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.
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) =
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
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