Homer gives mathematicians Patty and Selma each a different integer, not known to
the other or to you. Homer tells them, within each other’s hearing, that the number given to
Patty is the product $ab$ of the positive integers $a$ and $b$, and that the number given to Selma is the sum $a + b$ of the same numbers $a$ and $b$, where $b > a > 1.$ He doesn’t, however, tell Patty or Selma the numbers $a$ and $b.$ The following (honest) conversation then takes place:
Patty: “I can’t tell what numbers $a$ and $b$ are.”
Selma: “I knew before that you couldn’t tell.”
Patty: “In that case, I now know what $a$ and $b$ are.”
Selma: “Now I also know what $a$ and $b$ are.”
Supposing that Homer tells you (but neither Patty nor Selma) that neither $a$ nor $b$ is greater than 20, find $a$ and $b$, and prove your answer can result in the conversation above.
1. **Understanding the problem:**
   - Patty knows the product \( ab \).
   - Selma knows the sum \( a + b \).
   - Both \( a \) and \( b \) are integers such that \( b > a > 1 \) and \( a, b \leq 20 \).

2. **Analyzing Patty's statement:**
   - Patty says she cannot determine \( a \) and \( b \) from \( ab \). This implies that the product \( ab \) can be factored into more than one pair of integers \( (a, b) \) where \( b > a > 1 \).

3. **Analyzing Selma's statement:**
   - Selma knew that Patty couldn't determine \( a \) and \( b \). This implies that for the sum \( a + b \), every possible pair \( (a, b) \) that sums to \( a + b \) results in a product \( ab \) that can be factored into more than one pair of integers.

4. **Patty's realization:**
   - After hearing Selma, Patty now knows \( a \) and \( b \). This means that there is only one pair \( (a, b) \) that fits the conditions given the additional information from Selma.

5. **Selma's realization:**
   - After hearing Patty, Selma also knows \( a \) and \( b \). This confirms that the pair \( (a, b) \) is unique given the conditions.

6. **Finding possible sums \( a + b \):**
   - We need to find sums \( y = a + b \) such that for every possible pair \( (a, b) \) with \( a + b = y \), the product \( ab \) can be factored into more than one pair of integers.

7. **Listing possible sums and checking products:**
   - We list possible sums \( y \) and check the pairs \( (a, b) \) that sum to \( y \) and their corresponding products \( ab \).

   For \( y = 11 \):
   - Possible pairs: \( (2, 9), (3, 8), (4, 7), (5, 6) \)
   - Products: \( 18, 24, 28, 30 \)

   For \( y = 17 \):
   - Possible pairs: \( (2, 15), (3, 14), (4, 13), (5, 12), (6, 11), (7, 10), (8, 9) \)
   - Products: \( 30, 42, 52, 60, 66, 70, 72 \)

   For \( y = 23 \):
   - Possible pairs: \( (3, 20), (4, 19), (5, 18), (6, 17), (7, 16), (8, 15), (9, 14), (10, 13), (11, 12) \)
   - Products: \( 60, 76, 90, 102, 112, 120, 126, 130, 132 \)

   For \( y = 27 \):
   - Possible pairs: \( (4, 23), (5, 22), (6, 21), (7, 20), (8, 19), (9, 18), (10, 17), (11, 16), (12, 15), (13, 14) \)
   - Products: \( 92, 110, 126, 140, 152, 162, 170, 176, 180, 182 \)

   For \( y = 29 \):
   - Possible pairs: \( (5, 24), (6, 23), (7, 22), (8, 21), (9, 20), (10, 19), (11, 18), (12, 17), (13, 16), (14, 15) \)
   - Products: \( 120, 138, 154, 168, 180, 190, 198, 204, 208, 210 \)

   For \( y = 35 \):
   - Possible pairs: \( (6, 29), (7, 28), (8, 27), (9, 26), (10, 25), (11, 24), (12, 23), (13, 22), (14, 21), (15, 20), (16, 19), (17, 18) \)
   - Products: \( 174, 196, 216, 234, 250, 264, 276, 286, 294, 300, 304, 306 \)

   For \( y = 37 \):
   - Possible pairs: \( (7, 30), (8, 29), (9, 28), (10, 27), (11, 26), (12, 25), (13, 24), (14, 23), (15, 22), (16, 21), (17, 20), (18, 19) \)
   - Products: \( 210, 232, 252, 270, 286, 300, 312, 322, 330, 336, 340, 342 \)

8. **Identifying the unique product:**
   - From the above lists, the only product that appears only once is \( 52 \) from the sum \( 17 \).

9. **Determining \( a \) and \( b \):**
   - The pair that gives the product \( 52 \) and the sum \( 17 \) is \( (4, 13) \).

The final answer is \( \boxed{(4, 13)} \)