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## 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.