Find all sets of integers A, B, and C which satisfy the following:

1/A + 1/B + 1/C = 1.

Let us make a starting constraint, that A, B, and C must be written in increasing order (A <= B <= C). Any solution can be organized this way, so it won't affect the solutions.

A=0: Invalid.

A=1: 1+1/B+1/C=1. 1/B + 1/C must be 0, so B and C = infinity. No valid solution.

A=2: By the starting restraint, B cannot be 1.
_____B=2: 1/2+1/2+1/C=1. 1/C=0. C = infinity. Nope
_____B=3: 1/2+1/3+1/C=1. 1/C=1/6. C = 6.
_____B=4: 1/2+1/4+1/C=1. 1/C=1/4. C = 4.
_____B>4: The largest value, by the starting restraint, is 1/2+1/5+1/5=9/10.
Thus, there are two valid answers that have a lowest value of 2: (2,3,6) and (2,4,4)

A=3: By the starting restraint, B cannot be 1 or 2.
_____B=3: 1/3+1/3+1/C=1. 1/C=1/3. C = 3.
_____B>3: The largest value, by the starting restraint, is 1/3+1/4+1/4=5/6.
Thus, there is one valid answer that has a lowest value of 3: (3,3,3)

A>3: The largest value, by the starting restraint, is 1/4+1/4+1/4 = 3/4. No other solutions.

However, integers are considered any of the natural numbers (positive, zero, or negative). So what happens when we allow negatives?
B=1: 1/A+1/1+1/C=1 => 1/A+1/C=0 => 1/C=-1/A => C=-1*A
So the solution is all members of the set (-X,1,X)

Now we add that infinite list to the set. I'm not sure how to prove out all solutions involving one negative number (it can be shown that none involve two negatives or three negatives, though).

So the answer to the question appears to be:
(2,3,6), (2,6,3), (3,2,6), (3,6,2), (6,2,3), (6,3,2), (2,4,4), (4,2,4), (4,4,2), (3,3,3), and all (-X,1,X) (and its permutations)

Posted by Ender on 2003-07-18 03:59:27