RATIONAL SOLUTION get the greater root as 3. Hence the statement for the com- position is m=11 1=1 g=3 1=1 1=-2 i=-2 253 Here m multiplier (gunaka prakrti). Ilesser root (kaniştha-mula). g=greater root (jyestha-mala) and i-interpola- tor (kşepa). Here we have set down successively the lesser root, greater root and interpolator, and below them again set down the same (See Brahmagupta's Lemmas described by Bhaskara II). Now proceeding as before we obtain the roots for the additive 4: 1-6, g 20, (for) i=4. Then by the rule: "If the interpolator (of a varga-prakyti or Square-nature) divided by the square of an optional number be the interpolator (of another Square-nature), then the two roots (of the former) divided by that optional number will be the roots (of the other). Or, if the interpolator be multiplied, their roots should be multiplied."¹ are found the roots for the additive unity 1-3. g=10 (for) i=1. Whence by the Principle of Composition of Equals, we get the lesser and greater roots: 1-60, g=199 (for) i 1. In this way an infinite number of roots can be deduced. Alternative method:-Bhaskara II has given another method for finding the two roots for the additive unity : Or divide twice an optional number by the difference between the square of that optional number and the prakyti. This (quotient) will be the lesser root (of a Square-nature) when unity is the additive. From that (follows) the greater root." 1. इष्टवर्गहृतः वपः क्षपः स्वादिष्टभाजिते । मूले ते स्तोऽथवा चोपः क्षुण्णः तुरणे तदा पदे || 2. Siddhanta-sekhara, XIV. 32, Bijaganita II. 5.
