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Difference between revisions of "Talk:2007 AIME II Problems/Problem 14"
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]</math>. <math>f\left( x \right)f\left( {2x^2 } \right) = f\left( {2x^3  + x} \right) \Rightarrow \ldots \Rightarrow a_n = 1</math>. <math>f\left( { \pm i} \right)f\left( 2 \right) = f\left( { \mp i} \right) \Rightarrow f\left( { \pm i} \right) = 0 \Rightarrow \left. {\left( {x^2  + 1} \right)} \right|f\left( x \right)</math> or <math>f\left( x \right) \equiv 1</math>(impossible).
 
]</math>. <math>f\left( x \right)f\left( {2x^2 } \right) = f\left( {2x^3  + x} \right) \Rightarrow \ldots \Rightarrow a_n = 1</math>. <math>f\left( { \pm i} \right)f\left( 2 \right) = f\left( { \mp i} \right) \Rightarrow f\left( { \pm i} \right) = 0 \Rightarrow \left. {\left( {x^2  + 1} \right)} \right|f\left( x \right)</math> or <math>f\left( x \right) \equiv 1</math>(impossible).
 
Let <math>f_1 \left( x \right) = \frac{{f\left( x \right)}}{{x^2  + 1}}</math>.
 
Let <math>f_1 \left( x \right) = \frac{{f\left( x \right)}}{{x^2  + 1}}</math>.
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Then <math>f_1 \left( x \right)f_1 \left( {2x^2 } \right) = f_1 \left( {2x^3  + x} \right)</math> and the same thing got:<math>\[f_1 \left( x \right) \equiv 1]</math> or <math>\left. {\left( {x^2 + 1} \right)} \right|f_1 \left( x \right)</math>.
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Then <math>f_1 \left( x \right)f_1 \left( {2x^2 } \right) = f_1 \left( {2x^3  + x} \right)</math> and the same thing got:<math>[f_1 \left( x \right) \equiv 1]</math> or <math>\left. {\left( {x^2 + 1} \right)} \right|f_1 \left( x \right)</math>.
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Let <math>n</math> be an integer and <math>\[f_n \left( x \right) = \frac{{f\left( x \right)}}{{\left( {x^2 + 1} \right)^n }}\$ such that </math>\[deg f_n \left( x \right) = 0{\text{ or }}1\$.Then <math>\[f_n \left( x \right) = 1{\rm{ or }}x + 1]\$.Check if </math>f\left( 2 \right) + f\left( 3 \right) = 125<math> and we can easily get </math>n = 2<math> and </math>f_n \left( x \right) = 1<math> and </math>f\left( 5 \right) = \boxed{625}$.
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Let <math>n</math> be an integer and <math>[f_n \left( x \right) = \frac{{f\left( x \right)}}{{\left( {x^2 + 1} \right)^n }}</math> such that <math>\[deg f_n \left( x \right) = 0{\text{ or }}1</math>.Then <math>[f_n \left( x \right) = 1{\rm{ or }}x + 1]</math>.Check if <math>f\left( 2 \right) + f\left( 3 \right) = 125</math> and we can easily get <math>n = 2</math> and <math>f_n \left( x \right) = 1</math> and <math>f\left( 5 \right) = \boxed{625}</math>.

Revision as of 22:53, 15 August 2021

Here is a completed solution to 2007AIMEII-14. Let $f\left( x \right) = \sum\limits_{i = 0}^n {a_i x^i }$. $[f\left( 0 \right) = 1 \Rightarrow a_0 = 1 ]$. $f\left( x \right)f\left( {2x^2 } \right) = f\left( {2x^3 + x} \right) \Rightarrow \ldots \Rightarrow a_n = 1$. $f\left( { \pm i} \right)f\left( 2 \right) = f\left( { \mp i} \right) \Rightarrow f\left( { \pm i} \right) = 0 \Rightarrow \left. {\left( {x^2 + 1} \right)} \right|f\left( x \right)$ or $f\left( x \right) \equiv 1$(impossible). Let $f_1 \left( x \right) = \frac{{f\left( x \right)}}{{x^2 + 1}}$. Then $f_1 \left( x \right)f_1 \left( {2x^2 } \right) = f_1 \left( {2x^3 + x} \right)$ and the same thing got:$[f_1 \left( x \right) \equiv 1]$ or $\left. {\left( {x^2 + 1} \right)} \right|f_1 \left( x \right)$. Let $n$ be an integer and $[f_n \left( x \right) = \frac{{f\left( x \right)}}{{\left( {x^2 + 1} \right)^n }}$ such that $\[deg f_n \left( x \right) = 0{\text{ or }}1$ (Error compiling LaTeX. Unknown error_msg).Then $[f_n \left( x \right) = 1{\rm{ or }}x + 1]$.Check if $f\left( 2 \right) + f\left( 3 \right) = 125$ and we can easily get $n = 2$ and $f_n \left( x \right) = 1$ and $f\left( 5 \right) = \boxed{625}$.

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