2006 USAMO Problems/Problem 4
Contents
Problem
(Ricky Liu) Find all positive integers such that there are positive rational numbers satisfying .
Solutions
Solution 1
First, consider composite numbers. We can then factor into It is easy to see that , and thus, we can add 1s in order to achieve a sum and product of . For , which is only possible in one case, , we consider .
Secondly, let be a prime. Then we can find the following procedure: Let and let the rest of the be 1. The only numbers we now need to check are those such that . Thus, we need to check for . One is included because it is neither prime nor composite.
For , consider . Then by AM-GM, for . Thus, is impossible.
If , once again consider . Similar to the above, for since and . Obviously, is then impossible.
If , let . Again, . This is obvious for . Now consider . Then is obviously greater than . Thus, is impossible.
If , proceed as above and consider . Then and . However, we then come to the quadratic , which is not rational. For and we note that and . This is trivial to prove. If , it is obviously impossible, and thus does not work.
The last case, where , is possible using the following three numbers. shows that is possible.
Hence, can be any positive integer greater than with the exclusion of .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
- <url>viewtopic.php?t=84554 Discussion on AoPS/MathLinks</url>
2006 USAMO (Problems • Resources) | ||
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Followed by Problem 5 | |
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