History of Hindu Mathematics: A Source Book v2.c17.s17.t4: (iv) Solution of a²x² + by² + c = z

History of Hindu Mathematics: A Source Book v2.c17.s17.t4

Language: English

(iv) Solution of a²x² + by² + c = z

v2.c17.s17.t4.1

Let the coefficient of x² (or y²) be a square number. The equation is of the form $a²x² + by² + c = z².$

v2.c17.s17.t4.3

"If the *prakṛti* is a square, then obtain the roots by the rule: 'The interpolator divided by an optional number is set down at two places, etc.' "³

v2.c17.s17.t4.5

Thus, according to Bhâskara II, the solution of the above equation is $x = 1/2a((by² + c)/m) - m,$

v2.c17.s17.t4.7

$z = ½((by²+c)/m+m);$ where m is an arbitrary number.

v2.c17.s17.t4.8

Kamalâkara divides equations of this form into two classes according as c is or is not a square.¹

v2.c17.s17.t4.9

¹ *Vide* his gloss on *SiTVi*, xiii. 275.

v2.c17.s17.t4.10

(1) Let c be a square (= d², say). That is to say, we have to solve $a²x² + by² + d² = z².$

v2.c17.s17.t4.11

The solution of this particular case, says Kamalâkara, is given by $x = b/2ady².$

v2.c17.s17.t4.12

For, with this value, we have $z² &= (b²y⁴)/(4d²) + by² + d² / &= 1/(4d²) (by² + 2d²)².$

v2.c17.s17.t4.14

(2) When c is not a square, Kamalâkara first finds α, β such that $α² + c = β².$

v2.c17.s17.t4.15

He next obtains n such that the value of $(αβ n - b/2)/(a⁴. α/a)$ is also n; and then says that $x = ny² + (α)/a;$ whence will follow the value of z.

Since & & (a\beta n - b/2)/(a\alpha) &= n, \\ we get & & n &= b/(2a(\beta - \alpha)); \\ so that & & x &= (by²)/(2a(\beta - \alpha)) + (\alpha)/a. \end{aligned} Therefore z² = a²x² + by² + c &= (b²y⁴)/(4(\beta - \alpha)²) + \alpha² + (\alpha by²)/((\beta - \alpha)) + by² + c / &= (b²y⁴)/(4(\beta - \alpha)²) + (\beta by²)/((\beta - \alpha)) + \beta² / &= { (by²)/(2(\beta - \alpha)) + \beta }². Hence z = (by²)/(2(β - α)) + β.

v2.c17.s17.t4.17

*Example from Kamalâkara* : 4x² + 48y² + 20 = z². Since 4² + 20 = 6², and the solution of (12n - 24)/8 = n is n = 6, we get the required solution of the given equation as x &= 6y² + 2, / z &= 12y² + 6.

v2.c17.s17.t4.18

It may be noted that the solution stated by Kamalâkara follows easily from that of Bhâskara II, on putting therein m = β - α, where α² + c = β².

v2.c17.s17.t4.19

In particular, if we put c=0 and a=1, Bhâskara's solution reduces to

v2.c17.s17.t4.20

x &= 1/2(b/my²-m), \\ z &= 1/2(b/my²+m); \end{align*} where m is arbitrary.

v2.c17.s17.t4.21

If b=αβ, taking m=2β p², we easily arrive at \begin{align*} x &= \alpha q²-\beta p², \\ y &= 2pq, \\ z &= \alpha q²+\beta p², \end{align*} where p, q are arbitrary integers, as the solution in positive integers of z²=x²+by². This solution was given by A. Desboves (1879).¹

v2.c17.s17.t4.22

¹ Dickson, *Numbers*, II, p. 405.

v2.c17.s17.t4.23

Taking m=2ν², we can derive Matsunago's (*c.* 1735) solution of z²=x²+by², *viz.*, \begin{align*} x &= b\mu²-\nu², \\ y &= 2\mu\nu, \\ z &= b\mu²+\nu²; \end{align*} where μ, ν are arbitrary integers.

Linked texts (2)

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