mersenneforum.org  

Go Back   mersenneforum.org > Great Internet Mersenne Prime Search > Math

Reply
 
Thread Tools
Old 2019-06-16, 18:34   #1
wildrabbitt
 
Jul 2014

3·149 Posts
Default theorem?

If 2p+1 is prime then 2p + 1 divides 2^p-1, iff 2^1,2^2,2^3,....,2^p are a complete set of least possible residues modulo 2(2p+1).


If this were true (and I'm pretty sure it is), could it be useful?
wildrabbitt is offline   Reply With Quote
Old 2019-06-16, 21:47   #2
wildrabbitt
 
Jul 2014

3×149 Posts
Default

I hope noone tried to find a counterexample as they would have wasted their time since it's no use finding out that's wrong.


What I should have wrote though is this



If 2p+1 is prime then 2p + 1 divides 2^p-1, iff the numbers 2^1,2^2,2^3,....,2^p are each uniquely congruent modulo 2(2p+1) to a number in the set of numbers 2, 4, 6, 8, 10,....2p . i.e there's a one-one mapping.
wildrabbitt is offline   Reply With Quote
Old 2019-06-16, 23:12   #3
Batalov
 
Batalov's Avatar
 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

942510 Posts
Default

Quote:
Originally Posted by wildrabbitt View Post
What I should have wrote though is this

If 2p+1 is prime then 2p + 1 divides 2^p-1, iff the numbers 2^1,2^2,2^3,....,2^p are each uniquely congruent modulo 2(2p+1) to a number in the set of numbers 2, 4, 6, 8, 10,....2p . i.e there's a one-one mapping.
How about p=3? Where is a one-to-one mapping to the value of 6?

Are you trying to say "...iff znorder(Mod(2,2*p+1)) == p" ?
Batalov is offline   Reply With Quote
Old 2019-06-17, 08:33   #4
wildrabbitt
 
Jul 2014

44710 Posts
Default

I need a bit of time to refine what I'm saying, Can someone remind me how to use latex in these posts. i.e. which tags to use.


I'll then be able to show you what I've found and ask a few things.
wildrabbitt is offline   Reply With Quote
Old 2019-06-17, 10:17   #5
Nick
 
Nick's Avatar
 
Dec 2012
The Netherlands

7·239 Posts
Default

Quote:
Originally Posted by wildrabbitt View Post
Can someone remind me how to use latex in these posts. i.e. which tags to use.
Put a backslash followed by an open bracket at the start of LaTeX mathematics.
Put a backslash followed by a close bracket at the end.
Use square brackets instead of ordinary brackets if you want it displayed on a separate line.
Nick is online now   Reply With Quote
Old 2019-06-17, 12:30   #6
wildrabbitt
 
Jul 2014

3·149 Posts
Default

What I've found is this :



\(2p+1\mid2^p-1\Longrightarrow2^p\prod^{p}_{k=1}\cos\bigg(\frac{2k\pi}{2p+1}\bigg)=1\)


What I was saying before wasn't thought through properly. I don't think the right to left implication is also true which was what I was saying before.


I thought that this is interesting anyway. I've got a proof.

Last fiddled with by wildrabbitt on 2019-06-17 at 12:32
wildrabbitt is offline   Reply With Quote
Old 2019-06-17, 13:25   #7
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

23·34·7 Posts
Default

Quote:
Originally Posted by wildrabbitt View Post
If 2p+1 is prime then 2p + 1 divides 2^p-1, iff 2^1,2^2,2^3,....,2^p are a complete set of least possible residues modulo 2(2p+1).


If this were true (and I'm pretty sure it is), could it be useful?
If q == 2*p + 1 is prime, there are q - 1 = 2*p (not p) nonzero residues (mod q). If 2^p == 1 (mod q) then 2 has multiplicative order at most p, so the residues 2^k (mod q), k = 1 to p, are at most only half the nonzero residues (mod q), not all of them. Likewise, they account for at most half the invertible elements (mod 2*q).

If p is odd, the prime q = 2*p + 1 can only be of the form 8*n + 7.

Since you didn't say that p has to be prime, I offer p = 15. Then, q = 2*p + 1 = 31, a prime. And 31 does divide 2^15 - 1. However, it also divides 2^5 - 1, so 2^k (mod 31) only defines 5 of the nonzero residues.

If you assume p is prime, then p = 4*n + 3 and q = 8*n + 7 are a pair of "Sophie Germain primes," and it is well known that in this case, q divides 2^p - 1.

Last fiddled with by Dr Sardonicus on 2019-06-17 at 13:27 Reason: Adding needed hypothesis
Dr Sardonicus is offline   Reply With Quote
Old 2019-06-17, 13:43   #8
wildrabbitt
 
Jul 2014

1BF16 Posts
Default

I realised that what my first post said was rubbish (to try and salvage some self respect).
I guess I've embarrassed myself again with a naive cranky attempt at being clever.
Oh well, thanks for letting me know it's not helpful :)
wildrabbitt is offline   Reply With Quote
Old 2019-06-17, 15:49   #9
wildrabbitt
 
Jul 2014

1BF16 Posts
Default

Actually, thanks very much for all of that Dr. Sardonicus. It's certainly added something to my interest in the matter.
wildrabbitt is offline   Reply With Quote
Old 2019-06-18, 14:33   #10
LaurV
Romulan Interpreter
 
LaurV's Avatar
 
Jun 2011
Thailand

25·5·59 Posts
Default

Making mistakes is not a minus, and recognizing when you made a mistake is a plus. Learning from it is a big plus.

There is no embarrassment (is this a word?) in it. This is how we learn. The one without sin should throw the stone first. That is not me...

Others come here with ignorance, but full of vanity and insolence and try to sell cucumbers to the gardeners. That is a sin.

Last fiddled with by LaurV on 2019-06-18 at 14:33
LaurV is offline   Reply With Quote
Old 2019-06-18, 17:30   #11
wildrabbitt
 
Jul 2014

6778 Posts
Default

Thanks.



I've been rereading Sardonicus' post again because it takes me a while to digest properly. This statement puzzles me



Quote:
If you assume p is prime, then p = 4*n + 3 and q = 8*n + 7 are a pair of "Sophie Germain primes,"

7 is prime 7=4*1 + 3 and q = 15 = 8*1 + 7 are not a pair of Sophie Germain Primes.
Also
11 is prime 11 = 4*2 + 3 and q = 25 = 8*2 + 7 aren't a pair of Sophie Germain primes either.


I'm not disputing anything because I don't think I understood the post in the way it was meant to be understood. A bit of clarification or whatever would be appreciated.
wildrabbitt is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Pocklington's Theorem bgbeuning Computer Science & Computational Number Theory 7 2015-10-13 13:55
Koebe's 1/4 Theorem wpolly Math 3 2008-11-12 12:31
Number Theorem herege Math 25 2006-11-18 09:54
Størmer's theorem grandpascorpion Factoring 0 2006-09-10 04:59
Lagrange's Theorem Numbers Math 2 2005-09-07 15:22

All times are UTC. The time now is 08:11.

Mon May 10 08:11:04 UTC 2021 up 32 days, 2:51, 0 users, load averages: 2.05, 1.96, 1.86

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.