Find all triples $(x, p, n)$ of non-negative integers such that $p$ is prime and $2x(x + 5) = p^n + 3(x - 1)$.
We start with the given equation:
\[ 2x(x + 5) = p^n + 3(x - 1) \]

1. **Rearrange the equation:**
   \[ 2x^2 + 10x = p^n + 3x - 3 \]
   \[ 2x^2 + 7x + 3 = p^n \]

2. **Factor the left-hand side:**
   \[ 2x^2 + 7x + 3 = (x + 3)(2x + 1) \]
   Thus, we have:
   \[ (x + 3)(2x + 1) = p^n \]

3. **Consider the case \( x = 0 \):**
   \[ (0 + 3)(2 \cdot 0 + 1) = 3 \cdot 1 = 3 \]
   \[ p^n = 3 \]
   Since \( p \) is a prime number, we have:
   \[ p = 3 \quad \text{and} \quad n = 1 \]
   Therefore, one solution is:
   \[ (x, p, n) = (0, 3, 1) \]

4. **Consider the case \( x > 0 \):**
   Let \( x + 3 = p^a \) and \( 2x + 1 = p^b \) for \( a, b \in \mathbb{N} \). Clearly, \( a \leq b \).

5. **Case \( a = b \):**
   \[ p^a = p^b \implies x + 3 = 2x + 1 \]
   \[ x + 3 = 2x + 1 \]
   \[ 3 - 1 = x \]
   \[ x = 2 \]
   Substituting \( x = 2 \) back into the original equation:
   \[ (2 + 3)(2 \cdot 2 + 1) = 5 \cdot 5 = 25 \]
   \[ p^n = 25 \]
   Since \( p \) is a prime number, we have:
   \[ p = 5 \quad \text{and} \quad n = 2 \]
   Therefore, another solution is:
   \[ (x, p, n) = (2, 5, 2) \]

6. **Case \( a < b \):**
   \[ p^b - 2p^a = -5 \]
   \[ p^b < 2p^a \]
   \[ p^a < p^b < 2p^a \]
   \[ 1 < p^{b-a} < 2 \]
   Since no natural number lies between 1 and 2 (both exclusive), this leads to a contradiction.

Therefore, the only solutions are:
\[ (x, p, n) = (0, 3, 1) \quad \text{and} \quad (2, 5, 2) \]

The final answer is \( \boxed{ (0, 3, 1) } \) and \((2, 5, 2)\).