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

INDETERMINATE EQUATIONS 225 quotient or phala. In later years, Mahāvīra has called the unknown number (x) as rāśi. Preliminary Operations in Kuṭṭaka-Karma Usually it has been suggested that in order that an equation of the form by−ax= ± c or by+ax= ± c may be amenable to solution, the two numbers a and b must not have a common divisor; for otherwise, the equation would be absurd, unless the number c had the same common divisor. So before the rules which we shall give hereafter, could be applied, the numbers a, b, c must be made prime (dṛḍha or firm; niccheda or having no divisor, or nirapavarta, meaning irreducible to each other. In this connection Bhāskara I writes : The dividend and divisor will become prime to each other on being divided by the residue of their mutual division. The operation of the pulveriser should be considered in relation to them.¹ Similarly we find in the writings of Brahmagupta : Divide the multiplier and the divisor mutually and find the last residue; those quantities being divided by the residue will be prime to each other.² Āryabhaṭa's Rule : Āryabhaṭa I is probably the first Indian writer on this subject, but the operation given by him is rather obscure. His disciple Bhāskara I has given the solution of inde- terminate equations of the first degree in more satisfactory langu- age. We shall give here the translation of Āryabhaṭa's verse from the Āryabhaṭīya, as rendered by Bibhutibhusan Datta, because other translations of this verse do very often confuse the sense : Divide the divisor corresponding to the greater remain- der by the divisor corresponding to the smaller remain-

1. भूदिनेष्टगणा-न्योन्य भक्तशेषेण भाजितौ । हारभाज्यौ दृढौ स्यातां कुट्टाकारं तयोर्विदुः —MBh, I. 41
2. हत्योः परस्परं यच्छेषं गुणकारभागहारकयोः । तेन हृतौ निश्छेदौ तावेव परस्परं हत्योः । —BrSpSi. XVIII. 9.

226 BRAHMAGUPTA AS AN ALGEBRAIST der. The residue (and the divisor corresponding to the smaller remainder) being mutually divided, the last resi- due should be multiplied by such an optional integer that the product being added(in case the number of quo- tients of the mutual division is even) or subtracted (in case the number of quotients is odd) by the difference of the remainders (will be exactly divisible by the last but one remainder. Place the quotients of the mutual division successively one below the other in a column; below them the optional multiplier and underneath it the quotient just obtained). Any number below . : . the penultimate) is multiplied by the one just above it and then added by that just below it. Divide the last number (obtained by doing so repeatedly) by the divisor corresponding to the smaller remainder; then multiply the residue by the divisor corresponding to the greater remainder and add the greater remainder. (The result will be) the number corresponding to the two divisors.¹ There is an alternative rendering of this passage also as follows : . Divide the divisor corresponding to the greater remain- der by the divisor corresponding to the smaller remain- der. The residue (and the divisor corresponding to the smaller remainder) being mutually divided (until the remainder becomes zero), the last quotient should be multiplied by an optional integer and then added (in case the number of quotients of the mutual division is even) or subtracted (in case the number of quotients is odd) by the difference of the remainders. (Place the other quotients of mutual division successively one below the other in a column; below them the result just obtained and underneath it the optional integer). Any

1. अधिकाग्रभागहारं छिन्यादूनाग्रभागहारेण । शेषपरस्परभक्तं मतिगुणमग्रान्तरे क्षिप्तम् ॥ अध उपरि गुणितमन्त्ययुगूनाच्छेद भाजिते शेषम् । अधिकाग्रच्छेदगुणं द्विच्छेदाग्रमधिकाग्रयुतम् । —Ārya. II. 32-33

PRELIMINARY OPERATIONS 227 number below (i.e. the penultimate) is multiplied by the one just above it and then added by that just below it. Divide the last number (obtained by doing so repeatedly) by the divisor corres- ponding to the smaller remainder; then multiply the residue ty the divisor corresponding to the greater remainder and add the greater remainder. (The result will be) the number corresponding to the two divisors. Āryabhaṭa’s problem may be enunciated thus : To find a number (N) which being divided by two given numbers (a, b) will leave two given remainders (R₁, R₂). This gives : N=ax+R₁=by+R₂ (where R₁ is a greater remainder and R₂ lesser remainder, and a is the divisor corresponding to greater remainder and b the divisor corresponding to the lesser remainder.) Denoting as before by c the difference between R₁, and R₂, we get (i) by=ax+c, if R₁>R₂ (ii) ax=by+c, if R₂>R₁ the equation being so written as to keep c always positive. Hence the problem now reduces to making either (ax+c)/b or (by+c)/a according as R₁>R₂ or R₂>R₁, a positive integer. So Āryabhaṭa says : Divide the divisor corresponding to the greater remainder etc.” Now we shall proceed with the details of the operation as proposed by Datta and Singh in his History of Hindu Mathema- tics, Part II. Algebra : Suppose R₁>R₂; then the equation to be solved will be ax+c=by ...(i) a, b being prime to each other.

228 BRAHMAGUPTA AS AN ALGEBRAIST Let b) a (q bq ─── r₁) b (q₁ r₁q₁ ─── r₂) r₁ (q₂ r₂q₂ ─── r₃ ... ────── rₘ₋₁) rₘ₋₂ (qₘ₋₁ rₘ₋₁ qₘ₋₁ ─────── rₘ) rₘ₋₁ (qₘ rₘqₘ ───── rₘ₊₁ Then we get (when a < b, we shall have q = 0, r₁ = a) a = bq + r₁ b = r₁q₁ + r₂ r₁ = r₂q₂ + r₃ r₂ = r₃q₃ + r₄ ... ... ... rₘ₋₂ = rₘ₋₁ qₘ₋₁ + rₘ rₘ₋₁ = rₘqₘ + rₘ₊₁ Now, substituting the value of a in the given equation (1), we get by = (bq + r₁)x + c Therefore y = qx + y₁ where by₁ = r₁x + c In other words, since a = bq + r₁, on putting y = qx + y₁ (ii) the given equation (i) reduces to by₁ = r₁x + c (iii) Again, since b = r₁q₁ + r₂

PRELIMINARY OPERATIONS 229 putting similarly x=q₁y₁+x₁ the equation (iii) can be further reduced to r₁x₁=r₂y₁—c (iv) and so on. Writing down the successive values and reduced equations in columns, we have (1) y=qx+y₁ (I.1) by₁=r₁x+c (2) x=q₁y₁+x₁ (I.2) r₁x₁=r₂y₁—c (3) y₁=q₂x₁+y₂ (I.3) r₂y₂=r₃x₁+c (4) x₁=q₃y₂+x₂ (I.4) r₃x₂=r₄y₂—c (5) y₂=q₄x₂+y₃ (I.5) r₄y₃=r₅x₂+c (6) x₂=q₅y₃+x₃ (I.6) r₅x₃=r₆y₃—c ......... ......... (2n-1) yₙ₋₁=q₂ₙ₋₂ xₙ₋₁+yₙ (I. 2n-1) r₂ₙ₋₂ yₙ=r₂ₙ₋₁ xₙ₋₁+c (2n) xₙ₋₁=q₂ₙ₋₁ yₙ+xₙ (I. 2n) r₂ₙ₋₁ xₙ=r₂ₙ yₙ—c (2n+1) yₙ=q₂ₙ xₙ+yₙ₊₁ (I. 2n+1) r₂ₙ yₙ₊₁=r₂ₙ₊₁ xₙ+c Now the mutual division can be continued either (i) to the finish or (ii) so as to get a certain number of quotients and then stopped. In either csse the number of quotients found, negle- cting the first one (q), as is usual with Āryabhaṭa, may be even or odd. Case (i) First suppose that the mutual division is continued until the zero remainder is obtained. Since a, b are prime to each other, the last one remainder is unity. Subcase (i.1.). Let the number of quotients be even. We then have r₂ₙ=1, r₂ₙ₋₁=0, q₂ₙ=r₂ₙ₋₁ The equations (1,2n) and (I.2n+1), therefore become yₙ=q₂ₙ xₙ+c and yₙ₊₁=c respectively. Giving an arbitrary integral value (t) to xₙ we get an integral value of yₙ. From that we can find the value of xₙ₋₁ by the equation (2n). Procceding backwards step by step we ultimately find the values of x and y in positive integers. So that the equation (I) is solved. Subcase (i. 2) : If the number of quotients be odd, we shall have r₂ₙ₋₁=1, r₂ₙ=0, q₂ₙ₋₁=r₂ₙ₋₂.

230 BRAHMAGUPTA AS AN ALGEBRAIST The equations (2n+1) and (I. 2n+1) will then be absent and the equations (I. 2n-1) and (I. 2n) will be reduced respectively to xₙ₋₁ = q₂ₙ₋₁ y - c and xₙ = - c Giving an arbitrary integral value (t') to yₙ we get an in- tegral value of xₙ₋₁. Then proceeding backwards as before we calculate the values of x and y. Case (ii) : Next suppose that the mutual division is stopped after having obtained an even or odd number of quotients. Subcase (ii.1) : If the number of quotients obtained be even the reduced form of the original equation is r₂ y + 1 = r₂ₙ + 1 xₙ + c or yₙ ₊ ₁ = (r₂ₙ ₊ ₁ xₙ + c) / r₂ Giving a suitable integral value (t) to xₙ as will make yₙ₊₁ = (r₂ₙ₊₁ t + c) / r₂ₙ = an integral number, we get an integral value for yₙ by (2n+1). The values of x and y can then be calculated by proceeding as before. Subcase (ii.2) : If the number of quotients be odd the reduc- ed form of the quotient is r₂ₙ₋₁ xₙ = r₂ₙ yₙ - c or xₙ = (r₂ₙ yₙ - c) / r₂ₙ₋₁ Putting yₙ = t', where t' is an integer, such that xₙ = (r₂ₙ t' - c) / r₂ₙ₋₁ = a whole number, we get an integral value of xₙ ₋ ₁ by (2n). Whence can be calculated the values of x and y in integers. If x = α and y = β be the least integral solution of ax + c = by, we shall have aα + c = bβ Therefore a(bm + α) + c = b (am + β), m being any integer. Therefore, in general, x = bm + α But we have calculated before that

BHASKARA I AND KUṬṬAKA OPERATION 231 x = q₁y₁ + x₁ ; ∴ q₁y₁ + x₁ = bm + α Thus it is found that the minimum value α of x is equal to the remainder left on dividing its calculated value by b. whence we can calculate the minimum value of N (=aα+R₁). This will explain the rationale of the operations described in the latter portion of the rule of Āryabhaṭa I. Bhāskara I and Kuṭṭaka Operation In Chapter I of the Mahābhāskarīya, Bhāskara I has descri- bed the preliminary operation to be performed on the divisor and dividend of a pulveriser. We shall quote it from the edition of K.S, Shukla : The divisor (which is "the number of civil days in a yuga) and the dividend (which is "the revolution num- ber of the desired planet") become prime to each other on being divided by the (last non-zero) residue of the mutual division of the number of civil days in a yuga and the revolution number of the desired planet. The operations of the pulveriser should be performed on them (i. e. on the abraded divisor and abraded dividend). So has been said.¹ An indeterminate equation of the first degree of the type (ax — c) / a = y (with x and y unknown) is known in Hindu mathematics by the name of "pulveriser"—kuṭṭakāra). In this equation, a is called the "dividend" (bhājya), b the "divisor" (bhāgahāra), c the interpolator (kṣepa), x the "multiplier" (guṇakāra), and y the "quotient" (labdha). In the pulveriser contemplated in the above stanza : a = revolution number of a planet.- b = civil days in a yuga, c = residue of the revolutions of the planet (Śeṣa)

1. भूदिनेष्टगणावन्योन्य भक्तशेषेण भाजितौ हारभाज्यौ दृढौ स्यातां कुट्टाकारं तयोर्विदुः । — MBh. I. 41

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

BHĀSKARA I AND KUṬṬAKA OPERATIONS 233 furnished by the above and applying the method of such quadratics, obtain the (simplest) solution by the substitution of 2, 3 etc. successively in the general solu- tion). Then calculate the ahargaṇa and the revolu- tions performed by Saturn and Mars in that time toge- ther with the number of solar years elapsed.¹ Let x and y denote the residues of the revolution of Mars and Saturn respectively. Then we have to find out two numbers x and y such that each of the expressions x+y, x-y and xy+1 may be a perfect square. Let x+y=4P² and x-y=4Q², so that x=2P²+2Q² y=2P²-2Q² and therefore xy+1=(2P²-1)²+4(P²-Q⁴) Hence the condition that xy+1 be a perfect square is that P²=Q⁴. Substituting these values, we have x=2(Q⁴+Q²) y=2(Q⁴-Q²) where Q may possess any of the values 2, 3, 4,......but not 1. (We neglect the case when x or y is zero).

1. भाज्यं न्यसेदुपरि हारमधश्च तस्य । खण्ड्यात्परस्परमथो विनिधाय लब्धम् । केनाऽऽहतोऽ

234 BRAHMAGUPTA AS AN ALGEBRAIST Putting Q=2, we get x=40 and y=24, which is the least solution. Assuming now that the residues of the revolution (maṇḍa- laja-śeṣa) of Saturn and Mars are 24 and 40 respectively, we have to obtain the ahargaṇa (which means the number of mean civil days elapsed since the beginning of Kaliyuga, or, in fact, any epoch). The revolution-number of Saturn is 146564, and the number of civil days in a yuga is 1,577,917,500. In the present problem, these are respectively the dividend and the divisor. Their H.C.F. is 4, so that dividing them out by 4 we get 36641 and 394,479,375 as the abraded dividend and abraded divisor respectively. We have, therefore, to solve the pulveriser 36641x–24 ────────── = y 394479375 where x and y denote the ahargaṇa and the revolutions respecti- vely made by Saturn. Mutually dividing 36641 and 394479375, we get 36641) 394479375 (10766 394477006 ───────── 2369) 36641 (15 35535 ───── 1106) 2369 (2 157) 1106 (7 1099 7) 157 (22 154 3) 7 (2 6 ── 1×27–24=3)3(1 3 ── 0 We have chosen here the number 27 as the optional number (mati). In fact, mati may be chosen at any stage after an even number of quotients are obtained.

BHĀSKARA I AND KUṬṬAKA OPERATIONS 235 Writing down the quotients one below the other as pres- cribed in the rule, we get the chain 10766 15 2 7 22 2 (mati) 27 1 Reducing the chain, we successively get 10766 10766 10766 10766 10766 10766 3108044439 (multiplier) 15 15 15 15 15 288689 288689 (quotient) 2 2 2 2 18665 18665 7 7 7 8714 8714 22 22 1237 1237 2 55 55 (mati) 27 27 1 (it would be seen in this reduction of chain that mati or 27 × 2 plus 1 is 55; 55 × 22 plus 27 is 1237; 1237 × 7 plus 55 is 8714; 8714 × 2 plus 1237 is 18665; 18665 × 15 plus 8714 is 288689; and finally 288689 × 10766 plus 18665 is 3108044439 which is the multiplier). Dividing 3108044439 by 394479375, and 288689 by 36641, we obtain 346688814 and 32202 respectively as remainders, (This division is performed only when the multiplier and quotient are greater than the divisor and dividend respectively). These are the minimum values of x and y satisfying the above equation. Therefore, the required ahargaṇa = 346688814. and the revolutions performed by Saturn = 32202. To obtain the ahargaṇa and the revolutions of Mars, one has to solve the equation : 191402 z — 40 ───────────── = w 131493125

