20170105, 00:57  #1 
"Sam"
Nov 2016
2^{2}·79 Posts 
Composite integers n satisfying prime exponents of Mersennes
I think this is the right subforum to post in for my question.
Given the following Theorem 1: If n is an odd prime, each divisor p of 2^n1 has the form 2nk+1 and 8r+1. Correction: If n is an odd number, each divisor p of 2^n1 has the form 8r+1. Proofs can be found on this page. Are there any composite integers n following Theorem 1? Is there a proof that there do or do not exist composite integers n with this property and also if infinitely exist? Are there any known examples? Thanks for answering. One case scenario is a prime n in https://oeis.org/A000043 being a Wieferich Prime. The composite number n^2 follows theorem 1. No known examples. The other part I know of is n not being a multiple of 2, 3, 5, 7, 11, 13, etc. Any other thoughts/improvements? 
20170105, 02:05  #2  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×3,041 Posts 
Quote:
We already know that for 2^p1 to be prime, p must prime. So we are not interested in nonprime odd n values in the GIMPS process. Quote:
n is an odd prime in Theorem 1. Theorem 1's backbone is the "if S1, then S2" structure. What you are trying to ask is "if not S1, then 1) is S2 always false? 2) is S2 always true? 3) is S2 sometimes true and sometimes false?" ...and the answer is "3". Counterexamples to both 1 and 2: take n=15, then 2^n1 has factors 7, 31, 151 and 31 is = 2nk+1, but 7 2nk+1 So your new Theorem 1like proposition comes down to this "If n is an odd composite, then anything can happen." That's not much of a statement, right? P.S. You choice of the subforum was exactly right! Thanks for getting into the right forum. 

20170105, 02:08  #3  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
etc. 

20170105, 02:14  #4  
"Sam"
Nov 2016
474_{8} Posts 
Quote:
Property 1: Integers n such that all divisors p of 2^n1 is of the form 2kn+1 and 8r+1. Are there any integers n following property 1 that are composite? 

20170105, 02:30  #5  
"Forget I exist"
Jul 2009
Dumbassville
20B1_{16} Posts 
Quote:
if 2kn+1 = 8r+1 then 2kn=8r so kn=4r so there would have to be a factor of 4 in there case 2: if 2kn+1 = 8r1 then 2kn+2 = 8r which when you factor out a 2 you get kn+1 = 4r so kn must be an odd number of form 4j+3 for j=r1. edit: and following the links at the bottom of the page you linked to earlier we get that n can't have any sophie germain factors as the mersennes of prime exponents dividing such a case would need the original k to have all the factors of n for it to be of form 2kn+1 for the new value n. Last fiddled with by science_man_88 on 20170105 at 02:55 

20170105, 03:56  #6  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9123_{10} Posts 
Quote:
But don't even think about suggesting that this is a primality test for n. Because in order to prove that a tiny n is prime, you will have to factor 2^n1 completely, then prove primality of prime factors ...which are much larger than n. (I.e. sell a car to buy a loaf of bread. Only you don't have a car, so you sell the house to buy a car, and so on.) 

20170105, 04:13  #7 
Jun 2003
2^{2}·11·107 Posts 
The primitive part of Mersenne numbers will obey the property

20170105, 04:36  #8  
"Sam"
Nov 2016
474_{8} Posts 
Quote:
To show that at least 1 composite k  n: If some other odd prime factor y which is not k divides all s_n1, and for all divisors v_n1 of 2^y1 the gcd of (v1, v_21..., v_31) is of the form 2^x*y*k*z, then 2^(ky)1 follows property 1 and ky is obviously composite so we are done with this case. Of course there are many more scenarios/case of a composite n occurring. These are a few basic ones. Last fiddled with by carpetpool on 20170105 at 04:44 

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