Determine all triples $(m , n , p)$ satisfying :
\[n^{2p}=m^2+n^2+p+1\]
where $m$ and $n$ are integers and $p$ is a prime number.
To determine all triples \((m, n, p)\) satisfying the equation:
\[ n^{2p} = m^2 + n^2 + p + 1 \]
where \(m\) and \(n\) are integers and \(p\) is a prime number, we will consider two cases: \(p = 2\) and \(p > 2\).

1. **Case \(p = 2\):**
   \[ n^{2 \cdot 2} = m^2 + n^2 + 2 + 1 \]
   \[ n^4 = m^2 + n^2 + 3 \]
   Rearranging the equation, we get:
   \[ n^4 - n^2 - 3 = m^2 \]
   We need to find integer solutions for \(n\) and \(m\). Let's analyze the inequality:
   \[ (n^2 - 1)^2 < n^4 - n^2 - 3 < (n^2)^2 \]
   Simplifying the inequalities:
   \[ n^4 - 2n^2 + 1 < n^4 - n^2 - 3 < n^4 \]
   \[ -2n^2 + 1 < -n^2 - 3 \]
   \[ n^2 > 4 \]
   This implies \(|n| \leq 2\). We check the possible values of \(n\):

   - For \(n = 2\):
     \[ 2^4 - 2^2 - 3 = 16 - 4 - 3 = 9 = 3^2 \]
     Thus, \((n, m, p) = (2, 3, 2)\) is a solution.
   - For \(n = -2\):
     \[ (-2)^4 - (-2)^2 - 3 = 16 - 4 - 3 = 9 = 3^2 \]
     Thus, \((n, m, p) = (-2, 3, 2)\) is a solution.
   - For \(n = 1\):
     \[ 1^4 - 1^2 - 3 = 1 - 1 - 3 = -3 \]
     No integer \(m\) satisfies this.
   - For \(n = -1\):
     \[ (-1)^4 - (-1)^2 - 3 = 1 - 1 - 3 = -3 \]
     No integer \(m\) satisfies this.
   - For \(n = 0\):
     \[ 0^4 - 0^2 - 3 = -3 \]
     No integer \(m\) satisfies this.

   Therefore, the solutions for \(p = 2\) are:
   \[ (n, m, p) = (\pm 2, \pm 3, 2) \]

2. **Case \(p > 2\):**
   \[ n^2(n^{2p-2} - 1) = m^2 + p + 1 \]
   Since \(p\) is a prime number greater than 2, we know that \(p \equiv 1 \pmod{4}\) implies \(p \mid m^2 + 1\). This means:
   \[ p \mid m^2 + 1 \]
   However, if \(p \equiv 1 \pmod{4}\), then \(p \mid m^2 + 1\) implies \(p \mid m^2 + 2\), which is a contradiction because \(p\) cannot divide both \(m^2 + 1\) and \(m^2 + 2\) simultaneously. Therefore, there are no solutions for \(p > 2\).

The final answer is \( \boxed{ (n, m, p) = (\pm 2, \pm 3, 2) } \)