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

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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.
 
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}{=}m-n</math> so of course <math>m-n\mid f(m)-f(n)</math>!
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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>!
  
 
==See also==
 
==See also==

Revision as of 09:56, 18 April 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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