An ordered triple $(p, q, r)$ of prime numbers is called [i]parcera[/i] if $p$ divides $q^2-4$, $q$ divides $r^2-4$ and $r$ divides $p^2-4$. Find all parcera triples.
1. **Case 1: All primes are equal**

   If \( p = q = r \), then we have:
   \[
   p \mid q^2 - 4 \implies p \mid p^2 - 4 \implies p \mid 4
   \]
   Since \( p \) is a prime number, the only prime that divides 4 is 2. Therefore, \( p = q = r = 2 \). This gives us the solution:
   \[
   (2, 2, 2)
   \]

2. **Case 2: All primes are distinct**

   Assume \( p, q, r \) are distinct primes. Without loss of generality, let \( r \) be the largest prime. We have:
   \[
   r \mid p^2 - 4 \implies r \mid (p-2)(p+2)
   \]
   Since \( r \) is a prime and larger than \( p-2 \), it must divide \( p+2 \). Therefore:
   \[
   r \mid p+2 \implies p+2 = kr \text{ for some integer } k
   \]
   Since \( p \) and \( r \) are distinct primes, \( k \) must be 1. Thus:
   \[
   p + 2 = r \implies p = r - 2
   \]

3. **Substitute \( p = r - 2 \) into the conditions**

   We also have:
   \[
   p \mid q^2 - 4 \implies r-2 \mid q^2 - 4
   \]
   Since \( r-2 \) is a prime, it must divide \( q+2 \) or \( q-2 \). We also have:
   \[
   q \mid r^2 - 4 \implies q \mid (r-2)(r+2)
   \]
   Since \( q \) is a prime and distinct from \( r \), it must divide \( r+2 \). Therefore:
   \[
   q \mid r+2 \implies r+2 = mq \text{ for some integer } m
   \]
   Since \( q \) and \( r \) are distinct primes, \( m \) must be 1. Thus:
   \[
   r + 2 = q \implies q = r - 4
   \]

4. **Check the distinctness and primality**

   We now have:
   \[
   p = r - 2, \quad q = r - 4
   \]
   Since \( p, q, r \) are distinct primes, we need to check the values of \( r \) such that \( r-2 \) and \( r-4 \) are also primes. The smallest prime \( r \) that satisfies this is 7:
   \[
   r = 7, \quad p = 7 - 2 = 5, \quad q = 7 - 4 = 3
   \]
   This gives us the solution:
   \[
   (5, 3, 7)
   \]

5. **Check permutations**

   Since the problem does not specify the order of the primes, we also have the permutations:
   \[
   (7, 5, 3), \quad (3, 7, 5)
   \]

Summing up, all the solutions are:
\[
\boxed{(2, 2, 2), (5, 3, 7), (7, 5, 3), (3, 7, 5)}
\]