Verified answer

Wilson's theoremstates that a natural number n > 1 is a prime number if and only if(n-1)!=-1(mod n). We will use the (=>) part:If p is a prime number, then(p-1)!=-1(mod p), or(p-1)!=m*p-1……(1)for some m. Our problem is (p-1)!≡p-1(mod1+2+3+...+(p-1))<= > (p-1)!≡p-1(mod (p-1)p/2).First:We prove (p-2)!=-1(mod p)Let (p-2)!=kp + a ……(2),where 0<=a<p.(p-1)!= (p-2)!*(p-1)=(kp+a)*(p-1)=(kp+a)*p-kp-a=(kp+a-k)*p-a=m*p-1 (from (1)).= > a=1.Substituting to (2), we have(p-2)!=kp + 1, i.e.Or (p-2)!-1=kp.Now prove the mean result:(p-1)!-(p-1)=(p-1)*(p-2)!-(p-1)=(p-1)*[(p-2)!-1]=(p-1)*kp=2k[(p-1)p/2].So (p-1)!-(p-1)=0 (mod (p-1)p/2)i.e.(p-1)!=(p-1) (mod (p-1)p/2).[[Done]]

2011-04-12 19:18:20 補充：

The above proof is bad, we give a more concise proof in the following:

Wilson's theorem states that a natural number n > 1 is a prime number if and only if

(n-1)!=-1(mod n).

We will use the (=>) part:

If p is a prime number, then

(p-1)!=-1(mod p).

2011-04-12 19:18:51 補充：

Our problem is

(p-1)!≡p-1(mod1+2+3+...+(p-1))

< = > (p-1)!≡p-1(mod (p-1)p/2).

First:

We will prove (p-2)!=1 (mod p).

[proof]

Let (p-2)!= a (mod p),

-1= (p-1)! (mod p)

= (p-2)!*(p-1) (mod p)

=a*(p-1) (mod p)

(for p-1=-1 (mod p)

=a*(-1) (mod p)

=-a (mod p)

= > a=1.

2011-04-12 19:19:17 補充：

Hence (p-2)!= a=1 (mod p), or

(p-2)!-1=kp for some k.

Now prove the mean result:

(p-1)!-(p-1)=(p-1)*(p-2)!-(p-1)

=(p-1)*[(p-2)!-1]

=(p-1)*kp=2k[(p-1)p/2].

So (p-1)!-(p-1)=0 (mod (p-1)p/2)

i.e.

(p-1)!=(p-1) (mod (p-1)p/2).

[[Done]]