Mathematics in Ancient and Medieval India c7.s5: Indeterminate Equations of the First Degree

Mathematics in Ancient and Medieval India c7.s5

Language: English

Indeterminate Equations of the First Degree

c7.s5.1

The subject of indeterminate analysis (first degree) is designated in Indian mathematics by the term *kuṭṭaka*. *Kuṭṭ* means 'to break' or 'to pulverize'. The name has been given on account of the process of continued division that is adopted for the solution. Indian interest in the indeterminate problems is first found in the *Śulbasūtras*, Detailed rules of solutions of the type : by = ax ± c were given by scholars from Āryabhaṭa I (476 A. D.) onwards Colebrooke², Chasles³ and Sen⁴ gave credit to Indians for attaining a general solution of the problem. Some scholars favoured Greeks and the Chinese for the origin. They are of opinion that Indians got inspiration in this subject from the foreign countries. In the foregoing discussion it can be seen that while the career of indeterminate (first degree) is doubtful in other cultural areas, in India it is based on strong and solid foundation.

c7.s5.2

² Colebrooke, H. T. *Algebra with Arithmetic and Mensuration from the Sanskrit of Brahmagupte, and Bhāscara*, London, 1817.

c7.s5.3

³ Chasles, M. *Apercu historique sur l'origine et le developpement des methodes en geometric*, 2e ed. Paris, 1875.

c7.s5.4

⁴ Sen, S. N. 'Study of Indeterminate analysis in Ancient India' *Bulletin of History of science Congress*, held at Ithaca p. 493-97, 1962.

c7.s5.5

The problems of altar construction in the *Śulbasūtras* necessitated the evolution of simultaneous indeterminate equations of the type : i ) x + y = 21 x/(p²) + y/(q²) = 1 ii ) x + y + z + u = 200 x/m + y/n + z/p + u/q = 7½ etc.

c7.s5.6

A few arbitrary solutions were given by Baudhāyana (c. 600 B. C.), Āpastamba (c. 550 B. C.), and their commentators. The rational solutions of such equations were given by Datta¹ and Kamalamma.²

c7.s5.7

¹ Datta, B. *The Science in the Śulba*, p. 180-86, Calcutta University, 1932.

c7.s5.8

² Bulletin of the Calcutta Math. Society (BCMS), **40**, p. 140-44, 1948.

c7.s5.9

Āryabhaṭa I (b. 476 A. D.) first gave a rule for obtaining the general solution of the problem by = ax ± c. His rule was obviously intended for the solution of the following problems :

c7.s5.10

(1) To find the number (N), which when divided by a given number (a) will leave a remainder (r₁) and when divided by another given number (b) will leave a remainder (r₂).

c7.s5.11

(2) To find a number (N) which being divided severally by the given numbers a₁, a₂, a₃, _n leaves remainders as r₁, r₂, r₃_n respectively.

c7.s5.12

In modern notation, this can be written as, ( 1 ) N = ax + r₁ = by + r₂ ( 2 ) N = a₁x₁ + r₁ = a₂x₂ + r₂ = a₃x₃ + r₃ = = a_nx_n + r_n.

c7.s5.13

*Nature of Solution ( Āryabhaṭa I )* Āryabhaṭa I keeps c = r₁ - r₂ always positive, consequently his rule is directed to solve an equation of the form : i ) y = (ax + c)/b if r₁ > r₂ and ii ) x = (by + c)/a if r₂ > r₁. Where a, b = divisors, *bhāgahāra*, *bhājaka*, *cheda* etc. r₁, r₂ = remainders, *agra*, *śeṣa* etc.

c7.s5.14

Āryabhaṭa I's rule¹ : *adhikāgra bhāgahāraṃ chindyādūnāgra bhāgahāreṇa ।* *śeṣaparasparabhaktaṃ matiguṇamagrāntare kṣiptam ॥* *adhaupariguṇitamantyayugūnāgracchedabhājite śeṣam ।* *adhikāgracchedaguṇaṃ dvicchedāgramadhikāgrayutam ॥*

c7.s5.15

¹ *Āryabhaṭīya*, Gaṇita, vss. 32-33.

c7.s5.16

*English Translaton :* We follow Datta who based his translation on that of Bhāskara I ( 600 A. D. ), the first commentator on Āryabhaṭa I's work. The translation runs as follows : "Divide the divisor corresponding to the greater remainder by the divisor corresponding to the smaller remainder, being mutually divided, the last residue should be multiplied by an integer at our choice such that the product on being added to ( if the number of quotients in the division process is odd ) or subtracted ( if the number of quotients is even ) by the difference of the remainders, will be exactly divisible by the penultimate remainder. Place the quotients of the mutual division successively one below the other in a column, below them the optional multiplier and below it the quotient just obtained. The penultimate is multiplied by the one just above it and added to that just below it. Repeating this process, divide the last number obtained by the divisor corresponding to the smaller remainder. Then multiply the residue by the divisor corresponding to the greater remainder and add greater remainder. The result will be a number corresponding to the divisors.'¹

c7.s5.17

¹ *Bulletin of the Calcutta Mathematical Society*, 24,p.19–36,1933.

c7.s5.18

*Explanation of the Rule* :

c7.s5.19

i ) Solution of y=(ax+c)/b.........(1) in positive integers when a and b are prime to each other.

