20060625, 17:46  #1 
Jun 2003
Pa.,U.S.A.
2^{2}×7^{2} Posts 
Fermats theorem and defining a 'full set' for any prime.
This thread may have to be delegeted to the miscellanea as was my thread on an indefinitely continuing sequence.
I believe Fermats theorem can be read ,or would cover the following: if an odd number p is prime, then 2^(p1)1=K(f)*p where K(f) is a positive natural number. I have stressed at times the importance of defining all base 2 numbers, (2^x) that add to any given positive integer, and wish to demonstrate how this can be used to add a complete set and be consistent to Fermat. Given p, let 2^(x1) be the greatest power of two less then p.Then 2^x is greater than p. My question is ,is there some way I can add all lesser base 2, powers of 2 to 2^(x1) to obtain 2^x1,as would be consistent to 2^(p1)1, as given for a prime. Step 1:[ 2^(p1)1]/p = K(f) Step 2: Let p' be defined as the complement of p in 2^x1. That is ,p+p' = 2^x1 Step 3: Let k(s) be K(f)/(p+p') The result is , k(s)(p+p')=K(f), and hence , as p is prime, (p)(k(s)(p+p'))= 2^(p1)1. As examples of this occuring, one might take all the primes between 2^5 and 2^6, where x1 is 2^5 and 2^61=63=p+p' . Notice 63=2^0+2^1+2^2+2^3+2^4+2^5, as the full set asked for. Again , this in its form is an if statement alone, that is given a prime. Furthermore , as given a prime , Wilson;s theorem is consistent, and one does not have to worry about pseudoprimes. 
20060626, 01:15  #2  
Feb 2005
11×23 Posts 
Quote:
Quote:
Last fiddled with by maxal on 20060626 at 01:16 

20060626, 20:26  #3 
Jun 2003
Pa.,U.S.A.
2^{2}·7^{2} Posts 
Thanks for the reminder.
Thanks for the reminder.Using progression language in an adjusted Fermat and primes might be interesting.
Eons ago, on this topic, I dropped using 'series' type language because of convergence problems in (a)^x where a is not between 0 and 1. Even mentioning that every Mersenne would have to be twice raised to the base 2, would have become prohibitive from scratch. 
20060724, 02:18  #4  
May 2004
2^{2}×79 Posts 
Quote:
A.K.Devaraj 

20061126, 20:45  #5 
Jun 2003
Pa.,U.S.A.
2^{2}×7^{2} Posts 
A little more of the same.
Please read the attached.
While working with Mersennes or primes alone it might be worth having at the elementary level something like this included. By intuition , the complement might have properties of primeness that are important, (take a look at 2^891), or one might want to use calculus of variation methods with other constraints, or simply shorten observations such as p(p1)! for compositeness. John Hill 
20061128, 12:12  #6  
Feb 2005
11·23 Posts 
Quote:
(p1)! is divisible by 2*peak2 = 2*(2^m  1) simply because both 2 and 2^m  1 are smaller than p1 and thus both divide (p1)!. 

20061129, 07:06  #7 
Jun 2003
Pa.,U.S.A.
2^{2}·7^{2} Posts 
True in what you said, except that this begins with p, and from such is the correct factorial for p ,as (p1)! and for instance to allow as in the example,
to go from a k with the p, to a larger number with a smaller k ,multiplying all inclusive terms. Much merely as a definition. This with an automatic return to the 'fragmented' prime as such. Hence when looking as 2^p1, which IS a full set, it gives SOMETHING for one to look at equivalently, as p falls short (in addition)of being a 2^(smaller prime)1, itself. Hope this definition will give a little alternate insite, eventially,of primes cropping up and NOT going Mersenne, as well as Mersenne. 
20061203, 21:27  #8 
Jun 2003
Pa.,U.S.A.
2^{2}×7^{2} Posts 
Follow up example
Following the last post,
an example of using the approach, a Mersenne prime might be simply defined as a number with complement '1', that is prime. J.H. 
20061204, 04:26  #9 
Feb 2005
11·23 Posts 
David,
I do not see any practical implications from what you are saying. It all may be entertaining but useless. I will be grad if I'm wrong. 
20061205, 05:31  #10 
Jun 2003
Pa.,U.S.A.
2^{2}×7^{2} Posts 
one hypothetical use
given an enormous prime, is it possibly mersenne?
rest assured if its complement is not '1', it is not. just a quick way(or manner of thinking or speaking) of fitting 2^x1, and sorting out given primes. 
20061205, 12:04  #11  
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
11145_{10} Posts 
Quote:
This thread is getting ever sillier ... Paul 

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