236 BRAHMAGUPTA AS AN ALGEBRAIST where z and w denote the ahargaṇa and the revolutions per- formed by Mars respectively. The general solution of this equation is z = 131493125 s + 118076020 w = 191402 s + 171872 where s = 0, 1, 2, 3......When s = 0, we have the least solution. Brahmagupta’s Rules Concerning Indeterminate Analysis of the First Degree For the solution of Āryabhaṭa’s problem, Brahmagupta gives the following rule : What remains when the divisor corresponding to the greater remainder is divided by the divisor correspond- ing to the smaller remainder—that (and the latter divisor) are mutually divided and the quotients are severally set down one below the other. The last residue (of the reciprocal division after an even number of quo- tients has been obtained) is multiplied by such an optio- nal integer that the product being added with the differ- ence of the (given) remainders will be exactly divisible (by the divisor corresponding to that residue). That optional multiplier and then the (new) quotient just obtained should be set down (underneath the listed quotients). Now, proceeding from the lower-most number (in the column), the penultimate is multiplied by the number just above it and then added by the number just below it. The final value thus obtained (by repeating the above process) is divided by the divisor corresponding to the smaller remainder. The residue being multiplied by the divisor corresponding to the greater remainder and added to the greater remainder will be the number in view.¹

1. अधिकाग्रभागहारादूनाग्रच्छेद भाजिताच्छेषम् । यत् तत् परस्परहृतं लब्धमधोधः पृथक् स्थाप्यम् ॥ शेषं तदिष्टगुणितं यथाऽनयोरन्तरेण संयुक्तम् । शुध्यति गुणकः स्थाप्यो लब्धं चान्त्यादुपान्त्यगुणः ॥ स्वोर्ध्वोऽन्त्य युतोऽग्रान्तो हीनाग्रच्छेदभाजितः शेषम् । अधिकाग्रच्छेदहत मग्रिकाप्रयुतं भवत्यग्रम् ॥ —BrSpSi. XVIII, 3-5

BRAHMAGUPTA'S RULES OF ANALYSIS 237. Brahmagupta further observes : Such is the process when the quotients (of mutual division) are even in number. But if they be odd, what has been stated before as negative should be made as positive, or as positive should be made negative.¹ Regarding the direction for dividing the divisor corres- ponding to the greater number by the divisor corresponding to the smaller remainder, Pṛthūdaka Svāmī (860A.D.) observes that it is not absolute, rather optional; so that the process may be conducted in the same way by starting with the division of the divisor corresponding to the smaller remainder by the divisor corresponding to the greater remainder. But in this case of in- version of the process, he continues, the difference of the remain- ders, must be negative. That is to say, the equation by=ax+c can be solved by transforming it first to the form ax=by-c so that we shall have to start with the division of b by a. For the details of the "Theory of the pulveriser" as applied to the problems in Astronomy, the reader is referred to the writ- ings of Bhaṭṭa Govind, translated by K.S. Shukla, and given as an Appendix to the edition of the Laghu-Bhāskarīya. For the rationale of the rules in relation to kuṭṭaka or the pulveriser operation, one may also refer to the chapters by Datta and Singh in the History of Hindu Mathematics: Algebra. Solution of by=ax ± 1. This simple indeterminate equation has a special use in astronomical calculations and therefore, Indian algebraists have paid special attention to it. In fact, this equation is solved exactly in the same way as the equation by=ax ± c; it is a parti-

1. एवं समेषु विषमेष्वृणं धनं धनमृणं यदुक्तं तत् । ऋणधनयोर्व्यस्तत्वं गुण्य प्रक्षेपयोः कार्यम् ॥ —BrSpSi. XVIII. 13.

