<s>[INST] How to prove that $\frac{(5m)!(5n)!}{(m!)(n!)(3m+n)!(3n+m)!}$ is a natural number? 
How to prove that  $$\frac{(5m)! \cdot (5n)!}{m! \cdot n! \cdot (3m+n)! \cdot (3n+m)!}$$ is a natural number $\forall m,n\in\mathbb N$ , $m\geqslant 1$ and $n\geqslant 1$?

If $p$ is a prime, then the number of times $p$ divides $N!$ is $e_p(N)=\sum_{k=1}^\infty\left\lfloor\frac{N}{p^k}\right\rfloor$. So I need
$$e_p(5m)+e_p(5n)\geq e_p(m)+e_p(n)+e_p(3m+n)+e_p(3n+m).$$
What to do next?  Thanks in advance.
 [/INST] I became aware of this question due to some recent edits. The shorter proofs seem to rely on
$$
\lfloor5x\rfloor+\lfloor5y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\lfloor3x+y\rfloor+\lfloor3y+x\rfloor
$$
My answer also uses this inequality. I have tried to provide a complete, yet shorter, proof of it.

Define
$$
f(x,y)=\lfloor5x\rfloor+\lfloor5y\rfloor-\lfloor x\rfloor-\lfloor y\rfloor-\lfloor3x+y\rfloor-\lfloor3y+x\rfloor\tag1
$$
Note that $f(x+1,y)=f(x,y+1)=f(x,y)$. Thus, we only need worry about $x,y\in[0,1)$.
Let $x\in\left[\frac j5,\frac{j+1}5\right)$ and $y\in\left[\frac k5,\frac{k+1}5\right)$, where $j,k\in\{0,1,2,3,4\}$. Then
$$
\lfloor x\rfloor=\lfloor y\rfloor=0\tag{2a}
$$
$$
\lfloor5x\rfloor=j\qquad\text{and}\qquad\lfloor5y\rfloor=k\tag{2b}
$$
$$
\lfloor3x+y\rfloor\le\left\lfloor\frac{3j+k+3}5\right\rfloor
\qquad\text{and}\qquad
\lfloor x+3y\rfloor\le\left\lfloor\frac{j+3k+3}5\right\rfloor
\tag{2c}
$$
Apply $(2)$ to $(1)$:
$$
\begin{align}
f(x,y)
&=\lfloor5x\rfloor+\lfloor5y\rfloor-\lfloor x\rfloor-\lfloor y\rfloor-\lfloor3x+y\rfloor-\lfloor3y+x\rfloor\\
&\ge j+k-\left\lfloor\frac{3j+k+3}5\right\rfloor-\left\lfloor\frac{j+3k+3}5\right\rfloor\\
&=g(j,k)\tag3
\end{align}
$$
Compute the values of $g(j,k)$ for $j,k\in\{0,1,2,3,4\}$:
$$
\begin{array}{c|cc}
g&0&1&2&3&4\\\hline
0&0&0&0&0&0\\
1&0&0&0&1&0\\
2&0&0&0&1&1\\
3&0&1&1&0&1\\
4&0&0&1&1&2
\end{array}\tag4
$$
Inequality $(3)$, table $(4)$, and the lattice periodicity of $f$, show that $f(x,y)\ge0$ for all $x,y$. That is,
$$
\lfloor5x\rfloor+\lfloor5y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\lfloor3x+y\rfloor+\lfloor3y+x\rfloor\tag5
$$
Inequality $(5)$, when combined with Legendre's Formula, yields
$$
m!\,n!\,(3m+n)!\,(m+3n)!\mid(5m)!\,(5n)!\tag6
$$
</s>