History of Hindu Mathematics: A Source Book v2.c17.s25.20

v2.c17.s25.20

In general, let us assume, as directed by Bhâskara II, x = (m² + n²)z², y = 2mnz², which will make x ± y squares. We have, therefore, only to make 2mn(m² + n²)z⁴ = a cube. Let & 2mn(m² + n²)z⁴ = p³z³. / Then & z = (p³)/(2mn(m² + n²)). / Therefore & x = ((m² + n²)p⁶)/({2mn(m² + n²)}²), / & y = (2mnp⁶)/({2mn(m² + n²)}²), where m, n, p are arbitrary.