History of Hindu Mathematics: A Source Book v2.c17.s15.27

v2.c17.s15.27

Hence x = 5967, y = 48842 is a solution of (i). $(ii) 6Ix² + I = y².$ Here we start with the auxiliary equation $6I.I² + 3 = 8².$ By the Lemma, we have $6I((m+8)/3)² + (m²-6I)/3 = ((8m+6I)/3)². (I)$ Now the solution of $(m+8)/3 = an integer,$ is m = 3t + I. Putting t = 2, we get the value m = 7 which makes | m²-6I | least. On substituting this value in (1), it becomes $6I.5² - 4 = 39².$ Dividing out by 4, we get $6I(5/2)² - I = (39/2)². (2)$ By the Principle of Composition of Equals, we have $6I(2.5/2.39/2)² + I = (39/2)² + 6I(5/2)²²,$ $or 6I(195/2)² + I = (1523/2)². (3)$ Combining (2) and (3), $6I(3805)² - I = (29718)².$ Composing this with itself, we get $6I(226I53980)² + I = (I7663I9049)².$ Hence x = 226I53980, y = I7663I9049 is a solution of (ii).