History of Hindu Mathematics: A Source Book v2.c17.s15.29

v2.c17.s15.29

Their solutions are given substantially as follows : For (*iii*) we have the auxiliary equation $103.1² - 3 = 10².$ By the Lemma, we get $103((m+10)/(-3))² + (m²-103)/(-3) = ((10m+103)/(-3))².$ The general solution of $(m+10)/(-3) = an integer,$ is m = -3t + 2. Putting t = -3, we get m = 11. Then $103.7² - 6 = 71².$ Again, by the Lemma, $103((7n+71)/(-6))² + (n²-103)/(-6) = ((71n+103.7)/(-6))².$ The solution of $(7n+71)/(-6) = a whole number,$ is n = -6t + 1. Taking t = -1, we get $103.20² + 9 = 203².$ Next, we have $103((20p+203)/9)² + (p²-103)/9 = ((203p+103.20)/9)².$ Now, (20p+203)/9 = an integral number for p = 9t + 2. When t = 1, p = 11. On taking this value we find $103.47² + 2 = 477².$