## A property of twin primes, only?

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• Let p q consecutive prime numbers p
Message 1 of 2 , Jul 4 8:03 AM
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Let p q consecutive prime numbers p<q.
Let z=sqrt[(p^2+q^2)/2-1]

Conjecture: p&q ares twin primes IF AND ONLY IF z is
integer.

This is not a trivial result.

Sincerely
Sebastian Martin Ruiz

See you below:

The "only if" is trivial:
If p and q are twin primes, then q = p+2, so z =
(p+1).

As for the "if":
The conjecture is true for p = 2, so assume p >= 3.

If we write
q = p+2*d
then
z^2 = (p+d)^2 + (d^2-1)

If (p+d)^2 + (d^2-1) < (p+d+1)^2, then z^2 is not a
perfect square.
This condition is:
q < p + 2*(sqrt(2*p+3)+1)
which is easily verified to hold for p <= 10^9.
It is believed (but has never been proved) to hold for
all p >= 3.

-Mike Oakes

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• I dont spoke about you. There are other e-mails in this list that said this result is trivial. You don t understand me correctly. Sorry my english is very
Message 2 of 2 , Jul 4 10:14 AM
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I dont spoke about you. There are other e-mails in
this list that said this result is trivial.
You don't understand me correctly. Sorry my english is
very little.

Your e-mail is very intersting. I put it by a correct
example.

I am sorry that I have been understood badly.
--- mikeoakes2@... escribió:

> Sebastian
>
> I didn't say it was trivial.
>
> I said
> "p&q are twin primes => z is integer"
> is trivial.
>
> Isn't it??
>
> Mike Oakes
>
>
> -----Original Message-----
> From: Sebastian Martin <sebi_sebi@...>
> To: lista de primos <primenumbers@yahoogroups.com>
> Sent: Tue, 4 Jul 2006 17:03:44 +0200 (CEST)
> Subject: [PrimeNumbers] A property of twin primes,
> only?
>
>
> Let p q consecutive prime numbers p<q.
> Let z=sqrt[(p^2+q^2)/2-1]
>
> Conjecture: p&q ares twin primes IF AND ONLY IF z is
> integer.
>
> This is not a trivial result.
>
> Sincerely
> Sebastian Martin Ruiz
>
> See you below:
>
> The "only if" is trivial:
> If p and q are twin primes, then q = p+2, so z =
> (p+1).
>
> As for the "if":
> The conjecture is true for p = 2, so assume p >= 3.
>
> If we write
> q = p+2*d
> then
> z^2 = (p+d)^2 + (d^2-1)
>
> If (p+d)^2 + (d^2-1) < (p+d+1)^2, then z^2 is not a
> perfect square.
> This condition is:
> q < p + 2*(sqrt(2*p+3)+1)
> which is easily verified to hold for p <= 10^9.
> It is believed (but has never been proved) to hold
> for
> all p >= 3.
>
> -Mike Oakes
>
>
> ______________________________________________
> LLama Gratis a cualquier PC del Mundo.
> Llamadas a fijos y móviles desde 1 céntimo por
> minuto.
> http://es.voice.yahoo.com
>
>
>

______________________________________________
LLama Gratis a cualquier PC del Mundo.
Llamadas a fijos y móviles desde 1 céntimo por minuto.
http://es.voice.yahoo.com
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