Hi,

How to show that $\displaystyle (n-1)!\equiv-1[n] \Longrightarrow n \in \mathbb{P}$????

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- May 1st 2010, 03:21 AMbhitroofen01Prime number
Hi,

How to show that $\displaystyle (n-1)!\equiv-1[n] \Longrightarrow n \in \mathbb{P}$???? - May 1st 2010, 03:37 AMDeadstar
I take it $\displaystyle \mathbb{P}$ are the prime numbers? This is Wilsons theorem.

For any $\displaystyle a \in \{1, 2, \dots, p-1\}$ there's an $\displaystyle a' \in \{1, 2, \dots, p-1\}$ with $\displaystyle aa' \equiv 1$ (mod p). Further, $\displaystyle a' = a$ iff $\displaystyle a = 1$ or $\displaystyle p-1$. Hence, in $\displaystyle (p-1)!$, the numbers forming this product can be cancelled in pairs, apart from $\displaystyle 1$ and $\displaystyle p-1$. Hence result. - May 1st 2010, 03:41 AMtonio
- May 1st 2010, 03:47 AMtonio
- May 1st 2010, 03:53 AMDeadstar