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,
, so
has infinitely many roots. Since
is a polynomial (thus of finite degree) and
is nonzero,
has no real roots.
Note that is not constant. 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |