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 2021-03-13, 17:29 #23 PAT291   Feb 2017 Belgium 3·7 Posts Thank you very much for your response. That’s a bit of my problem. I don’t know where to go to, to put things in to music and if it’s worth the effort. That’s why I came here for some feedback or help. Can such a proof be useful? Can it help the forum? Or computations in some way? It will take time to put things in to music. Maybe it’s worth writing a paper on it. But I am not into that kind of writing business nor have I the time or the skills to write things scrupulously down from a to z in a paper. I found something (hobbywise) but maybe it’s not of use. I thought people here would be interested or know whether it’s worthwhile asking somebody to verify or whether it can be of any use. If my assumption stands it would be a sufficient test to say Mp is prime.
2021-03-13, 23:10   #24
PAT291

Feb 2017
Belgium

101012 Posts

Quote:
 Originally Posted by paulunderwood 3 is special because kronecker(3,Mp)==-1 for all odd p. Lehmer's test is an efficient implementation of Mod(Mod(x+2,Mp),x^2-3)^2^p==1
Indeed 3 seems to be very special. Apparently it was already noticed by yourself and others J

 2021-03-14, 07:31 #25 PAT291   Feb 2017 Belgium 3×7 Posts fyi I found everything by myself, not knowing any theorems. It was just some spielerei with numbers. I am not familiar with most of the notations (f.e. kronecker notation). That means that we don’t speak the same language yet. Which will lead almost certainly to some misunderstandings and miscommunications. Sorry in advance for this.
2021-03-15, 18:02   #26
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

240608 Posts

Quote:
 Originally Posted by PAT291 I claim that if (3^(Mn-1)/2 + 1)==0 (mod Mn) then Mn is prime. Mn being a Mersenne number.
It seems that for the most of the mersenne numbers, this test says nothing. If Mn=7, you have 3^6=1, 1/2=4, 4+1=5 (mod 7). So, it doesn't tell me that 7 is prime or not. Or maybe you screwed up the parentheses?
In fact, for any odd prime number q, mersenne or not, and any base b, b^(q-1)=1 (mod 2), so the result is always 1/2+1 (mod q). Or (q+3)/2. Always. Never zero. So, this is never true for primes.

Last fiddled with by LaurV on 2021-03-15 at 18:08

2021-03-15, 20:19   #27
Dr Sardonicus

Feb 2017
Nowhere

18F016 Posts

Quote:
Originally Posted by LaurV
Quote:
 Originally Posted by PAT291 I claim that if (3^(Mn-1)/2 + 1)==0 (mod Mn) then Mn is prime. Mn being a Mersenne number.
It seems that for the most of the mersenne numbers, this test says nothing. If Mn=7, you have 3^6=1, 1/2=4, 4+1=5 (mod 7). So, it doesn't tell me that 7 is prime or not. Or maybe you screwed up the parentheses?
In fact, for any odd prime number q, mersenne or not, and any base b, b^(q-1)=1 (mod 2), so the result is always 1/2+1 (mod q). Or (q+3)/2. Always. Never zero. So, this is never true for primes.
The outer parentheses may be superfluous, and perhaps confusing, but eliminating them leaves

3^(Mn-1)/2 + 1 == 0 (mod Mn)

There is no question the OP meant

3^((Mn-1)/2) + 1 == 0 (mod Mn)

Last fiddled with by Dr Sardonicus on 2021-03-15 at 20:20 Reason: fignix stopy

 2021-03-22, 09:42 #28 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 24·643 Posts I know that you know...

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