Let $x_1,x_2, ..., x_n$ be non-negative real numbers. Solve the system of equations:
$$x_k+x_{k+1}=x^2_{k+2}\,\,,\,\,\, (k =1,2,...,n),$$
where $x_{n+1} = x_1$, $x_{n+2} = x_2$.
1. Given the system of equations:
   \[
   x_k + x_{k+1} = x_{k+2}^2 \quad \text{for} \quad k = 1, 2, \ldots, n,
   \]
   with the conditions \( x_{n+1} = x_1 \) and \( x_{n+2} = x_2 \).

2. By adding all the equations, we get:
   \[
   (x_1 + x_2) + (x_2 + x_3) + \cdots + (x_n + x_1) = x_3^2 + x_4^2 + \cdots + x_2^2.
   \]
   Simplifying, we have:
   \[
   2 \sum_{i=1}^n x_i = \sum_{i=1}^n x_i^2.
   \]

3. By the Cauchy-Schwarz inequality:
   \[
   \left( \sum_{i=1}^n x_i^2 \right) n \geq \left( \sum_{i=1}^n x_i \right)^2.
   \]
   Substituting \( \sum_{i=1}^n x_i^2 = 2 \sum_{i=1}^n x_i \), we get:
   \[
   (2 \sum_{i=1}^n x_i) n \geq \left( \sum_{i=1}^n x_i \right)^2.
   \]
   Simplifying, we have:
   \[
   2n \sum_{i=1}^n x_i \geq \left( \sum_{i=1}^n x_i \right)^2.
   \]
   Let \( S = \sum_{i=1}^n x_i \). Then:
   \[
   2nS \geq S^2 \implies S^2 - 2nS \leq 0 \implies S(S - 2n) \leq 0.
   \]
   Therefore:
   \[
   0 \leq S \leq 2n.
   \]

4. Let \( a_j \) be the minimum of \( x_1, x_2, \ldots, x_n \). Then:
   \[
   x_{j-2} + x_{j-1} = x_j^2 \leq \left( \frac{x_{j-2} + x_{j-1}}{2} \right)^2.
   \]
   Simplifying, we get:
   \[
   (x_{j-2} + x_{j-1})^2 \geq 4(x_{j-2} + x_{j-1}).
   \]
   If \( x_{j-2} + x_{j-1} = 0 \), then \( x_j = 0 \). This implies \( x_{j-1} = x_{j-2} = 0 \), and by induction, all \( x_i = 0 \). Thus, \( (0, 0, \ldots, 0) \) is a solution.

5. If \( x_{j-2} + x_{j-1} \neq 0 \), then:
   \[
   x_j^2 = x_{j-2} + x_{j-1} \geq 4.
   \]
   Therefore, the minimum value of \( x_j \) is \( 2 \). Hence, \( x_1, x_2, \ldots, x_n \geq 2 \). This implies:
   \[
   x_1 + x_2 + \cdots + x_n \geq 2n.
   \]
   Since \( x_1 + x_2 + \cdots + x_n \leq 2n \), we have:
   \[
   x_1 + x_2 + \cdots + x_n = 2n.
   \]
   Therefore, \( x_1 = x_2 = \cdots = x_n = 2 \). Thus, \( (2, 2, \ldots, 2) \) is another solution.

The final answer is \( \boxed{ (0, 0, \ldots, 0) } \) and \((2, 2, \ldots, 2)\).