[Math] Proof for new deterministic primality test

Question

Claim:

Let $p$ be a positive prime. Let $n \in \left\{1, 2, 3, …\right\}$. Then $N =
p\cdot 2^n+1$ is prime, if and only if it holds the congruence $3^{(N-1)/2} \equiv \pm1\ ($mod $N)$.

If the claim is true, we would have a fast deterministic test for numbers of the form $p\cdot2^n + 1$. That means, with small $p$ and large $n$, we could generate huge prime numbers, similar to Mersenne primes or Fermat primes.

A proof is needed. Thanks for Your attention.

Answer 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 $N=p\cdot2^n + 1$.