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Difference between revisions of "1972 IMO Problems/Problem 3"

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First, let's look at <math>f(m,n-1)</math>:
 
First, let's look at <math>f(m,n-1)</math>:
  
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<math>f(m,n-1)=\frac{(2m)!(2n-2)!}{m!(n-1)!(m+n-1)!}=\frac{(2m)!(2n)!(n-1)(m+n-1)}{m!n!(m+n)!(2n-1)(2n-2)}=f(m,n) \frac{(n-1)(m+n)}{(2n-1)(2n-2)}=f(m,n) \frac{m+n}{2(2n-1)}</math>
+
<math>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) \frac{(n-1)(m+n)}{(2n-1)(2n-2)}=f(m,n) \frac{m+n}{2(2n-1)}</math>
  
 
Second, let's look at <math>f(m+1,n-1)</math>:
 
Second, let's look at <math>f(m+1,n-1)</math>:

Revision as of 19:54, 20 November 2018

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

Solution

Let $f(m,n)=\frac{(2m)!(2n)!}{m!n!(m+n)!}$. 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 $f(m,n-1)$:

$f(m,n-1)=\frac{(2m)!(2n-2)!}{m!(n-1)!(m+n-1)!}=\frac{(2m)!(2n)!(n-1)(m+n)}{m!n!(m+n)!(2n-1)(2n-2)}=f(m,n) \frac{(n-1)(m+n)}{(2n-1)(2n-2)}=f(m,n) \frac{m+n}{2(2n-1)}$

Second, let's look at $f(m+1,n-1)$:

$f(m+1,n-1)=\frac{(2m+2)!(2n-2)!}{(m+1)!(n-1)!(m+n)!}=\frac{(2m)!(2n)!(2m+1)(2m+2)(n-1)}{m!n!(m+n)!(m+1)(2n-1)(2n-2)}=f(m,n) \frac{(2m+1)(2m+2)(n-1)}{(m+1)(2n-1)(2n-2)}=f(m,n) \frac{(2m+1)}{2n-1}$

Combining,

$4f(m,n-1)-f(m+1,n-1)=f(m,n) (\frac{4(m+n-1)}{2(2n-1)}-\frac{2m+1}{2n-1})=f(m,n) \frac{2m+2n-2-2m+1}{2n-1}=f(m,n)$.

Therefore, we have found the recurrence relation $f(m,n)=4f(m,n-1)-f(m+1,n-1)$.

We can see that $f(m,0)=\frac{(2m)!}{m!m!}$ is integral because the RHS is just $_{2m} C _m$, which we know to be integral for all $m \geq 0$.

So, $f(m,1)=4f(m,0)-f(m+1,0)$ must be integral, and then $f(m,2)=4f(m,1)-f(m+1,1)$ must be integral, etc.

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

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

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