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History of Hindu Mathematics: A Source Book v2.c17.s25: 27. Multiple Equations

History of Hindu Mathematics: A Source Book v2.c17.s25

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27. Multiple Equations

v2.c17.s25.1

There are some very elegant problems in which three or more functions, linear or quadratic, of the unknowns have to be made squares or cubes. The following example occurs in the *Laghu-Bhâskarîya* of Bhâskara I¹ (522):

v2.c17.s25.2

¹ *LBh*, viii. 17.

v2.c17.s25.3

*Example 1.* To find two numbers x and y such that the expressions x + y, x - y, xy + 1 are each a perfect square.

v2.c17.s25.4

Brahmagupta gives the following solution :

v2.c17.s25.5

"A square is increased and diminished by another. The sum of the results is divided by the square of half their difference. Those results multiplied (severally) by this quotient give the numbers whose sum and difference are squares as also their product together with unity."²

v2.c17.s25.6

Thus the solution is : \begin{align*} x &= P(m² + n²), \\ y &= P(m² - n²), \end{align*} where P = ((m² + n²) + (m² - n²))/([½(m² + n²) - (m² - n²)]²), m, n being any rational numbers.

v2.c17.s25.7

² *BrSpSi*, xviii. 72.

v2.c17.s25.8

Nârâyaṇa (1357) says :

v2.c17.s25.9

"The square of the square of an optional number is set down at two places. It is decreased by the square (at one place) and increased (at another), and then doubled. The sum and difference of the results are squares and so also their product together with unity."³

v2.c17.s25.10

³ *GK*, i. 47.

v2.c17.s25.11

That is, \begin{align*} x &= 2(p⁴ + p²), \\ y &= 2(p⁴ - p²), \end{align*} where p is any rational number.

v2.c17.s25.12

Suppose $x = 2z²(m² + n²), y = 2z²(m² - n²),$ so that x ± y are already squares. We have, therefore, only to make $xy + 1 = 4z⁴(m² + n²)(m² - n²) + 1 = a square.$ Now $4z⁴(m⁴ - n⁴) + 1 = (2z²m² - 1)² + 4z²(m² - z²n⁴).$ Hence, in order that xy + 1 may be a square, one sufficient condition is $m² = z²n⁴.$ Therefore 2z² = (2m²)/(n⁴) = ((m² + n²) + (m² - n²))/([½(m² + n²) - (m² - n²)]²).

v2.c17.s25.13

Again x = 2z²(m² + n²) = 2(z⁴n⁴ + z²n²), or x = 2(p⁴ + p²), if p = zn. Therefore y = 2(p⁴ - p²).

v2.c17.s25.14

*Example 2.* "If thou be expert in mathematics, tell me quickly those two numbers whose sum and difference are squares and whose product is a cube."¹

v2.c17.s25.15

¹ *BBi*, p. 56.

v2.c17.s25.16

That is, solve $. x ± y = squares, / xy = a cube.$

v2.c17.s25.17

Bhâskara II says :

v2.c17.s25.18

"Here let the two numbers be 5z², 4z². *They are assumed such as will make their sum and difference both squares.* Their product is 20z⁴. This must be a cube. Putting it equal to the cube of an optional multiple² of 10z and removing the common factor z³ from the sides as before, (we shall ultimately find) the numbers to be 10000, 12500."

v2.c17.s25.19

² *G.K*, i. 49.

v2.c17.s25.20

In general, let us assume, as directed by Bhâskara II, x = (m² + n²)z², y = 2mnz², which will make x ± y squares. We have, therefore, only to make 2mn(m² + n²)z⁴ = a cube. Let & 2mn(m² + n²)z⁴ = p³z³. / Then & z = (p³)/(2mn(m² + n²)). / Therefore & x = ((m² + n²)p⁶)/({2mn(m² + n²)}²), / & y = (2mnp⁶)/({2mn(m² + n²)}²), where m, n, p are arbitrary.

v2.c17.s25.21

This general solution has been explicitly stated by Nârâyaṇa thus:

v2.c17.s25.22

“The square of the cube of an optional number is divided by the square of the product of the two numbers stated above and then severally multiplied by those numbers. (Thus will be obtained) two numbers whose sum and difference are squares and whose product is a cube.”¹

v2.c17.s25.23

¹ *GK*, i. 49.

v2.c17.s25.24

The two numbers stated above² are m² + n² and 2mn whose sum and difference are squares.

v2.c17.s25.25

² The reference is to rule i. 48.

v2.c17.s25.26

In particular, putting m = 1, n = 2, p = 10, Nârâyaṇa finds x = 12500, y = 10000. With other values of m, n, p he obtains the values (3165/16, 625/4), (62500/117, 250000/507), (15625/1872, 15625/2028); and observes: “thus by virtue of (the multiplicity of) the optional numbers many values can be found.”

