1

organized by

�व�ान भार�तand

Dr. KV Sharma Indological Research Institute,Chennai(Date:28th and 29th January 2015)

Venue:Vivekananda Hall, P. S. Higher Secondary School,215. R. K. Mutt Road, Mylapore, Chennai – 600 004

900th Birth Anniversary of Bhaskara-II (Born1114 AD):

Workshopon

Bhaskarscharya’s Contributions to Mathematics and Astrnomy

2

Compiled by: Venkatesha Murthy, Hon. Head, Vedic Math’s Group, National Institute of Vedic Sciences, # 58, Raghavendra Colony, Chamarajapet, Bangalore-18

900th Birth Anniversary of Bhaskara-II (Born1114 AD):

Solving Indeterminate EquationsfromLilavati of Bhaskara-II.

3

Introduction: - Indian Mathematicianswere basically astronomers,

and they stated a set of stepsto solve in which A, B and C

are known integers, and x & y are unknowns, whose integral values

are required.

Method for solving such equations in Indian Mathematics is named

“Kuttakaor Vallika Kuttakara”

Problems onLinear Indeterminate Equations are in several ancient

texts of Indian Astronomy.Aryabhata-I(5th c. AD) gavea general rule

for solving

Bhascara I (6thc.AD), Brahmagupta(8thcenturyAD),Mahaveeracarya

(9th century AD)and others explained general methods to solve them.

A C

B

xy

+=

A C

B

xy

+=

A C

B

xy

+=A

B

But the method ofreputed mathematician Bhascara-II (born1114 AD),

solved througha single recurring relationbased onpartial

quotientsof simple continued fraction for integral solutions.

4

I. Kuttaka (कु�क) from Lilavati of Bhascara-II, (a section ofSiddhanta-siromani):

Step 1: -भायो हारः �ेपक�चापव�य�ः केना�यादौ सभंवे कु�काथ�म ् ।येन ि#छ%नो भायहारौ न तेन �ेप�चत& द'ुटम)ु*'टमेव॥248||

[Ref. 1 (p.102)]

Purport: - HCF of A and B must divideC for solving the equation. Otherwise the problem is improper. (x and y cannot benon-zero integers).

In an indeterminate linear equation of the form;

A, dividend (भाय); B, divisor (भाजक); and C, addend (�ेपक); are

known integers.

x, a multiplier (गुणक) andy, a quotient (लि3ध) are to be found.

Lilavati states a methodto find the least integral valuesfor

x (गुणक) and y (लि3ध) of the equation in six steps.

A C

B

xy

+=

5

Step 2: - पर5परं भािजतयोय�योय�ः शेष5तयोः 5यादपवत�नं सः ।तेनापव�त8न �वभािजतौ यौ तौ भायहारौ 9ढसञंकौ 5तः ||249||

[Ref. 1 (p.103)]

Purport: -Find the HCF (k)by mutually dividingA and B. Divide A,

B and C by k (HCF) to get thereduced dividend(a); reduceddivisor

(b); and the reduced addend (c). Then the reduced equation is; A C

B

ax c xy

b

+ + = =

Step 3: - <मथोभजे�तौ 9ढभायहारौ याव& �वभाये भवतीय >पम ्।फला%यधोध5तदधो �नवे�य �ेप5ततः शू%यमपुाि%ततेन ॥250||

[Ref. 1 (p.103)]

Purport: - Divide the reduced dividend(a) and reduced divisor(b)

mutually(as in the case of finding their HCF) until the remainder1 is

obtained.Place the quotientsso obtainedone below the otherand

below themthe reduced addend (c) and thenzero (0).

6

Mathematically: Mutual divisionof dividend (a) and divisor (b) is;

a = q1 b + r1 ; 0 ≤ r1 < |b|, and,

b = q2 a + r2 ; 0 ≤ r2 < r1;

and so on, untilthe remainder rn = 1 is obtained.

In general: Mutual division ofa and b would be of the form;

r(t-2) = qt r(t-1) + r t ; 0 ≤ r t < r(t-1) , when r(-1) = a and ra = |b|

for t = 1, 2, 3, . . . (n-1).

When r(n-1) = 1, write thequotients q1, q2, q3, . . . q(n-1) one below

the other, and thenwrite the reduced addend (c) and thenzero.

