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Difference between revisions of "1972 IMO Problems/Problem 3"
 
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Let m and n be arbitrary non-negative integers. Prove that
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Let <math>m</math> and <math>n</math> be arbitrary non-negative integers. Prove that
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<cmath>\frac{(2m)!(2n)!}{m!n!(m+n)!}</cmath>
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is an integer. (<math>0! = 1</math>.)
  
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<math>((2m)!(2n)!)/mn!(m+n)!</math>
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== Solution 1 ==
  
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is an integer. (0! = 1.)
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Denote the given expression as <math>f(m,n)</math>. We intend to show that <math>f(m,n)</math> is integral for all <math>m,n \geq 0</math>. To start, we would like to find a recurrence relation for <math>f(m,n)</math>. First, let's look at <math>f(m,n-1)</math>:
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<cmath>\begin{align*}
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    f(m,n-1) &=\frac{(2m)!(2n-2)!}{m!(n-1)!(m+n-1)!}\\
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    &=\frac{(2m)!(2n)!(n-1)(m+n)}{m!n!(m+n)!(2n-1)(2n-2)}\\
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    &=f(m,n) \cdot \frac{(n-1)(m+n)}{(2n-1)(2n-2)}\\
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    &=f(m,n) \cdot \frac{m+n}{2(2n-1)}
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\end{align*}</cmath>
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Second, let's look at <math>f(m+1,n-1)</math>:
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<cmath>\begin{align*}
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    f(m+1,n-1) &=\frac{(2m+2)!(2n-2)!}{(m+1)!(n-1)!(m+n)!}\\
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    &=\frac{(2m)!(2n)!(2m+1)(2m+2)(n-1)}{m!n!(m+n)!(m+1)(2n-1)(2n-2)}\\
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    &= f(m,n) \cdot  \frac{(2m+1)(2m+2)(n-1)}{(m+1)(2n-1)(2n-2)}\\
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    &=f(m,n) \cdot \frac{(2m+1)}{2n-1}
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\end{align*}</cmath>
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Combining,
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<cmath>\begin{align*}
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    4f(m,n-1)-f(m+1,n-1) &=f(m,n)\cdot \Bigg(\frac{4(m+n)}{2(2n-1)}-\frac{2m+1}{2n-1}\Bigg)\\
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    &=f(m,n) \cdot \frac{2m+2n-2m-1}{2n-1}\\
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    &=f(m,n).
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\end{align*}</cmath>
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Therefore, we have found the recurrence relation <cmath>f(m,n)=4f(m,n-1)-f(m+1,n-1).</cmath>
  
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== Solution ==
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Note that <math>f(m,0)</math> is just <math>\binom{2m}{m}</math>, which is an integer for all <math>m \geq 0</math>. Then <cmath>f(m,1)=4f(m,0)-f(m+1,0),</cmath> so <math>f(m,1)</math> is an integer, and therefore <math>f(m,2)=4f(m,1)-f(m+1,1)</math> must be an integer, etc.
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By induction, <math>f(m,n)</math> is an integer for all <math>m,n \geq 0</math>.
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Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln723.html
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== Solution 2 ==
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Let p be a prime, and n be an integer. Let <math>V_p(n)</math> be the largest positive integer <math>k</math> such that <math>p^k|n</math>
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WTS: For all primes <math>p</math>, <math>V_p((2m)!)+V_p((2n)!) \ge V_p(m!)+V_p(n!)+V_p((m+n)!)</math>
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We know <cmath>V_p(x!)=\sum_{a=1}^{\infty} \left\lfloor{\frac{x}{p^a}}\right\rfloor</cmath>
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Lemma 2.1: Let <math>a,b</math> be real numbers. Then <math>\lfloor{2a}\rfloor+\lfloor{2b}\rfloor\ge\lfloor{a}\rfloor+\lfloor{b}\rfloor+\lfloor{a+b}\rfloor</math>
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Proof of Lemma 2.1: Let <math>a_1=\lfloor{a}\rfloor</math> and <math>b_1=\lfloor{b}\rfloor</math>
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<math>\lfloor{2a}\rfloor+\lfloor{2b}\rfloor=2(a_1+b_1)+\lfloor2\{a\}\rfloor+\lfloor2\{b\}\rfloor</math>
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 +
On the other hand, <math>\lfloor{a}\rfloor+\lfloor{b}\rfloor+\lfloor{a+b}\rfloor=a_1+b_1+(a_1+b_1)+\lfloor\{2(a+b)\}\rfloor</math>
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It is trivial that <math>\lfloor2\{a\}\rfloor+\lfloor2\{b\}\rfloor\ge\lfloor\{2(a+b)\}\rfloor</math> (Triangle Inequality)
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 +
Apply Lemma 2.1 to the problem: and we are pretty much done.
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 +
Note: I am lazy, so this is only the most important part. I hope you can come up with the rest of the solution. This is my work, but perhaps someone have come up with this method before I did.
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== See Also == {{IMO box|year=1972|num-b=2|num-a=4}}

Latest revision as of 23:26, 6 December 2022

Let $m$ and $n$ be arbitrary non-negative integers. Prove that \[\frac{(2m)!(2n)!}{m!n!(m+n)!}\] is an integer. ($0! = 1$.)

