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

Revision as of 20:46, 25 April 2025

Problem

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.

Remark about, and correction to Solution 2

The last step seems mistaken. For example [1.5]+[1.5]<[1.5+1.5]. Triangle inequality is about absolute value |x| not floor function [x]. Following is a rectified solution.

Lemma: [a+b]<=[a]+[b]+1, a,b>=0.

Proof: a<[a]+1, b<[b]+1, so a+b<[a]+[b]+2, [a+b]<=a+b<[a]+[b]+2, so [a+b]<=[a]+[b]+1.

Let (x,2x,…,nx)=n, p is any prime, V2(9!)=V2(1•2•3•4•5•6•7•8•9) =(2,4,6,8)+(4,8)+(8)=[9/2]+[9/4]+[9/8]=7.

Generalize it and we have, Vp(m!)=Vp(1•2•….•m)=[m/p]+[m/p^2]+…+[m/p^k], k=1~∞, or,

Vp(m!)=∑([m/p^k],k=1~∞) $\ \ \ \ (1)$ .

Obviously when p^k>m, k>=1, [m/p^k]=0.

Note the target statement is equivalent to:

for any prime p, Vp((2m)!(2n)!)>=Vp(m!n!(m+n)!), or,

Vp((2m)!)+Vp((2n)!)>=Vp(m!)+Vp(n!)+Vp(m+n)! $\ \ \ \ (2)$ .

Let m/p^k=a(k), n/p^k=b(k), k=1~∞. Combine (1), we note that (2) is equivalent to:

for any k in (1), [2a(k)]+[2b(k)]>=[a(k)]+[b(k)]+[a(k)+b(k)] $\ \ \ \ (3)$ .

For convenience of viewing, let a(k)=a, b(k)=b, a,b>=0, (3) is then rewritten as:

[2a]+[2b]>=[a]+[b]+[a+b] $\ \ \ \ (4)$ .

We prove it under different scenarios:

-If a<[a]+1/2, b<[b]+1/2, Then 2[a]<=[2a]<=2a<2[a]+1, [2a]=2[a], likewise, [2b]=2[b], so (4) is equivalent to [a]+[b]>=[a+b]. Per current condition, a+b<[a]+[b]+1, so [a+b]<[[a]+[b]+1]=[a]+[b]+1, also obviously [a+b]>=[a]+[b], so [a+b]=[a]+[b], proved;

-If a<[a]+1/2, b>=[b]+1/2, then [2a]=2[a], [2b]=2[b]+1, so (4) is equivalent to [a]+[b]+1>=[a+b], as per lemma, proved;

-If a>=[a]+1/2, b>=[b]+1/2, so [2a]=2[a]+1, [2b]=2[b]+1, so (4) is equivalent to [a]+[b]+2>=[a+b], as per lemma, proved;

Therefore (4) is proved, so is (2), as every step from (2) to (4) is reversible. Therefore (2m)!(2n)!/(m!n!(m+n)!) is an integer. Proof completed.

Remarks (added by pf02, April 2025)

1. The first solution has a few errors in the computation, but they cancel out, and the proof is essentially correct. Below I will give the corrections, to make the solution flawless.

2. As pointed out by a contributor, the second solution is incorrect. The correction given is good, but very sloppily written and hard to read (even after I edited the spacing, to make it legible). Below, I will repeat the argument in a simplified form, to make it easier to follow.

3. I will give a third solution below, inspired by the papers mentioned in the next paragraph.

4. The problem is a classical result, attributed to Eugene Catalan, who stated it in a paper published in 1874. The numbers $S(m, n) = \frac{(2m)!(2n)!}{m!n!(m+n)!}$ are sometimes called the super-Catalan numbers. (They are mentioned at the end of https://en.wikipedia.org/wiki/Catalan_number).

There are a few papers published about them. See for example the paper "The Super Catalan Numbers $S(m, m + s)$ for $s \le 3$ and Some Integer Factorial Ratios" by Xin Chen & Jane Wang (https://www-users.cse.umn.edu/~reiner/REU/ChenWang2012.pdf), and "Super Ballot Numbers" by Ira M. Gessel (especially section 6) (https://www.sciencedirect.com/science/article/pii/0747717192900342/pdf).

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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