【問題】一個證明題???

[FTICR 10]

設n為質數

則滿足:

(n-1)!+1=nk

ex:

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

(3-1)!+1=3=3*1

(5-1)!+1=25=5*5

(7-1)!+1=721=7*103

(11-1)!+1=3628801=11*329891

(13-1)!+1=479001601=13*36846277

(17-1)!+1=20922789888001=17*1230752346353

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請問這該如何證明對所有質數成立???

[00 10]

這個定理叫wilson定理，google搜尋就可以找到證明了。

[Xiang 10]

有一個...不是很好的證法XD

(1)

Let p is a prime number, A={2,3,...,p-2}.

For all x in A, there exists one and only one y in A, such that xy=1 (mod p) and x≠y.

Proof

Let a,m,n in A, m>n.

Suppose a^m-a^n=0 (mod p).

a^m-a^n=a^n(a^(m-n)-1)=0 (mod p)

Hence, a^n≠0 (mod p) since a≠0 (mod p).

So, a^(m-n)=1 (mod p).

By Fermat's little theorem, m-n=p-1 since 1<a<p. (contradiction)

Fix a in A.

For all m,n in A, m≠n, then a^m≠a^n (mod p).

Let a^i=b_i (mod p), i in A, then b_i in A.

|{b_i|i in A}|=|A|, so for every x in A, there exists i in A such that x=b_i.

By Fermat's little theorem, a^(p-1)=1 (mod p).

So, for all x=b_i in A, there exists y=b_(p-1-i) in A such that xy=1.

Suppose x,y,z in A, y≠z, xy=xz=1 (mod p).

Hence, x(y-z)=0 (mod p).

y=z (mod p) since x≠0 (mod p). (contradiction)

So, there exists one and only one y in A such that xy=1 (mod p) and x≠y.

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以上沒有很重要XD, 寫的有點隨便, 也有可能錯

重點在下面

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(2)

By (1), we know that for all x in {2,3,...,p-2}, there exists one and only one y in {2,3,...,p-2}, such that xy=1 (mod p) and x≠y.

So, (p-1)!=(p-1)*1*1*...*1=p-1 (mod p).

(p-1)!+1=(p-1)+1=p=0 (mod p)

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基本上就是這樣(汗

竟然用了費馬小定理>"<, 應該有好很多的證法

其實可以直接用Z_p是finite field去證...

不過我忘了怎麼用代數證Z_p是finite field XDDD

[FTICR 10]

感謝以上兩位

我有在wiki上找到了(只是有個地方看不太懂@@)

本來不知道這叫wilson定理