1997 USAMO Problems/Problem 3
Problem
Prove that for any integer , there exists a unique polynomial with coefficients in such that .
Solution
The problem implies that . We define . Then, after expanding, the coefficient of is where . The constant term is . Since this has to be within the set , must be unique according to . The linear term is determined by , which also must be within the set . Since is determined uniquely by , in order to keep the linear coefficient in the set , is also determined uniquely. With and determined uniquely, we move on to the general coefficient.
Note that is in the set . We define to be a function determining the remainder when is divided by . Thus, for all , . Note also that , since is in the range of . Therefore, , which results in for all . Thus, is determined uniquely, since all coefficients where are uniquely determined, which in turn determines a unique polynomial .
See Also
1997 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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