c7.s5.20

Repeated divisions following H. C. F. process may be placed as follows :—

c7.s5.21

{rll} b) a & (a_1 & / ba_1 & & a = ba_1 + r_1 / r_1) & b (a_2 & b = r_1a_2 + r_2 / & r_1a_2 & r_1 = r_2a_3 + r_3 / & r_2) & r_1 (a_3 & r_2 = r_3a_4 + r_4 / & & r_2a_3 & and so on. / & & r_3) r_2 (a_4 & / & & r_3a_4 & / & & r_4 & / & & \dots & / & & \dots & / & & \dots & / & & r*{n-1}) r*{n-2} (a_n / & & r_{n-1}a_n / & & r_n

c7.s5.22

$then a/b = a₁ + 1/(a₂ +) 1/(a₃ +) 1/(a₄ +) 1/(a_n-1 +) 1/a_n$

c7.s5.23

$Let p₁/q₁, p₂/q₂, p₃/q₃, p_n/q_n be the succe-$ $ssive convergents of a/b, then :$

c7.s5.24

$p₁/q₁ = a₁$ $p₂/q₂ = a₁ + 1/a₂ = (a₁a₂ + 1)/a₂$ $p₃/q₃ = a₁ + 1/(a₂ +) 1/a₃$ $= a₁ + 1/(a₂ + 1/a₃)$ $= a₁ + a₃/(a₂a₃ + 1) = (a₁(a₂a₃ + 1) + a₃)/(a₂a₃ + 1)$ $p₄/q₄ = a₁ + 1/(a₂ +) 1/(a₃ +) 1/a₄$ $= a₁ + 1/(a₂ + 1/(a₃ + 1/a₄))$ $= a₁ + 1/(a₂ + a₄/(a₃a₄ + 1)) = a₁ + (a₃a₄ + 1)/(a₂(a₃a₄ + 1) + a₄)$ $= (a₁a₂(a₃a₄ + 1) + a₄ + a₃a₄ + 1)/(a₂(a₃a₄ + 1) + a₄)$ $and so on, and the last convergent p_n/q_n = a/b$

c7.s5.25

Evidently, p_n. q_n-1 - q_n. p_n-1 = ± 1 according as n = even or odd.

c7.s5.26

This result was perhaps known to Āryabhaṭa I for he always used the result p_n-q_n. u=-c (whether n is even or odd), the two numbers u and v were obtained by following an ingenious technique discussed later.

c7.s5.27

*Case I : When n=even i.e. the number of partial quotients for finding u and v is odd* (vide Table I)

c7.s5.28

The cryptic Sanskrit terms *agrāntare kṣiptam* used by Āryabhaṭa I in the rule has been explained by his commentator Bhāskara I (c.600 A.D.) in his *Āryabhaṭīyabhāṣya* as *sameṣu kṣiptam, visameṣu, śodhyam*, i.e. add (the *kṣepa* quantity, c) when n=even, and subtract when n=odd." Since the ratio u/v was always used to determind the (n-1) th convergent, the *kṣepa* quantity c was to be added or subtracted according as the number of partial quotents arranged in Table I & II are odd and even. Suppose the process of division was stopped at remainder r₃

c7.s5.29

then (r₃t+c)/r₂=q₁ say, where t=(r₂ q₁-c)/r₃

c7.s5.30

The quotients a₁, a₂, a₃ with *mati* t, and final quotient q₁ are placed one after another in the first column (vide Table I) and the final numbers u and v were obtained as follows:—

c7.s5.31

*Table 1*

c7.s5.32

\begin{array}{c|c|c|c} a_1 & a_1 & a_1 & a_1 v+\alpha_1= \\ a_2 & a_2 & a_2 \alpha_1+t=v, say & u, say \\ a_3 & a_3 t+q_1=\alpha_1, say & \alpha_1 & \\ t & t & & \\ q_1 & & &

c7.s5.33

Now α₁ = a₃ t + q₁ {ll} = a_3 ( (r_2\ q_1 - c)/r_3 ) + q_1 & \because t = (r_2\ q_1 - c)/r_3 / = (q_1\ (\ r_2\ a_3 + r_3\ ) - a_3\ c)/r_3 / = (q_1\ r_1 - a_3\ c)/r_3 & ( p. 196 ) v = a₂ α₁ + t {ll} = a_2\ {\ (q_1\ r_1 - a_3\ c)/r_3\ } + (r_2\ q_1 - c)/r_3 / = (q_1\ (a_2 r_1 + r_2) - c\ (a_2 a_3 + 1))/r_3 / = (q_1 b - c\ (a_2 a_3 + 1))/r_3 & ( p. 196 ) u = a₁ v + α₁ {l} = a_1\ {\ (q_1\ b - c\ (a_2\ a_3 + 1))/r_3\ } + (q_1\ r_1 - a_3\ c)/r_3 / = (q_1\ (a_1\ b + r_1) - c\ {a_1\ (a_2\ a_3 + 1) + a_3\ })/r_3 / = (q_1\ a - c\ {a_1\ (a_2\ a_3 + 1) + a_3})/r_3 {ll} u/v & = (q_1\ a - c\ {a_1\ (a_2\ a_3 + 1) + a_3})/(q_1\ b - c\ (a_2\ a_3 + 1)) / & = (q_1\ a - cp_3)/(q_1\ b - cq_3) If there be (n-1) number of partial quotients in the Table I, then u/v = (q₁ a - p_n-1)/(q₁ b - q_n-1) The last n th convergent is always a/b hence p_n/q_n = a/b.

c7.s5.34

Āryabhaṭa 1 knew, $p_n. v - q_n. u = -c [ n = even] (2)$ For. p_n (q₁ b - cq_n-1) - q_n (q₁ a - c p_n-1) $= -c (p_n q_n-1 - q_n p_n-1) ( p_n = a, q_n = b)$ $= - c$

c7.s5.35

*Case II : When n = odd i.e, the number of partial qutient*¹ *used for finding u and v is even.*

c7.s5.36

¹ Shukla, Kripa Shankar, *Mahābhāskarīya*, edited and translated into English with notes, p. 30–33, Lucknow University, 1960.

c7.s5.37

Suppose the repeated division was stopped at the remainder r₄. Following rule, let us put, $(r₄ t - c)/r₃ = q₂, where t = (r₃ q₂ + c)/r₄$

c7.s5.38

Now we place below the quotients a₁, a₂, a₃, a₄, then *mati*, t and c and obtain u and v following a process of multiplication and addition as directed in the rule.

