Any solutions is either in the following set, or is some permutation of an element in this set:

It is a little tricky to prove that these are the only solutions, though.

Here is a sketch of a proof that if

, where

, then

:

Suppose

and

. Then suppose

divides any two of the numbers. Without loss of generality, assume

and

. Then since

, we know

. But then since

divides both

, we have some integer

such that

. Then

implies

. Hence, if a number divides two of the numbers, it must also divide the third.

Hence, any solution where

are all positive must be a multiple of some solution

where all three are pairwise coprime. But then

. Since

, it must be that

. Since these numbers are pairwise coprime, we can only have equality if

or

. Since

, it cannot be that

. Hence,

, so

. Then

. Since

, it must be

, so

. Then

implies

. Hence, the only solution with all three numbers positive and coprime is

, so the set of all solutions with all three numbers positive is

. Since

is a solution, we can include

.