PlanetMath: Wilson's theorem for prime powers

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Wilson's theorem for prime powers

(Theorem)

For every natural number , let $\left(n\underline{!}\right)_p$ denote the product of numbers $1 \le m \le n$ with .

For prime and $s \in \mathbb{N}$

$\begin{displaymath}\left(p^s\underline{!}\right)_p \equiv \left( \begin{array}{l... ... \ge 3 \ -1 & \mbox{otherwise} \end{array}\right. \pmod{p^s}.\end{displaymath}$

Proof: We pair up all factors of the product $\left(p^s\underline{!}\right)_p$ into those numbers where $m \not\equiv m^{-1} \pmod{p^s}$ and those where this is not the case. So $\left(p^s\underline{!}\right)_p$ is congruent (modulo ) to the product of those numbers where $m \equiv m{-1} \pmod{p^s} \leftrightarrow m^2 \equiv 1 \pmod{p^s}$ .

Let be an odd prime and $s \in \mathbb{N}$ . Since $2 \not\vert p^s$ , $p^s \vert (m^2 -1)$ implies $p^s \vert (m +1)$ either or $p^s \vert (m-1)$ . This leads to

$\displaystyle \left(p^s\underline{!}\right)_p \equiv -1 \pmod{p^s}$

for odd prime

and any $s \in \mathbb{N}$ .

Now let and $s \ge 2$ . Then

$\displaystyle \left(1 +t.2^{s-1}\right)^2 \equiv 1 \pmod{2^s}, t =\stackrel{-}{+}1.$

Since

$\displaystyle \left(2^{s -1} +1\right)\left(2^{s-1} -1\right) \equiv -1 \pmod{2^s},$

we have

$\displaystyle \left(p^s\underline{!}\right)_p \equiv (-1).(-1) =1 \pmod{p^s}$

For $p=2, s \ge 3$ , but

for $s=1,2. \square$

"Wilson's theorem for prime powers" is owned by Thomas Heye.

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Cross-references: implies, odd, congruent, factors, prime, numbers, product, natural number
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This is version 5 of Wilson's theorem for prime powers, born on 2003-01-18, modified 2003-01-18.
Object id is 3899, canonical name is WilsonsTheoremForPrimePowers.
Accessed 2104 times total.

Classification:

AMS MSC:	11A07 (Number theory :: Elementary number theory :: Congruences; primitive roots; residue systems)
	11A41 (Number theory :: Elementary number theory :: Primes)

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