ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

210 BRAHMAGUPTA AS AN ALGEBRAIST Similar solutions have been offered by the other Indian algebraists who followed Brahmagupta like Śrīpati, Bhāskara II and Nārāyaṇa. Here again, we take a problem proposed by Brahmagupta in this connection : Problem : Tell the number of elapsed days for the time when four times the twelfth part of the residual degrees increased by one, plus eight will be equal to the residual degrees plus one.¹ Pṛthūdaka Svāmī has solved this problem as follows : Here the residual degrees are (put as) yāvat-tāvat, yā; increased by one, yā 1 rū 1; twelfth part of it, (yā 1 rū 1) / 12 ; four times this, (yā 1 rū 1) / 3 ; plus the absolute quantity eight, (yā 1 rū 25) / 3 . This is equal to the residual degrees plus unity. The statement of both sides tripled is yā 1 rū 25 yā 3 rū 3 This difference between the coefficients of the unknown is 2. By this the difference of the absolute terms, namely 22, being divided, is produced the residual of the degrees of the Sun, 11. These residual degrees should be known to be irreducible. The elapsed days can be deduced then, (proceeding) as before. If put in the modern notations, it means the solution of the equation : 4/12 (x+1)+8=x+1, from which we have x+25=3x+3 or 2x=22 or x=11. Rule of Concurrence or Saṁkramaṇa Brahmagupta has included this rule in algebra, whereas other Indian mathematicians included it in arithmetic. Saṁ-

1. सैकादंशकशेषाद् द्वादशभागश्चतुर्गुणोऽष्टयुतः । सैकांशशेषतुल्यो यदा तदाऽहर्गणं कथय ॥ —BrSpSi. XVIII. 46