Suppose $n \ge 0$ is an integer and all the roots of $x^3 + 
\alpha x + 4 - ( 2 \times 2016^n) = 0$ are integers. Find all possible values of $\alpha$.
1. Let \(a, b, c\) be the integer roots of the polynomial equation \(x^3 + \alpha x + 4 - (2 \times 2016^n) = 0\). By Vieta's formulas, we have:
   \[
   a + b + c = 0,
   \]
   \[
   ab + bc + ca = \alpha,
   \]
   \[
   abc = 2 \times 2016^n - 4.
   \]

2. Suppose \(n \ge 1\). We start by using the identity for the sum of cubes:
   \[
   (a + b + c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a).
   \]
   Since \(a + b + c = 0\), this simplifies to:
   \[
   0 = a^3 + b^3 + c^3 - 3abc.
   \]
   Therefore,
   \[
   a^3 + b^3 + c^3 = 3abc.
   \]

3. Substituting \(abc = 2 \times 2016^n - 4\) into the equation, we get:
   \[
   a^3 + b^3 + c^3 = 3(2 \times 2016^n - 4).
   \]
   Simplifying, we have:
   \[
   a^3 + b^3 + c^3 = 6 \times 2016^n - 12.
   \]

4. Using the polynomial equation, we also have:
   \[
   a^3 + b^3 + c^3 = -\alpha(a + b + c) + 3abc + 4 - 2 \times 2016^n.
   \]
   Since \(a + b + c = 0\), this simplifies to:
   \[
   a^3 + b^3 + c^3 = 3abc + 4 - 2 \times 2016^n.
   \]
   Substituting \(abc = 2 \times 2016^n - 4\), we get:
   \[
   a^3 + b^3 + c^3 = 3(2 \times 2016^n - 4) + 4 - 2 \times 2016^n.
   \]
   Simplifying, we have:
   \[
   a^3 + b^3 + c^3 = 6 \times 2016^n - 12 + 4 - 2 \times 2016^n.
   \]
   \[
   a^3 + b^3 + c^3 = 4 \times 2016^n - 8.
   \]

5. Equating the two expressions for \(a^3 + b^3 + c^3\), we get:
   \[
   6 \times 2016^n - 12 = 4 \times 2016^n - 8.
   \]
   Simplifying, we have:
   \[
   2 \times 2016^n - 4 = 0.
   \]
   \[
   2 \times 2016^n = 4.
   \]
   \[
   2016^n = 2.
   \]

6. Since \(2016^n\) is an integer, the only possible value for \(n\) that satisfies this equation is \(n = 0\). Therefore, we must have \(n = 0\).

7. Substituting \(n = 0\) into the original polynomial equation, we get:
   \[
   x^3 + \alpha x + 4 - 2 = 0.
   \]
   Simplifying, we have:
   \[
   x^3 + \alpha x + 2 = 0.
   \]

8. Using Vieta's formulas again, we have:
   \[
   a + b + c = 0,
   \]
   \[
   ab + bc + ca = \alpha,
   \]
   \[
   abc = -2.
   \]

9. Since \(a, b, c\) are integers and their product is \(-2\), the possible sets of values for \((a, b, c)\) are permutations of \((1, -1, -2)\).

10. For \((a, b, c) = (1, -1, -2)\), we have:
    \[
    a + b + c = 1 - 1 - 2 = -2 \neq 0.
    \]
    This set does not satisfy \(a + b + c = 0\).

11. For \((a, b, c) = (1, -2, 1)\), we have:
    \[
    a + b + c = 1 - 2 + 1 = 0.
    \]
    This set satisfies \(a + b + c = 0\).

12. For \((a, b, c) = (-1, 2, -1)\), we have:
    \[
    a + b + c = -1 + 2 - 1 = 0.
    \]
    This set satisfies \(a + b + c = 0\).

13. For both valid sets, we have:
    \[
    ab + bc + ca = 1 \cdot (-2) + (-2) \cdot 1 + 1 \cdot 1 = -2 - 2 + 1 = -3.
    \]
    Therefore, \(\alpha = -3\).

Conclusion:
\[
\boxed{\alpha = -3}
\]