c7.s5.39

*Table II*

c7.s5.40

| a₁ | a₁ | a₁ | a₁ | & a₁ v + β₂ = u / & ( say ) |
| a₂ | a₂ | a₃ | a₂ β₂ + β₁ = v | |
| a₃ | a₂ | a₃ β₁ + t = β₂ | β₂ | |
| a₄ | a₄ t + q₂ = β₁ | β₁ | | |
| t | t | | | |
| q₂ | | | | |

c7.s5.41

If we calculate β₁, β₂, v, and u proceeding as before we will find, u/v = (q₂ a + c p₄)/(q₂ b + c q₄)

c7.s5.42

If there be (n-1) number of partial quotients in the Table II, then : $u/v = (q ₂a + cp_n-1)/(q ₂b + cq_n-1)$ Āryabhaṭa I knew that : $p_n · v - q_n · u = -c [n = odd] (3)$ For p_n · v - q_n · u $= p_n (q₂b + cq_n-1) - q_n (q ₂a + cp_n-1)$ $= c (p_nq_n-1 - q_n p_n-1) [ p_n = a, q_n = b]$ $= -c [n = odd]$

c7.s5.43

The equations (2) & (3) are identical. This was purposely done by Āryabhaṭa I, by selecting (r_nt ± c)/(r_n-1) = q₁ or q₂ according as the number of partial quotients in the Table are odd or even (chosen by Āryabhaṭa I.)

c7.s5.44

From (3), q_n u = p_n · v + c or bu = av + c (4) Evidently y = u and x = v when u = am + y, and v = bm + x, the equation (4) reduces to $by₁ = ax₁ + c$ ∴ y y₁ ( mod a ) and x = x₁ ( mod b ) hence ( x₁, y₁ ) is the least solution of (1)

c7.s5.45

ii) Solution of x = (by + c)/a in positive integer when a and b are prime to each other.

c7.s5.46

The method of solution is exactly similar, only select (r_n t ± c)/(r_n-1) = q₁ or q₂ according as the number of partial quotients are odd or even.

c7.s5.47

then x = u = bm + x₁ or x ≡ x₁ ( mod b ). and y = v = am + y₁ ≡ y₁ ( mod a )

c7.s5.48

( 2 ) *Solution of linear simultaneous indeterminate equation* To solve N = a₁x + r₁ = a₂y + r₂ = a₃z + r₃ = ... = ant + rₙ.

c7.s5.49

Then the least value of N = a₁α + r₁

c7.s5.50

General value = a₁ a₂n + a₁α + r₁ where n is any integer. Thus we find that a₁α + r₁ is the remainder left on dividing N by the product a₁ and a₂. This is exactly what has been stated by Āryabhaṭa I as follows : "( The result will be ) the residue corresponding to the product of the two divisors".

c7.s5.51

Let N' = a₁ a₂n + (a₁α + r₁) = a₃z + r₃

c7.s5.52

then N' = a₁ a₂ a₃t + a₁α + r₁ which shows that N' when divided severally by a₁ a₂ and a₃ leaves as remainder a₁ α and r₃ respectively and repeat the process and so on.

c7.s5.53

*Illustrations :*

c7.s5.54

(1) 45x + 7 = 29y Now, 29) 45 (1 29
16) 29 (1 16
13) 16 (1 13
3) 13 (4 12 1

c7.s5.55

Here no. of quotients omitting the first is odd.

c7.s5.56

Therefore, (1 × 10 - 7)/3 = 1

c7.s5.57

| | 1 | 1 | 1 | 1 | 143 |
| Now | 1 | 1 | 1 | 92 | 92 |
| | 1 | 1 | 51 | 51 | |
| | 4 | 41 | 41 | | |
| | 10 (*mati*) | 10 | | | |
| | 1 | | | | |

c7.s5.58

Now y = 143 = 45 × 3 + 8 = 8 ( mod 45 ) and x = 5 ( mod 29 )......

c7.s5.59

( 2 ) N = 12x + 5 = 31y + 7 To find x, y, and N. Now 31y + 2 = 12x

c7.s5.60

{r@{ }l} 12)\phantom31 & (2 / 24 & / 7)\phantom12 & (1 / \phantom7 & / 5)\phantom7 & (1 / 5 & / 2 &

c7.s5.61

Here the no. of quotients excepting the first = even.

c7.s5.62

$Here (2 × 4 + 2)/5 = 2$

c7.s5.63

| *Table* | 2 | 2 | 2 | 26 |
| | 1 | 1 | 10 | 10 |
| | 1 | 6 | 6 | |
| | 4 (*mati*) | 4 | | |
| | 2 | | | |

c7.s5.64

Now x = 26, y = 10 ∴ N = 31y + 7 = 317

c7.s5.65

*Bhāskara I*: Bhāskara I (600A.D.) stated clearly that a and b should be prime to each other. He put the equation always in the form (ax - c)/b = y which is same as Āryabhaṭa I’s (by + c)/a = x. His rule is similar to that of Āryabhaṭa I and was given in his *Mahāsiddhānta* (ch. 1, v. 41–44). He for the first time used this equation to astronomy to compute the *ahargaṇas* and revolutions of the planets. He gave also the following results :

c7.s5.66

( i ) If (x₁, y₁) be solutions of ax + 1 = by, then (cx₁, cy₁) is solution of ax + c = by. (*Mahābhāskarīya*, ch. 1, v. 45–46i ).

c7.s5.67

( ii ) If x = α, y = β be a solution of Ax - C = by, then x = α, y = ma + β be a solution of (ax - C)/b = y, where a = mb + A. (*Mahābhāskarīya*, ch. 1. v. 47).

c7.s5.68

( iii ) If cx₁ x'₁ ( mod b ), cy₁ y'₁ ( mod a ), then (x'₁ y'₁) is the solution of ax + C = by. (*Mahābhāskarīya* ch. 1, v. 50).

c7.s5.69

Various problems were given by Bhāskara I, a few of which are given here :

c7.s5.70

( 1 ) “What is that number which being divided by 12, leaves the remainder 5; again when divided by 31 has the remainder 7 ?” (*Āryabhaṭīya bhāṣya*). *Hints* : To solve N = 12x + 5 = 31y + 7. Here N = 317, x = 31, y = 10.

c7.s5.71

( 2 ) "Find what is that number which divided by 8 is known to leave a remainder 5; by 9 a remainder 4; and by 7 a remainder 1". ( *Āryabhaṭīya bhāṣya* ).

