April 20th, 2018, 05:55 AM  #21 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 188 Thanks: 57 Math Focus: Algebraic Number Theory, Arithmetic Geometry  
April 29th, 2018, 07:36 AM  #22 
Newbie Joined: Mar 2018 From: United States Posts: 14 Thanks: 0 
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, 03:05 AM  #23  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 188 Thanks: 57 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, 05:38 AM  #24  
Newbie Joined: Mar 2018 From: United States Posts: 14 Thanks: 0  Quote:
How are you defining this for any $n$?  
May 1st, 2018, 06:06 AM  #25  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 188 Thanks: 57 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, 11:02 AM  #26  
Newbie Joined: Mar 2018 From: United States Posts: 14 Thanks: 0  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, 12:34 PM  #27  
Senior Member Joined: Aug 2012 Posts: 1,889 Thanks: 525  Quote:
https://en.wikipedia.org/wiki/Primitive_root_modulo_n  

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