2017 IMO Problems/Problem 1
Problem
For each integer , define the sequence for as Determine all values of such that there exists a number such that for infinitely many values of .
Solution
First we observe the following:
When we start with , we get , , and the pattern repeats.
When we start with , we get , , and the pattern repeats.
When we start with , we get , , and the pattern repeats.
When we start with , we get , ,..., , , , and the pattern repeats.
When this pattern repeats, this means that there exists a number such that for infinitely many values of and that number is either or .
When we start with any number , we don't see a repeating pattern.
Therefore the claim is that where is a positive integer and we need to prove this claim.
When we start with , the next term if it is not a square is , then and so on until we get where is an integer and where is an integer. Then the next term will be and the pattern repeats again when or when or .
In order for these patterns to repeat, any square in the sequence need to be a multiple of 3.
To try the other two cases where , we can try then the next terms will be in the form .
When is a square, it will not be a multiple of because is not a multiple of and because and would have to be which is not an integer even if is a multiple of .
Therefore the pattern doesn't repeat for any of the other cases where and only repeats when
So, the answer to this problem is and or .
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2017 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |