# How to prove that $m!n!(m+n)!|(2m)!(2n)!$ for every pair of positive integers $(m,n)$? [duplicate]

I want to prove that

For every pair of positive integers $(m,n)$, the statement $m!n!(m+n)!|(2m)!(2n)!$ holds.

which is equivalent to

For every pair of positive integers $(m,n)$, the statement $$\sum_{i=1}^{\infty}\left\lfloor\frac m{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac n{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac {m+n}{p^i}\right\rfloor \le\sum_{i=1}^{\infty}\left\lfloor\frac {2m}{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac {2n}{p^i}\right\rfloor$$ holds for every prime number $p$.

I have tried setting $m=ap+c,n=bp+d$ with $a,b,c,d$ nonnegative integers and $c,d<p$. From that, the inequality becomes $$a+\sum_{i=1}^{\infty}\left\lfloor\frac a{p^i}\right\rfloor+ b+\sum_{i=1}^{\infty}\left\lfloor\frac b{p^i}\right\rfloor+ a+b+\left\lfloor\frac{c+d}{p}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac{a+b}{p^i}+\frac {c+d}{p^{i+1}}\right\rfloor \\\le 2a+\left\lfloor\frac{2c}{p}\right\rfloor+\sum_{i=1}^{\infty}\left\lfloor\frac {2a}{p^i}+\frac{2c}{p^{i+1}}\right\rfloor\\+ 2b+\left\lfloor\frac{2d}{p}\right\rfloor+\sum_{i=1}^{\infty}\left\lfloor\frac {2b}{p^i}+\frac{2d}{p^{i+1}}\right\rfloor$$ $$\iff \sum_{i=1}^{\infty}\left\lfloor\frac a{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac b{p^i}\right\rfloor+ \left\lfloor\frac{c+d}{p}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac{a+b+\frac {c+d}{p}}{p^i}\right\rfloor \\\le \left\lfloor\frac{2c}{p}\right\rfloor+\sum_{i=1}^{\infty}\left\lfloor\frac {2a+\frac{2c}{p}}{p^i}\right\rfloor\\+ \left\lfloor\frac{2d}{p}\right\rfloor+\sum_{i=1}^{\infty}\left\lfloor\frac {2b+\frac{2d}{p}}{p^i}\right\rfloor\text.$$

After that, I tried splitting cases based on whether $2c\ge p$, whether $2d\ge p$, and whether $c+d\ge p$. The case where $2c<p$, $2d\ge p$, and $c+d\ge p$ becomes $$\sum_{i=1}^{\infty}\left\lfloor\frac a{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac b{p^i}\right\rfloor+ 1+\sum_{i=1}^{\infty}\left\lfloor\frac{a+b+1}{p^i}\right\rfloor \le \sum_{i=1}^{\infty}\left\lfloor\frac {2a}{p^i}\right\rfloor+ 1+\sum_{i=1}^{\infty}\left\lfloor\frac {2b+1}{p^i}\right\rfloor$$ $$\iff \sum_{i=1}^{\infty}\left\lfloor\frac a{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac b{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac{a+b+1}{p^i}\right\rfloor \le \sum_{i=1}^{\infty}\left\lfloor\frac {2a}{p^i}\right\rfloor+ \sum_{i=1}^{\infty}\left\lfloor\frac {2b+1}{p^i}\right\rfloor$$ $$\iff \sum_{i=1}^{\infty}\left( \left\lfloor\frac a{p^i}\right\rfloor+ \left\lfloor\frac b{p^i}\right\rfloor+ \left\lfloor\frac{a+b+1}{p^i}\right\rfloor\right) \le \sum_{i=1}^{\infty}\left( \left\lfloor\frac {2a}{p^i}\right\rfloor+ \left\lfloor\frac {2b+1}{p^i}\right\rfloor\right)\text.$$

How to carry on from there?

## marked as duplicate by bof, Jaideep Khare, Community♦Feb 6 '18 at 6:14

Let $x=\frac{m}{p^i}$ and $y=\frac{n}{p^i}.$
Thus, by your work it's enough to prove that $$[2x]+[2y]\geq[x]+[y]+[x+y],$$ which is not so hard.
• By $[x]$, do you mean the floor function? – user_194421 Feb 6 '18 at 5:09
• $[2\{x\}]+[2\{y\}]\geq\{x+y\},$ doesn't hold if ${x}={y}=\frac14$. – user_194421 Feb 6 '18 at 5:16
• In your case $\{x+y\}=0.$ I can write a full proof if you want. It remains to write also three cases. – Michael Rozenberg Feb 6 '18 at 5:17
• What is $x$ and $y$? And why does $\{x+y\}=0$? – user_194421 Feb 6 '18 at 6:00