Subscribe to the weekly news from TrueShelf
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: .
From the second equation, is even.
From the third equation, if , then ; if , then is odd, which implies that .
Hence (so ), , and .
Hence is 2 or 4, and equals or .
Thus the solutions for are , or .
Case 2: .
Since , we have .
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.