 International Mathematical Olympiad 2015 Problem 2

Determine all triples of positive integers such that each of the numbers is a power of 2.

We may assume that , such that   , with .

Note that since otherwise , which is impossible.

Hence , i.e., and are positive.

Observe that if , we get , so and are (even and) powers of .

Hence is odd and .

Hence is also a power of , which implies .

But is not a solution; hence is infeasible.

Now let's consider the remaining cases.

Case 1: .

We have From the second equation, is even.

From the third equation, if , then ; if , then is odd, which implies that .

Hence (so ), , and .

Hence .

Hence is 2 or 4, and equals or .

Thus the solutions for are  or .

Case 2: .

Since , we have .

Hence  Hence is not divisible by , and is not divisible by for .

Adding and subtracting and , we get  From the latter equation, is divisible by .

Hence is not divisible by , which implies that is a multiple of .

Hence and Consider , which implies   .

Hence , or .

Hence   and .

Finally, consider   .

Hence . But implies and implies .

Hence there are no solutions with .

We obtain as the only solution with .

The solutions for are    , and all permutations of these triples.