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Old 2003-04-01, 14:22   #1
1260
 
Feb 2003

408 Posts
Default twin prime generator

while toying with twin primes, i noticed that following:

if p and p+2 are twin primes, then there exists an integer B such that
if q = (B^2 - p)^2 - B, then q and q+2 are also twin primes greater than p.

i have also noticed that in some cases, if r=q mod p, then r and r+2 are also twin primes.

does a proof for this already exist or are there counterexamples?
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Old 2003-04-01, 17:11   #2
ewmayer
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Sep 2002
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Default Re: twin prime generator

Quote:
Originally Posted by 1260
while toying with twin primes, i noticed that following:

if p and p+2 are twin primes, then there exists an integer B such that
if q = (B^2 - p)^2 - B, then q and q+2 are also twin primes greater than p.

i have also noticed that in some cases, if r=q mod p, then r and r+2 are also twin primes.

does a proof for this already exist or are there counterexamples?
You don't provide any bounds on B. For instance, if I take p=3, then for B<3 q is negative. Did you intend only positive q? For the same example, B=3 gives q=33, which is not prime.

Even saying there exists at least ONE such value of B for any twin-prime pair
(p,p+2) is a very strong statement, since it implies that the twin prime conjecture is true. While this is generally believed to be the case, it has never been proven, and some of the best mathematicians in history have worked on it. Are you saying you have a proof, or just some numerical examples?
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Old 2003-04-02, 11:14   #3
1260
 
Feb 2003

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Quote:
You don't provide any bounds on B.
Yes. B can be any positive or negative integer.

Quote:
if I take p=3, then for B<3 q is negative. Did you intend only positive q?
If p=3 and B=-1, q=5.

Yes, I only intended q to be positve.

The following are some negative values of B for p=3.
[code:1]
B q
-10 9419
-28 609989
-223 2472675299[/code:1]

Quote:
Are you saying you have a proof, or just some numerical examples?
All I have are still numerical examples. I haven't yet come up with a proof. Maybe a distributed effort might arrive at a proof faster.
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Old 2003-04-04, 05:40   #4
Kevin
 
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Aug 2002
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I don't think a distributed effort would necessarily lead to a proof, but it may give us an idea on where to start one. I've taken up interest in this, and have been doing some work on. Due to a small error in my program, I have a list of all positive B up to around a million for p=11 that make q and q+2 prime. I don't plan to go this far for the rest of the numbers, I'll probably restrict it to about B=1000. I'll try and get as much done as I can now, since I'm on Spring Break. Once I get back to school, I can set up a small cluster of TI-82's to do some work during Doc Block (AP Physics C followed AP Calculus BC, 3 hours long). I'm going to intentionally avoid B being negative, partially because it doesn't seem right to include negative numbers for this kinda thing, and I also see it more at positive values of B for q=(B^2-p)+B. Another interesting thing to look at might be to see how often B and -B both yield twin primes.

P.S. My only programming experience is with TI calculators ops: . Running stuff on a TI-89 emulator at full throttle my computer works good, but if anyone makes a better program, I'd like to know about it.
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Old 2003-04-06, 22:42   #5
Khemikal796
 
Feb 2003

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Off topic, but do you have any links/info on this cluster of ti-82s u talk about? I have a couple 82/83 sitting around, and if i could get them working together it could help on some of my progs that ive got crunching during class while i sleep.
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Old 2003-04-06, 23:43   #6
Kevin
 
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Aug 2002
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Oh, i didn't mean a cluster working together at once. I meant you'd have the 1st one testing for B with p= 11, then the second testing for B with p=17...etc. I probably mis-used "cluster" ops: .
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Old 2003-04-07, 02:36   #7
Khemikal796
 
Feb 2003

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Dang I got all excited about nothing heh. Oh well, guess I still have something to dream about.
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Old 2003-04-14, 11:02   #8
1260
 
Feb 2003

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Here's another thing I've noticed:

If the equation is rearranged like this:

(B^2 - p)^2 = B + q

then let A be another integer such that:

A^2 = B + q
A = B^2 - p

then:

B = A^2 - q
B^2 = A + p

resulting in:

(A^2 - p)^2 = A + p

They look symmetric but I haven't found a case where A = B.
I've also noticed that |A - B| = 0 mod 3.
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Old 2003-04-23, 12:55   #9
1260
 
Feb 2003

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Anyone interested in breaking the world twin prime record?

According to Chris Caldwell (http://www.utm.edu/research/primes/l...op20/twin.html)
the largest twin prime pair to date is 33218925*2^169690+-1, found in 2002 by Papp using Proth.exe.

Let p=33218925*2^169690-1.
then (B^2 - p)^2 - B is another twin prime greater than p.

Can anyone try B for 11 to 100? or -1 to -100?
I'm presently testing B for 1 to 10 using PFGW (PrimeForm).

Does anyone know of other programs (windows-based) that can test an expression for primality? Is there a version of PRP that implements this? (I guess, I should have asked this in The Software section.)
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Old 2003-06-10, 04:23   #10
1260
 
Feb 2003

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I tested B for values from -1999 to 2000 and didn't find any twin primes. I guess, its larger than 2000 or smaller than -1999.

By the way, B is always of the form 3n - 1 where n is any integer.
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Old 2003-06-13, 07:44   #11
loop quantum gravity
 

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Quote:
Originally Posted by 1260
Here's another thing I've noticed:

If the equation is rearranged like this:

(B^2 - p)^2 = B + q

then let A be another integer such that:

A^2 = B + q
A = B^2 - p

then:

B = A^2 - q
B^2 = A + p

resulting in:

(A^2 - p)^2 = A + p

They look symmetric but I haven't found a case where A = B.
I've also noticed that |A - B| = 0 mod 3.
it should be (A^2-q)^2=A+p
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