Find the roots $ r_i \in \mathbb{R}$  of the polynomial  \[ p(x) \equal{} x^n \plus{} n \cdot x^{n\minus{}1} \plus{} a_2 \cdot x^{n\minus{}2} \plus{} \ldots \plus{} a_n\] satisfying \[ \sum^{16}_{k\equal{}1} r^{16}_k \equal{} n.\]
1. **Given Polynomial and Condition:**
   We are given the polynomial:
   \[
   p(x) = x^n + n \cdot x^{n-1} + a_2 \cdot x^{n-2} + \ldots + a_n
   \]
   and the condition:
   \[
   \sum_{k=1}^{16} r_k^{16} = n
   \]
   where \( r_i \in \mathbb{R} \) are the roots of the polynomial.

2. **Applying Cauchy-Schwarz Inequality:**
   We apply the Cauchy-Schwarz inequality to the sequence of roots \( r_i \):
   \[
   \left( \sum_{i=1}^n r_i^{16} \right) \left( \sum_{i=1}^n 1 \right) \ge \left( \sum_{i=1}^n r_i^8 \right)^2
   \]
   Since there are \( n \) roots, the inequality becomes:
   \[
   \left( \sum_{i=1}^n r_i^{16} \right) n \ge \left( \sum_{i=1}^n r_i^8 \right)^2
   \]

3. **Using the Given Condition:**
   Given that:
   \[
   \sum_{i=1}^{16} r_i^{16} = n
   \]
   we substitute this into the inequality:
   \[
   n \cdot n \ge \left( \sum_{i=1}^n r_i^8 \right)^2
   \]
   Simplifying, we get:
   \[
   n^2 \ge \left( \sum_{i=1}^n r_i^8 \right)^2
   \]
   Taking the square root of both sides, we obtain:
   \[
   n \ge \sum_{i=1}^n r_i^8
   \]

4. **Equality Case of Cauchy-Schwarz:**
   For the equality case of the Cauchy-Schwarz inequality to hold, the sequences must be proportional. This implies:
   \[
   r_i^{16} = c \cdot 1 \quad \text{for some constant } c
   \]
   Given that \( \sum_{i=1}^n r_i^{16} = n \), we have:
   \[
   n \cdot c = n \implies c = 1
   \]
   Therefore:
   \[
   r_i^{16} = 1 \implies r_i = \pm 1
   \]

5. **Determining the Roots:**
   Since the polynomial has real coefficients and the sum of the roots raised to the 16th power is \( n \), the roots must be \( -1 \) to satisfy the polynomial structure. Thus, all roots are \( -1 \):
   \[
   r_1 = r_2 = \ldots = r_n = -1
   \]

6. **Forming the Polynomial:**
   If all roots are \( -1 \), the polynomial can be written as:
   \[
   p(x) = (x + 1)^n
   \]

The final answer is \( \boxed{ r_1 = r_2 = \ldots = r_n = -1 } \)