History of Hindu Mathematics: A Source Book v2.c17.s25.63

v2.c17.s25.63

That is to say, if *w* is any rational number, we assume $u = w ± α,$ where α is an absolute number which may be 0. Then $v = (w ± α) + √(k'/k).$ Now x² - y² + k' = (x - y)(x + y) + k' $= (u² - k)(v² - k) + k'$ $= u²v² - k(u² + v²) + k² + k'.$ One sufficient condition that the right-hand side may be a square is $k(v - u)² = k',$ or $v = u + √(k'/k),$ which is stated in the rule. Therefore, $x - y = (w ± α)² - k,$ $x + y = (w ± α + √(k'/k))² - k.$ Hence x = ½ (w ± α)² + (w ± α + √(k'/k))² - 2k, y = ½ (w ± α + √(k'/k))² - (w + α)². Now, if γ denotes √(k'/k), we get $x² + y² = u⁴ + 2γ u³ + (3γ² - 2k)u²$ $+ 2γ(γ² - k)u + ½k² + ½(γ² - k)².$ So it now remains to solve $u⁴ + 2γ u³ + (3γ² - 2k)u² + 2γ(γ² - k)u$ $+ ½k² + ½(γ² - k)² + k″ = t²,$ which is an indeterminate equation in *u*.