BRAHMAGUPTA AS AN ALGEBRAIST Let us solve the first example 8x²+1=y. We assume the optional number to be 3. Its square is 9; the prakṛti of multiplier is 8, their difference is 9-8-1. Dividing by this twice the optional number (2x3, i.e. 6), namely 6, we get the lesser root for the addi- tive unity as 6. Whence proceeding as before, we get the greater to be 17. Thus here x-6 and p=17. 254 Let us use this method for the equation 11x²+1=p². Let the optional number be 3. Its square is 9: multiplier or prakti is 11; the difference is 11-9-2; dividing by this twice the optional number (2x3), namely 6, we get 6/2-3, which is the lesser root. Consequently the greater root would be 10, Thus for this equation x=3 and p=10. Solution in Positive Integers The Indian algebraists usually aimed at obtaining solutions of the varga-prakyti or Square-nature in positive integers or abhinna. The tentative methods of Brahmagupta and Sripati always did not furnish solutions in positive integers. These auth- ors, however, discovered that if the interpolator of auxiliary equa- tion in the tentative method bet1l, 2 or 4, an integral solu- tion of the equation Nx³+1-y² can always be found. Thus Śripati says: If 1, 2 or 4 be the additive or subtractive (of the auxi- liary equation), the lesser and greater roots will be integral (abhinna)". (1) 1f k=+1, then the auxiliary equation will be No²+1=3 where a and ß are intergers. Then by Brahmagupta' Corollary we get x=2aß and y-8²+Na² as the required first solution in positive integers of the equation Nx²+1=² 1. इष्टवर्ग प्रकृत्योर्यविर तेन वा भजेत् । द्विघ्नमिष्टं कनिष्ठं तद् पदं स्यादेक संयुतौ । ततो ज्येष्ठमिहानन्त्यं भावनाभिस्तथेष्टतः || Bijaganita, Varga-Prakrti, 5-6
