Browse Groups

• ## RE A property of twin primes, only?

(2)
• NextPrevious
• ... From: Sebastian Martin Let p q consecutive prime numbers p
Message 1 of 2 , Jul 5, 2006
View Source
----- Original Message -----
From: "Sebastian Martin" <sebi_sebi@...>

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.

==============================

Some random (useless?) facts:

2(z^2 +1) = p^2 + q^2
2(z+i)(z-i) = (p+q·i)(p-q·i)

Since z+i <> p+q·i, z+i <> p-q·i and p,q >1,
z+i and z-i can't be gaussian primes.
Then z^2+1 is not a prime.

As 2(z^2+1) is a sum of two squares, it must be a product of factors of 2 and primes of
the form 4n+1.
But since we can consider p,q > 2, then p^2 + q^2 = = 2 (mod 4), it only can have one
factor of 2, so
z^2+1 is a product of primes of the form 4n+1.

2(z^2+1) = p^2 + q^2
2(z^2+1) = = q^2 (mod p)
2(z^2+1) = = p^2 (mod q)

2(z^2+1) is a quadratic residue in Zp and in Zq.

Moreover, if q = p+2d, then

z^2+1 = = 2d^2 (mod p) and
z^2+1 = = 2d^2 (mod q)

2d^2-1 = = z^2 (mod p)
2d^2 -1 = = z^2 (mod q)

So, 2d^2 -1 is a quadratic residue in Zp and Zq

In brief:
a) z^2+1 is a product of (at least two) primes of the form 4n+1.
b) 1 = Legendre(2(z^2+1),p) = Legendre(2(z^2+1),q) = Legendre(2d^2-1,p) =
Legendre(2d^2-1,q)
c) z^2+1 = = 2d^2 in Zp and in Zq.

A related question can then be:
If x is a quadratic residue of p,q primes and 2·x-1 is also a qr of p,q... is it
necessarily x = 1?

I think the answer is probably "no", but it is a necessary condition, so if the answer is
"yes", then the problem is solved.

Regards. Jose Brox
Your message has been successfully submitted and would be delivered to recipients shortly.
• Changes have not been saved
Press OK to abandon changes or Cancel to continue editing
• Your browser is not supported
Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View.