Exercise 2.7.13* ($(mp)! \equiv m! p!^m \pmod {p^{m+3}}$, if $p\geq 5$)

Question

Show that if $p\geq 5$ then $(mp)! \equiv m! p!^m \pmod {p^{m+3}}.$

Richard Ganaye · Accepted Answer

\begin{proof} 
As in Problem 12,
\begin{align}
f(x) &= (x-1)(x-2)\cdots(x-p+1)
\end{align}
We show in the solution of this problem (see (5)) that, under the hypothesis $p\geq 5$, 
\begin{align}
f(kp) \equiv f(p) \pmod{p^3}
\end{align}
for all $k \geq 1$.
Now, since 
$$[\![ 1, mp]\!] = \bigcup_{k=1}^m [\![ (k-1)p + 1, kp]\!],$$
(disjoint union), then
\begin{align*}
(mp)! &= \prod_{k=1}^m \prod_{i=(k-1)p+1} ^{kp} i\
&=  \prod_{k=1}^m \left[ kp \prod_{i=(k-1)p+1} ^{kp - 1} i ight],
\end{align*}
where
\begin{align*}
\prod_{i=(k-1)p+1} ^{kp - 1} i &= (kp - 1)(kp - 2)\cdots (kp - p + 1)\
&= f(kp).
\end{align*}
Therefore
\begin{align*}
(mp)! &= \prod_{k=1}^m \left[\,  kp\,   f(kp)ight]\
&= m! p^m \prod_{k=1}^m f(kp).
\end{align*}
By (2), 
\begin{align*}
\prod_{k=1}^m f(kp) &\equiv f(p)^m = [(p-1)!]^m \pmod {p^3}.
\end{align*}
Thus
\begin{align*}
(mp)! &\equiv m!p^m [(p-1)!]^m \pmod {p^{m+3}}\
\end{align*}
This shows that, for $p \geq 5$, and $m \geq 1$,
$$(mp)! \equiv m!(p!)^m \pmod {p^{m+3}}.$$
(The property is also true for $m = 0$.)
\end{proof}

Exercise 2.7.13* ($(mp)! \equiv m! p!^m \pmod {p^{m+3}}$, if $p\geq 5$)

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Exercise 2.7.13* ($(mp)! \equiv m! p!^m \pmod {p^{m+3}}$, if $p\geq 5$)

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