Find all $4$-tuple of real numbers $(x, y, z, w)$ that satisfy the following system of equations:
$$x^2 + y^2 + z^2 + w^2 = 4$$
$$\frac{1}{x^2} +\frac{1}{y^2} +\frac{1}{z^2 }+\frac{1}{w^2} = 5 -\frac{1}{(xyzw)^2}$$
1. **Given Equations:**
   We start with the given system of equations:
   \[
   x^2 + y^2 + z^2 + w^2 = 4
   \]
   \[
   \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} + \frac{1}{w^2} = 5 - \frac{1}{(xyzw)^2}
   \]

2. **Applying AM-HM Inequality:**
   By the Arithmetic Mean-Harmonic Mean (AM-HM) inequality, we have:
   \[
   \frac{x^2 + y^2 + z^2 + w^2}{4} \geq \frac{4}{\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} + \frac{1}{w^2}}
   \]
   Given \(x^2 + y^2 + z^2 + w^2 = 4\), we substitute:
   \[
   \frac{4}{4} \geq \frac{4}{\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} + \frac{1}{w^2}}
   \]
   Simplifying, we get:
   \[
   1 \geq \frac{4}{\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} + \frac{1}{w^2}}
   \]
   Therefore:
   \[
   \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} + \frac{1}{w^2} \geq 4
   \]

3. **Analyzing the Second Equation:**
   From the second equation:
   \[
   \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} + \frac{1}{w^2} = 5 - \frac{1}{(xyzw)^2}
   \]
   Since \(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} + \frac{1}{w^2} \geq 4\), we have:
   \[
   4 \leq 5 - \frac{1}{(xyzw)^2}
   \]
   Rearranging, we get:
   \[
   \frac{1}{(xyzw)^2} \leq 1
   \]
   Thus:
   \[
   (xyzw)^2 \geq 1
   \]

4. **Applying AM-GM Inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   \frac{x^2 + y^2 + z^2 + w^2}{4} \geq \sqrt[4]{x^2 y^2 z^2 w^2}
   \]
   Given \(x^2 + y^2 + z^2 + w^2 = 4\), we substitute:
   \[
   1 \geq \sqrt[4]{(xyzw)^2}
   \]
   Therefore:
   \[
   1 \geq (xyzw)
   \]
   Combining with \((xyzw)^2 \geq 1\), we get:
   \[
   (xyzw)^2 = 1
   \]
   Hence:
   \[
   xyzw = \pm 1
   \]

5. **Equality in AM-GM:**
   Equality in AM-GM holds if and only if all the terms are equal. Therefore:
   \[
   x^2 = y^2 = z^2 = w^2 = 1
   \]
   Thus:
   \[
   x, y, z, w \in \{1, -1\}
   \]

6. **Conclusion:**
   The 4-tuples \((x, y, z, w)\) that satisfy the given system of equations are:
   \[
   (1, 1, 1, 1), (1, 1, 1, -1), (1, 1, -1, 1), (1, -1, 1, 1), (-1, 1, 1, 1), (1, 1, -1, -1), (1, -1, 1, -1), (1, -1, -1, 1), (-1, 1, 1, -1), (-1, 1, -1, 1), (-1, -1, 1, 1), (1, -1, -1, -1), (-1, 1, -1, -1), (-1, -1, 1, -1), (-1, -1, -1, 1), (-1, -1, -1, -1)
   \]

The final answer is \( \boxed{ (1, 1, 1, 1), (1, 1, 1, -1), (1, 1, -1, 1), (1, -1, 1, 1), (-1, 1, 1, 1), (1, 1, -1, -1), (1, -1, 1, -1), (1, -1, -1, 1), (-1, 1, 1, -1), (-1, 1, -1, 1), (-1, -1, 1, 1), (1, -1, -1, -1), (-1, 1, -1, -1), (-1, -1, 1, -1), (-1, -1, -1, 1), (-1, -1, -1, -1) } \)