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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.
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Let <math>f^r(x)</math> denote the result when <math>f</math> is applied to <math>f^{r-1}(x)</math>, where <math>f^1(x)=f(x)</math>.
 
<math>\hfill \break \hfill \break</math>
 
<math>\hfill \break \hfill \break</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)</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)</math>
Line 15: Line 15:
 
<math>\implies p=q</math> since <math>p,q>0</math>.
 
<math>\implies p=q</math> since <math>p,q>0</math>.
  
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Therefore, <math>f</math> is injective.
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Therefore, <math>f</math> is injective. It follows that <math>f^r</math> is also injective.
  
  
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Lemma 1: If <math>f^r(b)=a</math> and <math>f(a)=a</math>, then b=a.
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Lemma 1: If <math>f^r(b)=a</math> and <math>f(a)=a</math>, then <math>b=a</math>.
  
  
 
Proof:
 
Proof:
  
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Otherwise, set <math>a</math>, <math>b</math>, and <math>r</math> to a counterexample of the lemma, such that <math>r</math> is minimized. By injectivity, <math>f(a)=a\implies f(b)\neq a</math>, so <math>r\neq1</math>. If <math>f^n(b)=a</math>, then <math>f^n-1(f(b))=a</math> and <math>f(b)\neq a</math>, a counterexample that contradicts our assumption that <math>r</math> is minimized, proving Lemma 1.
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<math>f^r(b)=a=f^r(a)</math> which implies <math>b=a</math> by injectivity of <math>f^r</math>.
  
  
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Since <math>k\neq0</math>, <math>f^{f(k)}(k)=k</math>
 
Since <math>k\neq0</math>, <math>f^{f(k)}(k)=k</math>
  
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<math>f^m(k)=k</math>
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<math>\implies f^m(k)=k</math>
  
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<math>f^{gcd(m, 2)}=k</math>
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<math>\implies f^{gcd(m, 2)}(k)=k</math>
  
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<math>m=f(k)=k</math>
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<math>\implies f(k)=k</math>
  
 
This proves Lemma 2.
 
This proves Lemma 2.
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Otherwise, let <math>m</math> be the least counterexample.
 
Otherwise, let <math>m</math> be the least counterexample.
  
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Since <math>f^2(m)\cdot f^{f(m)}(m)</math>, either
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Since <math>f^2(m)\cdot f^{f(m)}(m)=m^2</math>, either
  
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<math>(1) f^2(m)=k<m</math>, contradicted by Lemma 1 since <math>f(k)=k</math>.
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<math>(1) f^2(m)=k<m</math>, contradicted by Lemma 1 since <math>k</math> is odd and <math>f^2(k)=k</math>.
  
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<math>(2) f^{f(m)}(m)=k<m</math>, also contradicted by Lemma 1.
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<math>(2) f^{f(m)}(m)=k<m</math>, also contradicted by Lemma 1 by similar logic.
  
 
<math>(3) f^2(m)=m</math> and <math>f^{f(m)}(m)=m</math>, which implies that <math>f(m)=m</math> by Lemma 2.
 
<math>(3) f^2(m)=m</math> and <math>f^{f(m)}(m)=m</math>, which implies that <math>f(m)=m</math> by Lemma 2.

Latest revision as of 17:04, 5 April 2021

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

Let $f^r(x)$ denote the result when $f$ is applied to $f^{r-1}(x)$, where $f^1(x)=f(x)$. $\hfill \break \hfill \break$ If $f(p)=f(q)$, then $f^2(p)=f^2(q)$ and $f^{f(p)}(p)=f^{f(q)}(q)$

$\implies p^2=f^2(p)\cdot f^{f(p)}(p)=f^2(q)\cdot f^{f(q)}(q)=q^2$

$\implies p=\pm q$

$\implies p=q$ since $p,q>0$.

Therefore, $f$ is injective. It follows that $f^r$ is also injective.


Lemma 1: If $f^r(b)=a$ and $f(a)=a$, then $b=a$.


Proof:

$f^r(b)=a=f^r(a)$ which implies $b=a$ by injectivity of $f^r$.


Lemma 2: If $f^2(m)=f^{f(m)}(m)=m$, and $m$ is odd, then $f(m)=m$.


Proof:

Let $f(m)=k$. Since $f^2(m)=m$, $f(k)=m$. So, $f^2(k)=k$. $\newline f^2(k)\cdot f^{f(k)}(k)=k^2$.

Since $k\neq0$, $f^{f(k)}(k)=k$

$\implies f^m(k)=k$

$\implies f^{gcd(m, 2)}(k)=k$

$\implies f(k)=k$

This proves Lemma 2.


I claim that $f(m)=m$ for all odd $m$.

Otherwise, let $m$ be the least counterexample.

Since $f^2(m)\cdot f^{f(m)}(m)=m^2$, either

$(1) f^2(m)=k<m$, contradicted by Lemma 1 since $k$ is odd and $f^2(k)=k$.

$(2) f^{f(m)}(m)=k<m$, also contradicted by Lemma 1 by similar logic.

$(3) f^2(m)=m$ and $f^{f(m)}(m)=m$, which implies that $f(m)=m$ by Lemma 2. This proves the claim.


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.

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.



The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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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