Difference between revisions of "2007 AIME II Problems/Problem 14"
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== Solution == | == Solution == | ||
− | + | 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>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. | |
− | The polynomial is | + | The polynomial is thus in the form of <math>f(x) = (x^2 + 1)g(x)</math>. Substituting into the given expression, we have |
+ | |||
+ | <cmath>(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)</cmath> | ||
+ | <cmath>(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)</cmath> | ||
+ | |||
+ | Thus either <math>4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)</math> is 0 for any <math>x</math>, or <math>g(x)</math> satisfies the same constraints as <math>f(x)</math>. Continuing, by infinite descent, <math>f(x) = (x^2 + 1)^n</math> for some <math>n</math>. | ||
+ | |||
+ | Since <math>f(2)+f(3)=125=5^n+10^n</math> for some <math>n</math>, we have <math>n=2</math>; so <math>f(5) = 676</math>. | ||
== See also == | == See also == |
Revision as of 13:46, 12 March 2009
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |