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Wilson's theorem for prime powers (Theorem)

For every natural number $n$ , let $\pfac{n}$ denote the product of numbers $1 \le m \le n$ with $gcd(m,p) =1$ .

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

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

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

Let $p$ 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$$\pfac{p^s} \equiv -1 \pmod{p^s$$ for odd prime $p$ and any $s \in \mathbb{N}$ .

Now let $p=2$ and $s \ge 2$ . Then$$\left(1 +t.2^{s-1}\right)^2 \equiv 1 \pmod{2^s}, t =\stackrel{-}{+}1$$ Since$$\left(2^{s -1} +1\right)\left(2^{s-1} -1\right) \equiv -1 \pmod{2^s}$$ we have$$\pfac{p^s} \equiv (-1).(-1) =1 \pmod{p^s$$ For $p=2, s \ge 3$ , but $-1$ 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, proof, 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 2517 times total.

Classification:
AMS MSC11A07 (Number theory :: Elementary number theory :: Congruences; primitive roots; residue systems)
 11A41 (Number theory :: Elementary number theory :: Primes)

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