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

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==Solution==
 
==Solution==
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Note that <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>. We always have
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<cmath>
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n\cdot n!=(n+1)!-n!|f(n+1)!-f(n)!
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</cmath>
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(1)
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 +
Now if <math>f(n)=1</math> for any <math>n\ge 2</math>, then <math>f(k)=1</math> for all <math>k\ge n</math>. This follows because then <math>f(n+1)!\equiv 1 \mod n\cdot n!</math>, which is only possible if <math>f(n+1)=1</math>, and the rest follows by induction.
 +
We now divide into cases:
 +
 +
'''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>
 +
3-1|f(3)-f(1)
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</cmath>
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so <math>2|-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 bases cases are already shown. If <math>f(k)=k</math>, then by (1) we have
 +
<cmath>
 +
f(k+1)\equiv k! \mod k\cdot k!
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</cmath>
 +
Which gives <math>f(k+1)<2k</math> (otherwise <math>f(k+1)!\equiv 0 \mod k\cdot k!</math>). Also we have
 +
<cmath>
 +
k+1-1|f(k+1)-f(1)
 +
</cmath>
 +
so <math>f(k+1)\equiv 1 \mod k</math>. This gives the solutions <math>f(k+1)=1</math> (obviously impossible) and <math>f(k+1)=k+1</math>. Then by induction, this always holds. Note that this also satisfies the requirements.
 +
 +
'''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
 +
<cmath>
 +
f(k+1)\equiv 2 \mod k\cdot k!
 +
</cmath>
 +
Which gives <math>f(k+1)<2k</math> similarly. Also note that
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<cmath>
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k+1-1|f(k+1)-2
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</cmath>
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<cmath>
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k+1-2|f(k+1)-2
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</cmath>
 +
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 this always holds, and note that this satisfies the requirements.
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 +
The solutions are <math>\boxed{f(n)=1, f(n)=2, f(n)=n}</math>.
  
 
==See also==
 
==See also==

Revision as of 13:12, 26 April 2012

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

Note that $f(1)=f(1!)=f(1)!$ and $f(2)=f(2!)=f(2)!$, so $f(1)=1$ or $2$ and similarly for $f(2)$. We always have \[n\cdot n!=(n+1)!-n!|f(n+1)!-f(n)!\] (1)

Now if $f(n)=1$ for any $n\ge 2$, then $f(k)=1$ for all $k\ge n$. This follows because then $f(n+1)!\equiv 1 \mod n\cdot n!$, which is only possible if $f(n+1)=1$, and the rest 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 \[3-1|f(3)-f(1)\] so $2|-1$, a contradiction.

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

We claim $f(n)=n$ always by induction. The bases cases are already shown. If $f(k)=k$, then by (1) we have \[f(k+1)\equiv k! \mod k\cdot k!\] Which gives $f(k+1)<2k$ (otherwise $f(k+1)!\equiv 0 \mod k\cdot k!$). Also we have \[k+1-1|f(k+1)-f(1)\] so $f(k+1)\equiv 1 \mod k$. This gives the solutions $f(k+1)=1$ (obviously impossible) and $f(k+1)=k+1$. Then by induction, this always holds. Note that 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 \[f(k+1)\equiv 2 \mod k\cdot k!\] Which gives $f(k+1)<2k$ similarly. Also note that \[k+1-1|f(k+1)-2\] \[k+1-2|f(k+1)-2\] so $f(k+1)\equiv 2 \mod k(k-1)$. Then the only possible solution is $f(k+1)=2$. By induction this always holds, and note that this satisfies the requirements.

The solutions are $\boxed{f(n)=1, f(n)=2, f(n)=n}$.

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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