History of Hindu Mathematics: A Source Book v2.c17.s22: 24. Double Equations of the First Degree

History of Hindu Mathematics: A Source Book v2.c17.s22

Language: English

24. Double Equations of the First Degree

v2.c17.s22.1

The earliest instance of the solution of the simultaneous indeterminate quadratic equation of the type $.x ± a &= u², / x ± b &= v²,$ is found in the Bakhshâlî treatise. The portion of the manuscript containing the rule is mutilated. The example given in illustration can, however, be restored as follows :

v2.c17.s22.2

"A certain number being added by five {becomes capable of yielding a square-root} ; the same number {being diminished by} seven becomes capable of yielding a square-root. What is that number is the question."∗

v2.c17.s22.3

∗ *BMs*, Folio 59, recto.

v2.c17.s22.4

That is to say, we have to solve $√(x + 5) = u, √(x - 7) = v.$ The solution given is as follows :

v2.c17.s22.5

"The sum of the additive and subtractive is | 12 | ; its half | 6 | ; minus two | 4 | ; its half is | 2 | ; squared | 4 |. 'Should be increased by the subtractive' ; {the subtractive is} | 7 | ; adding this we get | 11 |. This is the number (required)."

v2.c17.s22.6

From this it is clear that the author gives the solution of the equations $x + a = u², x - b = v²;$ as $x = ½ ( (a + b)/m - m ) ² + b,$ where m is any integer.¹

v2.c17.s22.7

¹ In the above solution m is taken to be 2.

v2.c17.s22.8

Brahmagupta (628) gives the solution of the general case. He says:

v2.c17.s22.9

"The difference of the two numbers by the addition or subtraction of which another number becomes a square, is divided by an optional number and then increased or decreased by it. The square of half the result diminished or increased by the greater or smaller (of the given numbers) is the number (required)."²

v2.c17.s22.10

i.e., x &= { 1/2 ( (a - b)/m ± m ) }² \mp a, / or x &= { 1/2 ( (a - b)/m \mp m ) }² \mp b, where m is an arbitrary integer.

v2.c17.s22.11

The *rationale* is very simple. Since $u² = x ± a, v² = x ± b,$ we have $u² - v² = ± a b.$ Therefore $u - v = m,$ and $u + v = (± a b)/m,$ where m is arbitrary. Hence $u = ½ ( (± a b)/m + m ) = ± ½ ( (a - b)/m ± m ).$

v2.c17.s22.12

² *BrSpSi*, xviii. 74.

v2.c17.s22.13

Since it is obviously immaterial whether u is taken as positive or negative, we have $u = ½((a - b)/m ± m).$

v2.c17.s22.14

Similarly $v = ½((a - b)/m m).$

v2.c17.s22.15

Therefore $x = ½((a - b)/m ± m)² a,$ $or x = ½((a - b)/m m)² b,$ where m is an arbitrary number.

v2.c17.s22.16

Brahmagupta gives another rule for the particular case : $x + a = u²,$ $x - b = v².$

v2.c17.s22.17

"The sum of the two numbers the addition and subtraction of which make another number (severally) a square, is divided by an optional number and then diminished by that optional number. The square of half the remainder increased by the subtractive number is the number (required)."¹

v2.c17.s22.18

¹ *BrSpŚi*, xviii. 73.

v2.c17.s22.19

*i.e.*, $x = ½((a + b)/m - m)² + b.$

v2.c17.s22.20

Nârâyaṇa (1357) says :

v2.c17.s22.21

"The sum of the two numbers by which another number is (severally) increased and decreased so as to make it a square is divided by an optional number and then diminished by it and halved; the square of the result added with the subtrahend is the other number."²

v2.c17.s22.22

² *GK*, i. 52.

v2.c17.s22.23

He further states :

v2.c17.s22.24

“The difference of the two numbers by which another number is increased twice so as to make it a square (every time), is increased by unity and then halved. The square of the result diminished by the greater number is the other number.”

v2.c17.s22.25

$i.e., x = ((a - b + 1)/2)² - a$ is a solution of $x + a = u², x + b = v², a > b.$

v2.c17.s22.26

“The difference of the two numbers by which another number is diminished twice so as to make it a square (every time), is decreased by unity and then halved. The result multiplied by itself and added with the greater number gives the other.”

v2.c17.s22.27

$i.e., x = ((a - b - 1)/2)² + a$ is a solution of $x - a = u², x - b = v², a > b.$

v2.c17.s22.28

The general case $.ax + c = u², / bx + d = v²,$ has been treated by Bhâskara II. He first lays down the rule :

v2.c17.s22.29

² *GK*, i. 54.

