Determine all triples $(p, q, r)$ of prime numbers for which $p^q + p^r$ is a perfect square.
1. **Case 1: \( q = r \)**

   If \( q = r \), then the expression becomes:
   \[
   p^q + p^r = p^q + p^q = 2p^q
   \]
   For \( 2p^q \) to be a perfect square, \( p \) must be 2 (since 2 is the only even prime number). Therefore:
   \[
   2 \cdot 2^q = 2^{q+1}
   \]
   For \( 2^{q+1} \) to be a perfect square, \( q+1 \) must be even, implying \( q \) is odd. Thus, the solutions in this case are:
   \[
   (p, q, r) = (2, k, k) \quad \text{where } k \text{ is any odd prime}
   \]

2. **Case 2: \( q \neq r \)**

   Without loss of generality, assume \( q > r \). Then the expression becomes:
   \[
   p^q + p^r = p^r(p^{q-r} + 1)
   \]
   For \( p^r(p^{q-r} + 1) \) to be a perfect square, \( p^r \) must be a perfect square and \( p^{q-r} + 1 \) must also be a perfect square. Since \( p \) is a prime, \( p^r \) is a perfect square only if \( r \) is even. Let \( r = 2 \), then:
   \[
   p^2(p^{q-2} + 1) = a^2
   \]
   This implies:
   \[
   p^{q-2} + 1 = b^2
   \]
   for some integer \( b \). Therefore:
   \[
   p^{q-2} = b^2 - 1 = (b-1)(b+1)
   \]
   Since \( p \) is a prime, the factors \( b-1 \) and \( b+1 \) must be powers of \( p \). Let:
   \[
   b-1 = p^a \quad \text{and} \quad b+1 = p^b
   \]
   Then:
   \[
   p^b - p^a = 2
   \]
   The only solutions to this equation for prime \( p \) are:
   \[
   (p, a, b) = (3, 0, 1) \quad \text{or} \quad (3, 1, 2)
   \]
   Substituting back, we get:
   \[
   (p, q, r) = (3, 3, 2) \quad \text{or} \quad (3, 2, 3)
   \]

Combining both cases, the solutions are:
\[
(p, q, r) = (2, k, k) \quad \text{where } k \text{ is any odd prime}, \quad (3, 3, 2), \quad \text{and} \quad (3, 2, 3)
\]

The final answer is \( \boxed{ (2, k, k) \text{ where } k \text{ is any odd prime}, (3, 3, 2), (3, 2, 3) } \).