20211129, 21:46  #12  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,423 Posts 
Quote:
Your original thread title was false and misleading: 'Potential new primality test for Mersenne numbers'. One moderator implemented your request to change 'test' to 'conjecture'. Another added '(unproven)', so that potential readers could save themselves 10+ minutes to read by deciding not to not even being confused by title if what's being offered here. If I found a thread with this current title I wouldn't have even entered. Now that I have, I want my time back, but at least I can save it for someone else. 

20211129, 22:08  #13 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·2,423 Posts 
Also, suggestion for simplification of your statement.
Your a=1+√2 (mod p)  much simpler to look at. Are you saying that \(a^{{p+1}\over 2} = a\) (mod p) ? That's the same* as a^{p} = a (mod p) which is the aPRP test with a peculiar a value. _______ *except for implying the + sign of the "square root of the Fermal litttle test". Can the sign be '' for any odd p values? If you can follow the BerrizbeitiaIskra path (test for GaussianMersenne numbers, proven) and actually prove it, then maybe you will have something. 
20211129, 23:06  #14 
Apr 2020
593_{10} Posts 
As I'm sure you already know:
For general p, yes it can (eg p=17). For p a Mersenne prime, the answer is "no" for p greater than 7. By Euler's criterion, what we need to show is that a is a square mod p. This is not hard to do using quadratic reciprocity and other basic properties of Jacobi symbols; I will leave this as an exercise for the reader. In the process of solving it you will discover why the test fails for p=7. 
20211129, 23:19  #15 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,423 Posts 
Thank you for answering the rhetorical question.
The '' sign happens exactly at q=3. (Which makes aPRP still work, even for q=3, because you square both sides for the last time in it.) 
20211130, 13:53  #16  
Feb 2017
Nowhere
1010011110001_{2} Posts 
Quote:
In theory, yes AFAIK. In practice, I wouldn't hold my breath. Quote:
If q = 2*k*p + 1 divides M_{p}, and k does not divide 2^{p1}  1 (for example, if k is even), then M_{p} is not a Carmichael number. [Note: If k is even, it is also divisible by 4.] If M_{p} is completely factored, and has evenly many factors, at least one factor q = 2*k*p + 1 will have k divisible by 4, so M_{p} is not a Carmichael number. If a PRP test says M_{p} is composite, it is not a Carmichael number. If p is large, a PRP test will AFAIK not "officially" be run on M_{p} if a factor has been found, though the cofactor will be PRP tested. For "small" primes p, M_{p} has been PRP tested to the base 3 whether any factors are known or not, with no composite M_{p} "passing" the test. So M_{p} is definitely not a Carmichael number for p up to whatever limit this has been done. I don't know how many M_{p} among the "first tested" remain standing as possible candidates for being Carmichael numbers. My guess is, "Not many." Last fiddled with by Dr Sardonicus on 20211130 at 13:54 Reason: add qualifier 

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