7

Step 4: - 5वोAव8 हतेऽ%�येन युते तद%�यं�यजे%महुुः 5या)द�त रा<शयCुमम ् ।

ऊAवE �वभायेन 9ढेन त'टः फलं गणुो 5यादधरो हरेण ॥251||

[Ref. 1 (p.103)]

Purport: - Multiply the last quotient by the term below the quotient and, to the product add the last term, and write the sum by the side of the quotient. Reject the last term and repeat the process till a pair of results from the top is obtained. The first term from the top thus obtainedis to be dividedby thereduced dividendto get the remainder(the quotient y0, लि3ध) and, 2nd term from the top thus

obtained is to be dividedby the reduced divisorto get the remainder(the multiplier x0, गुणक).[The process described above is namedविJलका कु�क्ार as it movesfrom bottom to

top as a creeper(विJलका). It is shown in the chart]

8

9

Last line ofstep 3: -Write the quotients q1, q2, q3, . . . q (n-1), qn one below the otherand below themthe reduced addend (c) and thenzero, in the first column.

Step 4: - (Modified Vallika process)1. Multiply the last quotient qn by addend c below the quotientand to the product add zeroto get F1

[= qn(c) + 0], and write the sum F1 in the column Ft against the quotient qn.

2. Multiply the quotientq(n-1) by F1, and to

the product add the addend cto get

F2 [= q(n-1) (F1) + c], and write the sum

F2 aboveF1 against the quotient q(n-1).

Repeat the process in arecurring relationof the form; F(n- t+1) = qt F(n- t) + F(n-t -1) ; whereF0 = c and F(-1) = 0for [t = n, (n-1), . . . , 2, 1]until a pair of resultsFn and F(n-1)

are obtained from the top,in the column F(n-t+1) .

10

Fn is to be dividedby the reduced dividend(a) to get the remainder(the quotient y0 is लि3ध) [Fn = a t + y0], and F(n-1) is to be dividedby the reduced divisor(b) to get the remainder(the multiplier x0 is गुणक) [F(n-1) = b t + x0].

Step 5: -एवं तदैवाO यदा समा5ताः 5युल�3धय�चे&�वषम5तदानीम ्।यदागतौ लि3धगणुौ �वशोAयौ 5वत�णा#छेष<मतौ तु तौ 5तः ॥252||

[Ref. 1 (p.103)]Purport: - (i) When the numberof partial quotients n is even, and addend c is positivethe remainders are the least integral values ofy = y0 (लि3ध) and of x = x0 (गुणक).

(ii) When the number of partial quotientsn is oddand c is positive, ORwhen n is evenand c is negative, the above remaindersare to be subtracted fromthe respective divisorsto get their least integralvaluesof y [= (a - y0) (लि3ध)] andof x [ = (b - x0) (गुणक)].

11

Example 1: एक�वशं�तयुतं शत&वयं य&गणुं गणक पPचषि'टयुक् ।पPचविज�तशत&वयोQृतं शु�Qमे�त गणुकं वदाऽऽशु तम ् ॥252||

[Ref. 1 (p.103)]

Purport: - O, mathematician, tell me the numberwhen multiplied by

221and increased by65 is divisible by 195?

Answer: - The problem is to solve;

According to step 1, HCF of 221 and 195is 13.HCF 13divides the addend65.Therefore, the problem has integral solutionsfor x and y.According to step 2, dividing dividend 221, divisor 195and addend 65by the HCF (15)we getreduced dividend (17); reduced divisor (15); and reduced addend (5).

221 65

195

xy

+=

12

According to step 2, dividing dividend221, divisor 195 and addend 65 by the HCF (15)we get reduced dividend(17); reduced divisor (15); and reduced addend (5).

Then the reduced equation is;. 221 65 17 5

195 15

x xy

+ += =

According to step 3, when the reduced dividend17 and reduced divisor15 aremutually divided (as in the case of finding their HCF);

the remainder1 and quotients1 and 7are obtained .

13

The quotients1 and 7are placed one below the otherand below themthe reduced addend 5 of the equation and then zero.According to Step 4, apply the Modified Vallika processthus ;

F1 = 35and F2 = 40are obtained.

According to Step 5 (i);F2 = 40 = 17 x 2 + 6,hence the least integral value of y is; y0 = 6 and F1 = 35 = 15 x 2 + 5,hence the least integral value of x is; x0 = 5.

