June 17th, 2018, 01:44 PM  #31  
Newbie Joined: Mar 2018 From: United States Posts: 19 Thanks: 0  Quote:
. The following identities are mod ($2p+1$): . Case 1: If $2p+1$ divides $2^p1$, . $2^p ≡ 1$ . $2^{p1}$ ≡ $p+1$ . $(2^{(p1)/2})^2$ ≡ $p+1$ . Since a mod 4 square can only be 0 or 1, the only available solution is $p$ mod $4$ ≡ $3$. Example: $p$ = 11 . Case 2: If $2p+1$ divides $2^p+1$, . $2^p$ ≡ $2p$ . $2^{p1}$ ≡ $p$ . $(2^{(p1)/2})^2$ ≡ $p$ . Since a mod 4 square can only be 0 or 1, the only available solution is $p$ mod $4$ ≡ $1$ Example: $p$ = 5  
June 18th, 2018, 01:17 AM  #32 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 204 Thanks: 60 Math Focus: Algebraic Number Theory, Arithmetic Geometry  
July 18th, 2018, 04:45 AM  #33  
Newbie Joined: Mar 2018 From: United States Posts: 19 Thanks: 0  Quote:
For a given odd $n$, let the cycle generated by $2$, {$2^i$ mod $n$}, be represented by $< n >_2$. For example, $< 9 >_2$ = {$2, 4, 8, 7, 5, 1$}. We can call this the Mersenne Cycle or Mersenne Group for $n$. We can call the size or length of this group the Mersenne Order or Mersenne Length and designate it by $[n]_2$. Then $[9]_2$ = $6$. The Mersenne Group is a subgroup (not necessarily proper) of the full multiplicative group mod $n$, so its length must divide the size of the full group. In the case of $9$, they are the same. In the case of $7$, $< 7 >_2$ = {$2, 4, 1$} and $[7]_2$ = $3$. The full multiplicative group mod $7$ = {$1, 2, 3, 4, 5, 6$} and has size $6$. We see that $3$ divides $6$. The full multiplicative group mod $n$ has size $n1$ if and only if $n$ is a prime number. In any case, $n$ is a divisor of the Mersenne Number $2^{[n]_2}$  $1$, which is the smallest Mersenne Number that $n$ divides. For example, $9$ is a divisor of $2^6  1$. This is where I will stop for now, to not make this post too long.  
July 19th, 2018, 05:03 PM  #34  
Newbie Joined: Mar 2018 From: United States Posts: 19 Thanks: 0  Quote:
Finally, we need to know that if $r$ is a divisor of $s$, then $2^r  1$ divides $2^s1$. This can be seen from $2^s  1$ = ($2^r  1$)($2^{s  r} + 2^{s  2r} + . . . + 2^r + 1$) So $2n + 1$ divides $2^m  1$ which divides $2^{2n}  1$ = ($2^n  1$)($2^n + 1$), which implies the desired conclusion. As an example of the above, $43 = 2*21 + 1$ is a prime. Then $[43]_2 = 14$, so $43$ divides $2^{14}  1$ which divides $2^{42}  1 = (2^{21}  1)(2^{21} + 1) $. In this case, $43$ divides $2^{21} + 1$.  

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