由费马小定理可以得到p | 2^(p-1) - 1所以p | 2^(p-1) - 1-p = 2^(p-1) - (p+1)所以设n = k(p^2-1)那么2^n = [2^(p^2-1)]^k = [2^(p-1)]^(k(p+1)) = (-1)^(k(p+1)) = 1 (mod p)所2^n - n = 1 - k(p^2-1) = 1 + k (mo...