# How to prove this $p^{j-\left\lfloor\frac{k}{p}\right\rfloor}\mid c_{j}$

Let $p$ be a prime number and $g\in \mathbb{Z}[x]$.

Let $$\binom{x}{k}=\dfrac{x(x-1)(x-2)\cdots(x-k+1)}{k!} \in \mathbb{Q}[x]$$ for every $k \geq 0$.

Fix an integer $k$. Write the integer-valued polynomial $\binom{g(px)}{k}$ in the form $$\binom{g(px)}{k}=\sum_{i=0}^{\infty}c_{i}\binom{x}{i}$$ for some fixed $c_0, c_1, c_2, \ldots \in \mathbb{Z}$, where all but finitely many $i\geq 0$ satisfy $c_i \neq 0$.

show that $$p^{j-\left\lfloor\frac{k}{p}\right\rfloor}\mid c_{j},\text{where } j=0,1,2,\ldots$$

• Here is a lemma that is probably helpful. We define $S$ to be the set of all power series $f \in \left(\mathbb{Z}\left[x\right]\right)\left[\left[t\right]\right]$ such that, if we write $f$ in the form $f = \sum_{k\geq 0} \left(\sum_{i\geq 0} c_{i,k} \dbinom{x}{i}\right) t^k$, then the $c_{i,k}$ are integers and every $i \geq 0$ and $k \geq 0$ satisfy $p^{i - \left\lfloor k/p\right\rfloor} \mid c_{i,k}$. (If the left hand side of this divisibility is not an integer, then it means nothing.) Then, $S$ is a subring ... – darij grinberg Oct 19 '15 at 4:04
• ... of $\left(\mathbb{Z}\left[x\right]\right)\left[\left[t\right]\right]$. (Hint for the proof of this lemma: For any nonnegative integers $a$ and $b$, we have $\dbinom{x}{a} \dbinom{x}{b} = \sum_{i=a}^{a+b} \dbinom{i}{a} \dbinom{a}{a+b-i} \dbinom{x}{i}$.) Moreover, this subring $S$ is closed as a subset of the topological ring $\left(\mathbb{Z}\left[x\right]\right)\left[\left[t\right]\right]$. Hence, if $a \in \left(\mathbb{Z}\left[x\right]\right)\left[\left[t\right]\right]$ is a power series whose constant coefficient ... – darij grinberg Oct 19 '15 at 4:08
• ... (with respect to $t$) is $1$, and if $a \in S$, then $a^{-1} \in S$ as well. Now, your problem translates into the statement that $\left(1+t\right)^{g\left(px\right)} \in S$. By our lemma, it suffices to show that $\left(1+t\right)^{\left(px\right)^v} \in S$ for every nonnegative integer $v$. Since it is easy to check that $1+t \in S$ (indeed, every formal power series in $t$ whose coefficients are constants -- i.e., which does not depend on $x$ at all -- lies in $S$), I suspect that we can prove the stronger claim ... – darij grinberg Oct 19 '15 at 4:09
• ... that $h^{px} \in S$ for every $h \in S$ with constant term $1$. Volunteers? – darij grinberg Oct 19 '15 at 4:12
• @darijgrinberg,Hello,can you post your solution? – network o Oct 20 '15 at 2:47