20210322, 17:12  #12 
"Viliam Furík"
Jul 2018
Martin, Slovakia
1253_{8} Posts 

20210322, 17:18  #13 
"Viliam Furík"
Jul 2018
Martin, Slovakia
683 Posts 
The pattern you observed is fairly trivial I would say. If the prime exponent you're testing modulo p is itself a Mersenne prime, then you are testing the Mersenne prime with a smaller exponent. For all Mersenne primes, S(x) (for x >= p) is equal to 2, because S(p2) = 0 (mod p), S(p1) = 2 (mod p), and S(p) = 2 (mod p).
So if there can be such a test for Mersenne numbers modulo p, it won't work for double Mersennes. 
20210323, 12:39  #14 
Feb 2017
Nowhere
5010_{10} Posts 
The fact that S(P2) mod p doesn't tell you anything about whether 2^P  1 may be prime when P = 2^p  1 is prime, disproves the existence of a general "S(p2) mod p" test for Mersenne primes.
More important than this specific case IMO is the fact that S(p2) == trace(u^e) (mod p), where u = Mod(x+2,x^2  3) and e = 2^(p2)%m, where m = p  1 if p == 1 or 11 (mod 12) m = p + 1 if p == 5 or 7 (mod 12). We know that u^m == 1 (mod p) so the exact multiplicative order of u (mod p) is a divisor of m. I have not investigated when it might be known to be a proper divisor. If there were some pattern in S(p2) (mod p) that indicated whether 2^p  1 was prime, presumably it would somehow be manifest in the "reduced" exponent e = 2^(p2)%m. I am not aware of any convenient formula for this remainder, but I don't know that there isn't one. So far the "trivial" pattern when p is a Mersenne prime is the only consistent one I am aware of, and in that case S(p2) mod p utterly fails to indicate whether 2^p  1 is prime. 
20210411, 16:06  #15 
"Viliam Furík"
Jul 2018
Martin, Slovakia
683 Posts 
Update
I have been trying to find something and came across the Wikipedia page for the LL test, where it says, that the sign of the S(p3) is (1)^{(p1)/2} (except for p = 5), for starting value of 2/3, which is the Wagstaff number with exponent p, i.e. (2^{p}+1)/3. This turns out to be 1 (mod p). Thus LL test with starting value of 2/3, and done modulo p is entirely composed of terms p1, always. Using the property of the sign I tried to devise a checking test which under the assumption of the Mp being prime checks whether the conditions fit according to the known sign of the S(p3).
S(p2) = X * (2^{p}1) S(p2) = X (mod p) S(p3) = K * (2^{p}1) + sign * 2^{(p+1)/2} S(p3) = K + sign * 2^{(p+1)/2} (mod p) S(p2) = S(p3)^{2}  2 = K^{2} + sigma * 2 * K * 2^{(p+1)/2} + 2^{p+1}  2 = K^{2} + sigma * K * 2^{(p+3)/2} + 2 = K(K + sigma * 2^{(p+3)/2}) + 2 using the property that all terms after the first one are 1 (mod p) S(p2) = S(p3) = 1 (mod p) S(p2) = K(K + sigma * 2^{(p+3)/2}) + 2 = 1 (mod p) K(K + sigma * 2^{(p+3)/2}) = 3 (mod p) I observed that for all primes the 2^{(p+3)/2} is either 4 or 4 (mod p). I later realised that it's because 2^{p+3} is 16 (mod p), or equivalent if p < 16, thus the square root of 16 is 4 or 4. This means that the equation becomes K(K + 4) = 3 (mod p) For the plus case, the solutions for K are 3 and 1, of which only 3 seems to satisfy the condition from the fourth row. For the minus case, the solutions are 1 and 3, and again, only 1 seems to ever satisfy the condition. That would all be nice unless ALL primes p satisfied these conditions, which they do. Is there any other way to modify this, so that we can use this test at all, or does it mean something in the sense of all Mersenne numbers being something like PRP modulo p (thus signifying impossibility of such test)? I also tried something with starting value 4, and 10, because there is a relation between signs of S(p3) for these starting values, but it only gives the product of the signs. 
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