238- BRAHMAGUPTA AS AN ALGEBRAIST cular case only of the more general latter equation. Of course, there is a little justification also for treating it separately, since both the types of equations represent two different physical conditions of the astronomical problems. In the case of by=ax±c, the conditions are such that the value of either y or x, more particu- larly of the latter, has to be found and the rules for solution formulated with that objective. But in the case of the equation by=ax±1, the physical conditions require the values of both y and x. The equation by=ax±1 is usually known by the name sthira-kuṭṭaka, literally meaning the 'constant pulveriser' Pṛthū- daka Svāmī also names it as dṛḍha-kuṭṭaka meaning firm-pulveri- ser. Later on this term dṛḍha-was confined to another sense, equivlent to nicched (having no divisor) or nirapavarta (irreduci- ble). The origin of the name sthira-kuṭṭaka or constant pulveriser has been explained by Pṛthūdaka Svāmī as being due to the fact that the interpolator (±1) is here invariable. For the solution of this equation, we shall quote Bhāskara I's rule and the rule by Brahmagupta. Bhāskara I writes in this connection as follows : The method of the pulveriser is applied also after subtracting unity. The multiplier and quotient are respectively the numbers above and underneath. Multi- plying those quantities by the desired number divide by the reduced divisor and dividend; the residues are in this case known to be the (elapsed) days and (resid- ues of) revolutions respectively¹. The pulveriser ax - c -------- = y ... (1) b may be written as aX - 1 -------- = Y ... (2) b where x=cX and y=cY. If X=α, Y=β is a solution of (2), then x=cα, y=cβ will be a solution of (1). Hence the above rule.

1. रूपमेकमपास्यापि कुट्टाकारः प्रसाध्यते । गुणकारोऽथ लब्धं च राशी स्यातामुपर्यधः ॥ —MBh. I. 45

BRAHMAGUPTA'S RULES OF ANALYSIS 239 Brahmagupta's Rule in this connection is as follows : Solution of by=ax-1 : Divide them (i.e., the abraded coefficient of the multi- plier and the divisor) mutually and set down the quo- tients one below the other. The last residue (or the reciprocal division after an even number of quotients has been obtained) is multiplied by an optional integer such that the product being diminished by unity will be exactly divisible (by the divisor corresponding to that residue). The (optional) multiplier and then this quotient should be set down (underneath the listed quotients). Now proceeding from the lower most term to the uppermost, by the penultimate multiply the term just above it and then add the lowermost number. (The uppermost number thus calculated being divided by the reduced divisor, the residue (is the quan- tity required. This is the method of the constant pul- veriser¹. Solution of by+ax=±c Indian algebraists usually transformed this equation as by=-ax+c, so that it appeared as a particular case of by=ax+c, in which a was negative. Brahmagupta has been the first person to solve this equation, but the rule given by him is obscure : The reversal of the negative and positive should be made of the multiplier and interpolator.² Pṛthudaka Svāmī has tried to explain it, but he too is not very clear. He says :

1. हृतयोः परस्परं यच्छ्रेषं गुणकार भागहारकयोः । तेन हृतौ निश्छेदौ तावेव परस्परं हृतयोः ॥ लब्धमधोऽधः स्थाप्यं तथेष्ट गुणकारसङ्गुणं शेषम् । शुद्धस्यति यथैकहीनं गुणकः स्थाप्यः फलं चान्त्यम् ॥ अश्रान्तमुपान्त्येन स्वोर्ध्वो गुणितोऽन्त्य संयुतो भक्तम् । निःशेषभागहारेणैव स्थिरकुट्टकः शेषम् ॥ —BrSpSi. XVIII. 9-11
2. एवं समेषु विषमेष्वृणं धनं धनमृणं यदुक्तं तत् । ऋणधनयोर्व्यस्तत्वं गुण्यप्रक्षेपयोः कार्यम् ॥ BrSpSi XVIII. 13

