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Difference between revisions of "Talk:2012 USAMO Problems/Problem 4"
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7. For <math>n\ge 3</math>, <math>f(n)=C_1(n-1)+1=C_2(n-2)+2</math> implies that <math>f(n)>2</math>, otherwise <math>C_1=C_2=0</math> and that <math>f(n)=1=2</math>, which is a contradiction. Since <math>f(n)>2</math>, now we have <math>C_2\ge 1</math>, so <math>f(n)\ge (n-2)+2=n</math>. Now we have a lower bound. What is more difficult is to find an upper bound of <math>f(n)</math>.  
 
7. For <math>n\ge 3</math>, <math>f(n)=C_1(n-1)+1=C_2(n-2)+2</math> implies that <math>f(n)>2</math>, otherwise <math>C_1=C_2=0</math> and that <math>f(n)=1=2</math>, which is a contradiction. Since <math>f(n)>2</math>, now we have <math>C_2\ge 1</math>, so <math>f(n)\ge (n-2)+2=n</math>. Now we have a lower bound. What is more difficult is to find an upper bound of <math>f(n)</math>.  
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(I know from the main Page it is not too hard to get the upper bound, but to be honest it takes me a while to figure it out.)
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'''(I know from the main Page it is not too hard to get the upper bound, but to be honest it takes me quite a while to figure it out.)'''
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8. Therefore <math>n|f(n)!</math>. Now <math>f( (n+1)!) = f(n!) + C( n \cdot n!)</math>, therefore if <math>f(n)=n</math>, then suppose <math>f(n+1)=D(n+1)</math>, then we have
 
8. Therefore <math>n|f(n)!</math>. Now <math>f( (n+1)!) = f(n!) + C( n \cdot n!)</math>, therefore if <math>f(n)=n</math>, then suppose <math>f(n+1)=D(n+1)</math>, then we have
 
<cmath> (D(n+1))! = n! + Cn \cdot n!</cmath>
 
<cmath> (D(n+1))! = n! + Cn \cdot n!</cmath>

Revision as of 21:27, 3 May 2012

Though not as elegant as the inductive proof, my proof for this problem is quite different from the one posted in the Page, so I would like to paste it here just for reference.

1. First, $f(1)$, $f(2)$ equal to $1$ or $2$.

2. Second, suppose that for all $n$, $f(n)\le 2$. Then for $n\ge 3$,

$2\ge f(n!) = f(1) + C(n!-1)$ ($C\ge 0$ is an integer.) $\ge -2+5C$, therefore $5C\le 4$ and so $C=0$ is the only possibility. Hence $f(n)!=f(n!)=f(1)$. Similar argument yields that $f(n)!=f(n!)=f(2)$, so $f$ is a constant function, which can only be $f=1$ or $f=2$.

3. From now on, suppose that there exists $n_0\ge 3$ such that $f(n_0)>2$.

4. By $2| (n_0!-2) | f(n_0)!-f(2)$ and that $2|f(n_0)!$ we know that $2|f(2)$, or $f(2)=2$.

5. By $f(3)!=f(6)=4C+f(2)=4C+2$ we know that $f(3)\le 3$, and by $3|f(n_0)!-f(3)$ and $3|f(n_0)!$ we know that $3|f(3)$, therefore $f(3)=3$.

6. Then by $2|f(3)-f(1)$ we know that $f(1)$ is odd, so $f(1)=1$.

7. For $n\ge 3$, $f(n)=C_1(n-1)+1=C_2(n-2)+2$ implies that $f(n)>2$, otherwise $C_1=C_2=0$ and that $f(n)=1=2$, which is a contradiction. Since $f(n)>2$, now we have $C_2\ge 1$, so $f(n)\ge (n-2)+2=n$. Now we have a lower bound. What is more difficult is to find an upper bound of $f(n)$.

(I know from the main Page it is not too hard to get the upper bound, but to be honest it takes me quite a while to figure it out.)

8. Therefore $n|f(n)!$. Now $f( (n+1)!) = f(n!) + C( n \cdot n!)$, therefore if $f(n)=n$, then suppose $f(n+1)=D(n+1)$, then we have \[(D(n+1))! = n! + Cn \cdot n!\]

, which is not divisible by $n\cdot n!$ and so $D=1$. By induction we have $f(n)=n$ for all $n\ge 3$.

--Lightest 22:26, 3 May 2012 (EDT)

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