Difference between revisions of "Talk:2007 AIME II Problems/Problem 14"

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Here is a completed solution to 2007AIMEII-14.
 
Here is a completed solution to 2007AIMEII-14.
Let <math>f\left( x \right) = \sum\limits_{i = 0}^n {a_i x^i }</math>.<math>\[f\left( 0 \right) = 1 \Rightarrow a_0  = 1
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Let <math>f\left( x \right) = \sum\limits_{i = 0}^n {a_i x^i }</math>. <math>[f\left( 0 \right) = 1 \Rightarrow a_0  = 1
\]</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).
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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).
 
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>.
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
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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>\[f_1 \left( x \right) \equiv 1]</math> or <math>\left. {\left( {x^2  + 1} \right)} \right|f_1 \left( x \right)</math>.
\]</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}$.
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 23:52, 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]$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.) 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$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)\[deg f_n \left( x \right) = 0{\text{ or }}1$.Then $\[f_n \left( x \right) = 1{\rm{ or }}x + 1]$.Check if$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)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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