Find all the prime numbers $p, q$ and $r$ such that $p^2 + 1 = 74 (q^2 + r^2)$.
1. **Analyze the equation modulo 4:**
   \[
   p^2 + 1 = 74(q^2 + r^2)
   \]
   Since \( p \) is a prime number, \( p \) can be either 2 or an odd prime. Let's consider both cases:
   - If \( p = 2 \):
     \[
     2^2 + 1 = 4 + 1 = 5
     \]
     \[
     74(q^2 + r^2) = 5
     \]
     This is not possible since \( 74(q^2 + r^2) \) is always even, and 5 is odd.
   - If \( p \) is an odd prime:
     \[
     p^2 \equiv 1 \pmod{4}
     \]
     Therefore:
     \[
     p^2 + 1 \equiv 1 + 1 \equiv 2 \pmod{4}
     \]
     Now consider the right-hand side:
     \[
     74(q^2 + r^2) \equiv 2(q^2 + r^2) \pmod{4}
     \]
     Since \( q \) and \( r \) are primes, \( q^2 \) and \( r^2 \) are either 1 or 0 modulo 4. Thus:
     \[
     q^2 + r^2 \equiv 1 \pmod{2}
     \]
     This means one of \( q \) or \( r \) must be 2 (since \( 2^2 \equiv 0 \pmod{4} \) and any odd prime squared is 1 modulo 4).

2. **Assume without loss of generality \( q = 2 \):**
   \[
   p^2 + 1 = 74(2^2 + r^2)
   \]
   \[
   p^2 + 1 = 74(4 + r^2)
   \]
   \[
   p^2 + 1 = 296 + 74r^2
   \]
   \[
   p^2 = 295 + 74r^2
   \]

3. **Analyze the equation modulo 9:**
   - \( p^2 \) modulo 9 can be 0, 1, 4, or 7 (since \( p \) is a prime).
   - \( 295 \equiv 7 \pmod{9} \)
   - \( 74 \equiv 2 \pmod{9} \)
   - Therefore:
     \[
     p^2 \equiv 7 + 2r^2 \pmod{9}
     \]
   - Possible values for \( r^2 \) modulo 9 are 0, 1, 4, 7.
   - Check each case:
     - If \( r^2 \equiv 0 \pmod{9} \):
       \[
       p^2 \equiv 7 \pmod{9}
       \]
       Possible.
     - If \( r^2 \equiv 1 \pmod{9} \):
       \[
       p^2 \equiv 7 + 2 \equiv 9 \equiv 0 \pmod{9}
       \]
       Possible.
     - If \( r^2 \equiv 4 \pmod{9} \):
       \[
       p^2 \equiv 7 + 8 \equiv 15 \equiv 6 \pmod{9}
       \]
       Not possible.
     - If \( r^2 \equiv 7 \pmod{9} \):
       \[
       p^2 \equiv 7 + 14 \equiv 21 \equiv 3 \pmod{9}
       \]
       Not possible.

4. **Check the possible values:**
   - If \( r^2 \equiv 0 \pmod{9} \), then \( r \equiv 0 \pmod{3} \). Since \( r \) is a prime, \( r = 3 \).
   - If \( r^2 \equiv 1 \pmod{9} \), then \( r \equiv \pm 1 \pmod{3} \). Since \( r \) is a prime, \( r = 2 \) or \( r = 5 \).

5. **Verify the solutions:**
   - For \( r = 3 \):
     \[
     p^2 = 295 + 74 \cdot 9 = 961
     \]
     \[
     p = 31
     \]
     Check:
     \[
     31^2 + 1 = 961 + 1 = 962
     \]
     \[
     74(2^2 + 3^2) = 74(4 + 9) = 74 \cdot 13 = 962
     \]
     This is a valid solution.

   - For \( r = 2 \) or \( r = 5 \):
     \[
     p^2 = 295 + 74 \cdot 4 = 591
     \]
     \[
     p = \sqrt{591}
     \]
     Not an integer, so not a solution.

The final answer is \( \boxed{ (p, q, r) = (31, 2, 3) } \)