@Igor Rivin

I will answer Your question here. I have done a research about safe primes, and I have found a new deterministic primality test for safe primes. This test goes as follows: We have two statements:

1.) Let $p=3$ (mod $4$) be prime. $2p+1$ is also prime if and only if $2p+1$ divides $2^p−1$.

 

2.) Let $p=1$ (mod $4$) be prime. $2p+1$ is also prime if and only if $2p+1$ divides $2^p+1$.

(Statement 1. is proven by Lagrange 1775, and statement 2. is proven by Batominovsky 2015)

So if a number $N=2\cdot p+1$ holds the congruence $2^p\equiv \pm1\ ($mod $N)$ then it is definitely prime.

From this point I went one step further to $N=p\cdot2^n + 1$.

@Igor Rivin

I will answer Your question here. I have done a research about safe primes, and I have found a new deterministic primality test for safe primes. This test goes as follows: We have two statements:

1.) Let $p=3$ (mod $4$) be prime. $2p+1$ is also prime if and only if $2p+1$ divides $2^p−1$.

2.) Let $p=1$ (mod $4$) be prime. $2p+1$ is also prime if and only if $2p+1$ divides $2^p+1$.

(Statement 1. is proven by Lagrange 1775, and statement 2. is proven by Batominovsky 2015)

So if a number $N=2\cdot p+1$ holds the congruence $2^p\equiv \pm1\ ($mod $N)$ then it is definitely prime.

From this point I went one step further to $N=p\cdot2^n + 1$.

@Igor Rivin

I will answer Your question here. I have done a research about safe primes, and I have found a new deterministic primality test for safe primes. This test goes as follows: We have two statements:

1.) Let $p=3$ (mod $4$) be prime. $2p+1$ is also prime if and only if $2p+1$ divides $2^p−1$.

2.) Let $p=1$ (mod $4$) be prime. $2p+1$ is also prime if and only if $2p+1$ divides $2^p+1$.

(Statement 1. is proven by Lagrange 1775, and statement 2. is proven by Batominovsky 2015)

So if a number $N=2\cdot p+1$ holds the congruence $2^p\equiv \pm1\ ($mod $N)$ then it is definitely prime.

From this point I went one step further to $N=p\cdot2^n + 1$.

@Igor Rivin

I will answer Your question here. I have done a research about safe primes, and I have found a new deterministic primality test for safe primes. This test goes as follows: We have two statements:

1.) Let $p=3$ (mod $4$) be prime. $2p+1$ is also prime if and only if $2p+1$ divides $2^p−1$.

2.) Let $p=1$ (mod $4$) be prime. $2p+1$ is also prime if and only if $2p+1$ divides $2^p+1$.

(Statement 1. is proven by Lagrange 1775, and statement 2. is proven by Batominovsky 2015)

So if a number $N=2\cdot p+1$ holds the congruence $2^p\equiv \pm1\ ($mod $N)$ then it is definitely prime.

From this point I went one step further to $N=p\cdot2^n + 1$.