History of Hindu Mathematics: A Source Book v2.c17.s23.44

v2.c17.s23.44

Putting n = 1 in (a') and (a″), we have the integral solutions $.x &= 2m², & u &= 2m² + 1, / y &= 2m, & v &= 2m² - 1; (a'.1)[¹]$ and $.x &= 2m⁴(16m² + 3), / y &= 2m(16m⁴ + 1), / u &= (16m⁴ + 1)(2m² + 1), / v &= (16m⁴ + 1)(2m² - 1). (a″.1)$