History of Hindu Mathematics: A Source Book v2.c17.s24.10

v2.c17.s24.10

Now, from the first equation, we have $(m² + n²)z⁴ = u³.$ Assume u = rz; then $z = (r³)/(m² + n²).$ Hence we get $x = (mr⁶)/((m² + n²)²), y = (nr⁶)/((m² + n²)²);$ where r is any integer and m, n are such that $m³ + n³ = a square.$ One obvious solution¹ of this equation is m = 1, n = 2. Hence we get the solution $x = (r⁶)/25, y = (2r⁶)/25.$