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History of Hindu Mathematics: A Source Book v2.c17.s13: 15. Simultaneous Indeterminate Equations of the First Degree

History of Hindu Mathematics: A Source Book v2.c17.s13

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15. Simultaneous Indeterminate Equations of the First Degree

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**Śrîpati’s Rule.** We have described before the rule of Brahmagupta for the solution of simultaneous equations of the first degree.³ In the latter portion of that rule there are hints for the solution of simultaneous indeterminate equations by the application of the method of the pulveriser. Similar rules have been given by later Hindu algebraists. Thus Śrîpati (1039) says:

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³ See pp. 54f.

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“Remove the first unknown from any one side of an equation leaving the rest, and remove the rest from the other side. Then find the value of the first by dividing the other side by its coefficient. If there be found thus several values (of the first unknown), the same (operations) should be made again (by equating two and two of those values) after reducing them to a common denominator. (Proceed thus repeatedly) until there results a single value for an unknown. Now apply the method of the pulveriser; and from the values (determined in this way) the other unknowns will be found by proceeding backwards. In the pulveriser the multiplier will be the value of the unknown associated with the dividend and the quotient, of that with the divisor."¹

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¹ *SiŚe*, xiv. 15-6.

Bhâskara II's Rule

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Bhâskara II (1150) writes:

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"Remove the first unknown from the second side of an equation and the others as well as the absolute number from the first side. Then on dividing the second side by the coefficient of the first unknown, its value will be obtained. If there be found in this way several values of the same unknown, from them, after reduction to a common denominator and then dropping it, values of another unknown should be determined. In the final stage of this process, the multiplier and quotient obtained by the method of the pulveriser will be the values of the unknowns associated with the dividend and the divisor (respectively). If there be several unknowns in the dividend, their values should be determined after assuming values of all but one arbitrarily. Substituting these values and proceeding reversely, the values of the other unknowns can be obtained. If on so doing there results a fractional value (at any stage), the method of the pulveriser should be employed again. Then determining the (integral) values of the latter unknowns accordingly and substituting them, the values of the former unknowns should be found proceeding reversely again."²

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² *BBi*, p. 76.

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A similar rule has been given by Jñânarâja.

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*Example from Bhâskara II:*

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"(Four merchants), who have horses 5, 3, 6 and 8 respectively; camels 2, 7, 4 and 1; whose mules are 8, 2, 1 and 3; and oxen 7, 1, 2 and 1 in number; are all owners of equal wealth. Tell me instantly the price of a horse, etc."¹

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¹ *BBi*, p. 79.

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If *x*, *y*, *z*, *w* denote respectively the prices of a horse, a camel, a mule and an ox, and *W* be the total wealth of each merchant, we have $r@ l 5x + 2y + 8z + 7w = W & (1) / 3x + 7y + 2z + w = W & (2) / 6x + 4y + z + 2w = W & (3) / 8x + y + 3z + w = W & (4)$ Then $r@ l@ l x & = ½(5y - 6z - 6w), & from (1) and (2) / & = ⅓(3y + z - w), & from (2) and (3) / & = ½(3y - 2z + w), & from (3) and (4)$

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From the first and second values of *x*, we get $y = 1/9(20z + 16w);$ and from the second and third values, we have $y = ⅓(8z - 5w).$ Equating these two values of *y* and simplifying, $20z + 16w = 24z - 15w.$ Therefore $z = 31w/4.$ Take $w = 4t; then$ $z = 31t, y = 76t, x = 85t.$

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**Special Rules.** Bhâskara II observes that the physical conditions of problems may sometimes be such that the ordinary method of solving simultaneous indeterminate equations of the first degree, which has been just explained, will fail to give the desired result. One such problem has been described by him as follows :

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“Tell quickly, O algebraist, what number is that which multiplied by 23 and severally divided by 60 and 80 leaves remainders whose sum is 100.”¹

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¹ *BBi*, p. 91.

