# 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 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 $m,n \geq 0$.

## 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.