c7.s5.72

*Solution* : N = 8x + 5 = 9y + 4 = 7z + 1 ( say ).
Let N' = 8x + 5 = 9y + 4.

c7.s5.73

The least value of N is found to be 13 ( by the method of pulverizer ). This remainder is left by dividing the number N' by the product 8.9 = 72. Hence in general N = 72t + 13 where t is any arbitrary integer. Again, N = 72t + 13 = 7z + 1. Applying the same method again, least value of N = 85 which satisfies the three given divisors.

c7.s5.74

( 3 ) Find a number N which leaves the remainder 1, when divided by 2, 3, 4, 5 or 6 but is exactly divisible by 7". ( *Āryabhaṭīya bhāṣya* ).

c7.s5.75

This problem has been treated by Ibn-al-Haitam ( c. 1000 A. D. ), Leonardo Fibonacci ( c. 1202 A. D. ) and two writers of the 17th century.

c7.s5.76

*Hints* : N = 2x + 1 = 3y + 1 = 4z + 1 = 5u + 1
= 6v + 1 = 7t + 0.
Here N = 721.

c7.s5.77

( 4 ) "The residue of the revolutions of Saturn is 24, find the *ahargaṇa* and the revolutions performed by Saturn" ( *Laghubhāskarīya*, ch. 8, v. 17 )

c7.s5.78

*Hints* : The revolution number of Saturn = 146564.
No. of civil days in a yuga = 1577917500.
These two figures have a common factor 4.
Hence dividing them by 4, we get 36641

c7.s5.79

and 394479375 respectively. Therefore one has to solve the equation.

c7.s5.80

$36641x - 24 = 394479375y.$

c7.s5.81

where x = *ahargaṇa* and y = revolutions performed by Saturn.

c7.s5.82

Here x = 346688814, y = 32202.

c7.s5.83

( 5 ) “The mean ( position ) of the Sun has been observed by me at sun-rise to be in the sign Leo in the middle of the *navāṃśa* Sagittarius.¹ Calculate the *ahargaṇa* according to the ( Ārya )bhaṭaśāstra and also the revolutions performed by the Sun since the beginning of Kaliyuga”. ( *Mahābhāskarīya*, ch. 2, v. 32–33 )

c7.s5.84

¹ One *aṃśa* = 3°20'. Hence middle of the *navamaṃśa* = 3°20' × 8 + 1°40' = 28°20'.

c7.s5.85

*Hints* : The mean longitude of the Sun = 4 signs 28°20' = 8900'. The abraded revolution number of the Sun = 576. The abraded number of civil days in a yuga = 210389. Hence the residue² of the revolution = 86688.

c7.s5.86

² Residue of the revolutions = ( Mean longitude × abraded number of civil days in a yuga ) ÷ Number of divisions ( in zodiac ). No. of zodiacal divisions would be 12, 360, and 21600 accordingly as the mean longitude is expressed in terms of signs and degrees, signs and degrees and minutes respectively.

c7.s5.87

Now to solve the equation (576x - 86688)/210389 = y, where x = *ahargaṇa*, y = no. of revolution performed by the Sun.

c7.s5.88

Here x = 105345, y = 288.

c7.s5.89

General solution : x = 210389α + 105345. y = 576α + 288

c7.s5.90

*Scholars from Brahmagupta to Āryabhaṭa II* :

c7.s5.91

The method was subsequently discussed by Brahmagupta (c. 628 A. D.), Govindasvāmī (c. 850 A. D.) Mahāvīra (850 A. D.), Pṛthudakasvāmī (c. 861 A. D.) and others. The method of solution was further reduced by Āryabhaṭa II (950 A. D.). He continued the mutual division till the remainder becomes 1. Then the table (*valli*) was made with quotients and with 1 and 0 attached at the end as follows.

c7.s5.92

| a₁ | a₁ | a₁ | a₁ (a₂ a₃ + 1) ≡ u say |
| a₂ | a₂ | a₂ a₃ + 1 | a₂ a₃ + 1 ≡ v, say |
| a₃ | a₃ · 1 + 0 = a₃ | a₃ | |
| 1 | 1 | | |
| 0 | | | |

c7.s5.93

then u/v = (a₁ (a₂ a₃ + 1) + a₃)/(a₂ a₃ + 1) = p₃/q₃

c7.s5.94

Similarly if there be (n – 1) partial quotients in the table, then u/v = (p_n-1)/(q_n-1)

c7.s5.95

Since the mutual division was not continued upto the remainder zero but upto the one i.e., one operation less in the continued fraction.

c7.s5.96

i. e. (p_n)/(q_n) = a/b Āryabhaṭa II, similar to Aryabhaṭa I, knew that p_n. v-q_n. u=± 1 according as total number of partial quotients are even or odd.

c7.s5.97

or a v-b u=± 1 according as the number of partial quotients in the above Table or odd or even.

c7.s5.98

When the number of partial quotients in the table are even, a v-b u=-1 or b u=a v+1, then u and v is the solution of by=ax+1. If u=at+β, and v=bt+a, then ( a, β) is the least solution of by=ax+1.

c7.s5.99

When the number of partial quotients in the table are odd, then a v-b u=+1 or b u=a v-1, then ( u, v) is a solution of b y=ax-1.

c7.s5.100

If u=at+a-β and v=bt+b-α, then the equation becomes bβ=aα+1 i.e. (α, β) is the least solution of by=ax+1

c7.s5.101

This gives the solutton of by=ax± 1......(1). Then if α and β be the solutions of (1) then ca and cβ are solution of by=ax.

c7.s5.102

Here two cases should be considered depending on the number of quotients as follows :

c7.s5.103

*Case 1* : When the number of quotients are even, the method gives the solution of ax+c=by.

c7.s5.104

*Case 2* : When the number of quotients are odd, the method gives the solution of ax-c=by. Then x=(b-α)+bt, y=(a-β)+at is the general solution of ax+c=by.

c7.s5.105

Āryabhaṭa II eliminated the common factors among a, b and c and suggested also the following steps when the numbers ( a, c) or ( b, c) have a common factors (*Mahāsidhānta*, ch. 18, v. 1–2).

