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

232 BRAHMAGUPTA AS AN ALGEBRAIST x=ahargaṇa , and y=complete revolutions performed by the planet. The text says that as a preliminary operation to the solu- tion of this pulveriser, a and b. i.e., civil days in yuga and revo- lution-number of the planet, should be made prime to each other by dividing them out by their greatest common factor, That is to say, in solving a pulveriser, one should always make use of abraded divisor and abraded dividend. The interpolator, i.e., the residue, should also be divided out by the same factor. (This instruction is not given in the text, but it is implied that the residue should be computed for the abraded dividend and abraded divisor). Set down the dividend above and the divisor (hāra) below that. Divide them mutually and write down the quotients (labdha) of division one below the other (in the form of a chain). (When an even number of quotients is obtained) think out by what number the (last) remainder be multiplied so that the product being diminished by the (given) residue be exactly divisible (by the divisor corresponding to that remain- der). Put down the chosen number called mati below the chain and then the new quotient underneath it. Then by the chosen number multiply the number which stands just above it, and to the product add the quoti- ent (written below the chosen number). (Replace the upper number by the resulting sum and cancel the number below). Proceed afterwards also in the same way (until only two numbers remain). Divide the upper number (called the "multiplier") by the divisor by the usual process and the lower one (called the "quotient") by the dividend : the remainders (thus obtained) will respectively be the ahargaṇa and the revolutions etc. or what one wants to know.¹ We shall illustrate the operation by taking a problem from the Laghu-Bhāskarīya (VIII. 17) : The sum, the difference, and the product increased by one, of the residues of the revolution of Saturn and Mars—each is a perfect square. Taking the equations