History of Hindu Mathematics: A Source Book v2.c17.s17.t3: (iii) Solution of ax² + by² + c = z

History of Hindu Mathematics: A Source Book v2.c17.s17.t3

Language: English

(iii) Solution of ax² + by² + c = z

v2.c17.s17.t3.1

Bhâskara II followed several devices for the solution of the equation $ax² + by² + c = z². 3$ In every case his object was to transform the equation into the form of the Square-nature. He says :

"In such cases, where squares of two unknowns with (or without) an absolute number are present, supposing either of them optionally as the *prakṛti*, the rest (of the terms) should be considered as the interpolator. Then the roots should be investigated in the way described before. If there be more equations than one (the process will be especially helpful)."¹

v2.c17.s17.t3.3

¹ *BBi*, pp. 105f.

v2.c17.s17.t3.4

He then explains further :

v2.c17.s17.t3.5

"Where on finding the square root of the first side, there remain on the other side squares of two unknowns with or without an absolute number, there consider the square of one of the unknowns as the *prakṛti*; the remainder will then be the interpolator. Then by the rule: 'An optionally chosen number is taken as the lesser root, etc.,'² the unknown in the interpolator multiplied by one, etc., and added with one, etc., or not, according to one's own sagacity, should be assumed for the lesser root; then determine the greater root."³

v2.c17.s17.t3.6

² The reference is to the rule for solving the Square-nature (*vide supra* p. 144) (*BBi*, p. 33).

v2.c17.s17.t3.7

³ *BBi*, p. 106.

v2.c17.s17.t3.8

There are thus indicated two artifices for solving the equation (3). They are :

v2.c17.s17.t3.9

(i) Set x = my; so that equation (3) transforms into $z² = (am² + b) y² + c$ $= α y² + c,$ where α = am² + b. Hence the required solution of ax² + by² + c = z² is $x = my = m(rq ± ps),$ $y = rq ± ps,$ $z = sq ± α pr;$ where s² = α r² + c and q² = α p² + 1.

v2.c17.s17.t3.10

*(ii)* Set x = my ± n; then the equation reduces to $z² = α y² ± 2amny + γ$ where α = am² + b and γ = an² + c.

v2.c17.s17.t3.11

Completing the square on the right-hand side of this, we get $α z² - β = w²,$ where w = α y ± amn and β = γα - a² m² n² = a(bn² + cm²) + bc.

v2.c17.s17.t3.12

If z = s, w = r be a solution of this equation, another solution will be $z = sq ± rp,$ $w = rq ± α sp;$ where q² = α p² + 1. Hence the solution of ax² + by² + c = z² is $x = m/(α) (rq ± α sp amn) ± n,$ $y = 1/(α) (rq ± α sp amn),$ $z = sq ± rp;$ where q² = α p² + 1, r² = α s² - β, α = am² + b and β = a(bn² + cm²) + bc.

v2.c17.s17.t3.13

In working certain problems, Bhâskara II is found to have occasionally followed other artifices also for the solution of the equation (3). For instance :

v2.c17.s17.t3.14

(*iii*)¹ Set w² = by² + c. Then equation (3) becomes $z² - w² = ax².$

v2.c17.s17.t3.15

¹ See *BBi*, p. 108.

v2.c17.s17.t3.16

Whence $z = ½ (a/m + m)x,$

v2.c17.s17.t3.17

and $w = ½ (a/m - m)x;$ where m is an arbitrary number. Therefore $x = 2mw/(a - m²),$ $z = ((a + m²)/(a - m²))w.$

v2.c17.s17.t3.18

Now, if y = l, w = r be a solution of $w² = by² + c,$ another solution of it will be $y = lq ± pr,$ $w = rq ± blp;$ where ap² + 1 = q². Therefore, the solution of (3) will be $x = 2m/(a - m²) (rq ± blp),$ $y = lq ± pr,$ $z = (a + m²)/(a - m²)(rq ± blp);$ where ap² + 1 = q² and bl² + c = r².

v2.c17.s17.t3.19

(*iv*)² Suppose c = 0; then the equation to be solved will be $ax² + by² = z².$

v2.c17.s17.t3.20

² *BBi*, p. 106.

