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.

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

Jose Ramón Brox

... From: Jose Ramón Brox ... Sorry, I should have tried to verify it before sending; I just saw that finding a counterexample is

Message 2 of 2
, Jul 5, 2006

0 Attachment

----- Original Message -----
From: "Jose Ramón Brox" <ambroxius@...>

>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?

Sorry, I should have tried to verify it before sending; I just saw that finding a
counterexample is simple:

5 is a qr in Z11 because 4^2 = = 16 = = 5 (mod 11)
2·5-1 = 9 is a qr in Z11 because 3^2 = = 9 (mod 11)

5 is also a qr in Z19 because 9^2 = = 81 = = 5 (mod 19)
9 is also a qr in Z19 because 3^2 = = 9 (mod 19)

We will need a better condition.

Jose

Your message has been successfully submitted and would be delivered to recipients shortly.