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 2019-06-16, 18:34 #1 wildrabbitt   Jul 2014 3·149 Posts 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?
 2019-06-16, 21:47 #2 wildrabbitt   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^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.
2019-06-16, 23:12   #3
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

942510 Posts

Quote:
 Originally Posted by wildrabbitt 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" ?

 2019-06-17, 08:33 #4 wildrabbitt   Jul 2014 44710 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.
2019-06-17, 10:17   #5
Nick

Dec 2012
The Netherlands

7·239 Posts

Quote:
 Originally Posted by wildrabbitt 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.

 2019-06-17, 12:30 #6 wildrabbitt   Jul 2014 3·149 Posts 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
2019-06-17, 13:25   #7
Dr Sardonicus

Feb 2017
Nowhere

23·34·7 Posts

Quote:
 Originally Posted by wildrabbitt 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

 2019-06-17, 13:43 #8 wildrabbitt   Jul 2014 1BF16 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 :)
 2019-06-17, 15:49 #9 wildrabbitt   Jul 2014 1BF16 Posts Actually, thanks very much for all of that Dr. Sardonicus. It's certainly added something to my interest in the matter.
 2019-06-18, 14:33 #10 LaurV Romulan Interpreter     Jun 2011 Thailand 25·5·59 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 2019-06-18 at 14:33
2019-06-18, 17:30   #11
wildrabbitt

Jul 2014

6778 Posts

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.

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