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> | + | 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). | |
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> | + | 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>. |
− | + | 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>. | |
− | Let <math>n</math> be an integer and <math> | ||
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Latest revision as of 23:54, 15 August 2021
Here is a completed solution to 2007AIMEII-14.
Let .
.
.
or
(impossible).
Let
.
Then
and the same thing got:
or
.
Let
be an integer and
such that
.Then
.Check if
and we can easily get
and
and
.