14

Number of partial quotients,n = 2 (an even number) and addend c is positive. Therefore, x = x0 = 5 and y = y0 = 6 are the least integral values of xand y satisfying,

And its infinite integral values for x and y are; x = 15t + 5and y = 17t + 6.

The required values of them are obtained by substituting any natural number to t.

17 5 221 65

15 195

x xy

+ += =

15

According to step 3, the dividend60 and divisor 13 are co-primes. When they are mutually divided (as in the case of finding their HCF); thequotientsare 4, 1, 1, 1, and 1 and the remainderis 1.

Example 2: य&गणुा �यषि'टरि%वता विज�ता च य)द वा [O<भ5ततः ।5या�Oयोदश\ता �नर]का5तं गणुं गणक मे पथृCवद ॥ CCLVII || [Ref. 1 (p.170)]

Purport: - O, mathematician, tell me the numberwhen multiplied by

Minus 60and increased or decreased by3 is divisible by 13?

Answer: - The problem is; 60 3

13

xy

− ± = Let us solve;60 3

13

xy

+ =

16

Number of quotientsn = 5 (an odd number).Quotients 4, 1, 1, 1 and1 are placed one below the other and below themthe addend 3 and then zero.According to Step 4, applying the Modified Vallika process;

As per Step 5(ii)F5 = 69 = 60 x1 + 9 ;the integral value of y is; y0 = (60-9) = 51

F4 = 15 = 13 x 1 + 2; the integral value of x is; x0 = (13-2) = 11.

The integral values x = x0 = 11 andy = y0 = 51satisfy;60 3

13

xy

+ =

17

According to step 3, the dividend(-60) and divisor 13 are co-primes. When they are mutually divided (as in the case of finding their HCF); thequotients(-5), 2, 1 and 1, and the remainder1 are obtained .

Example 2: (can also be solved thus) ;य&गणुा �यषि'टरि%वता विज�ता च य)द वा [O<भ5ततः ।5या�Oयोदश\ता �नर]का5तं गणुं गणक मे पथृCवद ॥ CCLVII || [Ref. 1 (p.170)]

Purport: - O, mathematician, tell me the numberwhen multiplied by

Minus 60and increased or decreased by3 is divisible by 13?

Answer: - The problem is; 60 3

13

xy

− ± = Let us solve; 60 3

13

xy

− + =

18

Number of quotientsn = 4, and quotients (-5), 2, 1 and1 are placed one below the other and below themthe addend 3 and then zero.According to Step 4, applying the Modified Vallika process;

F4 = -69 = (-60)x1 + (-9) ;the integral value of y is; y0 = -9

According Step 5(i);F3 = 15 = 13x1 + 2; the least integral value of x is; x0 = 2.

Thenx = x0 =2 andy = y0 = (-9) satisfy; 60 3

13

xy

− + =

19

To solve the above problem with negative dividend (-60); 60 3

13

xy

− + =

the integral value of x andy are;x0 = 13 – 11 = 2 and y0 = -9

the integral value of x andy are;x0 = 11 – 13 = -2 and y0 = 9

To solve the above problem with negative addend (-3); 60 3

13

xy

− − =

20

A problem is given in[Ref. 1 page १07 and 180];

येन स_गु̀ णताः पPच Oयो�वशं�तसयंुताः ।विज�ता वा [O<भभ�aता �नर]ाः 5युः स को गणुः ॥२५९॥

Purport: -259. What is the multiplier, by which fivebeing multiplied

and twenty-threeadded to the product, or subtracted from it, the sum

or difference may be divided by threewithout remainder?

Rule 2: - A ‘common quotient’ and its implication;

गणुल3Aयोः समं ]ाeयं धीमता ��णे फलं ॥२५८॥

Purport: -258. The intelligent calculator should take a like quotient

(of both divisions) in the abrading of the numbers for the multiplier

and quotient(sought). [Ref. 3. (p. 11-12) & 1. (p. १07 & 179)]

5 23

3

xy

± =

21

Answer: - Let us solve the equation;5 23

3

xy

+ =

As per step 3 & 4, the dividendis 5 and divisor is 3, and they are co-primes. When dividendis 5 and divisor is 3are mutually divided (as in the case of finding their HCF);

Then, quotients1 and 1, and the remainder1 are placed one below the other and below themthe addend 23 of the equation and then zero.As per Step 4, apply the Modified Vallika processthus;

Hence F2 = 46and F1 = 23.

