# Prime number

• May 1st 2010, 03:21 AM
bhitroofen01
Prime number
Hi,

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

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

I take it $\mathbb{P}$ are the prime numbers? This is Wilsons theorem.

For any $a \in \{1, 2, \dots, p-1\}$ there's an $a' \in \{1, 2, \dots, p-1\}$ with $aa' \equiv 1$ (mod p).

I'm afraid this doesn't work: this presumes you already know p is prime, which is precisly what you're tring to prove!

Further, $a' = a$ i ff $a = 1$ or $p-1$. Hence, in $(p-1)!$, the numbers forming this product can be cancelled in pairs, apart from $1$ and $p-1$. Hence result.

.
• May 1st 2010, 03:47 AM
tonio
Quote:

Originally Posted by bhitroofen01
Hi,

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

Suppose $n=dk\,,\,\,d>1\Longrightarrow d\mid (n-1)!\,\,\,and\,\,\,also\,\,\,d\mid n\Longrightarrow d\mid 1$ , contradiction!

Tonio
• May 1st 2010, 03:53 AM
Suppose $n=dk\,,\,\,d>1\Longrightarrow d\mid (n-1)!\,\,\,and\,\,\,also\,\,\,d\mid n\Longrightarrow d\mid 1$ , contradiction!
Oh yeah! I never noticed the =>. I misread it as prove Wilsons formula with $n \in$ prime numbers...