# Anton’s congruence

For every $n\in\mathbb{N}$ $\left(n\underline{!}\right)_{p}$ stands for the product  of numbers between $1$ and $n$ which are not divisible by a given prime $p$. And we set $\left(0\underline{!}\right)_{p}=1$.

The corollary below generalizes a result first found by Anton, Stickelberger, and Hensel:

Let $N_{0}$ be the least non-negative residue of $n\pmod{p^{s}}$ where $p$ is a prime number  and $n\in\mathbb{N}$. Then

 $\left(n\underline{!}\right)_{p}\equiv\left(\pm 1\right)^{\left\lfloor n/p^{s}% \right\rfloor}\cdot\left(N_{0}\underline{!}\right)_{p}\pmod{p^{s}}.$
###### Proof.

We write each $r$ in the product below as $ip^{s}+j$ to get

 $\displaystyle\left(n\underline{!}\right)_{p}$ $\displaystyle=$ $\displaystyle\prod\limits_{\begin{subarray}{c}1\leq r\leq n\\ p^{s}\not\div r\end{subarray}}r$ $\displaystyle=$ $\displaystyle\left(\prod\limits_{\begin{subarray}{c}0\leq i\leq\left\lfloor n/% p^{s}\right\rfloor-1\\ 1\leq j $\displaystyle\equiv$ $\displaystyle\prod\limits_{i=0}^{\left\lfloor n/p^{s}\right\rfloor-1}\prod% \limits_{\begin{subarray}{c}1\leq j $\displaystyle\equiv$ $\displaystyle\left(p^{s}\underline{!}\right)_{p}^{\left\lfloor n/p^{s}\right% \rfloor}\cdot\left(N_{0}\underline{!}\right)_{p}\pmod{p^{s}}.$

From Wilson’s theorem for prime powers it follows that

 $\left(n\underline{!}\right)_{p}\equiv\begin{cases}\left(N_{0}\underline{!}% \right)_{p}\text{if}&p=2,s\geq 3\\ (-1)^{\left\lfloor n/p^{s}\right\rfloor}\cdot\left(N_{0}\underline{!}\right)_{% p}&\text{otherwise.}\end{cases}\pmod{p^{s}}.$

Title Anton’s congruence    AntonsCongruence 2013-03-22 13:22:49 2013-03-22 13:22:49 Thomas Heye (1234) Thomas Heye (1234) 10 Thomas Heye (1234) Theorem msc 11A07 Factorial  