22

As per the above rule to Step 5 (i); F2 = 46 = 5 x 9 + 1 ; hence the least integral value of y is; y0 = 1 and F1 = 23 = 3 x 7 + 2 ; hence the least integral value of x is; x0 = 2.

5 23

3

xy

+ =

But when the least integral value of x; x0 = 2 is substituted in ;

we get y0 = 11 as the least integral value of yinstead of 1. The rule to Step 5 (i) fails.

To overcome this difficulty Bhaskara-II has stated the Rule 2. Rule 2: - ‘common quotient’ and its implication;

गणुल3Aयोः समं ]ाeयं धीमता ��णे फलं ॥२५८॥Purport: -258. The intelligent calculator should take a like quotient (of both divisions) in the abrading of the numbers for the multiplier and quotient (sought). [Ref. 1 (p. १07 & 179)]

F2 = 46 = 5 x 7 + 11 ; hence the least integral value of y is; y0 = 11and F1 = 23 = 3 x 7 + 2 ; hence the least integral value of x is; x0 = 2.Here the common quotient is 7.

23

Ref. 1 (p.107 & 179)]

To solve the above problem with negative addend (-23);

the least integral value of x is; x0 = 3 – 2 = 1and therefore y0 = -6

Rule 1: -To solve such problems with ‘negative addend’;हरत'टे धन�ेपे गणुल3Aयो तु पूव�वत ् ।

�ेपत�णलाभाgया लि3धः शुQौ तु विज�ता ||258||Purport: -The multiplier and quotient may be found as before, the

additive quantity being (first) abraded by the divisor ; the quotient, however, must have added to it the quotient obtained

in the abrading of the additive. But in the case of a subtractive quantity,

it (the quotient) is subtracted.

Applying the above rule to the problem ;

5 23

3

xy

− =

5 23

3

xy

− =

24

Joseph Louise Lagrange (1736-1813 AD) found the least integral values of x and y for the problems like ax+ by= c (AD.1770) in which a and b are co-primes. Lagrange, first solved the equation

ax+ by= 1 using the partial quotientsof a continued fraction

expansion in a pair of recurring

relations

The least integral valuesx = xL and y = yL satisfyax+ by= 1.

Then, Theleast integral valuesx = cxL and y = cyL satisfy

ax+ by= c.

[ ]1 2 3 n 1 nq , q , q , . . . ,q , qa

b −=

r r-1 r-2 r -1 0q ;where 1, 0,x x x x x+ = = =

r r-1 r-2 r -1 0q ;where 0, 1y y y y y+ = = =

25

before solving for values x and y for5 3 23x y− = ∓

Answer: -Continued Fraction Expansionof

5 3 1x y− =Joseph Louise Lagrange first solves the equation

Here the partial quotients are;1, 1, 1, 1.Out of these the penultimatepartial quotient is1. he usesthe quotients1,1,1in a pair ofrecurring relationas shown below;

When x = 2 and y = 3 the equation 5(2) – 3(3) = 1is true.

26

To get the least integral valuesof x and y for 5x – 3y =23,

multiply both sides of the equation by23, then

23 (5x) – 23(3y) = 23x1

23 (5 x 2) – 23 (3 x 3) = 23,

so that (x, y) = (10,9) is a particular pair of values satisfying

5x - 3y = 23.

x = (10 + 3t) and y =(9 + 5t)are the general integral values,

for any integert, gives infinite number of integral valuesof

x and y satisfying the equation

5x - 3y = 23.

27

Summary: -

Joseph Louise Lagrange (1736-1813 AD) found the least integral values of x and y for the problems like ax+ by= c (AD.1770) in which a and b are co-primes. Lagrange, first solved the equation

ax+ by= 1 using the partial quotientsof a continued fraction

expansion in a pair of recurring

relations

The least integral values x = xL and y = yL satisfy ax+ by= 1.

Then, Theleast integral values x = cxL and y = cyL satisfy

ax+ by= c.

