2006 Cyprus MO/Lyceum/Problem 5
Problem
If both integers are bigger than 1 and satisfy , then the minimum value of is
Solution
Since is greater than and therefore not equal to zero, we can divide both sides of the equation by to obtain , or Since is an integer, we must have is an integer. So, we can start testing out seventh powers of integers.
doesn't work, since and are defined to be greater than . The next smallest thing we try is .
This gives , so . Thus, our sum is , and the answer is .
See also
2006 Cyprus MO, Lyceum (Problems) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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