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Let the number be denoted by *x*; the quotients by *u*, *v*; and the remainders by *s*, *t*. Then we have $(23x - s)/60 = u, (23x - t)/80 = v ;$ also $s + t = 100.$ Therefore $x = (60u + s)/23 = (80v + t)/23.$ Hence $x = (60u + 80v + s + t)/46,$ or $x = (30u + 40v + 50)/23.$

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For the solution of the above he observes : “Here, (although) there is more than one quotient (*u*, *v*) in the dividend, the value of any should not be arbitrarily assumed ; for on so doing the process will fail.”² “In a case like this,” continues he, “the (given) sum of the remainders should be so broken up that each remainder will be less than the divisor corresponding to it and further that impossibility will not arise ; then must be applied the usual method.”

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² *BBi*, p. 91f.

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In the present example we thus suppose s = 40, t = 60. Hence we have $160u + 40 = 80v + 60$ or $u = (80v + 20)/60 = (4v + 1)/3;$ whence by the method of the pulveriser, we get $v = 3w + 2, u = 4w + 3.$

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Therefore $x = (240w + 220)/23.$

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Again, applying the method of the pulveriser in order to obtain an integral value of x, we have $w = 23m + 1, x = 240m + 20.$

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If we take s = 30, t = 70, we shall find, by proceeding in the same way, another value of x as 240m + 90.

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**General Problem of Remainders.** One type of simultaneous indeterminate equations of the first degree is furnished by the general problem of remainders, *viz.*, to find a number N which being severally divided by a₁, a₂, a₃, ..., a_n leaves as remainders r₁, r₂, r₃, ..., r_n respectively.

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In this case, we have the equations $N = a₁x₁ + r₁ = a₂x₂ + r₂ = a₃x₃ + r₃ =$ $= a_nx_n + r_n.$

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The method of solution of these equations was known to Āryabhaṭa I (499). For this purpose the term *dvicchedâgraṁ* occurring in his rule for the pulveriser must be explained in a different way so that the last line of the translations given before (pp. 94-5) will have to be replaced by the following: "(The result will be) the remainder corresponding to the product of the two divisors."¹ This explanation is, in fact, given by Bhâskara I, the direct disciple and earliest commentator of Āryabhaṭa I. Such a rule is expressly stated by Brahmagupta.¹

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¹ See Bibhutibhusan Datta, *BCMS*, XXIV, 1932.

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¹ *BrSpSi*, xviii. 5.

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The *rationale* of this method is simple: Starting with the consideration 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 &= a_1(a_2t + \alpha) + r_1, / &= a_1a_2t + a_1\alpha + r_1, where t is an integer. Thus a₁α + r₁ is the remainder left on dividing N by a₁a₂, as stated by Âryabhata I and Brahmagupta. Now, taking into consideration the third condition, we have N = a₁a₂t + a₁α + r₁ = 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.

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Pṛthûdakasvâmî remarks: ·

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“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 measure.² When one divisor is exactly divisible by the other then the greater remainder is the (required) remainder and the divisor corresponding to

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² *i.e.*, if p be the L.C.M. of a₁ and a₂, the general value of N satisfying the above two conditions will be $N = pt + a₁α + r₁$ instead of $N = a₁a₂t + a₁α + r₁.$

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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.”

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*Examples from Bhâskara I* :

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(1) “Find that number which divided by 8 leaves 5 as remainder, divided by 9 leaves 4 as remainder and divided by 7 leaves 1 as remainder.”¹

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¹ See his commentary on *Â*, ii. 32-3.

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That is to say, we have to solve $N = 8x + 5 = 9y + 4 = 7z + 1.$ The solution is given substantially thus : The minimum value of N satisfying the first two conditions $N = 8x + 5 = 9y + 4$ is found by the method of the pulveriser to be 13. This is the remainder left on dividing the number by the product 8.9. Hence $N = 72t + 13 = 7z + 1.$ Again, applying the same method we find the minimum number satisfying all the conditions to be 85.

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(2) “Tell me at once, O mathematician, that number which leaves unity as remainder when divided by any of the numbers from 2 to 6 but is exactly divisible by 7.”

