- 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 - 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