The sequence $ \{ a_n \} _ { n \ge 0 } $ is defined by $ a_0 = 2 , a_1 = 4 $ and
\[ a_{n+1} = \frac{a_n a_{n-1}}{2} + a_n + a_{n-1} \]
for all positive integers $ n $. Determine all prime numbers $ p $ for which there exists a positive integer $ m $ such that $ p $ divides the number $ a_m - 1 $.
1. **Define a new sequence \( b_n \):**
   We start by defining a new sequence \( b_n \) such that \( b_n = a_n + 2 \). Given the initial conditions \( a_0 = 2 \) and \( a_1 = 4 \), we have:
   \[
   b_0 = a_0 + 2 = 2 + 2 = 4
   \]
   \[
   b_1 = a_1 + 2 = 4 + 2 = 6
   \]

2. **Derive the recurrence relation for \( b_n \):**
   Using the given recurrence relation for \( a_n \):
   \[
   a_{n+1} = \frac{a_n a_{n-1}}{2} + a_n + a_{n-1}
   \]
   we substitute \( a_n = b_n - 2 \) and \( a_{n-1} = b_{n-1} - 2 \):
   \[
   a_{n+1} = \frac{(b_n - 2)(b_{n-1} - 2)}{2} + (b_n - 2) + (b_{n-1} - 2)
   \]
   Simplifying the expression:
   \[
   a_{n+1} = \frac{b_n b_{n-1} - 2b_n - 2b_{n-1} + 4}{2} + b_n + b_{n-1} - 4
   \]
   \[
   a_{n+1} = \frac{b_n b_{n-1}}{2} - b_n - b_{n-1} + 2 + b_n + b_{n-1} - 4
   \]
   \[
   a_{n+1} = \frac{b_n b_{n-1}}{2} - 2
   \]
   Therefore:
   \[
   b_{n+1} = a_{n+1} + 2 = \frac{b_n b_{n-1}}{2}
   \]

3. **Express \( b_n \) in terms of powers of 2 and 3:**
   We observe that \( b_0 = 4 = 2^2 \cdot 3^0 \) and \( b_1 = 6 = 2^1 \cdot 3^1 \). We hypothesize that \( b_n \) can be expressed as \( b_n = 2^{F_{n-1} + 1} 3^{F_n} \), where \( F_n \) is the Fibonacci sequence.

4. **Verify the hypothesis:**
   We use induction to verify the hypothesis. For \( n = 0 \) and \( n = 1 \), the hypothesis holds:
   \[
   b_0 = 2^{F_{-1} + 1} 3^{F_0} = 2^{0 + 1} 3^0 = 2^1 = 4
   \]
   \[
   b_1 = 2^{F_0 + 1} 3^{F_1} = 2^{1 + 1} 3^1 = 2^2 \cdot 3 = 6
   \]
   Assume \( b_k = 2^{F_{k-1} + 1} 3^{F_k} \) and \( b_{k+1} = 2^{F_k + 1} 3^{F_{k+1}} \) hold for some \( k \). Then:
   \[
   b_{k+2} = \frac{b_{k+1} b_k}{2} = \frac{2^{F_k + 1} 3^{F_{k+1}} \cdot 2^{F_{k-1} + 1} 3^{F_k}}{2}
   \]
   \[
   b_{k+2} = \frac{2^{F_k + 1 + F_{k-1} + 1} 3^{F_{k+1} + F_k}}{2} = 2^{F_{k+1} + 1} 3^{F_{k+2}}
   \]
   Thus, the hypothesis holds for all \( n \).

5. **Determine the primes \( p \) for which \( p \) divides \( a_m - 1 \):**
   We need \( p \) to divide \( a_m - 1 = b_m - 3 \). Since \( b_m = 2^{F_{m-1} + 1} 3^{F_m} \), we have:
   \[
   a_m - 1 = 2^{F_{m-1} + 1} 3^{F_m} - 3
   \]
   For \( p > 3 \), by Fermat's Little Theorem, \( 2^{p-1} \equiv 1 \pmod{p} \) and \( 3^{p-1} \equiv 1 \pmod{p} \). We need \( p-1 \) to divide both \( F_{m-1} + 1 \) and \( F_m - 1 \).

6. **Cycle of Fibonacci pairs modulo \( p-1 \):**
   The pairs \( (F_{m-1}, F_m) \) modulo \( p-1 \) form a cycle. We need to show that the pair \( (-1, 1) \) lies on the same cycle as \( (F_0, F_1) = (0, 1) \). This is true because:
   \[
   (-1, 1) \to (1, 0) \to (0, 1)
   \]
   Thus, every prime \( p > 2 \) satisfies the condition.

The final answer is \( \boxed{ p > 2 } \).