2007 AIME II Problems/Problem 14
Problem
Let be a polynomial with real coefficients such that and for all , Find
Solution
Let be a root of . Then we have ; since is a root, we have ; therefore is also a root. Thus, if is real and non-zero, has infinitely many roots. Since is a polynomial (thus of finite degree) and is nonzero, has no real roots.
We then find two complex roots: . We find that , and that . This means that . Thus are roots of the polynomial, and so will be a factor of the polynomial.
The polynomial is thus in the form of . Substituting into the given expression, we have
Thus either is 0 for any , or satisfies the same constraints as . Continuing, by infinite descent, for some .
Since for some , we have ; so .
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AIME Problems and Solutions |