I show here that it is possible to find a same result for  p=1 [mod 4], with a new corresponding primality test, a generalization to Cunningham chains of second kind and primality tests of a new type ("in cluster").
 

Theorem 1:

    If p=1 [mod 4] is prime, q=2p+1 is also prime if and only if q divides 2^p + 1.

    It is also a primality test for Sophie Germain primes with p=4a+1.
    It can be used with Proth numbers p=k.2ⁿ+1.
 
 Demonstration :
    1) If p=1 [mod 4] and q=2p+1 are prime then q divides 2^p+1 :
        Indeed 2^2p-1 = (2^p-1)*(2^p+1)  and 2^2p-1 = 0 [mod q]  (Fermat th. for q).
        As p=1 [mod 4] ,  q=3 [mod 8] , therefore (2/q) = -1 , 2 <> m² [mod q]  or 2^p<> m^2p [mod q].
        But m^2p=1 [mod q] . Therefore q does not divide 2^p-1, it divides 2^p+1.
    2) If p is prime and q=2p+1 divides 2^p+1 then q is prime :
         Indeed we use the primality test  in n-1 : if for each prime factor f of n-1 there exists an integer a such that
a^n-1=1 [mod n] and  a^(n-1)/f<>1 [mod n] then n is prime.
        We have here n-1=2p , 2^2p= (2^p-1)*(2^p+1) +1, therefore 2^2p= 1 [2p+1] since 2p+1 divides 2^p+1
 2²<>1 [2p+1] and 2^p=-1<> 1 [mod 2p+1 ] , therefore q=2p+1 is prime.
 
 

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 Generalization to Cunningham chains of second kind :

Theorem 2:

    If p=3 [mod 4] and q=2p-1 are prime then q divides 2^p-1 + 1.
 
    It is also a pseudo-primality test for Cunningham chains of second kind with p=4a+3 :
    If p=3 [mod 4] is prime and q=2p-1 divides 2^p-1 + 1,  q is probably prime.

    Demonstration :
        2^2p-2-1 = (2^p-1-1)*(2^p-1+1)  and 2^2p-2-1 = 0 [mod q]  (Fermat th. for q).
        As p=3 [mod 4] ,  q=5 [mod 8] , therefore (2/q) = -1 , 2 <> m² [mod q]  or 2^p-1<> m^2p-2 [mod q].
        But m^2p-2=1 [mod q] . Therefore q does not divide 2^p-1-1, it divides 2^p-1+1.
 

 Theorem 3:

    If p=1 [mod 4] and q=2p-1 are  prime then q divides 2^p-1 - 1.
 
    It is also a pseudo-primality test for Cunningham chains of second kind with p=4a+1 :
    If p=1 [mod 4] is prime and q=2p-1 divides 2^p-1 - 1,  q is probably prime. 

    Demonstration :
        2^2p-2-1 = (2^p-1-1)*(2^p-1+1)  and 2^2p-2-1 = 0 [mod q]  (Fermat th. for q).
        As p=1 [mod 4] ,  q=1 [mod 8] , therefore (2/q) = 1 , 2 = m² [mod q]  or 2^p-1= m^2p-2 [mod q].
        But m^2p-2=1 [mod q] . Therefore q divides 2^p-1-1.

 
    But as p is prime, p divides also 2^p-1-1.
 

Theorem 3a :

    If p=1 [mod 4] and q=2p-1 are  prime then p*q divides 2^p-1 - 1.
 
    It is also a pseudo-primality test for Cunningham chains of second kind with p=4a+1 :
    If n=1 [mod 4] and n₁=2n-1  and if n*n₁ divides 2^n-1 - 1,  then n and n₁ are probably prime.

    The generalization of this test is even evident and immediate, since n=1 [mod 4], 2n-1=1[mod 4]  :
    Let p_i, p_i+1, p_i+2 prime such that p_i=1[mod 4] , p_i+1=2p_i-1, p_i+2=2p_i+1-1.
    Therefore we have 2^(p_i+1-1)-1=0 [mod p_i+1*p_i+2], but 2^(p_i+1-1)-1=2^(2(p_i-1))-1 and 2^(p_i-1)-1=0[mod p_i],
         2^(p_i+1-1)-1=0 [mod p_i*p_i+1*p_i+2].
    Therefore this relationship is necessary in order that p_i, p_i+1, p_i+2 are prime.

Pseudo test "in cluster" :

    If n=1 [mod 4] and n₁=2n-1,  n₂=2n₁-1, ..., n_i=2n_i-1-1, and if n*n₁*n₂*..*n_i divides 2^(n_i-1-1) - 1,
        then  n,n₁,n₂, .., n_i  are probably prime, otherwise one at least of numbers n,n₁,n₂, .., n_i is composite.
    This method that allows a pseudo test of several numbers at the same time can speed up considerably the research of Cunningham chains of second kind with any length. It is even well adapted to test numbers Proth n=k.2ⁿ+1 (form 4a+1) "in cluster", a complete test of each number being undertaken only if this first test passes.

Complements to these tests (soon ...)

• Prime Pages   by Chris Caldwell .
• Yves Gallot's Proth.exe and Cunningham Chains  by Warut Roonguthai.
• Psieve - Sieving exponents for k.2^n+-1  by Chris Nash.