1972 IMO Problems/Problem 3

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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1972 IMO Problems/Problem 3

Solution