Difference between revisions of "2019 USAMO Problems/Problem 1"
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Let <math>f^r(x)</math> denote the resulr when <math>f</math> is applied to <math>x</math> <math>r</math> times. | Let <math>f^r(x)</math> denote the resulr when <math>f</math> is applied to <math>x</math> <math>r</math> times. | ||
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If <math>f(p)=f(q)</math>, then <math>f^2(p)=f^2(q)</math> and <math>f^{f(p)}(p)=f^{f(q)}(q)\newline\implies p^2=f^2(p)\cdot f^{f(p)}(p)=f^2(q)\cdot f^{f(q)}(q)=q^2\newline\implies p=\pm q\newline\implies p=q</math> since <math>p,q>0</math>. | If <math>f(p)=f(q)</math>, then <math>f^2(p)=f^2(q)</math> and <math>f^{f(p)}(p)=f^{f(q)}(q)\newline\implies p^2=f^2(p)\cdot f^{f(p)}(p)=f^2(q)\cdot f^{f(q)}(q)=q^2\newline\implies p=\pm q\newline\implies p=q</math> since <math>p,q>0</math>. | ||
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Revision as of 21:48, 24 April 2019
Problem
Let be the set of positive integers. A function satisfies the equation for all positive integers . Given this information, determine all possible values of .
Solution
Let denote the resulr when is applied to times. If , then and since . \newline Therefore, is injective. Lemma 1: If and , then b=a. Proof: Otherwise, set , , and to a counterexample of the lemma, such that is minimized. By injectivity, , so . If , then and , a counterexample that contradicts our assumption that is minimized, proving Lemma 1. Lemma 2: If , and is odd, then . Proof: Let . Since , . So, . . Since , $\newlinef^{f(k)}(k)=k$ (Error compiling LaTeX. Unknown error_msg) This proves Lemma 2. I claim that for all odd . Otherwise, let be the least counterexample. Since , either (1) , contradicted by Lemma 1 since . (2) , also contradicted by Lemma 1. (3) and , which implies that by Lemma 2. This proves the claim. By injectivity, is not odd. I will prove that can be any even number, . Let , and for all other . If is equal to neither nor , then . This satisfies the given property. If is equal to or , then since is even and . This satisfies the given property.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |