which**
   \[
   \prod_{k=1}^n (r_k+1) = \prod_{k=1}^n (r_k^2+1) = \dots = \prod_{k=1}^n (r_k^m+1) = 1.
   \]

   **Proof of Necessity:**
   - Let \( S \) denote the set of products of elements in the set \( \{r_i\} \), not including 1.
   - Define the polynomial \( P(x) := \prod_{r \in S} (x-r) \).
   - Suppose some \( m \ge 2^n-1 \) works. Since the sums \( S_k := \sum_{r \in S} r^k \) are known to be 0 for \( 1 \le k \le m \) by the problem condition, it holds by Newton's sums that there is a unique solution set for \( S \), since \( m \ge \deg P \).
   - As \( (r_i)=0 \) works, it must be the only solution, a contradiction. Thus we have the upper bound \( m \le 2^n - 2 \).

   **Construction:**
   - Let \( r_k = e^{2\pi i/(2^n-1) \cdot 2^{k-1}} \) for each \( 1 \le k \le m = 2^n-2 \).
   - Then since the generating function identity
     \[
     (1 - x^{2^{\ell + 1}})/(1-x) = \prod_{0 \le j \le \ell} (1+x^{2^j})
     \]
     holds, we are done.

2. **For this value of \( m \), find all possible values of**
   \[
   \prod\limits_{k=1}^n (r_k^{m+1}+1).
   \]

   - The answer for \( m=2^n-2 \) is \( \boxed{\{2^n\}} \).
   - Consider \( P(x) \) and \( S \) from part (a). Note that \( S_1, \dots, S_m \) are all 0, and denote \( a := S_{m+1} + 1 \).
   - By Newton's sums, we inductively have \( [x^{m+1}] P(x) = 1 \), for each \( 1 \le k \le m \), \( [x^k]P(x)=0 \), and \( [x^0] P(x) = - \tfrac{m+1}{a-1} \).

   - That said, let \( P(x) = x^{m+1} - \alpha \). Note that there exists \( C \in \mathbb{C} \) such that multiplying the \( 2^n \)th roots of unity by \( C \) returns \( S \). As we saw in part (a), \( C \) being a \( 2^n \)th root of unity works, so it suffices to show that such \( C \) must be a \( 2^n \)th root of unity as it then follows that \( \alpha = 1 \), thus showing that \( 2^n \) is the only solution for \( a \).

   - Realize that
     \[
     \alpha = - \left( \prod r_i \right)^{2^{n-1}}.
     \]
     If we multiply the elements of \( S \) by \( C \), then \( \alpha = \alpha \cdot C^{2^{n-1}} \), meaning that \( C \) is a \( 2^{n-1} \)th root of unity. Thus we are done.