c7.s5.106

i ) If a = ka', c = kc' where k is an integer. The equation ax + c = by reduces to a'x + c' = by' where y' = y/k. Hence if ( x, y ) be solutions of the reduced equation, ( x, ky' ) is the solution of ax + c = by.

c7.s5.107

ii ) If c = kc', b = kb', the equation ax + c = by reduces to ax' + c' = b'y where x' = x/k. If ( x, y ) be a solution of ax' + c' = b'y, then ( kx', y ) is the solution of ax + c = by.

c7.s5.108

iii ) If a = ka', c = kc', again c' = lc" and b = lb", then the equation ax + c = by ... ... (1) becomes ka'x + klc" = lb"y.

c7.s5.109

&or (ka'x)/kl + (klc")/kl = (lb"y)/kl or a' · x/l + c" = b" · y/k / &or a'X + c" = b"Y ...\ ...\ (2) when X = x/l, / &\hphantomor a'X + c" = b"Y ...\ ...\ (2) when Y = y/k

c7.s5.110

Hence if ( x, y ) is solution of (2), then ( Xl, Yk ) is solution of (1).

c7.s5.111

Bhāskara II ( 1150 A. D. ) simply adopted the method of Āryabhaṭa II i.e. continued the mutual division till the remainder becomes one and prepared the table ( *valli* ) with quotients along with c and o. This shows that Bhāskara II directly calculated bu = al ± c (according as the number of partial quotients in the table is odd or even), when :

c7.s5.112

$u/l = (Cpn--l)/(Cqn--l)$

c7.s5.113

There are problems involving simultaneous equations of the more general type given by Mahāvīra, Āryabhaṭa II and Bhāskara II, Devarāja and others as follows :

c7.s5.114

by_1 & = a_1x ± c_1 / by_2 & = a_2x ± c_2 / by_3 & = a_3x ± c_3 / & .........

c7.s5.115

These problems are known as *saṃśliṣṭa kuṭṭaka* (the constant pulveriser). If α₁ and α₂ be the least values of x of the first and second equation and b₁t₁ + α₁ and b₂t₂ + α₂ are the corresponding general values of x satisfying the first and second equations respectively,

c7.s5.116

$then b₁t₁ + α₁ = b₂t₂ + α₂$

c7.s5.117

Solving t₁ and t₂ by the method of pulveriser, the general values of b₁t₁ + α₁ i.e., of x satisfying both the equations are obtained. Next this is equated to the general value of x satisfying the third equation and the process is repeated.

c7.s5.118

*Illustration :*

c7.s5.119

Five heaps of fruits together with two fruits were divided equally among 9 travellers; 6 heaps together with 4 fruits were divided among 8; 4 heaps together with 1 fruit were divided among 7. Find the number of fruits in each heap. (*Gaṇitasārasaṃgraha*, ch. 6, v. 129½).

c7.s5.120

*Solution :* To solve

c7.s5.121

9y_1 & = 5x + 2 & ... & ... & (1) / 8y_2 & = 6x + 4 & ... & ... & (2) / 7y_3 & = 4x + 1 & ... & ... & (3)

c7.s5.122

From equation (1), we get x 5 ( mod 9 ) or x = 9u + 5 ...... (4), Subst. in equation (2), we get 8y₂ = 6(9u + 5) + 4 or 4y₂ = 27u + 17 From this u 1 (mod 4) or u = 4v + 1. Subst. this value of u in equation (4), we get x = 9 (4v + 1) + 5 or x = 36v + 14 . . . . . . (5) Subst. (5) in equation (3), we get, 7y₃ = 4x + 1 = 4(36v + 14) + 1 or 7y₃ = 144v + 57 . . . . . . (6) From here v - 2(mod 7) or v = 7p - 2. Hence from (3) and (6), 4x + 1 = 1008p - 231 or x = 252p - 58. Least value = 252 - 58 = 194.

c7.s5.123

The problem of indeterminate equations was persued later by scholars like Nārāyaṇa, Kṛṣṇa, Kamalakara and various others.¹ The *Karaṇapaddhati* and *Yuktibhāṣā* have also discussed the rule.

c7.s5.124

¹ Bag, A. K. "The Method of Integral solution of Indeterminate Equations of the type : by = ax + c in Ancient and Medieval India", *Indian Journal of History of Science*, 12 No. 1, p. 1-16, 1977.

c7.s5.125

The *Karaṇapaddhati* furnishes interesting information on the calculation of approximations of *mahāhāra* : *mahāguṇa* ( i. e. *b* : *a* ) and has applied the both upward ( below-top ) and downward ( top-below ) techniques to calculate the successive approximations. The relevant verse of the *Karaṇapaddhati*, (ii, 5) runs as follows :

c7.s5.126

*anyonyam vibhajenmahāguṇahārau yāvadvibhakte'lpata*
*tāvallabdhaphalāni rūpamapi onnyasyedadho'dhaḥ kramāt*
*prakṣipyāntyamupāntimena guṇite svordhve tadantyam*
tyajed
*bhuyo'pyeṣa vidhirbhaved guṇahārau syātām tadordhvas-*
thitau ॥

c7.s5.127

*English Translation :*

c7.s5.128

“The *mahāguṇa* (a) and *hāra* (b) should be simultaneously divided till it becomes negligible (i. e. zero). The respective quotients and a number one are placed one below the other. Then multiply the last but one (*upānta*) by the number placed above it (*upānta ūrdha*) and the last is added to it, and then this last number is left. This is the process of obtaining the *guṇa* and *hāra* placed in two *urdha* positions.”

c7.s5.129

This method upward (below-top) technique is similar to that of Bhāskara I with the difference that the mutual division is repeated upto the remainder zero, like that of our modern method.

c7.s5.130

For example, $let b/a = 355/113 = 3 + 1/(7 + 1/16)$

c7.s5.131

The rule gives a process of calculating the final convergent p_n/q_n. Here for n = 3, p₃/q₃ = b/a = 355/113. This has been obtained from the quotients 3, 7, 16 and 1 placed one below another following the process discussed before.

