October 12th, 2015, 03:24 AM  #11 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
2^340 mod 341=1 I do not understand you. Either you read quickly what I wrote either you are talking about something else. I`m not talking about the Fermat numbers https://en.wikipedia.org/wiki/Fermat_number I`m talking about the Fermat pseudoprimes base 2 https://en.wikipedia.org/wiki/Fermat_pseudoprime 
October 12th, 2015, 06:28 AM  #12  
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47  Quote:
Now I see you want to treat general cases of pseudoprimality, like 341, using that other sequence, which I baptized as "Mobel Sequence" after you. Now I see what you want . I recall that I proved what term should be divisible by $p$ in those sequences, so I'll continue this discussion in that topic in regard to the proof as given in: Conjecture "2^n"  
October 12th, 2015, 07:05 AM  #13 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
I started by the sequence conjecture 2^n. Then I found a new (?) primality test which equivalent to Fermat test. Some numbers failed my test. Then come the question on filtering out those numbers. I have found the algorithm on how to filter out those numbers. I did not implement the algorithm so I do not know its algorithmic complexity. I`m not programmer. I use Excel plus paper and pen. That`s it. In any case I solved theoretically the problem. I have found today a particular link between 2 sequences. The question now on the table is : can we use my primality test to factorize odd numbers? 
October 12th, 2015, 07:29 AM  #14  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
You've reduced an easy problem to a very hard one. But it was already known that the factorization of p1 allows for a quick test of the primality of p. Further, in the 1970s Brillhart, Lehmer, and Selfridge improved this test so that you only needed a partial factorization of p1 to prove the primality of p, so there's a much better way already.  
October 12th, 2015, 07:38 AM  #15  
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  Quote:
You are commenting this step but many things came after that. Nothing astronomical. Read first and then comment please.  
October 12th, 2015, 07:53 AM  #16 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  
October 13th, 2015, 07:57 AM  #17 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
There is no need to factorize n1. All you need to know is n1 odd or even If it is even then you split in 2. As long as it is even you continue splitting in 2. n=341 n1=340 340 is even I split in 170170 I compute : 2^3401=(2^1701)*(2^170+1) I have two factors : f1=2^1701 and f2=2^170+1 I compute gcd(f1,n)=gcd(2^1701,n)=341 gcd(f2,2^170+1)=1 At this step I remove f2 because none of its factors divide 341 Hence I continue only with f1=2^1701 Now it 170 odd or even 170 is even 170=85+85 Hence : 2^1701=(2^851)*(2^85+1) f3=2^851 f4=2^85+1 I compute 2 gcd`s : gcd(2^851,341)=31 gcd(2^85+1,341)=11 Hence 341 is composite = 31*11 If you understand this algorithm then you will not need to know the factorization of n1 to filter out the Fermat pseudoprimes. If 2^t1 with t odd I have the solution too. Many Fermat pseudo prime numbers (like 341 and others) will be eliminated before reaching the step of odd exponents. Last edited by mobel; October 13th, 2015 at 08:14 AM. 
October 13th, 2015, 08:13 AM  #18 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
There is NO need to factorize n1. I repeat it and I will repeat it twice. Some primality tests require factorizing n1. When I say factorizing I mean finding all the primes dividing n1. I do not need that. I propose a binary research : You plit in 2. You compute 2 gcd`s. Either the 2 will show that n is composite Either one of them will be removed from the research and we continue with the other who gcd (n,fi)=n *fi is the dactor not in the sens of the particular divisor). As long you do not distinguish the 2 : the polynomial factor of the 2^s+1 or 2^s1) and the individual prime diving n1 (p1,p2,...pk) you could not understand my test. 
October 13th, 2015, 08:16 AM  #19 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
If this not a definitive solution of filtering out the Fermat pseudoprime numbers I will stop everything and go to spend my time with birds and donkeys.

October 13th, 2015, 08:52 AM  #20 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
When some primality tests require the factorization of n1 it is not a problem but when Mobel talks about the "factorization" of n1 it becomes astronomical and so on. The way I look to the factorization is DIFFERENT. n1 is a product of (2^k(*(odd number) I do not need to go on details of the factors (the final divisors). My conclusion is that no one read carefully what is written. Behaving like robots is the main behaviour. Fermat? and the guy start talking about the Fermat prime numbers instead of talking about Fermat pseudoprime numbers. 561 as I indicate does not FAIL to my test but it fails to Fermat test. He said I was wrong but no one corrected it. Factorization? what? what? it is ASTRONOMICAL!!!!!!!!!!! Because I`m Mobel. But if others use it then it is NOT astronomical it is doable. Better for me to stop posting. 

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