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Difference between revisions of "2012 USAMO Problems/Problem 4"

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==Solution==
 
==Solution==
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By the first condition we have <math>f(1)=f(1!)=f(1)!</math> and <math>f(2)=f(2!)=f(2)!</math>, so <math>f(1)=1</math> or <math>2</math> and similarly for <math>f(2)</math>.  By the second condition, we have
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<cmath>
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n\cdot n!=(n+1)!-n! \mid f(n+1)!-f(n)! \qquad \qquad (1)
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</cmath>
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for all positive integers <math>n</math>.
  
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==See also==
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Suppose that for some <math>n \geq 2</math> we have <math>f(n) = 1</math>.  We claim that <math>f(k)=1</math> for all <math>k\ge n</math>. Indeed, from Equation (1) we have <math>f(n+1)!\equiv 1 \mod n\cdot n!</math>, and this is only possible if <math>f(n+1)=1</math>; the claim follows by induction.
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*[[USAMO Problems and Solutions]]
 
  
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We now divide into cases:
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'''Case 1:''' <math>f(1)=f(2)=1</math>
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This gives <math>f(n)=1</math> always from the previous claim, which is a solution.
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'''Case 2:''' <math>f(1)=2, f(2)=1</math>
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This implies <math>f(n)=1</math> for all <math>n\ge 2</math>, but this does not satisfy the initial conditions. Indeed, we would have
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<cmath>
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3-1 \mid f(3)-f(1)
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</cmath>
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and so <math>2\mid -1</math>, a contradiction.
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'''Case 3:''' <math>f(1)=1</math>, <math>f(2)=2</math>
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We claim <math>f(n)=n</math> always by induction. The base cases are <math>n = 1</math> and <math>n = 2</math>. Fix <math>k > 1</math> and suppose that <math>f(k)=k</math>.  By Equation (1) we have that
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<cmath>
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f(k+1)! \equiv k! \mod k\cdot k! .
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</cmath>
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This implies <math>f(k+1)<2k</math> (otherwise <math>f(k+1)!\equiv 0 \mod k\cdot k!</math>). Also we have
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<cmath>
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(k+1)-1  \mid  f(k+1)-f(1)
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</cmath>
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so <math>f(k+1)\equiv 1 \mod k</math>. This gives the solutions <math>f(k+1)=1</math> and <math>f(k+1)=k+1</math>. The first case is obviously impossible, so <math>f(k + 1) = k + 1</math>, as desired.  By induction, <math>f(n) = n</math> for all <math>n</math>.  This also satisfies the requirements.
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'''Case 4:''' <math>f(1)=f(2)=2</math>
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We claim <math>f(n)=2</math> by a similar induction. Again if <math>f(k)=2</math>, then by (1) we have
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<cmath>
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f(k+1)\equiv 2 \mod k\cdot k!
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</cmath>
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and so <math>f(k+1)<2k</math>.  Also note that
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<cmath>
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k+1-1  \mid  f(k+1)-2
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</cmath>
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and
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<cmath>
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k+1-2  \mid  f(k+1)-2
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</cmath>
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so <math>f(k+1)\equiv 2 \mod k(k-1)</math>. Then the only possible solution is <math>f(k+1)=2</math>. By induction, <math>f(n) = 2</math> for all <math>n</math>, and this satisfies all requirements.
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In summary, there are three solutions: <math>\boxed{f(n)=1, f(n)=2, f(n)=n}</math>.
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==Summary==
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Three functions: <math>f(x)=x</math> since <math>x!=x!</math>, <math>1!=1</math> and <math>2!=2</math>, fixed points on the <math>x!</math> function.
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No elegant<math>\mod</math> argument needed; <math>f(m)-f(n)\textcolor{red}{\textbf{ = }}m-n</math> so of course <math>m-n\mid f(m)-f(n)</math>!
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==See Also==
 
{{USAMO newbox|year=2012|num-b=3|num-a=5}}
 
{{USAMO newbox|year=2012|num-b=3|num-a=5}}
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{{MAA Notice}}
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[[Category:Olympiad Algebra Problems]]
 +
[[Category:Functional Equation Problems]]

Latest revision as of 07:18, 19 July 2016

Problem

Find all functions $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ (where $\mathbb{Z}^+$ is the set of positive integers) such that $f(n!) = f(n)!$ for all positive integers $n$ and such that $m - n$ divides $f(m) - f(n)$ for all distinct positive integers $m$, $n$.

Solution

By the first condition we have $f(1)=f(1!)=f(1)!$ and $f(2)=f(2!)=f(2)!$, so $f(1)=1$ or $2$ and similarly for $f(2)$. By the second condition, we have \[n\cdot n!=(n+1)!-n! \mid f(n+1)!-f(n)! \qquad \qquad (1)\] for all positive integers $n$.

Suppose that for some $n \geq 2$ we have $f(n) = 1$. We claim that $f(k)=1$ for all $k\ge n$. Indeed, from Equation (1) we have $f(n+1)!\equiv 1 \mod n\cdot n!$, and this is only possible if $f(n+1)=1$; the claim follows by induction.

We now divide into cases:

Case 1: $f(1)=f(2)=1$

This gives $f(n)=1$ always from the previous claim, which is a solution.

Case 2: $f(1)=2, f(2)=1$

This implies $f(n)=1$ for all $n\ge 2$, but this does not satisfy the initial conditions. Indeed, we would have \[3-1 \mid f(3)-f(1)\] and so $2\mid -1$, a contradiction.

Case 3: $f(1)=1$, $f(2)=2$

We claim $f(n)=n$ always by induction. The base cases are $n = 1$ and $n = 2$. Fix $k > 1$ and suppose that $f(k)=k$. By Equation (1) we have that \[f(k+1)! \equiv k! \mod k\cdot k! .\] This implies $f(k+1)<2k$ (otherwise $f(k+1)!\equiv 0 \mod k\cdot k!$). Also we have \[(k+1)-1 \mid f(k+1)-f(1)\] so $f(k+1)\equiv 1 \mod k$. This gives the solutions $f(k+1)=1$ and $f(k+1)=k+1$. The first case is obviously impossible, so $f(k + 1) = k + 1$, as desired. By induction, $f(n) = n$ for all $n$. This also satisfies the requirements.

Case 4: $f(1)=f(2)=2$

We claim $f(n)=2$ by a similar induction. Again if $f(k)=2$, then by (1) we have \[f(k+1)\equiv 2 \mod k\cdot k!\] and so $f(k+1)<2k$. Also note that \[k+1-1 \mid f(k+1)-2\] and \[k+1-2 \mid f(k+1)-2\] so $f(k+1)\equiv 2 \mod k(k-1)$. Then the only possible solution is $f(k+1)=2$. By induction, $f(n) = 2$ for all $n$, and this satisfies all requirements.


In summary, there are three solutions: $\boxed{f(n)=1, f(n)=2, f(n)=n}$.

Summary

Three functions: $f(x)=x$ since $x!=x!$, $1!=1$ and $2!=2$, fixed points on the $x!$ function.

No elegant$\mod$ argument needed; $f(m)-f(n)\textcolor{red}{\textbf{ = }}m-n$ so of course $m-n\mid f(m)-f(n)$!

See Also

2012 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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