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History of Hindu Mathematics: A Source Book v2.c17.s18: 20. Rational Triangles

History of Hindu Mathematics: A Source Book v2.c17.s18

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English

Contents

20. Rational Triangles

v2.c17.s18.1

**Rational Right Triangles: Early Solutions.** The earliest Hindu solutions of the equation $x² + y² = z² 1$ are found in the *Sulba*. Baudhâyana (*c.* 800 B.C.), Âpastamba and Kâtyâyana (*c.* 500 B.C.)¹ give a method for the transformation of a rectangle into a square, which is the equivalent of the algebraical identity $mn = (m - (m-n)/2)² - ((m-n)/2)²,$ where *m*, *n* are any two arbitrary numbers. Thus we get $(√mn)² + ((m-n)/2)² = ((m+n)/2)².$ Substituting p², q² for *m*, *n* respectively, in order to eliminate the irrational quantities, we get $p²q² + ((p²-q²)/2)² = ((p²+q²)/2)²,$ which gives a rational solution of (1).

v2.c17.s18.2

¹ *BŚl*, i. 58 ; *ÂpŚl*, ii. 7; *KŚl*, iii. 2. For details of the construction see Datta, *Sulba*, pp. 83f, 178f.

v2.c17.s18.3

For finding a square equal to the sum of a number of other squares of the same size, Kâtyâyana gives a very elegant and simple method which furnishes us with another solution of the rational right triangle. Kâtyâyana says :

v2.c17.s18.4

"As many squares (of equal size) as you wish to combine into one, the transverse line will be (equal to) one less than that ; twice a side will be (equal to) one more than that ; (thus) form (an isosceles) triangle. Its arrow (*i.e.*, altitude) will do that."²

v2.c17.s18.5

² *KŚl*, vi. 5 ; Compare also its *Pariśiṣṭa*, verses 40-1.

v2.c17.s18.6

Thus for combining n squares of sides a each, we form the isosceles triangle ABC, such that AB=AC=(n+1)a/2,

v2.c17.s18.7

Fig. 2

v2.c17.s18.8

and BC=(n-1)a. Then AD²=na². This gives the formula $a²(√n)²+a²((n-1)/2)²=a²((n+1)/2)².$

v2.c17.s18.9

Putting m² for n in order to make the sides of the right-angled triangle free from the radical, we have $m²a²+((m²-1)/2)²a²=((m²+1)/2)a²,$ which gives a rational solution of (1).

v2.c17.s18.10

Tacit assumption of the following further generalisation is met with in certain constructions described by Āpastamba:¹

v2.c17.s18.11

¹ *ĀpŚl*, v. 3, 4. Also compare Datta, *Śulba*, pp. 65f.

v2.c17.s18.12

If the sides of a rational right triangle be increased by any rational multiple of them, the resulting figure will be a right triangle.

v2.c17.s18.13

In particular, he notes $3²+4²=5²,$ $(3+3·3)²+(4+4·3)²=(5+5·3)²,$ $(3+3·4)²+(4+4·4)²=(5+5·4)²;$

v2.c17.s18.14

$5² + 12² = 13²,$ $(5 + 5 . 2)² + (12 + 12 . 2)² = (13 + 13 . 2)².$

v2.c17.s18.15

Âpastamba also derives from a known right-angled triangle several others by changing the unit of measure of its sides and vice versa.¹ In other words, he recognised the principle that if (α, β, γ) be a rational solution of x² + y² = z², then other rational solutions of it will be given by (la, lβ, lγ), where l is any rational number. This is clearly in evidence in the formula of Kâtyâyana in which a is any quantity. It is now known that all rational solutions of x² + y² = z² can be obtained without duplication in this way.

v2.c17.s18.16

¹ Datta, *Śulba*, p. 179.

v2.c17.s18.17

**Later Rational Solutions.** Brahmagupta (628) says:

v2.c17.s18.18

“The square of the optional (*iṣṭa*) side is divided and then diminished by an optional number; half the result is the upright, and that increased by the optional number gives the hypotenuse of a rectangle.”²

v2.c17.s18.19

In other words, if m, n be any two rational numbers, then the sides of a right triangle will be $m, ½((m²)/n - n), ½((m²)/n + n).$

v2.c17.s18.20

² *BrSpSi*, xii. 35.

v2.c17.s18.21

The Sanskrit word *iṣṭa* can be interpreted as implying “given” as well as “optional”. With the former meaning the rule will state how to find rational right triangles having a given leg. Such is, in fact, the interpretation which has been given to a similar rule of Bhâskara II.³

v2.c17.s18.22

³ *Vide infra* p. 211; H. T. Colebrooke, *Algebra with Arithmetic and Mensuration from the Sanscrit of Brahmegupta and Bhascara*, London, 1817, (referred to hereafter as, Colebrooke, *Hindu Algebra*), p. 61 footnote.

