## Four Egyptian Fractions

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.

Solutions were received from Radu Ionescu, Joseph DeVincentis, Jayavel Sounderpandian, Al Zimmermann, Alan O'Donnell, Ravi Subramanian, Bernie Erickson, Henry Shin, Philippe Fondanaiche, and Susan Hoover.

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
```

Mail to Ken