2019 USAJMO Problems/Problem 2
Problem
Let be the set of all integers. Find all pairs of integers
for which there exist functions
and
satisfying
for all integers
.
Solution
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 for
let
and
, so
does work and are the only solutions, as desired.
-Stormersyle
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |