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History of Hindu Mathematics: A Source Book v2.c17.s17.t5: (v) Solution of ax²+bxy+cy²=z

History of Hindu Mathematics: A Source Book v2.c17.s17.t5

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(v) Solution of ax²+bxy+cy²=z

v2.c17.s17.t5.1

For the solution of the equation $ax²+bxy+cy²=z² 5$ Bhâskara II lays down the following rule :

v2.c17.s17.t5.2

"When there are squares of two unknowns together with their product, having extracted the square-root of one part, it should be put equal to half the difference of the remaining part divided by an optional number

v2.c17.s17.t5.3

and the optional number.”¹

v2.c17.s17.t5.4

¹ Y. Mikami, *The Development of Mathematics in China and Japan*, Leipzig, 1913, p. 231.

v2.c17.s17.t5.5

It has been again elucidated thus :

v2.c17.s17.t5.6

“Where there exists also the product of the unknowns (in addition to their squares), by the rule, ‘when there are squares etc.,’ the square-root of as much portion of it as affords a root, should be extracted. The remaining portion, divided by an optional number and then diminished by that optional number and halved, should be put equal to that square-root.”²

v2.c17.s17.t5.7

The above rule, in fact, contemplates a particular case of equation (5) in which a or c is a square number.

v2.c17.s17.t5.8

(*i*) Suppose a = p². The equation to be solved is then $p²x² + bxy + cy² = z². 5.1$

v2.c17.s17.t5.9

Therefore (px + by/2p)² + y²(c - (b²)/(4p²)) = z².

v2.c17.s17.t5.10

Putting px + by/2p = w, we get $z² - w² = y²(c - (b²)/(4p²)).$

v2.c17.s17.t5.11

Whence $z - w = λ,$ $z + w = (y²)/(λ)(c - (b²)/(4p²)),$ where λ is an arbitrary rational number. So $w = ½(y²)/(λ)(c - (b²)/(4p²)) - λ,$ as stated in the rule. Therefore, $x = 1/2p(y²)/(λ)(c - (b²)/(4p²)) - λ - by/(2p²),$

v2.c17.s17.t5.12

² *BBi*, p. 106.

v2.c17.s17.t5.13

$z = ½(y²)/(λ)(c - (b²)/(4p²)) + λ.$

v2.c17.s17.t5.14

Now, if we suppose y = m/n, where m, n are arbitrary integers, we get the solution of (5.1) as x &= 1/(8\lambda p²n²) {m²(4cp² - b²) - 4\lambda² p²n² - 4\lambda bmn}, / y &= m/n, / z &= 1/(8\lambda p²n²) {m²(4cp² - b²) + 4\lambda² p²n²}.

v2.c17.s17.t5.15

Since the given equation is homogeneous, any multiple of these values of x, y, z will also be its solution. Therefore, multiplying by 8λ p² n², we get the following solution of the equation p²x² + bxy + cy² = z² in integers : .x &= m² (4cp² - b²) - 4\lambda²p²n² - 4\lambda bmn, / y &= 8\lambda mnp², / z &= m² (4cp² - b²) + 4\lambda²p²n²,} (5.2) where m, n, are arbitrary integers.

v2.c17.s17.t5.16

In particular, putting a = b = c = 1, and λ = p = 1 in (5.2), we get x &= 3m² - 4n(n + m), / y &= 8mn, / z &= 3m² + 4n², as the solution of the equation x² + xy + y² = z².

v2.c17.s17.t5.17

Dividing out by 8n, the above solution can be put into the form $x = ½((3m²)/4n - n - m),$

v2.c17.s17.t5.18

$y = m,$ $z = 1/8n (3m² + 4n²) ;$ as has been stated by Nârâyaṇa :

v2.c17.s17.t5.19

"An arbitrary number is the first. Its square less by its (square's) one-fourth, is divided by an optional number and then diminished by the latter and also by the first. Half the remainder is the second number. The sum of their squares together with their product is a square."¹

v2.c17.s17.t5.20

¹ *GK*, i. 55.

v2.c17.s17.t5.21

It is noteworthy that in practice Nârâyaṇa approves of only integral solutions of his equation. For instance, he says :

v2.c17.s17.t5.22

" 'Any arbitrary number is the first.' Suppose it to be 12. Then with the optional number unity, are obtained the numbers (12, 95/2). For *integral values*, they are doubled (24, 95). With the optional number 2, are obtained (12, 20). It being possible, these are reduced by the common factor 4 to (3, 5). In this way, owing to the varieties of the optional number, an infinite number of solutions can be obtained."²

v2.c17.s17.t5.23

(*ii*) If neither *a* nor *c* be a square, the solution can be obtained thus :

v2.c17.s17.t5.24

Multiplying both sides of the equation (5) by *a* and then completing a square on the left-hand side, the equation transforms into $(ax + ½by)² + (ac - ¼b²)y² = az².$ Putting ax + ½by = w and β = ¼(b² - 4ac), we get w² = az² + β y². (5.3)

v2.c17.s17.t5.25

² See the example in illustration of the same.

v2.c17.s17.t5.26

The method of the solution of an equation of this form, according to Bhâskara II, has been described before.

v2.c17.s17.t5.27

Assume w = vy, z = uy; so that the values of u, v will be given by $v² = α u² + β. 5.4$

v2.c17.s17.t5.28

If u = m, v = n be a solution of (5.4), another solution will be $u = mq ± pn,$ $v = nq ± α mp;$ where α p² + 1 = q². Therefore, a solution of (5) is $x = y/2a 2(nq ± α mp) - b,$ $z = y(mq ± pn);$ where α p² + 1 = q² and α m² + β = n².

v2.c17.s17.t5.29

Put n = ar + ½b and y = s/t; then we have $x = s/2at q(2ar + b) ± 2α mp - b,$ $y = s/t,$ $z = s/2t 2mq ± p (2ar + b).$

v2.c17.s17.t5.30

Multiplying by 2at, we get the following solution of ax² + bxy + cy² = z² in integers : $x = s q (2ar + b) ± 2α mp - b,$ $y = 2as,$ $z = as2mq ± p(2ar + b);$ where α p² + 1 = q² and m² = ar² + br + c.

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