Find all triples of prime numbers $(p, q, r)$ such that $p^q + p^r$ is a perfect square.
To find all triples of prime numbers \((p, q, r)\) such that \(p^q + p^r\) is a perfect square, we start by assuming \(q \geq r\). This gives us the equation:

\[ p^q + p^r = p^r (p^{q-r} + 1) = x^2 \]

1. **Assume \(r = 2\)**:
   \[ p^q + p^2 = x^2 \]
   Dividing both sides by \(p^2\):
   \[ p^{q-2} + 1 = y^2 \]
   where \(y = \frac{x}{p}\). This implies:
   \[ p^{q-2} = (y-1)(y+1) \]

2. **Case 1: \(y - 1 = 1\)**:
   \[ y = 2 \]
   \[ p^{q-2} = 2 + 1 = 3 \]
   This implies \(p = 3\) and \(q - 2 = 1\), so \(q = 3\). Therefore, one solution is:
   \[ (p, q, r) = (3, 3, 2) \]

3. **Case 2: \(y - 1 = p^a\) and \(y + 1 = p^b\)**:
   \[ p^b - p^a = 2 \]
   \[ a + b = q - 2 \]
   Solving \(p^b - p^a = 2\) for prime \(p\):
   - If \(p = 2\), then:
     \[ 2^b - 2^a = 2 \]
     \[ 2^a (2^{b-a} - 1) = 2 \]
     This implies \(a = 1\) and \(b = 2\), so:
     \[ q - 2 = a + b = 1 + 2 = 3 \]
     \[ q = 5 \]
     Therefore, another solution is:
     \[ (p, q, r) = (2, 5, 2) \]

4. **General Case: \(q = r = 2n + 1\)**:
   If \(q = r = 2n + 1\), then:
   \[ p^q + p^r = p^{2n+1} + p^{2n+1} = 2p^{2n+1} \]
   For \(2p^{2n+1}\) to be a perfect square, \(p\) must be 2:
   \[ 2 \cdot 2^{2n+1} = 2^{2n+2} \]
   which is indeed a perfect square. Therefore, another family of solutions is:
   \[ (p, q, r) = (2, 2n+1, 2n+1) \]

Combining all the solutions, we have:
\[ (p, q, r) \in \{ (2, 5, 2), (2, 2, 5), (2, 3, 3), (3, 3, 2), (2, 2n+1, 2n+1) \} \]

The final answer is \( \boxed{ (p, q, r) \in \{ (2, 5, 2), (2, 2, 5), (2, 3, 3), (3, 3, 2), (2, 2n+1, 2n+1) \} } \)