2002 USAMO Problems/Problem 4
Problem
Let be the set of real numbers. Determine all functions such that
for all pairs of real numbers and .
Solutions
Solution 1
We first prove that is odd.
Note that , and for nonzero , , or , which implies . Therefore is odd. Henceforth, we shall assume that all variables are non-negative.
If we let , then we obtain . Therefore the problem's condition becomes
.
But for any , we may set , to obtain
.
(It is well known that the only continuous solutions to this functional equation are of the form , but there do exist other solutions to this which are not solutions to the equation of this problem.)
We may let , to obtain .
Letting and in the original condition yields
But we know , so we have , or
.
Hence all solutions to our equation are of the form . It is easy to see that real value of will suffice.
Solution 2
As in the first solution, we obtain the result that satisfies the condition
.
We note that
.
Since , this is equal to
It follows that must be of the form .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
2002 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.