Find all real numbers $x_1, \dots, x_{2016}$ that satisfy the following equation for each $1 \le i \le 2016$. (Here $x_{2017} = x_1$.)
\[ x_i^2 + x_i - 1 = x_{i+1} \]
1. **Summing the Equations:**
   Given the equation for each \(1 \le i \le 2016\):
   \[
   x_i^2 + x_i - 1 = x_{i+1}
   \]
   Summing all these equations from \(i = 1\) to \(i = 2016\), we get:
   \[
   \sum_{i=1}^{2016} (x_i^2 + x_i - 1) = \sum_{i=1}^{2016} x_{i+1}
   \]
   Since \(x_{2017} = x_1\), the right-hand side is a cyclic sum and equals the left-hand side:
   \[
   \sum_{i=1}^{2016} x_i^2 + \sum_{i=1}^{2016} x_i - 2016 = \sum_{i=1}^{2016} x_i
   \]
   Simplifying, we get:
   \[
   \sum_{i=1}^{2016} x_i^2 = 2016
   \]

2. **Considering \(x_i = -1\):**
   If any \(x_i = -1\), then substituting into the original equation:
   \[
   (-1)^2 + (-1) - 1 = -1
   \]
   This implies that all \(x_i = -1\). Therefore, \(\boxed{(-1, -1, \ldots, -1)}\) is a solution.

3. **Multiplying the Equations:**
   Multiplying all the given equations from \(i = 1\) to \(i = 2016\), we get:
   \[
   \prod_{i=1}^{2016} (x_i^2 + x_i - 1) = \prod_{i=1}^{2016} x_{i+1}
   \]
   Since \(x_{2017} = x_1\), the right-hand side is a cyclic product and equals the left-hand side:
   \[
   \prod_{i=1}^{2016} (x_i^2 + x_i - 1) = \prod_{i=1}^{2016} x_i
   \]
   Simplifying, we get:
   \[
   \prod_{i=1}^{2016} x_i = 1
   \]

4. **Applying AM-GM Inequality:**
   Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
   \[
   \frac{x_1^2 + x_2^2 + \cdots + x_{2016}^2}{2016} \geq \sqrt[2016]{(x_1^2 x_2^2 \cdots x_{2016}^2)}
   \]
   Given \(\sum_{i=1}^{2016} x_i^2 = 2016\), we have:
   \[
   \frac{2016}{2016} \geq \sqrt[2016]{(x_1 x_2 \cdots x_{2016})^2}
   \]
   Simplifying, we get:
   \[
   1 \geq \sqrt[2016]{1^2} = 1
   \]
   Equality holds if and only if \(x_1^2 = x_2^2 = \cdots = x_{2016}^2\).

5. **Analyzing \(x_i\):**
   Since \(x_i^2 = 1\), we have \(x_i = \pm 1\). Given \(\prod_{i=1}^{2016} x_i = 1\), the number of \(x_i = -1\) must be even. 

6. **Considering \(x_i = 1\):**
   If all \(x_i = 1\), substituting into the original equation:
   \[
   1^2 + 1 - 1 = 1
   \]
   This implies that all \(x_i = 1\). Therefore, \(\boxed{(1, 1, \ldots, 1)}\) is a solution.

7. **Combining Solutions:**
   The solutions are either all \(x_i = -1\) or all \(x_i = 1\).