2007 AIME II Problems/Problem 14
Contents
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. (Note: This requires the assumption that
. Clearly,
, because that would imply the existence of a real root.)
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
.
Comment:
The answer is clearly correct, but the proof has a gap, i.e. there is no reason that . Since
has no real roots, the degree must be even. Consider
. Then since
is non-zero,
. Now the function
applied repeatedly from some real starting value of x becomes arbitrarily large, and the limit of
as
approaches infinity is 1, so
=1 for all x, or
. Then
for some polynomial
, and
. Now suppose h has degree m. It is clearly monic. Assume that the next highest non-zero coefficient in h is k. Then, subtracting
from both sides of the equation yields a polynomial equality with degree
on the left and degree
on the right, a contradiction. So
, and
.
Solution 2
Let be
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 | ||
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