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2019 USAMO Problems/Problem 1

Revision as of 20:45, 24 April 2019 by Blizzardwizard (talk | contribs) (→‎Solution)

Problem

Let $\mathbb{N}$ be the set of positive integers. A function $f:\mathbb{N}\to\mathbb{N}$ satisfies the equation \[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]for all positive integers $n$. Given this information, determine all possible values of $f(1000)$.

Solution

\usepackage{amsmath}

$\newline$ Let $f^r(x)$ denote the resulr when $f$ is applied to $x$ $r$ times. $\newline\newline$ If $f(p)=f(q)$, then $f^2(p)=f^2(q)$ and $f^{f(p)}(p)=f^{f(q)}(q)\newline\implies p^2=f^2(p)\cdot f^{f(p)}(p)=f^2(q)\cdot f^{f(q)}(q)=q^2\newline\implies p=\pm q\newline\implies p=q$ since $p,q>0$. \newline Therefore, $f$ is injective. $\newline\newline$ Lemma 1: If $f^r(b)=a$ and $f(a)=a$, then b=a. $\newline\newline$ Proof: $\newline$ Otherwise, set $a$, $b$, and $r$ to a counterexample of the lemma, such that $r$ is minimized. By injectivity, $f(a)=a\implies f(b)\neq a$, so $r\neq1$. If $f^n(b)=a$, then $f^n-1(f(b))=a$ and $f(b)\neq a$, a counterexample that contradicts our assumption that $r$ is minimized, proving Lemma 1. $\newline\newline$ Lemma 2: If $f^2(m)=f^{f(m)}(m)=m$, and $m$ is odd, then $f(m)=m$. $\newline\newline$ Proof: $\newline$ Let $f(m)=k$. Since $f^2(m)=m$, $f(k)=m$. So, $f^2(k)=k$. $\newline f^2(k)\cdot f^{f(k)}(k)=k^2$.$\newline$ Since $k\neq0$, $\newlinef^{f(k)}(k)=k$ (Error compiling LaTeX. Unknown error_msg) $\newline$ $f^m(k)=k$ $\newline$ $f^{gcd(m, 2)}=k$ $\newline$ $m=f(k)=k$ $\newline$ This proves Lemma 2. $\newline\newline$ I claim that $f(m)=m$ for all odd $m$. $\newline$ Otherwise, let $m$ be the least counterexample. $\newline$ Since $f^2(m)\cdot f^{f(m)}(m)$, either $\newline$ (1) $f^2(m)=k<m$, contradicted by Lemma 1 since $f(k)=k$. $\newline$ (2) $f^{f(m)}(m)=k<m$, also contradicted by Lemma 1. $\newline$ (3) $f^2(m)=m$ and $f^{f(m)}(m)=m$, which implies that $f(m)=m$ by Lemma 2. This proves the claim. $\newline\newline$ By injectivity, $f(1000)$ is not odd. I will prove that $f(1000)$ can be any even number, $x$. Let $f(1000)=x, f(x)=1000$, and $f(k)=k$ for all other $k$. If $n$ is equal to neither $1000$ nor $x$, then $f^2(n)\cdot f^{f(n)}(n)=n\cdot n=n^2$. This satisfies the given property. $\newline$ If $n$ is equal to $1000$ or $x$, then $f^2(n)\cdot f^{f(n)}(n)=n\cdot n=n^2$ since $f(n)$ is even and $f^2(n)=n$. This satisfies the given property. $\newline$

$\newline$ Let $f^r(x)$ denote the resulr when $f$ is applied to $x$ $r$ times. $\newline\newline$ If $f(p)=f(q)$, then $f^2(p)=f^2(q)$ and $f^{f(p)}(p)=f^{f(q)}(q)\newline\implies p^2=f^2(p)\cdot f^{f(p)}(p)=f^2(q)\cdot f^{f(q)}(q)=q^2\newline\implies p=\pm q\newline\implies p=q$since $p,q>0$. \newline Therefore, $f$ is injective. $\newline\newline$ Lemma 1: If $f^r(b)=a$ and $f(a)=a$, then b=a. $\newline\newline$ Proof: $\newline$ Otherwise, set $a$, $b$, and $r$ to a counterexample of the lemma, such that $r$ is minimized. By injectivity, $f(a)=a\implies f(b)\neq a$, so $r\neq1$. If $f^n(b)=a$, then $f^n-1(f(b))=a$ and $f(b)\neq a$, a counterexample that contradicts our assumption that $r$ is minimized, proving Lemma 1. $\newline\newline$ Lemma 2: If $f^2(m)=f^{f(m)}(m)=m$, and $m$ is odd, then $f(m)=m$. $\newline\newline$ Proof: $\newline$ Let $f(m)=k$. Since $f^2(m)=m$, $f(k)=m$. So, $f^2(k)=k$. $\newline f^2(k)\cdot f^{f(k)}(k)=k^2$. $\newline$ Since $k\neq0$, $\newlinef^{f(k)}(k)=k$ (Error compiling LaTeX. Unknown error_msg) $\newline$ $f^m(k)=k$ $\newline$ $f^{gcd(m, 2)}=k$ $\newline$ $m=f(k)=k$ $\newline$ This proves Lemma 2. $\newline\newline$ I claim that $f(m)=m$ for all odd $m$. $\newline$ Otherwise, let $m$ be the least counterexample. $\newline$ Since $f^2(m)\cdot f^{f(m)}(m)$, either $\newline$ (1) $f^2(m)=k<m$, contradicted by Lemma 1 since $f(k)=k$. $\newline$ (2) $f^{f(m)}(m)=k<m$, also contradicted by Lemma 1. $\newline$ (3) $f^2(m)=m$ and $f^{f(m)}(m)=m$, which implies that $f(m)=m$ by Lemma 2. This proves the claim. $\newline\newline$ By injectivity, $f(1000)$ is not odd. I will prove that $f(1000)$ can be any even number, $x$. Let $f(1000)=x, f(x)=1000$, and $f(k)=k$ for all other $k$. If $n$ is equal to neither $1000$ nor $x$, then $f^2(n)\cdot f^{f(n)}(n)=n\cdot n=n^2$. This satisfies the given property. $\newline$ If $n$ is equal to $1000$ or $x$, then $f^2(n)\cdot f^{f(n)}(n)=n\cdot n=n^2$ since $f(n)$ is even and $f^2(n)=n$. This satisfies the given property. $\newline$


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See also

2019 USAMO (Problems • Resources)
First Problem Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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