(1-A to the power of two) (1-B to the power of two) + 4AB factorization

(1-A to the power of two) (1-B to the power of two) + 4AB factorization

（1-a²)(1-b²)+4ab
=1-b²-a²+a²b²+4ab
=a²b²+2ab+1-(a²-2ab+b²)
=(ab+1)²-(a-b)²
=(ab+1-a+b)(ab+1+a-b)

Prove that for any prime P > 3,2 * (P-3)! & # 8803; - 1 (mod p)
Hint: Wilson theorem can be used

(p-1)!-2*(p-3)!=（p-3）!（p^2 -3p）=（p-3）!×p（p-3）
So p | ((p-1)! - 2 * (P-3)!)
So according to Wilson's theorem:
2*(p-3)!≣(p-1)!≣-1(mod p)

If n is a positive integer greater than 2, we prove that there is at most one prime number between the n-th power-1 of 2 and the n-th power + 1 of 2

If 2 ^ n-1, 2 ^ n, 2 ^ n + 1 are divided by 3, the remainder must be 0, 1, 2 respectively (the order may be different)
However, 2 ^ n has only a prime factor of 2, and it can not be divided by 3 to make 0
Then at least one of 2 ^ n - 1 and 2 ^ n + 1 is divided by 3 to make 0. It is not a prime number
So at most one of 2 ^ n - 1 and 2 ^ n + 1 is prime
The proof is complete