[ ]1 2 3 n 1 nq , q , q , . . . ,q , qa

b −=

r r-1 r-2 r -1 0q ;where 1, 0,x x x x x+ = = =

r r-1 r-2 r -1 0q ;where 0, 1y y y y y+ = = =

28

Bhaskara-II (born 1114 AD) found the least integral values of

x and y for the problems like;

by dividing A and B mutually to find their HCF (k) to get

reduced dividend(a); reduceddivisor (b); reduced addend (c),

by dividing A, B and C by the HCF (k).

Thus the reduced equation in the form;

Afterwards, a and b are mutually divided using an operation;

r(t-2) = qt r(t-1) + r t ; 0 ≤ r t < r(t-1), in which r(-1) = a and r0 = |b|

for t = 1, 2, 3, . . . n. until r n = 1. Bhaskara II used the partial

quotients q1, q2, q3, . . . qn thus obtained in a processnamed

Vallika (creeper like).

A C

B

xy

± =

ax cy

b

± =

Bhaskara-II (born 1114 AD) found the least integral values of

x and y for the problems like;

by dividing A and B mutually to find their HCF (k) to get

reduced dividend(a); reduceddivisor (b); reduced addend (c),

by dividing A, B and C by the HCF (k).

Thus the reduced equation in the form; is obtained.

Afterwards, a and b are mutually divided using an operation;

r(t-2) = qt r(t-1) + r t ; 0 ≤ r t < r(t-1), in which r(-1) = a and r0 = |b|

for t = 1, 2, 3, . . . n. until r n = 1. Bhaskara II used the partial

quotients q1, q2, q3, . . . qn thus obtained in a processnamed

Vallika (creeper like) process.

ax cy

b

± =

A C

B

xy

± =

29

The Vallika process could be shown as a a single recurring relation

F(n- t +1) = qt F(n- t) + F(n-t-1) in which F0 = c and F(-1) = 0

for [t = n, (n-1), . . . , 2, 1] successively.

Fn and F(n-1) thus obtained give the least integral values of y and x

for the problems of the form;

Divide Fn by the reduced dividend(a) so that [Fn = at + y0].

The remainder y0 = लि3ध, is the least integral value of y.

The least integral value of is found as the remainder

Divide F(n-1) by the reduced divisor (b) so that [F(n-1) = bt + x0]

The remainder x0 = गणुक, least integral value of x.

The quotient in both the divisions must be same.

A C

B

xy

± =

30

Joseph Louise Lagrange (born 1736 AD)gives a Method for solving Indeterminate equations in two unknowns

to find their least integral values using the partial quotients of the Continued Fraction Expansion of a rational number

in a PAIR of recurring relations(from top to bottom).

Bhaskara II (born 1114 AD)gives a Method for solving Indeterminate equations in two unknowns

to find their least integral values using the partial quotients of the Continued Fraction Expansion of a rational number

in a SINGLE recurring relation(from bottom to top, creeper like).

31

Reference: -1. “COLEBROOKE’S TRANSLATION OF THE LILAVATI with notes”- by

H C Bannerji, Asian Educational Services Madras (1993). 2. “Lilavati of Bhaskaracharya: A Treatise of Mathematics of Vedic

Tradition”- S Patwardhan and others, Motilal Banarsidass PublishersPrivate Ltd. (2001).

3. “HINDU WORK ON LINEAR DIOPHANTINE EQUATIONS” : - VenkateshaMurthy, Dr. M S Rangachary, Dr. Baskaran): JOURNAL OF THE MADRAS NIVERSITY, [section B, Vol.48 (1985) pp 1-19].

4. “COMPARATIVE STUDY OF ChakravalaMCF & ICF METHODS- VenkateshaMurthy, MATHEMATICS NEWS-LETTER, [Vol.5, No.3, P.50-57 (1995)]. Published by Ramanajuanujan Mathematical Society.

5. “Continued Fractions” – C. D. Olds, The L. W. Singer Company, (1963) 6. KUTTAKA, BHAVANA AND CAKRAVALA’ (in Kannada) – Venkatesha Murthy,

Rashtriya Sanskrit Vidyapeeth (Deemed University) Tirupati. (2003). 7. ‘KUTTAKA – An Indian Method for solving Linear Indeterminate Equations for

Integral Solutions’– Venkatesha Murthy, Rashtriya Sanskrit Vidyapeeth (Deemed University) Tirupati. (2004).