BRAHMAGUPTA AS AN ALGEBRAIST Substituting the value of N in the right-hand side expres- sion from (i), we have 256 2 aß N. (-²/² ) ² + 1 = ( ³² / ³²) ² Composing (ii) and (iii), N {(²-1) Hence x aß, y=1 (B²--2); and xa(-1), y=15(²-3); are solutions of Nx²+1=y². +1= Na²-4-8² N(a/2)²-1(3/2)* (iii) { £ (P²-3) } If ß be even the first values of (x,y) are integral. If ß be odd, the second values are integral. (iv) Finally, suppose k=-4; the auxiliary equation is Na²-4 = 8² Then the required first solution in positive integers of Nx+1-y² is xaß(²+3) (²+1) v=(B²+2) {1(P²+3) (5²+1)-1}. 2 Brahmagupta says: In the case of 4 as subtractive, the square of the second is increased by three and by unity; half the product of these sums and that as diminished by unity (are obtained). The latter multiplied by the first sum less unity is the (required) second root; the former multi- plied by the product of the (old) roots will be the first root corresponding to the (new) second root.¹ The rationale of this solution. as given by Datta and Singh is as follows: Hence by Brahmagupta's Corollary. we get ´`N ( ²² ) ² + ¹ = ( + + N²^ ) ² 2 1. चतुरूनेऽन्त्यपद कृती त्र्येकयुते नषदलं पृथग्व्येकम् । व्येकाद्याइतमन्त्यं पदवध गुणमाद्यमान्त्यपदम् ॥ (i) BrSpSi. XVIII. 68
