Fermats theorem and defining a 'full set' for any prime.

[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.

[maxal]

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.

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.

[David John Hill Jr]

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.

[devarajkandadai]

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

[David John Hill Jr]

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

[maxal]

Originally Posted by David John Hill Jr

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)!.

[David John Hill Jr]

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.

[David John Hill Jr]

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.

[maxal]

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.

[David John Hill Jr]

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.

[xilman]

Originally Posted by David John Hill Jr

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 ...


Paul