Proving that $p^{\alpha + \beta + 1} \mid {n \choose k} p^{k\alpha}$ when $p^\beta \mid n$.

Question

Let $n,\alpha\in\mathbb{N},\beta\in\mathbb{N}_0$, and let $p$ be odd prime number s.t. $p^\beta|n$.

How do we prove that $p^{\alpha+\beta+1}|{n\choose k}p^{k\alpha}$ for every $k\in\{1,2,\ldots,n-1\}$?

Answer 1

We have the following formula for the largest power of $p$ that divides integer $m!$: $$\frac{m-s_p(m)}{p-1}$$ where $s_p(m)$ represents the digit sum of $m$ in base $p$. We write $p^x\|n$ to denote $p^x|n$ but $p^{x+1}\nmid n$.

For convenience, set $\gamma=1+\lfloor \log_p n\rfloor$, the number of digits of $n$ in base $p$. Because $p^\beta|n$, the last $\beta$ digits of $n$ (in base $p$) are all zero. Hence $s_p(n)\le (p-1)(\gamma-\beta)$, and $p^x\|n!$ where $$x\ge \frac{n-(p-1)(\gamma-\beta)}{p-1}=\frac{n}{p-1}+(\beta-\gamma)$$ Next, we consider the denominator of $\displaystyle {n\choose k}=\frac{n!}{k!(n-k)!}$. We have $p^y\|k!$, where $y\le \frac{k-1}{p-1}$, and $p^z\|(n-k)!$, where $z\le \frac{n-k-1}{p-1}$. Combining these three facts, we get $p^s|{n\choose k}$, where $$s\ge \frac{n+k-1+n-k-1}{p-1}+\beta-\gamma=2\frac{n-1}{p-1}+\beta-\gamma$$ The desired result would follow if we could show that $$\alpha+\beta+1\le 2\frac{n-1}{p-1}+\beta-\gamma+k\alpha$$ which rearranges as $$\gamma+1\le 2\frac{n-1}{p-1}+(k-1)\alpha$$ It suffices to prove this for $k=1$, i.e. $$\gamma+1\le 2\frac{n-1}{p-1}$$ or $$\lfloor \log_p n\rfloor \frac{p-1}{2}\le n$$

Answer 2

It`s false for $k = 1$: $p^{\alpha +\beta+1} | np^{\alpha}$, $n = p,\beta=1$.