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• Found no reference on the web. Will this statement hold up in the realm of huge numbers? In every occurrence of positive twin prime pairs P and Q=P+2, except
Message 1 of 5 , Aug 5, 2010
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Found no reference on the web. Will this statement hold up in the realm of huge numbers?

In every occurrence of positive twin prime pairs P and Q=P+2, except (3, 5), where A=(5*P^2-1)/4 and B=(5*Q^2-1)/4 are simultaneously prime, and each is a member of a twin prime set, A and B are always the smaller members of those twin prime sets.

Here are the first 5 occurrences of such twin pairs. The format is (P, Q, A, B, Is A a small twin?, Is A a large twin?, Is B a small twin?, Is B a large twin?).
(3, 5, 11, 31, 0, 1, 1, 0)
(5, 7, 31, 61, 1, 0, 1, 0)
(71 73, 6301, 6661, 1, 0, 1, 0)
(117539, 117541, 17269270651, 17269858351, 1, 0, 1, 0)
(384257, 384259, 184566802561, 184568723851, 1, 0, 1, 0)

I stopped calculating with P=33613859, Q=33613861, A=1412364396089851, B=1412364564159151. I found no exceptions to the above statement. Noteworthy is that in every case, the rightmost digits of A and B equaled 1. Anyone care to check me on this or comment?

Thanks folks.

Bill Sindelar
____________________________________________________________
SHOCKING: 2010 Honda Civic for \$1,732.09
SPECIAL REPORT: High ticket items are being auctioned for an incredible 90% off!
http://thirdpartyoffers.juno.com/TGL3131/4c5b0b256a18c5a0854st03duc

[Non-text portions of this message have been removed]
• Don t you mean that A and B are the larger members of those twin prime pairs? For any X of the form X = 6*k+/-1, note that (5*X^2-1)/4 will be of the form
Message 2 of 5 , Aug 5, 2010
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Don't you mean that A and B are the larger members of those twin prime
pairs?

For any X of the form X = 6*k+/-1, note that (5*X^2-1)/4 will be of the
form 6*m+1, so it can never be the smaller member of a twin prime pair.

w_sindelar@... wrote:
> Found no reference on the web. Will this statement hold up in the realm of huge numbers?
>
> In every occurrence of positive twin prime pairs P and Q=P+2, except (3, 5), where A=(5*P^2-1)/4 and B=(5*Q^2-1)/4 are simultaneously prime, and each is a member of a twin prime set, A and B are always the smaller members of those twin prime sets.
>
> Here are the first 5 occurrences of such twin pairs. The format is (P, Q, A, B, Is A a small twin?, Is A a large twin?, Is B a small twin?, Is B a large twin?).
> (3, 5, 11, 31, 0, 1, 1, 0)
> (5, 7, 31, 61, 1, 0, 1, 0)
> (71 73, 6301, 6661, 1, 0, 1, 0)
> (117539, 117541, 17269270651, 17269858351, 1, 0, 1, 0)
> (384257, 384259, 184566802561, 184568723851, 1, 0, 1, 0)
>
> I stopped calculating with P=33613859, Q=33613861, A=1412364396089851, B=1412364564159151. I found no exceptions to the above statement. Noteworthy is that in every case, the rightmost digits of A and B equaled 1. Anyone care to check me on this or comment?
>
> Thanks folks.
>
> Bill Sindelar
> ____________________________________________________________
> SHOCKING: 2010 Honda Civic for \$1,732.09
> SPECIAL REPORT: High ticket items are being auctioned for an incredible 90% off!
> http://thirdpartyoffers.juno.com/TGL3131/4c5b0b256a18c5a0854st03duc
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
>
>
>
>
• Also, with regard to the ending digit of 1, that s a consequence of: X == 6*k-1, (5*X^2-1)/4 == 15*k*(3*k-1) + 1 X == 6*k+1, (5*X^2-1)/4 == 15*k*(3*k+1) + 1
Message 3 of 5 , Aug 5, 2010
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Also, with regard to the ending digit of 1, that's a consequence of:

X == 6*k-1, (5*X^2-1)/4 == 15*k*(3*k-1) + 1
X == 6*k+1, (5*X^2-1)/4 == 15*k*(3*k+1) + 1

Noting that k*(3*k-1) and k*(3*k+1) are always even, it's easy to
see that if X is of the form 6*k+/-1, then (5*X^2-1)/4 is of the
form 30*m+1.

Jack Brennen wrote:
> Don't you mean that A and B are the larger members of those twin prime
> pairs?
>
> For any X of the form X = 6*k+/-1, note that (5*X^2-1)/4 will be of the
> form 6*m+1, so it can never be the smaller member of a twin prime pair.
>
>
>
> w_sindelar@... wrote:
>> Found no reference on the web. Will this statement hold up in the
>> realm of huge numbers?
>>
>> In every occurrence of positive twin prime pairs P and Q=P+2, except
>> (3, 5), where A=(5*P^2-1)/4 and B=(5*Q^2-1)/4 are simultaneously
>> prime, and each is a member of a twin prime set, A and B are always
>> the smaller members of those twin prime sets.
>>
>> Here are the first 5 occurrences of such twin pairs. The format is (P,
>> Q, A, B, Is A a small twin?, Is A a large twin?, Is B a small twin?,
>> Is B a large twin?).
>> (3, 5, 11, 31, 0, 1, 1, 0)
>> (5, 7, 31, 61, 1, 0, 1, 0)
>> (71 73, 6301, 6661, 1, 0, 1, 0)
>> (117539, 117541, 17269270651, 17269858351, 1, 0, 1, 0)
>> (384257, 384259, 184566802561, 184568723851, 1, 0, 1, 0)
>>
>> I stopped calculating with P=33613859, Q=33613861, A=1412364396089851,
>> B=1412364564159151. I found no exceptions to the above statement.
>> Noteworthy is that in every case, the rightmost digits of A and B
>> equaled 1. Anyone care to check me on this or comment?
>>
>> Thanks folks.
>>
>> Bill Sindelar
>> ____________________________________________________________
>> SHOCKING: 2010 Honda Civic for \$1,732.09
>> SPECIAL REPORT: High ticket items are being auctioned for an
>> incredible 90% off!
>> http://thirdpartyoffers.juno.com/TGL3131/4c5b0b256a18c5a0854st03duc
>>
>> [Non-text portions of this message have been removed]
>>
>>
>>
>> ------------------------------------
>>
>> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
>> The Prime Pages : http://www.primepages.org/
>>
>>
>>
>>
>>
>>
>
>
• Bill, Jack wrote: For any X of the form X = 6*k+/-1, note that (5*X^2-1)/4 will be of the form 6*m+1, so it can never be the smaller member of a twin prime
Message 4 of 5 , Aug 5, 2010
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Bill,