v2.c17.s25.27

*Example 3.* To find numbers such that each of them severally added to a given number becomes a square; and so also the product of every contiguous pair increased by another given number.

v2.c17.s25.28

For instance, let it be required to find *four* numbers such that x + \alpha &= p², & xy + \beta &= \xi², / y + \alpha &= q², & yz + \beta &= \eta², / z + \alpha &= r², & zw + \beta &= \zeta². / w + \alpha &= s²,

v2.c17.s25.29

The method for the solution of a problem of this kind is indicated in the following rule quoted by Bhâskara II (1150) from an earlier writer, whose name is not known:

v2.c17.s25.30

“As many multiple (*guṇa*) as the product-interpolator (*vadha-kṣepa*) is of the number-interpolator (*râśi-kṣepa*), with the square-root of that as the common difference are assumed certain numbers; these squared and diminished by the number-interpolator (severally) will be the unknowns.”¹

v2.c17.s25.31

¹ *BBi*, p. 68.

v2.c17.s25.32

In applying this method to solve a particular problem, to be stated presently, Bhâskara II observes by way of explanation:

v2.c17.s25.33

“In these cases, that which being added to an (unknown) number makes it a square is designated as the number-interpolator. The number-interpolator multiplied by the square of the difference of the squareroots pertaining to the numbers, is equal to the product-interpolator. For the product of those two numbers added with the latter certainly becomes a square. The products of two and two contiguous of the squareroots pertaining to the numbers diminished by the

v2.c17.s25.34

number-interpolator are the square-roots corresponding to the products of the numbers."-1

v2.c17.s25.35

-1 *BBi*, p. 67.

v2.c17.s25.36

Since x = p² - α, y = q² - α, we get $xy + β = (p² - α)(q² - α) + β$ $= (pq - α)² + β - α(q - p)².$

v2.c17.s25.37

In order that xy + β may be a square, a sufficient condition is $α(q - p)² = β,$ or q = p ± √(β/α) = p ± γ, where γ = √(β/α).

v2.c17.s25.38

Then xy + β = (pq - α)².

v2.c17.s25.39

Hence ξ = pq - α.

v2.c17.s25.40

Similarly r = q ± γ, s = r ± γ.

v2.c17.s25.41

Thus, it is found that the square-roots p, q, r, s form an A.P. whose common difference is γ (= √(β/α)).

v2.c17.s25.42

Further, we have $x = p² - α,$ $y = (p ± γ)² - α,$ $z = (p ± 2γ)² - α,$ $w = (p ± 3γ)² - α,$ as stated in the rule.

v2.c17.s25.43

These values of the unknowns, it will be easily found, satisfy all the conditions about their products. For $xy + β = p(p ± γ) - α²,$ $yz + β = (p ± γ)(p ± 2γ) - α²,$ $zw + β = (p ± 2γ)(p ± 3γ) - α².$

v2.c17.s25.44

Thus we have $ξ = p(p ± γ) - α,$ $η = (p ± γ)(p ± 2γ) - α,$ $ζ = (p ± 2γ)(p ± 3γ) - α;$ as stated by Bhâskara II.

v2.c17.s25.45

It has been observed by him that the above principle is well known in mathematics. But we do not find it in the works anterior to him, which are available to us.

v2.c17.s25.46

It is noteworthy that the above principle will hold even when all the β's are not equal. For, suppose that in the above instance the second set of conditions is replaced by the following : $xy + β₁ = ξ²,$ $yz + β₂ = η²,$ $zw + β₃ = ζ².$

v2.c17.s25.47

Then, proceeding in the same way, we find that $q = p ± √(β₁/α), r = q ± √(β₂/α), s = r ± √(β₃/α),$ and ξ = pq - α, η = qr - α, ζ = rs - α.

v2.c17.s25.48

It should also be noted that in order that xy + β or p²q² - α(p² + q²) + α² + β may be a square, there may be other values of q besides the one specified above, namely q = p ± √(β/α). We may, indeed, regard $p²q² - α(p² + q²) + α² + β = ξ²$ as an indeterminate equation in q. Since we know one solution of it, namely q = p ± γ, ξ = p(p ± γ) - α, we can find an infinite number of other solutions by the method of the Square-nature.

v2.c17.s25.49

Now, suppose that another condition is imposed on the numbers, *viz.*, $wx + β' = μ².$

v2.c17.s25.50

On substituting the values of x and w this condition transforms into $p⁴ ± 6γ p³ + (9γ² + 2α)p² ± 6αγ p + α² - 9β + β' = μ²,$ an indeterminate equation of the fourth degree in p.