v2.c17.s17.t3.21

In this case set x = uy, z = vy; so that u, v will be given by $au² + b = v²,$ which can be solved by the method of the Square-nature. Some of these devices were followed also by later Hindu algebraists, for instance, Jñânarâja (1503) and Kamalâkara (1658).¹

v2.c17.s17.t3.22

¹ *SiTVi*, xiii. 260-1.

v2.c17.s17.t3.23

*Example* from Kamalâkara:² $7x² + 8y² = z².$ This is one of a double equation by Bhâskara II.³

v2.c17.s17.t3.24

³ *BBi*, p. 106.

v2.c17.s17.t3.25

To solve ax² + by² + c = z², Kamalâkara observes:

v2.c17.s17.t3.26

² *SiTVi*, xiii. 258.

v2.c17.s17.t3.27

"In this case, suppose the coefficient of the square of the first unknown as the *prakṛti* and the coefficient of the square of the other unknown together with the absolute number as the interpolator to that. The two roots can thus be determined in several ways."⁴ And again:

v2.c17.s17.t3.28

⁴ *SiTVi*, xiii. 264.

v2.c17.s17.t3.29

"(Suppose) the coefficient of the square of one of the unknowns as the *prakṛti* and the rest comprising two terms, the square of an unknown and an absolute number, as the interpolator. Then assume the value of the lesser root to be equal to the other unknown together with an absolute term."⁵

v2.c17.s17.t3.30

⁵ *SiTVi*, xiii. 267 f.

v2.c17.s17.t3.31

He seems to have indicated also a slightly different method:

v2.c17.s17.t3.32

"Or assume the value of the lesser root to be equal to another unknown plus or minus an absolute number and similarly also the value of the greater root. The remaining operations should be performed by the intelligent in the way described by Bhâskara in his algebra."¹

v2.c17.s17.t3.33

¹ *SiTVi*, xiii. 265.

v2.c17.s17.t3.34

That is to say, assume $x = mw ± α, z = nw ± β.$

v2.c17.s17.t3.35

Substituting in the equation ax² + by² + c = z², we get $(am² - n²)w² ± 2w (ama nβ) + by²$ $+ (c + aa² - β²) = 0.$

v2.c17.s17.t3.36

Putting λ = am² - n², μ = ama nβ, ν = c + aa² - β², this equation can be reduced to $- λ by² + (μ² - νλ) = u²,$ where u = λ w ± μ.

v2.c17.s17.t3.37

Kamalâkara gives also some other methods which are applicable only in particular cases.

v2.c17.s17.t3.38

*Case i.* Suppose that b and c are of different signs.² Two sub-cases arise:

v2.c17.s17.t3.39

(1) *Form* ax² + by² - c = z².

v2.c17.s17.t3.40

First find u, v, says Kamalâkara, such that $au² - c = v².$

v2.c17.s17.t3.41

Assuming $x = √(b/ac) vy + u,$ we have $ax² + by² - c = b/c v²y² + 2√(ab/c) uvy + by²$ $+ (au² - c)$ $= b/c (au² - c) y² + 2√(ab/c) uvy + by² + v²$

v2.c17.s17.t3.42

² *SiTVi*, xiii. 285-7.

v2.c17.s17.t3.43

$= ab/c u² y² + 2 √(ab/c) uvy + v²$ $= (√(ab/c) uy + v)².$

v2.c17.s17.t3.44

Hence z = √(ab/c) uy + v.

v2.c17.s17.t3.45

The following illustrative example and its solution are given :¹ $5x² + 16y² - 20 = z².$ Its solution is $x = ⅖vy + u,$ $z = 2uy + v;$ where 5u² - 20 = v². An obvious solution of this equation is given by u = 3, v = 5. Hence, we get a solution of the given equation as $x = 2y + 3,$ $z = 6y + 5.$ Therefore (x, y) = (5, 1), (7, 2), (9, 3),

Linked texts (2)

Bījagaṇita — Bhāskara II, Sanskrit
Siddhāntaśiromaṇi — Bhāskara II, Sanskrit