c7.s5.132

The *Karaṇapāddhati*, (ii 6) has given an alternative (top-below) process of this rule for calculating the successive approximations of circumference : diameter, i. e. p₁/q₁, p₂/q₂, p₃/q₃ ... etc., in the following rule.

c7.s5.133

*anyonyāhṛtabhājyahārakaphalaṃ sarvaṃ tvadho'dho*
*nyasedekatrādyaphalena hīnamaparatraikam dvayaścapari|*
*kurjād valyupasaṃhṛtiṃ hyuparitaḥ pūrvapraṇāśaṃ vinā*
*tyājyaṃ tatprathamordhvagaṃ hāraguṇāśśiṣṭāśca vā*
*svecchayā||*

c7.s5.134

*English Translation:*

c7.s5.135

"The *hāra* and *bhājya* (*guṇa*) are simultaneously divided and the results are all placed systematically one below the other in one place (first place). Put the partial quotients without the first quotient in another place (second place) and place one over both the places. Perform the *valli* operations from the top and leave the number of the first place which was not destroyed in the operation. The *hāra*, *guṇa* and the remaining results (are obtained) as desired.

c7.s5.136

This is undoubtedly a reverse (top-below) technique of calculating p₁/q₁, p₂/q₂, p₃/q₃ ...etc., followed before. Here the *valli* operation is performed starting from top like that of our modern method of calculation as follows :

c7.s5.137

| 1st place | *hāra* | 2nd place | *guṇa* |
| 1 | | | |
| a₁ | a₁ | 1 | 1 |
| a₂ | a₁a₂+1 | a₂ | a₂ |
| a₃ | a₃a₁a₂+1+a₁ | a₃ | a₂a₃+1 |
| a₄ | a₄[a₃a₁a₂+1+a₁] | a₄ | a₄a₂a₃+1+a₂ |

c7.s5.138

Hence the successive approximations of circumference (*hāra*) to diameter (*guṇa*), i.e. p₁/q₁, p₂/q₂, p₃/q₃...etc. are : $p₁/q₁ = a₁/1, p₂/q₂ = (a₁a₂ + 1)/a₂, p₃/q₃ = (a₃a₁a₂ + 1 + a₁)/(a₂a₃ + 1)$ and so on.

c7.s5.139

The author of *Yuktibhāṣā* (vide appendix of the edited text) has used the same technique to calculate p₁/q₁, p₂/q₂, p₃/q₃... .

c7.s5.140

This shows that Indian scholars had a more or less distinct idea about the application of continued fraction and used the tool p_nq_n-1 - q_np_n-1 = ± 1 for the solution by = ax ± c, according as n is even or odd.

c7.s5.141

The discussion shows that Āryabhaṭa I (496 A. D.) first obtained a general solution of the problem by = ax ± c. But Āryabhaṭa I himself admitted that he was discussing the knowledge current in Kusumpura (founded in the 1st century B. C.). Further we find that the Indian interest began during the time of *Śulbasūtras*. The process thus has a long history behind it and did not appear all of a sudden in a fully developed form. The subject of *Kuṭṭaka* was considered among the Indian mathematicians as an important branch of their algebra.

c7.s5.142

Regarding Greek contribution, Nicomachus of Geresa¹ (1st century A. D.) gave an example on the problem of remainders involving linear indeterminate analysis. Diophantus¹ ( 3rd century A. D. ) discussed the indeterminate equations of the second degree and higher degrees but did not deal equations of the first degree. Preliminary notions of Greek geometry, according to Kaye², which is responsible for the evolution of the rule is not found common among Indian works as it has been traced in Greek works.

c7.s5.143

¹ Dickson, L. E. *History of the Theory of Numbers*, 2, p. 58, New York, 1934.

c7.s5.144

¹ Heath, T. L. *A History of Greek Mathematics*, Diophantus of Alexandria, 2, p. 440–517, Oxford, 1921.

c7.s5.145

² Kaye, G. R. "Notes on Indian Math.– No. 2 : Āryabhaṭa, *JASB*, 4, p 111–47, 1908.

c7.s5.146

The basis for Chinese contribution is placed on the following example found in the *Sun-Tzu Suan Ching*.³ "There are certain things whose number is unknown. Repeatedly divided by 3, the remainder is 2; by 5 the remainder is 3; and by 7, the remainder is 2. What will be the numbers ?". No process of finding them has been indicated. The solution can be obtained by inspection⁴. The application of the equation : by = ax ± c was first found in a calendrical work, *Ta-Yenli* of I-Hsing⁵ ( 683–727 A. D.). The problems of indeterminate analysis found in *Ta-Yenli* was solved by I-Hsing with a method *Ta-Yen-Shu*⁶ which was similar to the Indian method of *kuṭṭaka*.⁷ I-Hsing came to India in 673 A. D., became a Tantric-Buddhist monk and learnt Sanskrit. Hence it is quite possible that I–Hsing acquired the technique of solving indeterminate problems from Indian scholars and it is through his effort, the knowledge was carried to China.

c7.s5.147

³ Mikami, Y. *The development of Mathematics in China and Japan*, 1913; Needham, J. *Science and Civilisation in China*, p. 119–22, Cambridge, 1959.

c7.s5.148

⁴ Mazumdar, N. K. "On Chinese Indeterminate Analysis", *Bulletin of the Calcutta Mathematical Society* 5, p.9–11,1913-14.

c7.s5.149

⁵ Sarton, G. *Introduction to History of Science*, 1, "I-Hsing" p. 514.

c7.s5.150

⁶ Sarton, G. *Ibid*., 1, p. 514.

c7.s5.151

⁷ Needham, J. *Ibid*., 3, p. 119–22, Cambridge, 1959.

c7.s5.152

*Indeterminate Equations of Second Degree ( Vargaprakṛti )*

c7.s5.153

The term *vargaprakṛti* has been used by Indian scholars to designate the equations of the type Nx² ± c = y². The most fundamental equation of this type has been regarded as Nx² ± 1 = y². Brahmagupta ( 628 A. D. ) first gave solutions in rational integers of both of these types. Further refinements, clarifications, extensions were made by subsequent mathematicians such as Śrīpati, Bhāskara II, *Nārāyaṇa* and others including several commentators who rendered considerable service to this branch of algebra. Here N is termed as *guṇakaprakṛti*, x : kaniṣṭhapada, hrasvamūla, or *ādyamūla*, y : jyeṣṭhapada, or *anyamūla*; c : kṣepa, *prakṣepa* or *prakṣepaka*.

