April 20th, 2018, 06:55 AM  #21 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 282 Thanks: 85 Math Focus: Algebraic Number Theory, Arithmetic Geometry  
April 29th, 2018, 08:36 AM  #22 
Newbie Joined: Mar 2018 From: United States Posts: 27 Thanks: 2 
For clarification, if p is a prime then 1, 2, . . . . . . p1 is a multiplicative group mod p. The 2cycle of numbers generated by 2 of form $2^n$ mod p is a subgroup (not necessarily proper). The size of the subgroup must divide p1. For example, 1, 2, . . . . 6 is the multiplicative group mod 7 and the 2cycle 2, 4, 1 is a subgroup. The subgroup has size 3 which divides 6. Let's designate the length of the 2cycle for n by $[n]_2$ Then $[7]_2 = 3$. If we look at the prime number 11, $[11]_2 = 10$. Thm: If n > 1 is an odd number and [n]_2 = n1, then n is a prime number. Proof: ??? 
April 30th, 2018, 04:05 AM  #23  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 282 Thanks: 85 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
There's nothing special about $2$ here. If there is any integer $k$ such that $[n]_k = n1$, then $n$ is prime for the same reason.  
May 1st, 2018, 06:38 AM  #24  
Newbie Joined: Mar 2018 From: United States Posts: 27 Thanks: 2  Quote:
How are you defining this for any $n$?  
May 1st, 2018, 07:06 AM  #25  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 282 Thanks: 85 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
In the special case that $n$ is prime, every number from $1$ to $n1$ is coprime to $n$ (i.e. is invertible mod $n$), so the group consists of all of these numbers.  
May 13th, 2018, 12:02 PM  #26  
Newbie Joined: Mar 2018 From: United States Posts: 27 Thanks: 2  Quote:
If we look at mod 7, then {1, 2, 3, 4, 5, 6} is a group, and 2 generates {2, 4, 1) while 3 generates the full group. So $[7]_2 = 3$ does not allow us to draw conclusions about 7 being a prime whereas $[7]_3 = 6$ does. In general, how can we best use the prime divisors of $n1$ to extract information about $n$?  
May 13th, 2018, 01:34 PM  #27  
Senior Member Joined: Aug 2012 Posts: 2,076 Thanks: 593  Quote:
https://en.wikipedia.org/wiki/Primitive_root_modulo_n  
May 29th, 2018, 04:13 PM  #28 
Newbie Joined: Mar 2018 From: United States Posts: 27 Thanks: 2 
True or false? If $p$ is a prime and $2p+1$ is a prime, then $2p+1$ divides $2^p1$ or $2^p+1$ 
May 29th, 2018, 05:25 PM  #29  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 282 Thanks: 85 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
If you work a little harder, you can say when each case happens: if $p \equiv 3 \bmod 4$ then $2p + 1$ divides $2^p  1$; if $p \equiv 1 \bmod 4$ then $2p + 1$ divides $2^p + 1$. The only ways I can think of to prove this right now use a small amount of the theory of quadratic residues. If I think of a more elementary approach (at least one that isn't essentially spending a page building up the theory from scratch!) I'll come back and post it.  
June 10th, 2018, 11:14 AM  #30  
Newbie Joined: Mar 2018 From: United States Posts: 27 Thanks: 2  Quote:
Quote:
Given any n, if $2n+1$ is a prime, then $2n+1$ divides $2^n1$ or $2^n+1$  

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