mersenneforum.org > Math (unproven) Potential new primality conjecture for Mersenne numbers
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2021-11-29, 21:46   #12
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

11·881 Posts

Quote:
 Originally Posted by ETLem As an aside the thread title was changed to "unproven ... conjecture". That seems a tad redoundant ;-) How about "A new conjecture about Mersenne Primes".
What made you think that this is a 'new conjecture'?

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.

 2021-11-29, 22:08 #13 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100101110110112 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 ap = a (mod p) which is the a-PRP 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 Berrizbeitia-Iskra path (test for Gaussian-Mersenne numbers, proven) and actually prove it, then maybe you will have something.
2021-11-29, 23:06   #14
charybdis

Apr 2020

22×3×72 Posts

Quote:
 Originally Posted by Batalov Can the sign be '-' for any odd p values?
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.

 2021-11-29, 23:19 #15 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 11·881 Posts Thank you for answering the rhetorical question. The '-' sign happens exactly at q=3. (Which makes a-PRP still work, even for q=3, because you square both sides for the last time in it.)
2021-11-30, 13:53   #16
Dr Sardonicus

Feb 2017
Nowhere

23×233 Posts

Quote:
 Originally Posted by ETLem By the way is there known Carmichael number(s) that are also Mersenne number(s)?
Not AFAIK. Probably better to ask whether any Mersenne numbers are Carmichael numbers. I will here restrict this question to prime exponents, since Mn has algebraic factors so is obviously composite when n is composite.

In theory, yes AFAIK. In practice, I wouldn't hold my breath.
Quote:
 Or could it be proven that it exist Mersenne numbers that are also Carmichael numbers?
AFAIK none are known, and all the easy things to check are exclusionary.

If q = 2*k*p + 1 divides Mp, and k does not divide 2p-1 - 1 (for example, if k is even), then Mp is not a Carmichael number. [Note: If k is even, it is also divisible by 4.]

If Mp is completely factored, and has evenly many factors, at least one factor q = 2*k*p + 1 will have k divisible by 4, so Mp is not a Carmichael number.

If a PRP test says Mp is composite, it is not a Carmichael number. If p is large, a PRP test will AFAIK not "officially" be run on Mp if a factor has been found, though the cofactor will be PRP tested.

For "small" primes p, Mp has been PRP tested to the base 3 whether any factors are known or not, with no composite Mp "passing" the test. So Mp is definitely not a Carmichael number for p up to whatever limit this has been done.

I don't know how many Mp 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 2021-11-30 at 13:54 Reason: add qualifier

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