Difference between revisions of "2007 AIME II Problems/Problem 14"
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Let <math>r</math> be a root of <math>f(x)</math>. Then we have <math>f(r)f(2r^2)=f(2r^3+r)</math>; since <math>r</math> is a root, we have <math>f(r)=0</math>; therefore <math>2r^3+r</math> is also a root. Thus, if <math>r</math> is real and non-zero, <math>|2r^3+r|>r</math>, so <math>f(x)</math> has infinitely many roots. Since <math>f(x)</math> is a polynomial (thus of finite degree) and <math>f(0)</math> is nonzero, <math>f(x)</math> has no real roots. | Let <math>r</math> be a root of <math>f(x)</math>. Then we have <math>f(r)f(2r^2)=f(2r^3+r)</math>; since <math>r</math> is a root, we have <math>f(r)=0</math>; therefore <math>2r^3+r</math> is also a root. Thus, if <math>r</math> is real and non-zero, <math>|2r^3+r|>r</math>, so <math>f(x)</math> has infinitely many roots. Since <math>f(x)</math> is a polynomial (thus of finite degree) and <math>f(0)</math> is nonzero, <math>f(x)</math> has no real roots. | ||
− | We then find two complex roots: <math>r = \pm i</math>. We find that <math>f(i)f(-2) = f(-i)</math>, and that <math>f(-i)f(-2) = f(i)</math>. This means that <math>f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2) - 1) = 0</math>. Thus <math>\pm i</math> are roots of the polynomial, and so <math>(x - i)(x + i) = x^2 + 1</math> will be a factor of the polynomial. | + | We then find two complex roots: <math>r = \pm i</math>. We find that <math>f(i)f(-2) = f(-i)</math>, and that <math>f(-i)f(-2) = f(i)</math>. This means that <math>f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2)^2 - 1) = 0</math>. Thus <math>\pm i</math> are roots of the polynomial, and so <math>(x - i)(x + i) = x^2 + 1</math> will be a factor of the polynomial. |
The polynomial is thus in the form of <math>f(x) = (x^2 + 1)g(x)</math>. Substituting into the given expression, we have | The polynomial is thus in the form of <math>f(x) = (x^2 + 1)g(x)</math>. Substituting into the given expression, we have |
Revision as of 12:25, 18 August 2010
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.
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 |