240 BRAHMAGUPTA AS AN ALGEBRAIST If the multiplier be negative, it must be made positive; and the additive must be made negative : and then the method of the pulveriser should be employed. Pṛthūdaka Svāmī, however, does not indicate how to derive the solution of the equation. by = -ax+c ...(1) from that of the equation by = ax-c ...(2) The method, however, seems to have been this : Let x=α, y=β be the minimum solution of (2). Then we get bβ = a α-c or b(a-β) = -a(a-b)+c Hence x=a-b, y=a-β is the minimum solution of (1). This rule is very clearly indicated by Bhāskara II and others. We shall give two examples from Bhāskara II (Bījagaṇita) to illustrate the rule : Example I. 13y = -60x + 3 By the method described before, we find that the minimum solution of 13y = 60x+3 is x=11, y=51. Subtracting these values from their respective abraders, namely 13 and 60, we get 2 and 9. Then by the maxim : "In the case of the dividend and divisor being of differ- ent signs, the results from the operation of division should be known to be so", making the quotient negative we get the solu- tion of 13y = -60x+3 as x=2, y=-9. Subtracting these values again from their respec- tive abraders (13, 60), we get the solution of 13y = -60x-3 as x=11, y=-51. Example II. 11y = 18x+10

LINEAR EQUATIONS 241 Proceeding as before, we find the minimum solution of 11y=18x+10 to be x=8, y=14. These will also be the values of x and y in the case of the negative divisor but the quotient for the reasons stated before should be made negative. So the solution of -11y=18x+10 is x=8, y=-14. Subtracting these (i.e., their numerical values) from their respective abraders, we get the solution of -11y=18x-10 as x=3, y=-4, "When the divisor is positive or negative the numeri- cal values of the quotient and multiplier remain the same : when either the divisor or the dividend is negative, the quotient must always be known to be negative"¹. One Linear Equation in More Than Two Unknowns Whenever a linear equation involves more than two unkno- wn's the Indian algebraists used to assume arbitrary values for all the unknowns except two and then to apply the method of kuṭ- ṭaka or "pulveriser". In this connection, Brahmagupta says : The method of the pulveriser (should be employed if there be present many unknowns (in any equation)²,

1. Bhāskara II gives the following rule : "Those(the multiplier and quotient)cbtained for a positive divi- dend being treated in the same manner give the results corres- ponding to a negative dividend." The treatment alluded to in this rule is that of subtraction from the respective abraders. He has further elaborated it thus : The multiplier and quotient should be determined by taking the dividend, divisor and interpolator as positive. They will be the quantities for the additive interpolator. Subtracting them from their respective abraders, the quantities for a negative interpolator are found. If the dividend or divi- sor, be negative, the quotient should be stated as negative, the quotient should be stated as negative. —Bījagaṇita
2. आद्याद्वर्णादन्यान् वर्णान् प्रोह्याद्यमानमाद्यहृतम् । सदृशच्छेदावसकृद् द्वौ व्यस्तौ कुट्टको बहुषु ॥ — BrSpSi. XVIII. 51

242 BRAHMAGUPTA AS AN ALGEBRAIST We shall take up one of the problems posed by Brahmagupta concerning astronomy and leading to the equation :¹ 197x - 1644 y - z = 6302. Hence 1644 y + z + 6302 x = ------------------- 197 The commentator assumes z = 131. Then 1644 y + 6433 x = --------------- ; 197 hence by the usual method of the pulveriser x = 41; y = 1. General Problem of Remainders A certain type of simultaneous indeterminate equations of the first degree arise out of the general problem of remainders which may thus be stated : To find a number N which being severally divided by a₁, a₂, a₃.......aₙ , leaves as remainders r₁, r₂, r₃.........rₙ respectively. While dealing with such a case, we shall have the following series of equations : N = a₁x₁ + r₁ = a₂x₂ + r₂ = a₃x₃ + r₃ = ...... = aₙ xₙ + r . We have reasons to believe that the method of solution of these equations was known to Āryabhaṭa I. In the translation of the verse in the Āryabhaṭīya, II. 32-33 (the translation of which we have already given), the term dvicchedāgram should be translated as "the result will be the remainder corresponding to the product of the two divisors", instead of "the result will be the number corresponding to the two divisors." (the last line of the translation). This explanation is in fact given by Bhāskara I, the direct disciple and earliest commentator of Āryabhaṭa I. Such a rule is clearly stated by Brahmagupta².

