2019 USAJMO Problems/Problem 2
Let be the set of all integers. Find all pairs of integers for which there exist functions and satisfying for all integers .
We claim that the answer is .
Proof: and are surjective because and can take on any integral value, and by evaluating the parentheses in different order, we find and . We see that if then to as well, so similarly if then , so now assume .
We see that if then , if then , if then ... if then . This means that the -element collection contains all residues mod since is surjective, so . Doing the same to yields that , so this means that only can work.
For let and , and for let and , so does work and are the only solutions, as desired.
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