use the lagrange formula for the max power of $p$, p prime, dividing $n!$ and $\left\lfloor 2m/p\right\rfloor+\left\lfloor 2n/p\right\rfloor\geq \left\lfloor m/p\right\rfloor+\left\lfloor n/p\right\rfloor+\left\lfloor m/p+n/p\right\rfloor$ using that: $\left\lfloor 2m/p\right\rfloor-\left\lfloor m/p\right\rfloor=\left\lfloor m/p+1/2\right\rfloor$ and $\left\lfloor m/p+1/2\right\rfloor+\left\lfloor n/p+1/2\right\rfloor \geq \left\lfloor m/p+n/p\right\rfloor$

let $f(m,n) = \frac{(2m)!(2n)!}{m!n!(m+n)!}$. we have $f(m,n) = 4f(m,n-1) - f(m+1,n-1)$. see that we can use this many times to represent $f(m,n)$ by $\sum{f(m+x,0)}$, but $f(a,0) = C^a_{2a} \in Z$ so, the problem is solved.

orl wrote: Prove that $(2m)!(2n)!$ is a multiple of $m!n!(m+n)!$ for any non-negative integers $m$ and $n$. Suppse that $p$ be an arbitrary prime then we should prove that $[\frac{2m}{p}]+[\frac{2n}{p}]\geq [\frac mp]+[\frac np]+[\frac{m+n}{p}]$ (where math=inline]$x$[/math denote the greatest integer less than or equal to $x$) And we know $[\frac{2m}{p}]-[\frac mp]=[\frac mp+\frac 12]$ so that $[\frac mp+\frac 12]+[\frac np+\frac 12]\geq [\frac mp+\frac np]$ and we are done.

To prove $\lfloor \frac{2m}{p} \rfloor + \lfloor \frac{2n}{p} \rfloor \geq \lfloor \frac{m}{p} \rfloor + \lfloor \frac{n}{p} \rfloor + \lfloor \frac{m+n}{p} \rfloor$, it suffices to prove that $\lfloor 2a \rfloor + \lfloor 2b \rfloor \geq \lfloor a \rfloor + \lfloor b \rfloor + \lfloor a+b \rfloor$ for arbitrary real $a,b \in [0,1)$, which is obviously true by considering the four cases: $a \geq \frac{1}{2}$ & $b \geq \frac{1}{2}$, $a \geq \frac{1}{2}$ & $b < \frac{1}{2}$, $a < \frac{1}{2}$ & $b \geq \frac{1}{2}$, and $a < \frac{1}{2}$ & $b < \frac{1}{2}$. From the nature of the solution, I feel this problem should be in the Number Theory section instead

This is the only one case of Kummer's Theorem.

True. I'm just proving this particular case here, which is sufficient for this problem.

e.lopes wrote: we have $ f(m,n) = 4f(m,n - 1) - f(m + 1,n - 1)$. Very nice! Let $ G(x, y) = \sum_{m, n \ge 0} f(m, n) x^m y^n$. Then the recurrence you gave is equivalent to $ 4G(x, y) = \frac{G(x, y) - G(0, y)}{x} + \frac{G(x, y) - G(x, 0)}{y}$, or $ (x + y - 4xy) G(x, y) = y G(0, y) + x G(x, 0)$. But of course $ G(x, 0) = \sum_{m \ge 0} {2m \choose m} x^m = \frac{1}{ \sqrt{1 - 4x} }$ and similarly, which gives $ G(x, y) = \frac{ \frac{y}{ \sqrt{1 - 4y}} + \frac{x}{ \sqrt{1 - 4x} } }{x + y - 4xy}$. Questions: 1. Can anyone come up with a natural combinatorial description of $ f(m, n)$? Does it lead to the above in a more general way? (Such a description would trivially solve the original question.) 2. What other identities of the form "a quotient of factorials is an integer" follow from Hermite's identity? (Here several of you have used Hermite's identity with $ n = 2$.) 3. The hook length formula is the most general statement I know of this form; for example, it implies that the Catalan numbers are integers. Is there a generalization of the hook length formula that reduces to this question as a special case? 4. What is the most general statement of this form we can make? I made an obvious conjecture some time ago (the terms in the numerator should majorize the terms in the denominator) that is quickly verified to be wrong, so what is the appropriate generalization? For example, is $ \frac{(3m)!(3n)!}{m!n!(m+n)!^2}$ an integer?

By factorization, f(m,n) = (2m m)*(2n n)/(m+n m). I don't know if it will be useful(it isn't so natural), but I got a counting solution trough this form. It's a little hard to explain, and I'm very tired. Tomorrow I will try to explain my solution.

I'm not sure if this works, but could someone check it? Thank you in advance.

bokagadha wrote: We have 2m+2n people in total. Arbitrarily split them into two groups of size 2m and 2n. From a group of 2m people we want to pick m people to join our club. From another group of 2n people, we want to choose n people to get preferential treatment in our club. Later the rules were changed, and we decided not to give preferential treatment to n people from the group of n+m people. This expression counts the number of ways you can do this, and thus is an integer. Could you be more precise about it? The group of $ n+m$ people chosen can still be split uniquely into those with preferential treatment and those without, because the initial splitting is known.

I'm not very sure about my solution, and I can see why that last part might be sketchy at best. Here's another stab at a combinatorial interpretation: Represent the quantity as such: \[ \frac{m! \binom{2m}{m} n! \binom{2n}{n}}{(m+n)!}\] So take a group of 2m boys and 2n girls. The numerator counts the way we can take m boys and order them in terms of height and do the same with the girls. The denominator says that we don't care about the order and just want a group of m+n people in no arrangement whatsoever. Again, I'm not very sure about this argument, and any help/comments would be greatly appreciated.

bokagadha wrote: The numerator counts the way we can take m boys and order them in terms of height and do the same with the girls. The denominator want a group of m+n people in no arrangement whatsoever. I don't think it is good, because the denominator don't say sth about the gene/ they may be all men for example.

e.lopes wrote: we have $f(m,n) = 4f(m,n-1) - f(m+1,n-1)$. How you got $f(m,n) = 4f(m,n-1) - f(m+1,n-1)$ ?

Nevermind, I posted a wrong solution

WLOG let $n > m$; if the two variables are equal, the expression simply equals ${2m \choose m}$. Taking $\nu_p$, it suffices to show that $$\sum_{i=1}^\infty \left \lfloor \frac{2m}{p_i} \right \rfloor + \sum_{i=1}^\infty \left \lfloor \frac{2n}{p^i} \right \rfloor \geq \sum_{i=1}^\infty \left \lfloor \frac m{p^i} \right \rfloor + \sum_{i=1}^\infty \left \lfloor \frac n{p^i} \right \rfloor + \sum_{i=1}^\infty \left \lfloor \frac{m+n}{p^i} \right \rfloor.$$In other words, it suffices to show that $$\lfloor 2a \rfloor + \lfloor 2b \rfloor \geq \lfloor a \rfloor + \lfloor b \rfloor + \lfloor a+b \rfloor$$for all real numbers $a, b$; note that $b>a$. This follows by using the inequality $\lfloor a+b \rfloor \geq \lfloor a \rfloor + \lfloor b \rfloor$ on the following three expressions and summing: \begin{align*} \lfloor 2b \rfloor - \lfloor b-a \rfloor &\geq \lfloor a+b \rfloor \\ \lfloor 2a \rfloor &\geq 2\lfloor a \rfloor \\ \lfloor b-a \rfloor &\geq \lfloor b \rfloor - \lfloor a \rfloor. \end{align*}So we are done.