History of Hindu Mathematics: A Source Book v2.c17.s15.31

v2.c17.s15.31

For the solution of (*iv*) the auxiliary equation is $97 . 1² + 3 = 10².$ Therefore $97((m+10)/3)² + (m² - 97)/3 = ((10m + 97)/3)².$ The solution of $(m+10)/3 = an integer,$ is m = 3t + 2. Taking t = 3, we have m = 11. Then $97 . 7² + 8 = 69.²$ Next, we have $97((7n+69)/8)² + (n² - 97)/8 = ((69n+97 . 7)/8)².$ The solution of $(7n+69)/8 = an integer,$ is n = 8t + 5. Taking t = 1, that is, n = 13, we get $97 . 20² + 9 = 197².$ Whence $97((20p + 197)/9)² + (p² - 97)/9 = ((197p + 97 . 20)/9)².$ The solution of $(20p + 197)/9 = a whole number,$ is p = 9t + 5. Putting t = 1, we get p = 14. With this value of p we have $97·53² + 11 = 522².$ Whence $97 ((53q + 522)/11)² + (q² - 97)/11 = ((522q + 97·53)/11)².$ The solution of $(53q + 522)/11 = an integer,$ is q = 11t + 8. The appropriate value of q is given by t = 0. So, taking q = 8, we have $97·86² - 3 = 847².$ Next, we find $97 ((86r + 847)/(-3)) + (r² - 97)/(-3) = ((847r + 97·86)/(-3))².$ The solution of $(86r + 847)/(-3) = a whole number,$ is r = 3t + 1. Putting t = -3, we get r = 10. Taking this value, we have $97·569² - 1 = 5604².$ By the Principle of Composition of Equals, we find $97 (6377352)² + 1 = (62809633)².$ Hence x = 6377352, y = 62809633 is a solution of (*iv*).