Jack wrote:

"For any X of the form X = 6*k+/-1, note that (5*X^2-1)/4 will be of the
form 6*m+1, so it can never be the smaller member of a twin prime pair. "

Here's how one would go about determining this:
First note that all primes except 2 and 3 (not just twin primes) are of the
form 6*k+/-1. That's because of all the possible residues, mod 6, namely
-2, -1, 0, 1, 2 and 3, residues -2, 0 and 2 are divisible by 2 (so can't be
prime) and residues 0 and 3 are divisible by 3 (so can't be prime). That
leaves only residues -1 and 1. All primes (except 2 and 3) must therefore
be equal to 6*k +/- 1, for some k.

So, first let's look at the bigger of the twin primes:
Let x = 1 mod 6. (this means x = 6*k+1 for some k)
Then x^2 = 1 mod 6.
then 5*x^2 = 5 mod 6.
Then 5*x^2-1 = 4mod 6
Then (5*x^2-1)/4 = 1 mod 6.

For the smaller of the twin primes:
Let x = -1 mod 6. (this means x = 6*k-1 for some k)
Then x^2 = 1 mod 6.
then 5*x^2 = 5 mod 6.
Then 5*x^2-1 = 4mod 6
Then (5*x^2-1)/4 = 1 mod 6.

So both twin primes, when put through the (5x^2-1)/4 formula will be of the
form 6k+1 for some k.

And to answer your question, yes, this will hold up in the realm of huge
numbers. :-)

Hope this helps.

On Thu, Aug 5, 2010 at 2:18 PM, Jack Brennen <jfb@...> wrote:

>
>
> Don't you mean that A and B are the larger members of those twin prime
> pairs?
>
> For any X of the form X = 6*k+/-1, note that (5*X^2-1)/4 will be of the
> form 6*m+1, so it can never be the smaller member of a twin prime pair.
>
>
> w_sindelar@... <w_sindelar%40juno.com> wrote:
> > Found no reference on the web. Will this statement hold up in the realm
> of huge numbers?
> >
> > In every occurrence of positive twin prime pairs P and Q=P+2, except (3,
> 5), where A=(5*P^2-1)/4 and B=(5*Q^2-1)/4 are simultaneously prime, and each
> is a member of a twin prime set, A and B are always the smaller members of
> those twin prime sets.
> >
> > Here are the first 5 occurrences of such twin pairs. The format is (P, Q,
> A, B, Is A a small twin?, Is A a large twin?, Is B a small twin?, Is B a
> large twin?).
> > (3, 5, 11, 31, 0, 1, 1, 0)
> > (5, 7, 31, 61, 1, 0, 1, 0)
> > (71 73, 6301, 6661, 1, 0, 1, 0)
> > (117539, 117541, 17269270651, 17269858351, 1, 0, 1, 0)
> > (384257, 384259, 184566802561, 184568723851, 1, 0, 1, 0)
> >
> > I stopped calculating with P=33613859, Q=33613861, A=1412364396089851,
> B=1412364564159151. I found no exceptions to the above statement. Noteworthy
> is that in every case, the rightmost digits of A and B equaled 1. Anyone
> care to check me on this or comment?
> >
> > Thanks folks.
> >
> > Bill Sindelar
> > __________________________________________________________
> > SHOCKING: 2010 Honda Civic for \$1,732.09
> > SPECIAL REPORT: High ticket items are being auctioned for an incredible
> 90% off!
> > http://thirdpartyoffers.juno.com/TGL3131/4c5b0b256a18c5a0854st03duc
> >
> > [Non-text portions of this message have been removed]
> >
> >
> >
> > ------------------------------------
>
> >
> > The Prime Pages : http://www.primepages.org/
> >
> >
> >
> >
> >
> >
>
>
>

[Non-text portions of this message have been removed]
• Jack Brennen wrote; Don t you mean that A and B are the larger members of those twin prime pairs? Yes, Jack that is what I meant. I mixed them up when I
Message 5 of 5 , Aug 6, 2010
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Jack Brennen wrote;
"Don't you mean that A and B are the larger members of those twin prime
pairs?"

Yes, Jack that is what I meant. I mixed them up when I wrote the post. Thank you for correcting me and also for why the rightmost digit must always be 1.
"Jack wrote:
For any X of the form X = 6*k+/-1, note that (5*X^2-1)/4 will be of the
form 6*m+1, so it can never be the smaller member of a twin prime pair.
Here's how one would go about determining this:"
Thank you Tom for your neat explanation. It seems so obvious now, but the fact that A and B will always be large twins with rightmost digits equal to 1 had me stumped.
Bill Sindelar
____________________________________________________________
Penny Stock Jumping 2000%