Let $p$  be a prime number, $p> 5$. Determine the non-zero natural numbers $x$  with the property that $5p + x$ divides $5p ^ n + x ^ n$, whatever $n \in N ^ {*} $.
To solve the problem, we need to determine the non-zero natural numbers \( x \) such that \( 5p + x \) divides \( 5p^n + x^n \) for any \( n \in \mathbb{N}^* \). 

1. **Initial Consideration:**
   Let's consider the greatest common divisor (GCD) of \( 5p \) and \( x \). We have three cases to consider:
   - \( \gcd(5p, x) = 1 \)
   - \( \gcd(5p, x) = 5 \)
   - \( \gcd(5p, x) = p \)
   - \( \gcd(5p, x) = 5p \)

2. **Case 1: \( \gcd(5p, x) = 1 \):**
   If \( \gcd(5p, x) = 1 \), then \( 5p + x \) must divide \( 5p^n + x^n \). We can write:
   \[
   5p + x \mid 5p^n + x^n - 5^n x
   \]
   Simplifying, we get:
   \[
   5p + x \mid x^n (1 - 5^{n-1})
   \]
   Since \( 5p + x \) must divide \( x^n (1 - 5^{n-1}) \) for all \( n \), it must also divide \( 1 - 5^{n-1} \). However, \( 1 - 5^{n-1} \) can take values such as \( -4, 24, -124 \), etc., which are not divisible by \( 5p + x \) for \( p > 5 \). Therefore, this case is impossible.

3. **Case 2: \( \gcd(5p, x) = 5 \):**
   Let \( x = 5k \) where \( \gcd(p, k) = 1 \). Then:
   \[
   5p + 5k = 5(p + k)
   \]
   We need \( 5(p + k) \mid 5p^n + (5k)^n \). Simplifying, we get:
   \[
   5(p + k) \mid 5p^n + 5^n k^n
   \]
   Dividing both sides by 5:
   \[
   p + k \mid p^n + 5^{n-1} k^n
   \]
   Since \( \gcd(p, k) = 1 \), \( p + k \) must divide \( 5^{n-1} k^n \). For this to hold for all \( n \), \( p + k \) must be a divisor of 5. The only possible values are 1 and 5. Since \( p > 5 \), \( p + k = 5 \) is impossible. Therefore, this case is also impossible.

4. **Case 3: \( \gcd(5p, x) = p \):**
   Let \( x = pk \) where \( \gcd(5, k) = 1 \). Then:
   \[
   5p + pk = p(5 + k)
   \]
   We need \( p(5 + k) \mid 5p^n + (pk)^n \). Simplifying, we get:
   \[
   p(5 + k) \mid 5p^n + p^n k^n
   \]
   Dividing both sides by \( p \):
   \[
   5 + k \mid 5p^{n-1} + p^{n-1} k^n
   \]
   Since \( \gcd(5, k) = 1 \), \( 5 + k \) must divide \( 5 + k^n \). For this to hold for all \( n \), \( k \) must be 1. Therefore, \( x = p \).

5. **Case 4: \( \gcd(5p, x) = 5p \):**
   Let \( x = 5pk \). Then:
   \[
   5p + 5pk = 5p(1 + k)
   \]
   We need \( 5p(1 + k) \mid 5p^n + (5pk)^n \). Simplifying, we get:
   \[
   5p(1 + k) \mid 5p^n + 5^n p^n k^n
   \]
   Dividing both sides by \( 5p \):
   \[
   1 + k \mid p^{n-1} + 5^{n-1} p^{n-1} k^n
   \]
   Since \( 1 + k \) must divide \( 1 + 5^{n-1} k^n \) for all \( n \), \( k \) must be 0. Therefore, \( x = 0 \), which is not a non-zero natural number.

Thus, the only solution is \( x = p \).

The final answer is \( \boxed{ x = p } \).