20120919, 16:06  #34 
Aug 2002
Buenos Aires, Argentina
5·269 Posts 
Based on the previous post: let n = (2^{2p} 2^{p}+1)/3 where p is a prime number, is it true that 2^{n1} = 1 (mod n) ?

20120919, 16:55  #35  
"Robert Gerbicz"
Oct 2005
Hungary
2×727 Posts 
Quote:
Observe that: (2^(2*p)2^p+1)/3  2^(6*p)1 so it is enough to prove that for p>2: 2^(n1)==1 mod 2^(6*p)1 it is equivalent to n1==0 mod (6*p) so (2^(2*p)2^p2)/3==0 mod (6*p) (18*p)2^p*(2^p1)2 Let p>3 then it is true mod p (use Fermat's little theorem), 2 also divides, this is trivial since 2^p is even. For 9 use: p=6k+1 then 2^p=={2,5} mod 9, for the two cases: 2^p*(2^p1)2=={0,18}==0 mod 9. Since 2,9,p are relative prime numbers for p>3 it means that the divisibility is also true for 2*9*p=18*p. For p=3: 2^(n1)==1 mod n is also true. 

20120919, 17:18  #36 
Aug 2002
Buenos Aires, Argentina
5·269 Posts 
Thanks.

20210320, 08:03  #37 
Mar 2021
1 Posts 
Hi everyone
I know I am a bit late.... BUT I have the answer to your problem: take N = 2^524287  1. Notice that 524287 = 2^19  1, this is a clue for the proof ;) Indeed, you just need to consider the sequence X(n+1) = 2^X(n)  1 with X(0) = 19 and proove that all terms of the sequence verify X(n)*X(n+1) divide 2^X(n)  2. Have a nice day! 
20210321, 16:14  #38 
Mar 2021
1 Posts 
Hi
I would like to know if a^(2^n) mod (2^n1) = a² when a is between 0 and sqrt(2^n1) and a is an integer when 2^n1 is prime and only if it's prime ? I tried this with some Mersenne exposant and it apparently works. It can't be used to eliminate nonMersenne exponant quickly ? Thanks :D 
20210321, 22:43  #39 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{3}·89 Posts 
In My Humble Opinion (IMHO) the even numbers are easier to keep track of

20210322, 03:03  #40  
Romulan Interpreter
Jun 2011
Thailand
22227_{8} Posts 
Quote:
Yes, it can be used to eliminate composite Mp's, but not "fast". Well, not faster than we are doing already. 

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