History of Hindu Mathematics: A Source Book v2.c17.s15.25

v2.c17.s15.25

(*i*) 67x² + 1 = y². We take the auxiliary equation $67.1² - 3 = 8².$ Then, by the Lemma, $67((1.m+8)/(-3)) + (m²-67)/(-3) = ((8m+67.¹)/(-3))². (1)$ By the method of the *Kuṭṭaka* the solution of $(m+8)/(-3) = an integer,$ is m = -3t + 1. Putting t = -2, we get m = 7 which makes |m² - 67| least. On substituting this value, the equation (1) reduces to $67.5² + 6 = 41².$ Again, by the Lemma, we have $67((5n + 41)/6)² + (n² - 67)/6 = ((41n + 67.5)/6)². (2)$ The solution of $(5n + 41)/6 = a whole number,$ is n = 6t + 5. |n² - 67| will be least for the value t = 0, that is, when n = 5. The equation (2) then becomes $67.11² - 7 = 90².$ Now, we form $67((11p + 90)/(-7))² + (p² - 67)/(-7) = ((90p + 67.11)/(-7))². (3)$ The solution of $(11p + 90)/(-7) = an integral number,$ is p = -7t + 2. Taking t = -1, we have p = 9; and this value makes |p² - 67| least. Substituting that in (3) we get $67.27² - 2 = 221².$ By the Principle of Composition of Equals, we get from this equation $67(2.27.221)² + 4 = (221² + 67.27²)²,$ $or 67(11934)² + 4 = (97684)².$ Dividing out by 4, we have $67(5967)² + 1 = (48842)².$