2019 USAMO Problems/Problem 1
Problem
Let be the set of positive integers. A function satisfies the equation for all positive integers . Given this information, determine all possible values of .
Solution
Let denote the result when is applied to , where . If , then and
since .
Therefore, is injective. It follows that is also injective.
Lemma 1: If and , then .
Proof:
which implies by injectivity of .
Lemma 2: If , and is odd, then .
Proof:
Let . Since , . So, . .
Since ,
This proves Lemma 2.
I claim that for all odd .
Otherwise, let be the least counterexample.
Since , either
, contradicted by Lemma 1 since is odd and .
, also contradicted by Lemma 1 by similar logic.
and , which implies that by Lemma 2. This proves the claim.
By injectivity, is not odd.
I will prove that can be any even number, . Let , and for all other . If is equal to neither nor , then . This satisfies the given property.
If is equal to or , then since is even and . This satisfies the given property.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |