Now, suppose that p ≡4 −1 and q ≡4 1. Then, again, x, y are of different
parity and z is even. If x = 2k, then we obtain 2pk = qy − 1. So 2pk ≡4 2 and
qy − 1 ≡4 0 which implies that 2pk 6≡4 qy − 1. Similarly, if y = 2l, then we get
2ql = px − 1. Since 2ql ≡6 2 and px − 1 ≡6 4, 2ql 6≡6 px − 1.