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By the same method, says Bhâskara I (522), the number is found to be 721. By a different method Sûryadeva Yajvâ obtains the number 301. It is interesting to find that this very problem was afterwards treated by Ibn-al-Haitam (*c.* 1000) and Leonardo Fibonacci of Pisa (*c.* 1202).²

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² L.E. Dickson, *History of the theory of Numbers*, Vol. II, referred to hereafter as Dickson, *Numbers II*, pp. 59, 60.

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To solve a problem of this kind Bhâskara II adopts two methods. One is identical with the method of Āryabhaṭa I and the other follows from his general rule for the solution of simultaneous indeterminate equations of the first degree. They will be better understood from his applications to the solution¹ of the following problem which, as Pṛthûdakasvâmî (860) observes,² was popular amongst the Hindus :

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¹ *BBi*, pp. 85f.

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² *Vide* his commentary on *BrSpSi*, xviii. 3-6.

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To find a number *N* which leaves remainders 5, 4, 3, 2 when divided by 6, 5, 4, 3 respectively.

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*i.e.*, N = 6x + 5 = 5y + 4 = 4z + 3 = 3w + 2.

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(1) We have $x = (5y - 1)/6, y = (4z - 1)/5, z = (3w - 1)/4.$

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Now by the method of the pulveriser, we get from the last equation $w = 4t + 3, z = 3t + 2,$ where *t* is an arbitrary integer. Substituting in the second equation, we get $y = (12t + 7)/5.$

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To make this integral, we again apply the method of the pulveriser, so that $t = 5s + 4, y = 12s + 11.$

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This value of *y* makes *x* a whole number. Hence we have finally $w = 20s + 19, z = 15s + 14, y = 12s + 11, x = 10s + 9.$ $N = 60s + 59.$

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(2) Or we may proceed thus :

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Since N = 6x + 5 = 5y + 4, we have $x = (5y - 1)/6.$ But x must be integral, so y = 6t + 5, x = 5t + 4. Hence $N = 30t + 29.$ Again $N = 30t + 29 = 4z + 3.$ $∴ t = (2z - 13)/15.$ Since t must be integral, we must have z = 15s + 14; hence t = 2s + 1. Therefore $N = 60s + 59.$

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The last condition is identically satisfied. Pṛthûdakasvâmî followed this second method to solve the above problem.

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**Conjunct Pulveriser.** The foregoing system of indeterminate equations of the first degree can be put into the form¹ .by_1 &= a_1x ± c_1 / by_2 &= a_2x ± c_2 / by_3 &= a_3x ± c_3 / \dots & \dots\dots\dots\dots} \tag

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¹ For, we have a₁x₁ + r₁ = a₂x₂ + r₂ = a₃x₃ + r₃ = = a_nx_n + r_n. Then \begin{aligned} a_2x_2 &= a_1x_1 + (r_1 - r_2), \\ a_2x_3 &= a_1a_2/a_3x_1 + a_2/a_3(r_1 - r_3), \\ a_2x_4 &= a_1a_2/a_4x_1 + a_2/a_4(r_1 - r_4), \\ \dots & \dots\dots\dots\dots\dots\dots\dots\dots

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On account of its important applications in mathematical astronomy this modified system has received special treatment at the hands of Hindu algebraists from Āryabhaṭa II (950) onwards. It is technically called *saṁśliṣṭakuṭṭaka* or the "conjunct pulveriser" (from, *kuṭṭaka* = pulveriser and *saṁśliṣṭa* = joined together, related).

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For the solution of the above system of equations Āryabhaṭa II lays down the following rule :

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"In the solution of simultaneous indeterminate equations of the first degree with a common divisor, the dividend will be the sum of the multipliers¹ and the interpolator the sum of the given interpolators."²

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¹ In the equations (1), a₁, a₂, ... are called multipliers.

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² *MSi*, xviii. 48.

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A similar rule is given by Bhâskara II. He says :

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"If the divisor be the same but the multipliers different then making the sum of the multipliers the dividend and the sum of residues the residue (of a pulveriser), the investigation is carried on according to the foregoing method. This true method of the pulveriser is called the conjunct pulveriser."³

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³ *BBi*, p. 33 ; *L*, p. 82.

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*Rationale.* If the equations (1) are satisfied by some value a of x, then the same value will satisfy the equation $b(y₁ + y₂ + ) = (a₁ + a₂ + )x + (c₁ + c₂ + ) (2).$

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Thus, if we can find the general value of x satisfying equation (2), one of these values, at least, will satisfy all the equations (1).

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To illustrate the application of the above Bhâskara II gives the following example :⁴ $. 63y₁ & = 5x - 7 / 63y₂ & = 10x - 14 (A)$

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⁴ *BBi*, p. 33 ; *L*, p. 82.

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Adding up the equations and dividing by the common factor 3, we get $21Y = 5x - 7,$ where Y = y₁ + y₂. By the method of the pulveriser the least positive value of x satisfying this equation is x = 14. This value of x is found to satisfy both the equations (A).

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**Generalised Conjunct Pulveriser.** A generalised case of the conjunct pulveriser is that in which the divisors as well as the multipliers vary. Thus we have $b₁ y₁ = a₁ x ± c₁,$ $b₂ y₂ = a₂ x ± c₂,$ $b₃ y₃ = a₃ x ± c₃,$ . . . . .

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Simultaneous indeterminate equations of this type have been treated by Mahâvîra (850) and Śrîpati (1039). Mahâvîra says :

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“Find the least solutions of the first two equations. Divide the divisor corresponding to the greater solution by the other divisor (and as in the method of the pulveriser find the least value of) the multiplier with the difference of the solutions as the additive. That multiplied by the divisor (corresponding to the greater solution) and then added by the greater solution (will be the value of the unknown satisfying the two equations).”¹

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¹ *GSS*, vi. 115½, 136½ (last lines).

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A similar rule is given by Śrîpati :

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“Find the least solutions of the first two equations. Dividing the divisor corresponding to the greater solution by the divisor corresponding to the smaller solution, the residue (and its divisor) should be mutually divided. Then taking the difference of the numbers as the additive, determine (the least value of) the multiplier of the divisor corresponding to the greater solution in the manner explained before. Multiply that value by the

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latter divisor and then add the solution (corresponding to it). The resulting number (severally) multiplied by the two multipliers and divided by the corresponding divisors will leave remainders as stated.”¹

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¹ *SiŚe*, xiv. 28.

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The *rationale* of these rules will be clear from the following:

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Taking the first two equations, we have $b₁ y₁ = a₁ x ± c₁,$ $b₂ y₂ = a₂ x ± c₂.$ Suppose α₁ to be the least value of x satisfying the first equation as found by the method of the pulveriser. Then b₁ m + α₁, where m is an arbitrary integer, will be the general value of x satisfying that equation. Similarly, we shall find from the second equation the general value of x as b₂ n + α₂. If the same value of x satisfies both the equations we must have $b₂ n + α₂ = b₁ m + α₁,$ $or b₂ n = b₁ m + (α₁ - α₂);$ supposing α₁ > α₂. Solving this equation, we can find the value of m and hence of b₁ m + α₁ of x satisfying both the equations. The general value of x derived from this may be equated to the value of x from the third equation and the resulting equation solved again, and so on.

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In illustration of his rule Mahâvîra proposed several problems. One of these has already been given (Part I, p. 233). Here are two others:

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(1) “Five (heaps of fruits) added with two (fruits) were divided (equally) between nine travellers; six (heaps) added with four (fruits) were divided amongst eight; four (heaps) increased by one (fruit) were divided amongst seven. Tell the number (of fruits in each heap).”¹

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¹ *GSS*, vi. 129½.

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This gives the equations : $9y₁ = 5x + 2, 8y₂ = 6x + 4, 7y₃ = 4x + 1.$

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(2) “The (dividends) are the sixteen numbers beginning with 35 and increasing successively by three ; divisors are 32 and others successively increasing by 2 ; and 1 increasing by 3 gives the remainders positive and negative. What is the unknown multiplier ?”²

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² *GSS*, vi. 138½.

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This gives the equations : $32y₁ = 35x ± 1, 34y₂ = 38x ± 4, 36y₃ = 41x ± 7,$

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**Alternative Method.** In four palm-leaf manuscript copies of the *Lîlâvatî* of Bhâskara II Sarada Kanta Ganguly discovered a rule describing an alternative method for the solution of the generalised conjunct pulveriser.³ There is also an illustrative example. The genuineness of this rule and example is accepted by him; but it has been questioned by A. A. Krishnaswami Ayyangar⁴ who attributes them to some commentator of the work. His arguments are not convincing.⁵ The chief points against the presumption, which have been noted also by Ganguly, are: (1) the rule and example in question have not been mentioned by the earlier commentators of the *Lîlâvatî* and (2) they have not been so far traced in any manuscript of the *Bîjagaṇita*, though the treatment of the pulveriser occurs nearly word for word in the two works. Still we are in favour of accepting Ganguly's conclusion.¹ The rule in question is this:

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³ S. K. Ganguly, “Bhâskarâcârya and simultaneous indeterminate equations of the first degree,” *BCMS*, XVII, 1926, pp. 89–98.

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⁴ A. A. Krishnaswami Ayyangar, “Bhâskara and samslishta Kuttaka,” *JIMS*, XVIII, 1929.

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⁵ For Ganguly’s reply to Ayyangar’s criticism see *JIMS*, XIX, 1931.

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¹ Of the four manuscripts containing the rule and example in question two are from Puri, in Oriya characters, with the commentary of Śrīdhara Mahâpâtra (1717); the other two, in Andhra characters and without any commentary, are preserved in the Oriental Libraries of Madras and Mysore. So these four manuscript copies do not appear to have been drawn from the same source. This is a strong point in favour of the genuineness of the rule and example.

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"If the divisors as well as the multipliers be different, find the value of the unknown answering to the first set of them. That value being multiplied by the second dividend and then added by the second interpolator will be the interpolator (of a new *kuṭṭaka*); the product of the second dividend and first divisor will be the dividend there and the divisor will be the second divisor. The value of the unknown multiplier determined from the *kuṭṭaka* thus formed being multiplied by the first divisor and added by the previous value of the unknown multiplier will be the value (answering to the two divisors). The dividend (for the next step) has been stated to be equal to the product of the two divisors. So proceed in the same way with the third divisor. And so on with the others, if there be many."

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The rationale of this rule is as follows: Let α₁ be the least value of x satisfying the first equation of the system, *viz*., $b₁ y₁ = a₁ x ± c₁.$ Hence the general value is x = b₁ t + α₁, where t is any integer. Substituting this value in the second equation, we get $b₂ y₂ = a₂ b₁ t + (a₂ α₁ ± c₂).$ If t = τ be a solution of this equation, a value of x satisfying both the equations will be α₂=b₁τ+α₁ as stated in the rule. Now the general value of t will be t=b₂m+τ, where m is an integer. Hence x=b₁t+α₁=b₁b₂m+b₁τ+α₁=b₁b₂m+α₂. Substituting this value in the third equation we can find the least value of m and hence a value of x answering to the three equations. And so on for the other equations.

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The example runs thus :

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“Tell me that number which multiplied by 7 and then divided by 62, leaves the remainder 3. That number again when multiplied by 6 and divided by 101 leaves the remainder 5 ; and when multiplied by 8 and divided by 17 leaves the remainder 9. Also (give) at once the process of the pulveriser for (finding) the number with the remainders all positive.”

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Symbolically, we have

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(1) 62y₁=7x-3, 101y₂=6x-5, 17y₃=8x-9;

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(2) 62y₁=7x+3, 101y₂=6x+5, 17y₃=8x+9.

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