梅森素数的证明

前提:

如果n是正整数,则代数恒等式成立:

 

a^{n_-bn_=(a-b)∑(ak_bn-1-k_{)     (0}}

 证明过程略.

则

  2^sp-1 = (2^s)^p-1^p=(2^s-1)^_∑(2s₎k  ( 0≤k≤p, k = k+1)

Theorem.

If for some positive integer n, 2ⁿ-1 is prime, then so is n.

Proof.

Let r and s be positive integers, then the polynomial x^rs-1 is x^s-1 times x^s(r-1) + x^s(r-2) + ... + x^s + 1.  So if n is composite (say r^.s with 1<s<n), then 2ⁿ-1 is also composite (because it is divisible by 2^s-1).

Corollary.

Let a and n be integers greater than one.  If aⁿ-1 is prime, then a is 2 and n is prime.