Find all quadrupels $(a, b, c, d)$ of positive real numbers that satisfy the following two equations:
\begin{align*}
ab + cd &= 8,\\
abcd &= 8 + a + b + c + d.
\end{align*}

1. We start with the given equations:
   \[
   ab + cd = 8
   \]
   \[
   abcd = 8 + a + b + c + d
   \]

2. We need to find all quadruples \((a, b, c, d)\) of positive real numbers that satisfy these equations. Let's first consider the inequality \(a + b + c + d \leq 8\).

3. Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   \frac{a + b + c + d}{4} \geq \sqrt[4]{abcd}
   \]
   Multiplying both sides by 4, we get:
   \[
   a + b + c + d \geq 4\sqrt[4]{abcd}
   \]

4. From the second equation \(abcd = 8 + a + b + c + d\), we substitute \(a + b + c + d\) with \(8\) (since \(a + b + c + d \leq 8\)):
   \[
   abcd = 8 + a + b + c + d \leq 16
   \]

5. Combining the inequalities, we get:
   \[
   4\sqrt[4]{abcd} \leq 8
   \]
   Simplifying, we find:
   \[
   \sqrt[4]{abcd} \leq 2
   \]
   Raising both sides to the fourth power, we get:
   \[
   abcd \leq 16
   \]

6. From the second equation \(abcd = 8 + a + b + c + d\), we know \(abcd \geq 8\). Therefore:
   \[
   8 \leq abcd \leq 16
   \]

7. Now, we use the first equation \(ab + cd = 8\). By AM-GM inequality:
   \[
   ab + cd \geq 2\sqrt{abcd}
   \]
   Given \(ab + cd = 8\), we have:
   \[
   8 \geq 2\sqrt{abcd}
   \]
   Simplifying, we get:
   \[
   4 \geq \sqrt{abcd}
   \]
   Squaring both sides, we find:
   \[
   16 \geq abcd
   \]

8. Combining all the inequalities, we have:
   \[
   abcd = 16
   \]

9. Substituting \(abcd = 16\) into the second equation:
   \[
   16 = 8 + a + b + c + d
   \]
   Simplifying, we get:
   \[
   a + b + c + d = 8
   \]

10. Given \(ab + cd = 8\) and using AM-GM inequality again:
    \[
    ab \leq \frac{(a+b)^2}{4} \quad \text{and} \quad cd \leq \frac{(c+d)^2}{4}
    \]
    Since \(a + b + c + d = 8\), we have:
    \[
    ab + cd \leq \frac{(a+b)^2}{4} + \frac{(c+d)^2}{4} \leq \frac{(a+b+c+d)^2}{4} = \frac{64}{4} = 16
    \]
    But \(ab + cd = 8\), so equality must hold in AM-GM, implying \(a = b\) and \(c = d\).

11. Let \(a = b\) and \(c = d\). Then:
    \[
    2a + 2c = 8 \implies a + c = 4
    \]
    \[
    ab = a^2 \quad \text{and} \quad cd = c^2
    \]
    \[
    a^2 + c^2 = 8
    \]

12. Since \(abcd = 16\), we have:
    \[
    a^2c^2 = 16 \implies (ac)^2 = 16 \implies ac = 4
    \]

13. Solving the system:
    \[
    a + c = 4
    \]
    \[
    ac = 4
    \]

14. The solutions to this system are \(a = c = 2\). Therefore, \(a = b = c = d = 2\).

Conclusion:
\[
(a, b, c, d) = (2, 2, 2, 2)
\]

The final answer is \( \boxed{ (2, 2, 2, 2) } \)