Wilson's theorem, infinite monkey's Theorem

Wilson's theorem proves that if p is a prime number, p can be divisible by (p-1 )! + 1

First: First test the strange Value

If p = 2, true;

If p = 3, true;

Then we will start to study the situation where p> = 5

Second: let's prove several conclusions first.

Set A = {2, 3, 4 ,......, P-2}

Note B = {a, 2a, 3a ,......, (P-1)}

[There will be no number of modulo p with the same remainder in B .......................................... Conclusion 1

Proof: Set b1a, b2a, B, b2, b1, [1, P-1] to different numbers.

Assume that b1a has b2a (mod p)

Then | b1-b2 | a limit 0 (mod p)

However | b1-b2 | ε [1, P-2]

So | b1-b2 | a, B

However, it is clear that no number in B can be divisible by p, so the question cannot be set.

So Conclusion 1 proves

Because B has a certain number of elements, the remainder of non-p-multiples modulo p is also only a certain number of elements. Therefore, the remainder of modulo p in B forms a set of C = {1, 2, 3 ,..., P-1 }]

In B, the number of the remainder 1 divided by p is ba, B, A, and B! = ....................... Conclusion 2

Proof: If B = 1, ba = a, because a is A, so! = 1, not true

If B is P-1, p-a (mod p) is a p-a (p). Because A is a, p-a is [2, P-2], not true.

If B = a, then a2 contains 1 (mod p), so (a + 1) (A-1) + mod p), Because p is a prime number, so (a + 1) and (A-1) have and only one is a multiple of p. because a in A, so (a + 1) in [P-1], (A-1) in [1, P-3], so they can not be a multiple of p

Conclusion 2: Evidence]

[Conclusion 2: If a is different, B is different .............................. Conclusion 3

Proof: Set a1! = A2, but all belong to A, and ba1 ba2 has 1 (mod p)

Yi Zhi ba1, ba2 ε B, but according to conclusion 1, it is not true

Conclusion 3: Evidence]

Third: If we have proved so many conclusions, this theorem has basically surfaced.

Conclusion 1 tells us that in B, each a has a unique bits, [1, P-1], so that AB branch 1 (mod p)

Conclusion 3 sublimation Conclusion 1: The range of B is reduced to B, [2, P-2], and a = B. that is to say, in [2, P-2], each number can find a unique inverse element (Note: p is obviously an even number, therefore, all integers in this range can be divided into a pair of children), and the reverse element belongs to [2, P-2].

P-1 )! + 1 second (p-1) + 1 second 0 (mod p)