c7.s5.154

The equation Nx² + 1 = y² is known by mistake now a days as Pell's equation ( c. 1668 A. D. ). The equation must be known as Brahmagupta–Bhāskara equation, for Bhāskara II gave a satisfactory solution of the nearly solved method of Brahmagupta. Lagrange gave a complete solution of the equation in 1767 A. D. with the help of continued fraction.

c7.s5.155

(A) *Solution of* Nx² + 1 = y² *(Brahmagupta)*–Find a set of integral values α and β of x and y and form an auxiliary equation of the form :

c7.s5.156

Na² ± k = β² for k = ± 1, ± 2 or ± 4.

c7.s5.157

From these an unlimited number of integral so'utions can be readily obtained by principle of composition (*samāsabhāvanā*) discovered by Brahmagupta (*Brāhmasphuṭasiddhānta*, ch. 18, p. 64–55) and applied by later mathematicians.¹ In algebraic symbo's the principle of *samāsabhāvanā* may be expressed as fol'ows : For conveniently chosen values of c₁ and c₂, if (a₁, b₁) and (a₂, b₂) be a set of solutions of Nx² + c₁ = y² and Nx² + c₂ = y² respectively, then x = a₁b₂ ± a₂b₁ and y = b₁b₂ ± Na₁a₂ wi'l satisfy the equation Nx² + c₁c₂ = y². The aim of the scholars like Brahmagupta and Śrīpati was to obtain solutions of Nx² + 1 = y² in positive integers. They however discovered that an integral solution of Nx² + k = y² can always be found if k = ± 1, ± 2 or ± 4. Having got the solution, an infinite number of solution can be obtained by repeated application of *samāsabhāvanā*. This is undoubtedly a remarkable feat when we realise that this was done in 628 A. D. The method would have been perfect if only a method other than trial be avilable to Brahmagupta for the solution. Bhāskara II did it by method known as *cakravāla* as following.

c7.s5.158

¹ Bhāṣkara II's *Bījagaṇita*, v. 71; *Bījagaṇitavataṃśa* of Nārāyaṇa ed. K. S. Shukla, p. 36–44.

c7.s5.159

(B) *Bhāskara II's Cakravāla Process* : $i. e. solution of Nx² + 1 = y²$

c7.s5.160

*hrasvajyeṣṭhapadakṣepānbhājyaprakṣepa bhājakān* ||
*kṛtvā kalpo guṇastatra tathā prakṛtitaścute* |
*guṇavarge prakṛtyone ‘thavālpaṃ śeṣakam yathā* ||
*tattu kṣepahṛtam kṣepo vyastaḥ prakṛtitaścute* |

c7.s5.161

*guṇalabdhiḥ padaṃ hrasvaṃ tato jyeṣṭhamato 'sakṛt* ||
*tyaktā pūrvapadakṣepāṃścakravālamidaṃ jaguḥ* |
*caturdvaikayutā vevamabhinne bhavataḥ pada* ||
*caturdvikṣepamūlābhyāṃ rūpakṣepārthabhāvanā* |¹
( *Bījagaṇita*, v. 75 )

c7.s5.162

¹ Bhāskara II's *Bījagaṇīta*—Vargaprakṛti, v. 2–4.

c7.s5.163

*English Translation :*

c7.s5.164

“Take the lesser root, greater root and interpolator of a square nature ) as divident, interpolator and divisor. Select an arbitrary number so that its square diminished by *prakṛti* or the *prakṛti* diminished by square of the chosen number, ( as the case may be ) is the least. This ( difference ) divided by the original interpolator is the ( new ) interpolator; it should be reversed in sign in case of the subtraction from *prakṛti*. The quotient corresponding to that value of the multiplier is the ( new ) lesser root; likewise is obtained the greater root. The same process should be followed repeatedly putting aside ( each time ) the previous roots and the interpolator. This process is called *cakravāla* ( or the cyclic method ). By this method there will appear two integral roots corresponding to an equation with ± 1, ± 2 or ± 4 as interpolator. In order to derive integral roots corresponding to an equation with the additive unity from those of the equation with the interpolator ± 2 or ± 4, the principle of composition ( should be applied ).”

c7.s5.165

Select Na² + k = b², for any suitable k,

c7.s5.166

and N. 1² + (m² - N) = m² then by *bhāvanā*, Bhāskara II obtained $N((am+b)/k)²+(m²-N)/k=((bm+Na)/k)²......(1)$ Next to choose the value of the m so that m²-N is numerically smaller and (am+b)/k becomes integer, since its value is determined by means of *kuṭṭaka*. $Let (am+b)/k=a₁, (m²-N)/k=k₁$ $(bm+Na)/k=b₁$

c7.s5.167

*Proposition I* :

c7.s5.168

When a₁ is an integer, b₁, k₁ are also integers. Then the equation (i) becomes, $Na₁²+k₁=b₁²,$

c7.s5.169

A new equation of the same kind, viz.

c7.s5.170

Na₂²+k₂=b₂² can be obtained, proceeding in a similar way.

c7.s5.171

*Proposition II* :

c7.s5.172

After an finite number of repetitions, the equations,

c7.s5.173

Nα²+t=β², when t=±1, ±2 or ±4.

c7.s5.174

is available.

c7.s5.175

In order to obtain an integral solution of the original equation Nx²+t=y² from those of the interpolators with ±2, ±4, the principle of composition is applied. The truth of these two propositions was perhaps understood by Bhāskara II in case of concrete instances, for no proof was given by him.

