Question

z=cos4xy-sen2x+pn2y+x^3y^2

Answer 1

x=|p2−3q2|,y=2pq,z=p2+3q2

x1=1−3m21+3m2,y1=2m1+3m2,

(1+3m2)x2+6m2x+(3m2−1)=0.

x^2 + 3y^2 = z^2 3y^2 = (z - x)(z + x) = 3rp^2 * rq^2 y = pqr 2x = r(q^2 - 3p^2) 2z = r(q^2 + 3p^2) (1) r = 1: 2x = q^2 - 3p^2, y = pq, 2z = q^2 + 3p^2, pq odd (2) r = 2: x = q^2 - 3p^2, y = 2pq, z = q^2 + 3p^2