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## Re: [PrimeNumbers] The Squaring of Twin Primes

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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
Message 1 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
> > __________________________________________________________
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> > [Non-text portions of this message have been removed]
> >
> >
> >
> > ------------------------------------
>
> >
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> >
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> >
> >
> >
> >
> >
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>
>

[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 2 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.