v2.c17.s25.51

In the following example and its solution from Bhâskara II we find the application of the above principle :

v2.c17.s25.52

*Example.* "What are those four numbers which together with 2 become capable of yielding squareroots ; also the products of two and two contiguous of which added by 18 yield square-roots ; and which are such that the square-root of the sum of all the roots added by 11 becomes 13. Tell them to me, O algebraist friend."[1]

v2.c17.s25.53

*Solution.* "In this example, the product-interpolator is 9 times the number-interpolator. The square-root of 9 is 3. Hence the square-roots corresponding to the numbers will have the common difference 3. Let them be $x, x + 3, x + 6, x + 9.$

v2.c17.s25.54

"Now the products of two and two contiguous of these minus the number-interpolator are the square-roots pertaining to the products of the numbers as increased by 18. So these square-roots are $x² + 3x - 2,$ $x² + 9x + 16,$ $x² + 15x + 52.$

v2.c17.s25.55

"The sum of these and the previous square-roots all together is 3x² + 31x + 84. This added with 11 becomes equal to 169. Hence $3x² + 31x + 95 = 0x² + 0x + 169.$

v2.c17.s25.56

¹ *BBi*, p. 67. It will be noticed that by virtue of the last condition the problem becomes, in a way, determinate.

v2.c17.s25.57

“Multiplying both sides by 12, superadding 961, and then extracting square-roots, we get $6x + 31 = 0x + 43.$ $∴ x = 2.$

v2.c17.s25.58

“With the value thus obtained, we get the values of the square-roots pertaining to the numbers to be 2, 5, 8, 11. Subtracting the number-interpolator from the squares of these, we have the (required) numbers as 2, 23, 62, 119.”

v2.c17.s25.59

*Example* 4. To find two numbers such that $x - y + k = u²,$ $x + y + k = v²,$ $x² - y² + k' = s²,$ $x² + y² + k″ = t².$

v2.c17.s25.60

Bhâskara II says :

v2.c17.s25.61

“Assume first the value of the square-root pertaining to the difference (of the numbers wanted) to be any unknown with or without an absolute number. The root corresponding to the sum will be equal to the root pertaining to the difference together with the square-root of the quotient of the interpolator of the difference of the squares divided by the interpolator for the sum or difference of the numbers. The squares of these two less their interpolator are the sum and difference of the numbers. From them the two numbers can be found by the rule of concurrence.”¹

v2.c17.s25.62

¹ *BBi*, pp. 111ff.

v2.c17.s25.63

That is to say, if *w* is any rational number, we assume $u = w ± α,$ where α is an absolute number which may be 0. Then $v = (w ± α) + √(k'/k).$ Now x² - y² + k' = (x - y)(x + y) + k' $= (u² - k)(v² - k) + k'$ $= u²v² - k(u² + v²) + k² + k'.$ One sufficient condition that the right-hand side may be a square is $k(v - u)² = k',$ or $v = u + √(k'/k),$ which is stated in the rule. Therefore, $x - y = (w ± α)² - k,$ $x + y = (w ± α + √(k'/k))² - k.$ Hence x = ½ (w ± α)² + (w ± α + √(k'/k))² - 2k, y = ½ (w ± α + √(k'/k))² - (w + α)². Now, if γ denotes √(k'/k), we get $x² + y² = u⁴ + 2γ u³ + (3γ² - 2k)u²$ $+ 2γ(γ² - k)u + ½k² + ½(γ² - k)².$ So it now remains to solve $u⁴ + 2γ u³ + (3γ² - 2k)u² + 2γ(γ² - k)u$ $+ ½k² + ½(γ² - k)² + k″ = t²,$ which is an indeterminate equation in *u*.

v2.c17.s25.64

*Applications.* We take an illustrative example with its solution from Bhâskara II.

v2.c17.s25.65

"O thou of fine intelligence, state a pair of numbers, other than 7 and 6, whose sum and difference (severally) added with 3 are squares; the sum of their squares decreased by 4 and the difference of the squares increased by 12 are also squares; half their product together with the smaller one is a cube; again the sum of all the roots plus 2 is a square."¹

v2.c17.s25.66

¹ *BBi*, p. 115.

v2.c17.s25.67

That is to say, if x > y, we have to solve $x - y + 3 = u²,$ $x + y + 3 = v²,$ $x² - y² + 12 = s²,$ $x² + y² - 4 = t²,$ $½xy + y = p³,$ $u + v + s + t + p + 2 = q².$

v2.c17.s25.68

This problem has been solved in two ways:

v2.c17.s25.69

*First Method.* As directed in the above rule, assume $u = w - 1.$ Then $x - y = (w - 1)² - 3 = w² - 2w - 2,$ $x + y = (w - 1 + 2)² - 3 = w² + 2w - 2.$ Therefore $x = w² - 2, y = 2w.$

