1992 USAMO Problems/Problem 5
Problem 5
Let be a polynomial with complex coefficients which is of degree and has distinct zeros. Prove that there exist complex numbers such that divides the polynomial .
Solution
Since the zeros of the polynomial are distinct, the polynomial divides the polynomial if and only if for every . So it is enough to show that for any complex numbers , there exist complex numbers , such that the polynomial satisfies for every . This can be generalized:
Lemma) Let be a positive integer, and be complex numbers. Then, there exist complex numbers such that the polynomial satisfies for every .
Proof: We use induction over .
For , the lemma is trivial, since , so we can take , and then the polynomial clearly satisfies .
Let be a positive integer. Assume that for , the lemma is true. For any complex numbers , there exist complex numbers such that the polynomial satisfies for every .
In fact, after our assumption, there exist complex numbers such that the polynomial satisfies for every . We want to construct our polynomial from this polynomial .
In fact, let for every , then let and . Then,
Hence, for , we have because and . Thus, holds for every , and this proves the lemma for . Hence, the induction step is complete, and the lemma is proven.
Resources
1992 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
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.