Find positive integer solutions to the equation 1/a + 1/b + 1/c + 1/d = 1 (a<=b<=c<=d). How many solutions exist?
Source: Alan O'Donnell.
Susan Hoover's solution aptly shows there are exactly 14 solutions:
a != 0. If a = 1, then 1/b = 1/c = 1/d = 0, which cannot be true for positive finite integers, so a > 1. If a >= 5, then b, c, and d are all >= 5, and the sum of (1/a + 1/b + 1/c + 1/d) is at most 4/5 < 1, so a < 5. Therefore a = 2, 3, or 4. Case A2. Let a = 2. 2 <= b <= c <= d, and 1/b + 1/c + 1/d = 1/2. If b = 2, then 1/c = 1/d = 0, so b != 2. If b > 6, then c and d are both >= 7, and the sum of (1/b + 1/c + 1/d) is at most 3/7 < 1/2, so b <= 6. Therefore, b = 3, 4, 5, or 6. Case A2B3. Let b = 3. 3 <= c <= d, and 1/c + 1/d = 1/6. Using reasoning similar to the above, we have 6 < c <= 12. Therefore, c = 7, 8, 9, 10, 11, or 12. Rearranging the reciprocal sum and solving for d, we have d = 6c / (c - 6). Plugging in the possibilities for c, we find the following solutions (none for c = 11): c = 7, d = 42 c = 8, d = 24 c = 9, d = 18 c = 10, d = 15 c = 12, d = 12 Case A2B4. Let b = 4. 4 <= c <= d, and 1/c + 1/d = 1/4. Reasoning as above, 4 < c <= 8. Therefore, c = 5, 6, 7, or 8. d = 4c / (c - 4). Solutions are (none for c = 7): c = 5, d = 20 c = 6, d = 12 c = 8, d = 8 Case A2B5. Let b = 5. 5 <= c <= d, and 1/c + 1/d = 3/10. Reasoning as above, 5 <= c < 7. Therefore, c = 5 or 6. d = 10c / (3c - 10). Solutions are (none for c = 6): c = 5, d = 10. Case A2B6. Let b = 6. 6 <= c <= d, and 1/c + 1/d = 1/3. Reasoning as above, 6 <= c <= 6. Therefore, c = 6. d = 3c / (c - 3). Solutions are: c = 6, d = 6 Case A3. Let a = 3. 3 <= b <= c <= d, and 1/b + 1/c + 1/d = 2/3. Reasoning as above, 3 <= b < 5. Therefore, b = 3 or 4. Case A3B3. Let b = 3. 3 <= c <= d, and 1/c + 1/d = 1/3. Reasoning as above, 4 <= c <= 6. Therefore, c = 4, 5, or 6. d = 3c / (c - 3). Solutions are (none for c = 5): c = 4, d = 12 c = 6, d = 6 Case A3B4. Let b = 4. 4 <= c <= d, and 1/c + 1/d = 5/12. Reasoning as above, 4 <= c < 5. Therefore, c = 4. d = 12c / (5c - 12). Solutions are: c = 4, d = 6 Case A4. Let a = 4. 4 <= b <= c <= d, and 1/b + 1/c + 1/d = 3/4. Reasoning as above, 4 <= b < 5. Therefore, b = 4. Case A4B4. Let b = 4. Solutions are: c = 4, d = 4 Summary of solutions: a = 2, b = 3, c = 7, d = 42 a = 2, b = 3, c = 8, d = 24 a = 2, b = 3, c = 9, d = 18 a = 2, b = 3, c = 10, d = 15 a = 2, b = 3, c = 12, d = 12 a = 2, b = 4, c = 5, d = 20 a = 2, b = 4, c = 6, d = 12 a = 2, b = 4, c = 8, d = 8 a = 2, b = 5, c = 5, d = 10 a = 2, b = 6, c = 6, d = 6 a = 3, b = 3, c = 4, d = 12 a = 3, b = 3, c = 6, d = 6 a = 3, b = 4, c = 4, d = 6 a = 4, b = 4, c = 4, d = 4