v2.c17.s25.70

Now, we find that $x² - y² + 12 = (w² - 4)²,$ $x² + y² - 4 = w⁴,$ $½xy + y = w³.$

v2.c17.s25.71

So all the equations except the last one are already satisfied. This remaining equation now reduces to $2w² + 3w - 2 = q².$

v2.c17.s25.72

Completing the square on the left-hand side of this equation, we get $(4w + 3)² = 8q² + 25.$

v2.c17.s25.73

By the method of the Square-nature its solutions are $.q &= 5 / 4w + 3 &= 15, .q &= 175 / 4w + 3 &= 495,$ Therefore w = 3, 123, Hence the values of (x, y) are (7, 6), (15127, 246),

v2.c17.s25.74

*Second Method.* Or assume¹ $x - y + 3 = w²,$ then x + y + 3 = w² + 4w + 4 = (w + 2)². Whence x = w² + 2w - 1, y = 2w + 2.

v2.c17.s25.75

¹ This is clearly equivalent to the supposition, u = w, v = w + 2.

v2.c17.s25.76

Now, we find that $x² - y² + 12 &= (w² + 2w - 3)², / x² + y² - 4 &= (w² + 2w + 1)², / ½xy + y &= (w + 1)³.$ Then the remaining condition reduces to $2w² + 7w + 3 = q².$ Completing the square on the left-hand side, we get $(4w + 7)² = 8q² + 25.$ Whence by the method of the Square-nature, we get $.q &= 5 / 4w + 7 &= 15, .q &= 175 / 4w + 7 &= 495,$ Therefore w = 2, 122, Hence (x, y) = (7, 6), (15127, 246),

v2.c17.s25.77

Another very interesting example which has been borrowed by Bhâskara II from an earlier writer is the following:² "Tell me quickly, O sound algebraist, two numbers, excepting 6 and 8, which are such that the cube-root of half the sum of their product and the smaller one, the square-root of the sum of their squares, the square-roots of the sum and difference of them (each) increased by 2, and of the difference of their squares plus 8, all being added together, will be capable of yielding a square-root."¹

v2.c17.s25.78

² The text is *kasyâpyudâharaṇam* ("the example of some one"). This observation appears to indicate that this particular example was borrowed by Bhâskara II from a secondary source; its primary source was not known to him.

v2.c17.s25.79

¹ *BBi*, p. 110.

v2.c17.s25.80

That is to say, if x > y, we have to solve $[3]½(xy+y)+√(x²+y²)+√(x²-y²+8)$ $+√(x+y+2)+√(x-y+2)=q².$

v2.c17.s25.81

In every instance of this kind, remarks Bhâskara II, "the values of the two unknown numbers should be so assumed in terms of another unknown that all the stipulated conditions will be satisfied." In other words, the equation will have to be resolved into a number of other equations all of which have to be satisfied simultaneously. Thus we shall have to solve $x-y+2=u²,$ $x+y+2=v²,$ $x²-y²+8=s²,$ $x²+y²=t²,$ $½(xy+y)=p³,$ $u+v+s+t+p=q².$

v2.c17.s25.82

The last equation represents the original one.

v2.c17.s25.83

There have been indicated several methods of solving these equations.

v2.c17.s25.84

(*i*) Set x=w²-1, y=2w; then we find that x-y+2=(w-1)², x+y+2=(w+1)², x² - y² + 8 &= (w² - 3)², \\ x² + y² &= (w² + 1)², \\ 1/2(xy + y) &= w³. \end{aligned}

v2.c17.s25.85

So all the equations except the last one are identically satisfied. This last equation now becomes $2w² + 3w - 2 = q².$

v2.c17.s25.86

Completing the square on the left-hand side, we get $(4w + 3)² = 8q² + 25.$

v2.c17.s25.87

Solutions of this are $.q &= 5 / 4w + 3 &= 15, .q &= 30 / 4w + 3 &= 85, .q &= 175 / 4w + 3 &= 495,$

v2.c17.s25.88

Therefore, we have the solutions of our problem as $(x, y) = (8, 6), (1677/4, 41), (15128, 246),$

v2.c17.s25.89

Or set (ii) &x = w² + 2w, / y = 2w + 2; / (iii) &x = w² - 2w, / y = 2w - 2; / or (iv) &x = w² + 4w + 3, / y = 2w + 4.

v2.c17.s25.90

In conclusion Bhâskara II remarks : "Thus there may be a thousandfold artifices ; since they are hidden to the dull, a few of them have been indicated here out of compassion for them."¹

v2.c17.s25.91

¹ *BBi*, p. 110.

v2.c17.s25.92

It will be noticed that in devising the various artifices noted above for the solution of the problem, Bhâskara II has been in each case guided by the result that if u = w ± α, then, v = w ± α + √(k'/k). He has simply taken different values of α in the different cases.

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