History of Hindu Mathematics: A Source Book v2.c17.s17.t3.18

v2.c17.s17.t3.18

Now, if y = l, w = r be a solution of $w² = by² + c,$ another solution of it will be $y = lq ± pr,$ $w = rq ± blp;$ where ap² + 1 = q². Therefore, the solution of (3) will be $x = 2m/(a - m²) (rq ± blp),$ $y = lq ± pr,$ $z = (a + m²)/(a - m²)(rq ± blp);$ where ap² + 1 = q² and bl² + c = r².