----- Original Message -----
From: "Ignacio Larrosa Cañestro" <ilarrosa@...>
>> ((B*2-1)^2+5)^2*5 == 4 (modulo p)
>> In order for this to have solutions, 5 must be a quadratic
>> residue modulo p, which eliminates primes p ending in 3 or 7.
Because the product of quadratic residues is an inner product and (quadratic residue) *
(quadratic non residue) = (quadratic non residue).
Let be p>3. Then 4 is a quadratic residue modulo p, since it's 2^2 (mod p). If p=3, then
4==1 = 1^2 (mod p), and if p=2, then 4 ==0 = 0^2 (mod p).
So we have a equation of the form [(quadratic residue)*(quadratic residue)]*5 = (quadratic
residue) (mod p) and therefore 5 must be a qr in order for the equation to have solutions.
Regards. Jose Brox