20190616, 18:34  #1 
Jul 2014
3·149 Posts 
theorem?
If 2p+1 is prime then 2p + 1 divides 2^p1, 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? 
20190616, 21:47  #2 
Jul 2014
3×149 Posts 
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^p1, 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 oneone mapping. 
20190616, 23:12  #3  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9425_{10} Posts 
Quote:
Are you trying to say "...iff znorder(Mod(2,2*p+1)) == p" ? 

20190617, 08:33  #4 
Jul 2014
447_{10} Posts 
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. 
20190617, 10:17  #5  
Dec 2012
The Netherlands
7·239 Posts 
Quote:
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. 

20190617, 12:30  #6 
Jul 2014
3·149 Posts 
What I've found is this :
\(2p+1\mid2^p1\Longrightarrow2^p\prod^{p}_{k=1}\cos\bigg(\frac{2k\pi}{2p+1}\bigg)=1\) 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 20190617 at 12:32 
20190617, 13:25  #7  
Feb 2017
Nowhere
2^{3}·3^{4}·7 Posts 
Quote:
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 20190617 at 13:27 Reason: Adding needed hypothesis 

20190617, 13:43  #8 
Jul 2014
1BF_{16} Posts 
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 :) 
20190617, 15:49  #9 
Jul 2014
1BF_{16} Posts 
Actually, thanks very much for all of that Dr. Sardonicus. It's certainly added something to my interest in the matter.

20190618, 14:33  #10 
Romulan Interpreter
Jun 2011
Thailand
2^{5}·5·59 Posts 
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 20190618 at 14:33 
20190618, 17:30  #11  
Jul 2014
677_{8} Posts 
Thanks.
I've been rereading Sardonicus' post again because it takes me a while to digest properly. This statement puzzles me Quote:
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. 

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