Determine a subset $ A\subset \mathbb{N}^*$ having $ 5$ different elements, so that the sum of the squares of its elements equals their product.
Do not simply post the subset, show how you found it.
To find a subset \( A \subset \mathbb{N}^* \) with 5 different elements such that the sum of the squares of its elements equals their product, we can follow these steps:

1. **Define the problem mathematically:**
   Let \( A = \{a, b, c, d, e\} \) be the subset. We need to find \( a, b, c, d, e \in \mathbb{N}^* \) such that:
   \[
   a^2 + b^2 + c^2 + d^2 + e^2 = abcde
   \]

2. **Consider the quadratic form:**
   Suppose we have a solution and consider it as a quadratic in \( a \). We have:
   \[
   a^2 - (bcde)a + (b^2 + c^2 + d^2 + e^2) = 0
   \]
   The discriminant of this quadratic equation must be a perfect square:
   \[
   \Delta = (bcde)^2 - 4(b^2 + c^2 + d^2 + e^2)
   \]
   For the discriminant to be a perfect square, we can set:
   \[
   (bcde - k)(bcde + k) = 4(b^2 + c^2 + d^2 + e^2)
   \]

3. **Simplify the equation:**
   Suppose \( bcde - k = 2 \) and \( bcde + k = 2(b^2 + c^2 + d^2 + e^2) \). Then:
   \[
   bcde = b^2 + c^2 + d^2 + e^2 + 1
   \]

4. **Consider it as a quadratic in \( b \):**
   Similarly, we can consider it as a quadratic in \( b \):
   \[
   b^2 - (cde)b + (c^2 + d^2 + e^2 + 1) = 0
   \]
   The discriminant of this quadratic equation must also be a perfect square:
   \[
   \Delta = (cde)^2 - 4(c^2 + d^2 + e^2 + 1)
   \]
   Again, we set:
   \[
   cde - k = 2 \quad \text{and} \quad cde + k = 2(c^2 + d^2 + e^2 + 1)
   \]
   Thus:
   \[
   cde = c^2 + d^2 + e^2 + 2
   \]

5. **Consider it as a quadratic in \( c \):**
   Now, consider it as a quadratic in \( c \):
   \[
   c^2 - (de)c + (d^2 + e^2 + 2) = 0
   \]
   The discriminant of this quadratic equation must be a perfect square:
   \[
   \Delta = (de)^2 - 4(d^2 + e^2 + 2)
   \]

6. **Choose a value for \( e \):**
   Let's choose \( e = 3 \). We need to find an integer \( d \) such that:
   \[
   5d^2 - 44 = k^2
   \]
   For \( d = 4 \), we have:
   \[
   5 \cdot 4^2 - 44 = 80 - 44 = 36
   \]
   Thus, \( k = 6 \).

7. **Solve for \( c \):**
   Returning to the previous equations, we find:
   \[
   12c = c^2 + 27
   \]
   Solving this quadratic equation:
   \[
   c^2 - 12c + 27 = 0
   \]
   The solutions are:
   \[
   c = \frac{12 \pm \sqrt{144 - 108}}{2} = \frac{12 \pm 6}{2} = 9 \quad \text{or} \quad 3
   \]

8. **Solve for \( b \):**
   Similarly, we find:
   \[
   (b - 1)(b - 107) = 0
   \]
   Thus, \( b = 1 \) or \( b = 107 \).

9. **Solve for \( a \):**
   If \( b = 1 \), we find:
   \[
   (a - 107)(a - 1) = 0
   \]
   Thus, \( a = 1 \) or \( a = 107 \).

   If \( b = 107 \), we find:
   \[
   (a - 1)(a - 11555) = 0
   \]
   Thus, \( a = 1 \) or \( a = 11555 \).

10. **Verify the solutions:**
    The subsets found are \( \{1, 3, 4, 9, 107\} \) and \( \{3, 4, 9, 107, 11555\} \). We need to verify that these subsets satisfy the original condition:
    \[
    1^2 + 3^2 + 4^2 + 9^2 + 107^2 = 1 \cdot 3 \cdot 4 \cdot 9 \cdot 107
    \]
    \[
    3^2 + 4^2 + 9^2 + 107^2 + 11555^2 = 3 \cdot 4 \cdot 9 \cdot 107 \cdot 11555
    \]

The final answer is \( \boxed{ \{1, 3, 4, 9, 107\} } \) and \( \{3, 4, 9, 107, 11555\} \).