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Old 2006-06-25, 17:46   #1
David John Hill Jr
 
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Default 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^(p-1)-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^(x-1) 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^(x-1) to obtain 2^x-1,as would be consistent to
2^(p-1)-1, as given for a prime.
Step 1:[ 2^(p-1)-1]/p = K(f)
Step 2: Let p' be defined as the complement of p in 2^x-1.
That is ,p+p' = 2^x-1
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^(p-1)-1.

As examples of this occuring, one might take all the primes between
2^5 and 2^6, where x-1 is 2^5 and 2^6-1=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.
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Old 2006-06-26, 01:15   #2
maxal
 
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Quote:
Originally Posted by David John Hill Jr
My question is ,is there some way I can add all lesser base 2, powers of 2 to 2^(x-1) to obtain 2^x-1
Strange question. As 2^0, 2^1, ..., 2^(x-1) form a geometric progression, their sum is 2^x - 1, exactly what you are asking for.
Quote:
Originally Posted by David John Hill Jr
Notice 63=2^0+2^1+2^2+2^3+2^4+2^5, as the full set asked for.
Nice illustration.

Last fiddled with by maxal on 2006-06-26 at 01:16
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Old 2006-06-26, 20:26   #3
David John Hill Jr
 
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Default 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.
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Old 2006-07-24, 02:18   #4
devarajkandadai
 
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Default

Quote:
Originally Posted by David John Hill Jr
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^(p-1)-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^(x-1) 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^(x-1) to obtain 2^x-1,as would be consistent to
2^(p-1)-1, as given for a prime.
Step 1:[ 2^(p-1)-1]/p = K(f)
Step 2: Let p' be defined as the complement of p in 2^x-1.
That is ,p+p' = 2^x-1
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^(p-1)-1.

As examples of this occuring, one might take all the primes between
2^5 and 2^6, where x-1 is 2^5 and 2^6-1=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.
Good;what about the corresponding Eulerphis?
A.K.Devaraj
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Old 2006-11-26, 20:45   #5
David John Hill Jr
 
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Default 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 2^89-1), or one might want to use calculus of variation methods with other constraints, or simply shorten observations
such as p|(p-1)! for compositeness.

John Hill
Attached Files
File Type: txt kp.txt (2.5 KB, 129 views)
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Old 2006-11-28, 12:12   #6
maxal
 
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Quote:
Originally Posted by David John Hill Jr View Post
Let me take Kp-1 and/or (p-1)! and divide by 2*peak-2 (it could be looked on as p+p'-2 to the 2*peak or 2^(peak+1). As p is prime and k a natural number, it is also a natural number.
If I understood you correctly, you define the "peak" as 2^m such that 2^m <= p < 2^(m+1). But then the described divisibility has nothing to do with primality of p.
(p-1)! is divisible by 2*peak-2 = 2*(2^m - 1) simply because both 2 and 2^m - 1 are smaller than p-1 and thus both divide (p-1)!.
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Old 2006-11-29, 07:06   #7
David John Hill Jr
 
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True in what you said, except that this begins with p, and from such is the correct factorial for p ,as (p-1)! 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^p-1, 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.
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Old 2006-12-03, 21:27   #8
David John Hill Jr
 
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Default 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.
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Old 2006-12-04, 04:26   #9
maxal
 
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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.
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Old 2006-12-05, 05:31   #10
David John Hill Jr
 
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Default 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^x-1,
and sorting out given primes.
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Old 2006-12-05, 12:04   #11
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Quote:
Originally Posted by David John Hill Jr View Post
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^x-1,
and sorting out given primes.
A much easier check is to add 1 to the prime given and use trial division to factor the result. If the largest prime needed in the factoring step is 2, the original was a Mersenne prime.

This thread is getting ever sillier ...


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