Is this an integer?

Solution 1:

If we consider darij's comment, it is enough to prove that $$ r(a, b, c) = \frac{(2a+2b+c)!(a+c)!}{(a+b)!b!(2a+c)!c!} $$ is an integer for all $a, b, c\geq 0$. (I just slightly changed the definition for convenience.) To prove this, we will prove the following lemma : $$ \lfloor 2x+2y+z\rfloor+\lfloor x+z\rfloor\geq \lfloor x+y\rfloor +\lfloor y\rfloor + \lfloor 2x+z \rfloor+\lfloor z \rfloor $$ for all $x, y, z\in \mathbb{R}$. If we can show this, our previous claim follows from the Lagrange's identity $$ \mathrm{ord}_{p}(n!)=\sum_{k=1}^{\infty}\bigg\lfloor\frac{n}{p^{k}}\bigg\rfloor. $$ By the way, since the equation in the lemma remains same if we change $x$ to $x+n$ for any $n\in \mathbb{Z}$ (and same for $y$ and $z$), we only need to check for $0\leq x, y, z<1$, which is an easy exercise.