Suppose that three prime numbers $p,q,$ and $r$ satisfy the equations $pq + qr + rp = 191$ and $p + q = r - 1$. Find $p + q + r$.

[i]Proposed by Andrew Wu[/i]
1. We start with the given equations:
   \[
   pq + qr + rp = 191
   \]
   \[
   p + q = r - 1
   \]

2. We will assume without loss of generality (WLOG) that \( p \leq q \leq r \). We will test small prime values for \( p \) to find a solution.

3. **Case \( p = 2 \)**:
   \[
   p = 2 \implies q + 2 = r - 1 \implies r = q + 3
   \]
   Substituting \( r = q + 3 \) into the first equation:
   \[
   2q + q(q + 3) + 2(q + 3) = 191
   \]
   Simplifying:
   \[
   2q + q^2 + 3q + 2q + 6 = 191
   \]
   \[
   q^2 + 7q + 6 = 191
   \]
   \[
   q^2 + 7q - 185 = 0
   \]
   Solving this quadratic equation using the quadratic formula \( q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   q = \frac{-7 \pm \sqrt{49 + 740}}{2} = \frac{-7 \pm \sqrt{789}}{2}
   \]
   Since \(\sqrt{789}\) is not an integer, \( q \) is not a prime number. Thus, \( p = 2 \) does not work.

4. **Case \( p = 3 \)**:
   \[
   p = 3 \implies q + 3 = r - 1 \implies r = q + 4
   \]
   Substituting \( r = q + 4 \) into the first equation:
   \[
   3q + q(q + 4) + 3(q + 4) = 191
   \]
   Simplifying:
   \[
   3q + q^2 + 4q + 3q + 12 = 191
   \]
   \[
   q^2 + 10q + 12 = 191
   \]
   \[
   q^2 + 10q - 179 = 0
   \]
   Solving this quadratic equation:
   \[
   q = \frac{-10 \pm \sqrt{100 + 716}}{2} = \frac{-10 \pm \sqrt{816}}{2}
   \]
   Since \(\sqrt{816}\) is not an integer, \( q \) is not a prime number. Thus, \( p = 3 \) does not work.

5. **Case \( p = 5 \)**:
   \[
   p = 5 \implies q + 5 = r - 1 \implies r = q + 6
   \]
   Substituting \( r = q + 6 \) into the first equation:
   \[
   5q + q(q + 6) + 5(q + 6) = 191
   \]
   Simplifying:
   \[
   5q + q^2 + 6q + 5q + 30 = 191
   \]
   \[
   q^2 + 16q + 30 = 191
   \]
   \[
   q^2 + 16q - 161 = 0
   \]
   Solving this quadratic equation:
   \[
   q = \frac{-16 \pm \sqrt{256 + 644}}{2} = \frac{-16 \pm \sqrt{900}}{2}
   \]
   \[
   q = \frac{-16 \pm 30}{2}
   \]
   \[
   q = 7 \quad \text{or} \quad q = -23
   \]
   Since \( q \) must be a prime number, we take \( q = 7 \).

6. Substituting \( q = 7 \) back into the equation for \( r \):
   \[
   r = q + 6 = 7 + 6 = 13
   \]

7. Verify the solution:
   \[
   p = 5, \quad q = 7, \quad r = 13
   \]
   \[
   pq + qr + rp = 5 \cdot 7 + 7 \cdot 13 + 5 \cdot 13 = 35 + 91 + 65 = 191
   \]
   \[
   p + q = 5 + 7 = 12 = r - 1 = 13 - 1
   \]

8. Therefore, the sum \( p + q + r \) is:
   \[
   p + q + r = 5 + 7 + 13 = 25
   \]

The final answer is \(\boxed{25}\)