c7.s5.176

*Examples :*

c7.s5.177

$(i) 61x² + 1 = y² / (ii) 67x² + 1 = y² [ Bījagaṇita, v. 76 ]$

c7.s5.178

*Solution :*

c7.s5.179

(i) 61x² + 1 = y² Select an auxiliary equation 61.1² + 3 = 8² (1) Let a₁ = 1, b₁ = 8, k₁ = 3 and N = 61 Then by Bhāskara II's lemma we get $N((a₁m + b₁)/k₁)² + (m² - N)/k₁ = ((b₁m + Na₁)/k₁)$ where m is an arbitrary chosen number. Here m is so chosen that m² - N is least and (a₁m + b₁)/k₁ is an integer Taking m = 7, we get (a₁m + b₁)/k₁ = 5, (m² - N)/k₁ = - 4. $and (b₁m + Na₁)/k₁ = 39$ Let a₂ = 5, b₂ = 39, k₂ = - 4 Hence from (1), we get 61.5² - 4 = 39² (2) Thus in one step, we get an equation Nx² + t = y² when t = - 4 From (2), 61.(5/2)² - 1 = (39/2)²

c7.s5.180

Apply the principle of *bhāvanā*, We get 195/2, 1523/2, 1 . . . . . (3)

c7.s5.182

Again applying *bhāvanā* between

c7.s5.184

We get 3805, 29718, -1 . . . . . (4) Finally applying *bhāvanā* in this with itself,

c7.s5.186

We get 226153980, 1766319049, 1 Hence x = 226153980 y = 1766319049 is the solution of 61x² + 1 = y²

c7.s5.187

Bhāskara II gave no proofs of Propositions I and II. Proofs are given by Datta and Singh¹, Hankel² and Krishnaswami Ayangar³.

c7.s5.188

¹ Datta, B and Singh, A. *Ibid.*, 2. , p. 157-166.

c7.s5.189

³ Krishnaswāmī Ayyangar, A. A. *Journal of the Indian Mathematical Society*, 18, first series, second part, p. 232 45.

c7.s5.190

(ii) Solution of : 67x² + 1 = y² We know, 67.1² - 3 = 8² . . . . . (1) Comparing with Na² + c = b², N = 67 a = 1, b = 8, c = -3. Now (am + b)/c = (1.m + 8)/(-3) = (m + 8)/(-3) is an integer when m = 7.

c7.s5.191

² Hankel, H. *Zur Geschichte der Mathematik*, p. 202, Leipzig, 1874.

c7.s5.192

Hence, taking m=7 $a₁=(am+b)/c=(7+8)/(-3)=-5$ $b₁=(bm+Na)/c=(8.7+67)/(-3)=(56+67)/(-3)=123/(-3)=-41$ $c₁=(m²-N)/c=(49-67)/(-3)=(-18)/(-3)=6$ $∴ 67.(-5)²+6=(-41)⁸$ $or 67.5²+6=41². (2)$

c7.s5.193

Hence we get, $67.5²+6=41²$

c7.s5.194

Now, a₂=(5p+41)/6 is an integer for p=5 $₂=66/6=11$ $b₂=(41.(+5)+67.5)/6=(+205+335)/6=540/6=90$ $c₂=(25-67)/6=(-42)/6=-7$

c7.s5.195

Hence, we get 67.11²-7=90²(3)

c7.s5.196

Now, a₃=(11t+90)/(-7) is an integer for t=9 $₃=(99+90)/(-7)=189/(-7)=-27$ $b₃=(90t+67.11)/(-7)=(90× 9+737)/(-7)=1547/(-7)$ $=-221$ $c₃=(81-67)/(-7)=14/(-7)=-2$

c7.s5.197

We get, 67.(-27)²-2=(-221)² $or 67(.27)²-2=(221)²(4)$

c7.s5.198

Applying the *samāsa* between : 27 & 221 & -2 / 27 & 221 & -2 We get X = 27 × 221 + 27 × 221 = 11934, Y = 221 × 221 + 67 × 27 × 27 = 97684, satisfying the equation, 67.X² + 4 = Y² or 67.(11934)² + 4 = (97684)² ... ... ... ...(5) Dividing both sides by 4, or 67.(5967)² + 1² = (48842)² (6) ∴ x = 5967 and y = 48842

c7.s5.199

From modern approach we know : \begin{align*} √67 &= 8 + 1/(5+) \ 1/(2+) \ 1/(\bar+) \ 1/(1+) \ 1/(7+) \ 1/(8+) \dots\dots\dots \\ &= 8, \ 41/5, \ 90/11, \ 221/27, \ 97684/11939 (= 48842/5967), \dots\dots\dots \end{align*}

c7.s5.200

For the solution of 67x² + 1 = y², if 1, m be the solution then according to Bhāskara II's method k = m² - N i.e. if (1, 8) be an an arbitrary solution, then k = 8² - 67 = -3 Hence the first set can be written as : \begin{align*} &1)\ 67.1² - 3 = 8². Likewise the other sets are \\ &2)\ 67.5² + 6 = (41)² \\ &3)\ 67.\ (11)² - 7 = (90)² \\ &4)\ 67.\ (27)² - 2 = (221)² \\ &5)\ 67.\ (11934)² + 4 = (97684)² \\ &and 6)\ 67.(5967)² + 1 = (48842)² \end{align*}

Linked texts (12)

Brāhmasphuṭasiddhānta — Brahmagupta, Sanskrit
Gaṇitasārasaṅgraha — Mahāvīra (mathematician), Sanskrit
Āpastambaśulbasūtra — Āpastamba, Sanskrit
Bījagaṇita — Bhāskara II, Sanskrit
Āryabhaṭīya — Āryabhaṭa, Sanskrit
Mahābhāskarīya — Bhāskara I, Sanskrit
Yuktibhāṣā (Gaṇitayuktibhāṣā) — Jyeṣṭhadeva 1520, Malayalam
Mahāsiddhānta (Āryasiddhānta) — Āryabhaṭa II, Sanskrit
Laghubhāskarīya — Bhāskara I, Sanskrit
Karaṇapaddhati — Putumana Somayājī 1660, Sanskrit
Pāṭīgaṇita — Śrīdhara (mathematician), Sanskrit
Bījagaṇitāvataṃsa — Nārāyaṇa Paṇḍita (mathematician), Sanskrit