2015 IMO Problems/Problem 2
Determine all triples of positive integers such that each of the numbers is a power of 2.
(A power of 2 is an integer of the form where is a non-negative integer ).
The solutions for are , , , , and permutations of these triples.
Throughout the proof, we assume , so 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. We consider the remaining cases as follows.
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 .
|2015 IMO (Problems) • Resources)|
|1 • 2 • 3 • 4 • 5 • 6||Followed by|
|All IMO Problems and Solutions|