1: अंशकशेषेण युतात् लिप्ताशेषात्तदन्तरादथवा । भानोर्भ'दिने द्युगणं यः कथयति कुट्टकज्ञः सः ॥ —BrSpSi. XVIII. 55 2: स्वोच्छेऽन्ययुतोऽग्रान्तो हीनाग्रच्छेदभाजितः शेषम् । अधिकाग्रच्छेदहतमधिकाग्रयुतं भवत्यग्रम् ॥ —BrSpSi. XVIII. 5

GENERAL PROBLEM OF REMAINDERS 243 The rationale of this method is not difficult. I shall quote it from the book of Datta and Singh: Starting with the considera- tion of the first two divisors, we have N = a₁x₁ + r₁ = a₂x₂ + r₂. By the method described before, we can find the minimum value α of x₁ satisfying this equation. Then the minimum value of N will be a₁α + r₁. Hence the general value of N will be given by N = a₁ (a₂t + α) + r₁ = a₁a₂ t + a₁α + r₁ where t is an integer. Thus a₁α + r₁ is the remainder left on dividing N by a₁ a₂ as stated by Āryabhaṭa I and Brahmagupta. Now taking into consideration the third condition, we have N = a₁a₂t + a₁αr₁ = a₃x₃ + r₃ which can be solved in the same way as before. Proceeding in this way successively, we shall ultimately arrive at a value of N satisfying all the conditions ; Pṛthūdaka Svāmī remarks : Wherever the reduction of two divisors by a common measure is possible, there 'the product of the divisors' should be understood as equivalent to the product of the divisor corresponding to the greater remainder and quotient of the divisor corresponding to the smaller remainder as reduced (i.e. divided) by the common mea- sure.¹ When one divisor is exactly divisible by the other, then the greater remainder is the (required) remainder and the divisor corresponding to the greater remainder is taken as 'the product of the divisors'. (The truth of) this may be investigated by an intelligent mathematician by taking several symbols. As an illustration we shall take up a problem quoted by Bhāskara II in his Bījagaṇita, and which in its solution follows the method of Āryabhaṭa 1. Pṛthūdaka Svāmī while commenting on serveral verses from Brahmagupta (BrSpSi. XVIII. 3-6)

1. i.e., if p be the L.C.M. of a, and a2, the general value of N satisfying the above two conditions will be N = pt + a₁a + r, instead of N = a₁a₂t + a₁a + r₁.

244 BRAHMAGUPTA AS AN ALGEBRAIST observes that such problems were very popular amongst the an- cient Indian mathematicians. Problem : To find a number N which leaves remainders 5, 4,3,2 when divided by 6,5,4,3 respectively. That is to solve the equations : N = 6x + 5 = 5y + 4 = 4z + 3 = 3w + 2. We have since N = 6x + 5 = 5y + 4, x = (5y - 1) / 6 But x must be integral, so y = 6t + 5, x = 5t + 4 Hence N = 30t + 29 Again N = 30t + 29 = 4z + 3 Therefore, t = (2z - 13) / 15 Since t must be integral, we must have z = 15s + 14; hence t = 2s + 1. Therefore N = 60s + 59. The last condition is identically satisfied. The method given here is the one followed by Pṛthūdaka Svāmī. Thus when N = 60s + 59 = 6x + 5 x = (60s + 54) / 6 = 10s + 9 ...(1) Again, when N = 60s + 59 = 5y + 4, y = (60s + 55) / 5 = 12s + 11 Again when N = 60s + 59 = 4z + 3 z = (60s + 56) / 4 = 15s + 14 Lastly, when N = 60s + 59 = 3w + 2. w = (60s + 57) / 3 = 20s + 19, Varga Prakṛti or Kṛti Prakṛti or Square-Nature The word varga-prakṛti (literally meaning 'square-nature') has been given by Indian algebraists to the indeterminate quadra- tic equation Nx² ± c = y²