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“In those cases where remains the (simple) unknown with an absolute number, there find its value by forming an equation with the square, etc., of another unknown plus an absolute number. Then proceed to the solution of the next equation comprising the simple unknown with an absolute number by substituting in it the root obtained before.”

v2.c17.s22.31

¹ *GK*, i. 53.

v2.c17.s22.32

³ *BBi*, pp. 117-8.

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He then proceeds to explain it further :

v2.c17.s22.34

“In those cases where on taking the square-root of the first side, there remains on the other side the (simple) unknown with or without an absolute number, find there the value of that unknown by forming an equation with the square of another unknown plus an absolute number. Having obtained the value of the unknown in this way and substituting that value (in the next equation) further operations should be proceeded with. If, however, on equating the root of the first with another unknown plus an absolute number, no further operations remain to be done, then the equation has to be made with the square, etc., of a known number.”

v2.c17.s22.35

(*i*) Set u = mw + α; then substituting in the first equation, we get $x = 1/a(m² w² + 2mwa + α² - c).$

v2.c17.s22.36

Substituting this value of x in the next equation, we have $b/a(m² w² + 2mwa + α² - c) + d = v², 1.1$ which can be solved by the method of the Square-nature.

v2.c17.s22.37

(*ii*) In the course of working out an example¹ Bhâskara II is found to have followed also a different procedure, which was subsequently adopted by Lagrange.²

v2.c17.s22.38

¹ *Vide infra*, p. 265.

v2.c17.s22.39

Eliminate x between the two equations. Then $bu² + (ad - bc) = av²,$ $or abu² + k = w², 1.2$ where w = av, k = a² d - abc.

v2.c17.s22.40

² *Addition to Euler's Algebra*, p. 547.

v2.c17.s22.41

If u = r, w = s be a solution of this transformed equation, another solution of it will be $u = rq ± ps,$ $w = qs ± abrp ;$ where abp² + 1 = q². Therefore, the general solution of (1) is $x = 1/a(rq ± ps)² - c/a,$ $u = rq ± ps,$ $v = 1/a(qs ± abrp) ;$ where abp² + 1 = q² and abr² + a²d - abc = s².

v2.c17.s22.42

Now, a rational solution of the equation abp² + 1 = q² is p = 2t/(t² - ab), q = (t² + ab)/(t² - ab), where t is any rational number. Therefore, the above general solution becomes .x &= 1/(a(t² - ab)²)l{ r(t² + ab) ± 2st r}² - c/a, / u &= 1/((t² - ab))l{ r(t² + ab) ± 2st r}, / v &= 1/(a(t² - ab))l{ s(t² + ab) ± 2abrt r}, } (1.3) where abr² + a²d - abc = s².

v2.c17.s22.43

(*iii*) Suppose c and d to be squares, so that c = α², d = β². Then we shall have to solve $ax + α² = u²,$ $bx + β² = v².$

v2.c17.s22.44

The auxiliary equation abr² + a²d - abc = s² in this case becomes abr² + (a²β² - aba²) = s². The same equation is obtained by proceeding as in case (*i*) with the assumption v = by + β. An obvious solution of it is r = α, s = aβ. Hence in this case the general solution (1.3) becomes x &= 1/(a(t² - ab)²){\alpha(t² + ab) ± 2a\beta t}² - (\alpha²)/c, / u &= 1/((t² - ab)){\alpha(t² + ab) ± 2a\beta t}, / v &= 1/((t² - ab)){\beta(t² + ab) ± 2bat}, where t is any rational number.

v2.c17.s22.45

Putting α = β = 1, t = a, and taking the positive sign only, we get a particular solution of the equations .ax + 1 = u² / bx + 1 = v²} as x = (8(a + b))/((a - b)²), u = (3a + b)/(a - b), v = (a + 3b)/(a - b).

v2.c17.s22.46

This solution has been stated by Brahmagupta (628):

v2.c17.s22.47

"The sum of the multipliers multiplied by 8 and divided by the square of the difference of the multipliers is the (unknown) number. Thrice the two multipliers increased by the alternate multiplier and divided by their difference will be the two roots."¹

Linked texts (4)

Brāhmasphuṭasiddhānta — Brahmagupta, Sanskrit
Bījagaṇita — Bhāskara II, Sanskrit
Bakhshālī Manuscript — Sanskrit
Gaṇitakaumudī — Nārāyaṇa Paṇḍita (mathematician), Sanskrit