2007 AIME II Problems/Problem 14: Difference between revisions

Revision as of 12:05, 18 August 2010

Problem

Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ , $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$

Solution

Let $r$ be a root of $f(x)$ . Then we have $f(r)f(2r^2)=f(2r^3+r)$ ; since $r$ is a root, we have $f(r)=0$ ; therefore $2r^3+r$ is also a root. Thus, if $r$ is real and non-zero, $|2r^3+r|>r$ , so $f(x)$ has infinitely many roots. Since $f(x)$ is a polynomial (thus of finite degree) and $f(0)$ is nonzero, $f(x)$ has no real roots.

Note that $f(x)$ is not constant. We then find two complex roots: $r = \pm i$ . We find that $f(i)f(-2) = f(-i)$ , and that $f(-i)f(-2) = f(i)$ . This means that $f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2)^2 - 1) = 0$ . Thus*, $\pm i$ are roots of the polynomial, and so $(x - i)(x + i) = x^2 + 1$ will be a factor of the polynomial.

The polynomial is thus in the form of $f(x) = (x^2 + 1)g(x)$ . Substituting into the given expression, we have

$(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)$ $(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)$

Thus either $4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)$ is 0 for any $x$ , or $g(x)$ satisfies the same constraints as $f(x)$ . Continuing, by infinite descent, $f(x) = (x^2 + 1)^n$ for some $n$ .

Since $f(2)+f(3)=125=5^n+10^n$ for some $n$ , we have $n=2$ ; so $f(5) = 676$ .

This requires the assumption that $f(x)\neq1$ . Clearly, $f(x)\neq-1$ , because that would imply the existence of a real root.

@@ Line 5: / Line 5: @@
 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.
-Note that <math>f(x)</math> is not constant. 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>. Assuming that <math>f(-2)\neq1</math>, <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.
+Note that <math>f(x)</math> is not constant. 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
@@ Line 15: / Line 15: @@
 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>.
+*This requires the assumption that <math>f(x)\neq1</math>. Clearly, <math>f(x)\neq-1</math>, because that would imply the existence of a real root.
 == See also ==

2007 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2007 AIME II Problems/Problem 14: Difference between revisions

Revision as of 12:05, 18 August 2010

Problem

Solution

See also