Solution 1

Denote the given expression as $f(m,n)$. We intend to show that $f(m,n)$ is integral for all $m,n \geq 0$. To start, we would like to find a recurrence relation for $f(m,n)$. First, let's look at $f(m,n-1)$: \begin{align*} f(m,n-1) &=\frac{(2m)!(2n-2)!}{m!(n-1)!(m+n-1)!}\\ &=\frac{(2m)!(2n)!(n-1)(m+n)}{m!n!(m+n)!(2n-1)(2n-2)}\\ &=f(m,n) \cdot \frac{(n-1)(m+n)}{(2n-1)(2n-2)}\\ &=f(m,n) \cdot \frac{m+n}{2(2n-1)} \end{align*} Second, let's look at $f(m+1,n-1)$: \begin{align*} f(m+1,n-1) &=\frac{(2m+2)!(2n-2)!}{(m+1)!(n-1)!(m+n)!}\\ &=\frac{(2m)!(2n)!(2m+1)(2m+2)(n-1)}{m!n!(m+n)!(m+1)(2n-1)(2n-2)}\\ &= f(m,n) \cdot \frac{(2m+1)(2m+2)(n-1)}{(m+1)(2n-1)(2n-2)}\\ &=f(m,n) \cdot \frac{(2m+1)}{2n-1} \end{align*} Combining, \begin{align*} 4f(m,n-1)-f(m+1,n-1) &=f(m,n)\cdot \Bigg(\frac{4(m+n)}{2(2n-1)}-\frac{2m+1}{2n-1}\Bigg)\\ &=f(m,n) \cdot \frac{2m+2n-2m-1}{2n-1}\\ &=f(m,n). \end{align*} Therefore, we have found the recurrence relation \[f(m,n)=4f(m,n-1)-f(m+1,n-1).\]

Note that $f(m,0)$ is just $\binom{2m}{m}$, which is an integer for all $m \geq 0$. Then \[f(m,1)=4f(m,0)-f(m+1,0),\] so $f(m,1)$ is an integer, and therefore $f(m,2)=4f(m,1)-f(m+1,1)$ must be an integer, etc.

By induction, $f(m,n)$ is an integer for all $m,n \geq 0$.

Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln723.html

Solution 2

Let p be a prime, and n be an integer. Let $V_p(n)$ be the largest positive integer $k$ such that $p^k|n$

WTS: For all primes $p$, $V_p((2m)!)+V_p((2n)!) \ge V_p(m!)+V_p(n!)+V_p((m+n)!)$

We know \[V_p(x!)=\sum_{a=1}^{\infty} \left\lfloor{\frac{x}{p^a}}\right\rfloor\]

Lemma 2.1: Let $a,b$ be real numbers. Then $\lfloor{2a}\rfloor+\lfloor{2b}\rfloor\ge\lfloor{a}\rfloor+\lfloor{b}\rfloor+\lfloor{a+b}\rfloor$

Proof of Lemma 2.1: Let $a_1=\lfloor{a}\rfloor$ and $b_1=\lfloor{b}\rfloor$

$\lfloor{2a}\rfloor+\lfloor{2b}\rfloor=2(a_1+b_1)+\lfloor2\{a\}\rfloor+\lfloor2\{b\}\rfloor$

On the other hand, $\lfloor{a}\rfloor+\lfloor{b}\rfloor+\lfloor{a+b}\rfloor=a_1+b_1+(a_1+b_1)+\lfloor\{2(a+b)\}\rfloor$

It is trivial that $\lfloor2\{a\}\rfloor+\lfloor2\{b\}\rfloor\ge\lfloor\{2(a+b)\}\rfloor$ (Triangle Inequality)

Apply Lemma 2.1 to the problem: and we are pretty much done.

Note: I am lazy, so this is only the most important part. I hope you can come up with the rest of the solution. This is my work, but perhaps someone have come up with this method before I did.

See Also

1972 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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