1987 USAMO Problems/Problem 3
Problem
is the smallest set of polynomials such that:
- 1. belongs to .
- 2. If belongs to , then and both belong to .
Show that if and are distinct elements of , then for any .
Solution
Let be an arbitrary polynomial in Then when Define for some and for some
If and we have for all with Therefore for any
For any , Let and for If for then for
Similarly, for any , Let and for If for then for
The proof is done by an induction.
J.Z.
See Also
1987 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.