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Old 2016-09-20, 15:08   #1
wildrabbitt
 
Jul 2014

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Default 2 holes in bizarre theorem about composite Mersenne Numbers

Hi again,

I've worked out a bizarre theorem about when Mersenne Numbers are composite.

There's a couple of holes left in it which I don't think mean it's wrong. It's just that I've worked so hard on the problem, I can't see the wood for the trees at the moment - ie what's obvious and what isn't,

I'm wondering if someone can help with the holes because I think someone can :

Hole 1 :

If q = 2kp + 1 divides 2^p - 1 then q cannot divide any number of the form

2^c - 1 where c is composite and less than p.

Hole 2 :

If 2^k = sq + r where q is factor of the Mersenne Number 2^p - 1
then r cannot be equal to p.

Other than these two things I've got a theorem which I'd be happy to share when I've written it up properly.
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Old 2016-09-20, 15:28   #2
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Quote:
Originally Posted by wildrabbitt View Post
Hi again,

I've worked out a bizarre theorem about when Mersenne Numbers are composite.

There's a couple of holes left in it which I don't think mean it's wrong. It's just that I've worked so hard on the problem, I can't see the wood for the trees at the moment - ie what's obvious and what isn't,

I'm wondering if someone can help with the holes because I think someone can :

Hole 1 :

If q = 2kp + 1 divides 2^p - 1 then q cannot divide any number of the form

2^c - 1 where c is composite and less than p.

Hole 2 :

If 2^k = sq + r where q is factor of the Mersenne Number 2^p - 1
then r cannot be equal to p.

Other than these two things I've got a theorem which I'd be happy to share when I've written it up properly.
well hole 1 is filled by the fact that all exponents in 2^n-1 that are coprime lead to coprime values. Hole 2 there are a few parts note that that all the powers k above p lead to an even remainder value. and then try to figure out for k<p this last part is where I'd have to think about it a bit more edit: except that if k<p then we definitely know r is not equal to 0 in these cases. edit2: oh and we know s and r are same parity.

Last fiddled with by science_man_88 on 2016-09-20 at 15:35
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Old 2016-09-20, 15:38   #3
wildrabbitt
 
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Massive thanks. Will only post again if I can't assimilate what you've expressed into my work.
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Old 2016-09-20, 15:42   #4
wildrabbitt
 
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Except for this one in which I want to make clear that I'll post the theorem soon.
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Old 2016-09-20, 15:46   #5
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Quote:
Originally Posted by wildrabbitt View Post
Except for this one in which I want to make clear that I'll post the theorem soon.
one thing to remember 2^p-1 is of form 2*k*p+1 as well in cases p prime.
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Old 2016-09-20, 16:29   #6
wildrabbitt
 
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I can't wait to get it written up properly.

I'm not sure if it's all that amazing or more importantly whether it's useful.

If someone could type it up in latex for me, that would be fab because then it would be easier to read and I could be sure that it has been understood properly.


The bizarre theorem ;

This is I suspect part of a much stronger result. This is for when

q = 2p + 1

q = 2p + 1 divides 2^p - 1

if and only if

(2^p) * (cos(pi/q)) * capital_pi( (k = 1 to p - 1), cos((2kpi)/q) = 1

Would be most interested to know what people think of it.
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Old 2016-09-20, 17:24   #7
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Quote:
Originally Posted by wildrabbitt View Post
I can't wait to get it written up properly.

I'm not sure if it's all that amazing or more importantly whether it's useful.

If someone could type it up in latex for me, that would be fab because then it would be easier to read and I could be sure that it has been understood properly.


The bizarre theorem ;

This is I suspect part of a much stronger result. This is for when

q = 2p + 1;<br />
<br />
 q = 2p + 1 | 2^p - 1 ,<br />
<br />
\text{iff}:<br />
<br />
(2^p)(cos(\frac{\pi}{q}))  \prod_{k = 1}^ {p - 1} cos((\frac{2k\pi}{q}) = 1

Would be most interested to know what people think of it.
I think is what you mean.
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Old 2016-09-20, 17:28   #8
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This is what you seem to mean:

Theorem. Let p be a prime number and let q = 2p + 1. Then q divides 2^p - 1 if and only if
\[
2^p \cdot \cos(\pi/q) \cdot \prod_{k = 1}^{p - 1} \cos((2k\pi)/q) = 1
\]

But it doesn't seem to be right. p = 79 and p = 83 both give -1, but only one of those has the required divisibility property. p = 5 gives 1 but doesn't have the property. Perhaps I've misinterpreted something though, or maybe my calculations have gone awry (though I checked my results with two programs).

Quote:
Originally Posted by wildrabbitt View Post
I'm not sure if it's all that amazing or more importantly whether it's useful.
Computationally it's worthless -- products of trig functions over huge ranges are much, much slower than naive modular exponentiation. There may be some aesthetic value to such a theorem, though. But you'll need to work the bugs out.

Last fiddled with by CRGreathouse on 2016-09-20 at 17:29
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Old 2016-09-20, 18:02   #9
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Quote:
Originally Posted by CRGreathouse View Post
Computationally it's worthless -- products of trig functions over huge ranges are much, much slower than naive modular exponentiation. There may be some aesthetic value to such a theorem, though. But you'll need to work the bugs out.
okay but it can be turned into a sum form:

http://www.purplemath.com/modules/idents.htm states:

cos(x)cos(y) = 1/2[cos(x-y)+cos(x+y)] could this narrow it down in complexity since p-1 is even ( at least assuming p is an odd prime) they could all be paired up inside the product to form a sum. admittedly we know which p can have 2p+1 divide them but it might be nice to explore a different angle of it.

Last fiddled with by science_man_88 on 2016-09-20 at 18:06
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Old 2016-09-20, 18:14   #10
wildrabbitt
 
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I was totally sure it was right after the holes were revealed to be trivial.

Now that it's been doubted I'm fairly sure it's true. I've got a full proof.

I'm open to the idea I made a sign error or took only a positive square root
when I should have taken both.

If it's true for 5 that's because like I said I think it's part of a stronger result.
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Old 2016-09-20, 18:15   #11
wildrabbitt
 
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Oh yeah, and thanks for typing out the latex for it.
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