v2.c17.s18.23

A similar rule is given by Śrîpati (1039):

v2.c17.s18.24

“Any optional number is the side; the square of that divided and then diminished by an optional number and halved is the upright; that added with the previous divisor is the hypotenuse of a right-angled triangle. For, so it has been explained by the learned in the matter of the rules of geometry.”¹

v2.c17.s18.25

¹ *SiŚe*, xiii. 41.

v2.c17.s18.26

Karavindasvâmî a commentator of the *Āpastamba Śulba*, finds the solution $m, ((n² + 2n)/(2n + 2))m, ((n² + 2n + 2)/(2n + 2))m,$ by generalising a rule of the *Śulba*.²

v2.c17.s18.27

**Integral Solutions.** Brahmagupta was the first to give a solution of the equation x² + y² = z² in integers. It is $m² - n², 2mn, m² + n²,$ m, n being any two unequal integers.³

v2.c17.s18.28

² *ĀpŚl*, 1.2 (*Com.*); also see Datta, *Śulba*, pp. 14-16.

v2.c17.s18.29

³ *BrSpSi*, xii. 33 ; *vide infra*, p. 222.

v2.c17.s18.30

Mahâvîra (850) says :

v2.c17.s18.31

“The difference of the squares (of two elements) is the upright, twice their product is the base and the sum of their squares is the diagonal of a *generated* rectangle.”⁴

v2.c17.s18.32

⁴ *GSS*, vii. 90½.

v2.c17.s18.33

He has re-stated it thus :

v2.c17.s18.34

“The product of the sum and difference of the elements is the upright. The *saṅkramaṇa*⁵ of their squares gives the base and the diagonal. In the operation of generating (geometrical figures), this is the process.”⁶

v2.c17.s18.35

⁵ For the definition of this term see pp. 43f.

v2.c17.s18.36

⁶ *GSS*, vii. 93½.

v2.c17.s18.37

"Twice the product of two optional numbers is the upright; the difference of their squares is the side; and the sum of their squares is the hypotenuse. (Each of these quantities is) rational (and integral)."¹

v2.c17.s18.38

¹ *L*, p. 36.

v2.c17.s18.39

It has been stated before that the early Hindus recognised that fresh rational right triangles can be derived from a known one by multiplying or dividing its sides by any rational number. The same principle has been used by Mahâvîra and Bhâskara II in their treatment of the solution of rational triangles and quadrilaterals. Gaṇeśa (1545) expressly states:

v2.c17.s18.40

"If the upright, base and hypotenuse of a rational right-angled triangle be multiplied by any arbitrary rational number, there will be produced another right-angled triangle with rational sides."

v2.c17.s18.41

Hence the most general solution of x² + y² = z² in integers is $(m² - n²)l, 2mnl, (m² + n²)l$ where m, n, l are integral numbers.

v2.c17.s18.42

**Mahâvîra's Definitions.** A triangle or a quadrilateral whose sides, altitudes and other dimensions can be expressed in terms of *rational* numbers is called *janya* (meaning generated, formed or that which is generated or formed) by Mahâvîra.² Numbers which are employed in forming a particular figure are called its *bîja-saṁkhyâ* (element-numbers) or simply *bîja* (element or seed). For instance, Mahâvîra has said: "Forming O friend! the generated figure from the *bîja* 2, 3,"¹ "forming another from half the base of the figure (rectangle) from the *bîja* 2, 3,"² etc. Thus, according to Mahâvîra, "forming a rectangle from the *bîja* *m*, *n*" means taking a rectangle with the upright, base and diagonal as m² - n², 2mn, m² + n² respectively. It is noteworthy that Mahâvîra's mode of expression in this respect very closely resembles that of Diophantus who also says, "Forming now a right-angled triangle from 7, 4," meaning "taking a right-angled triangle with sides 7² - 4², 2.7.⁴, 7² + 4² or 33, 56, 65."³ It should also be noted that Mahâvîra never speaks of "right-angled triangle." What Diophantus called "forming a right-angled triangle from *m*, *n*," Mahâvîra calls "forming a longish quadrilateral or rectangle from *m*, *n*."

v2.c17.s18.43

² *GSS*, introductory line to vii. 90½. The section of Mahâvîra's work devoted to the treatment of rational triangles and quadrilaterals bears the sub-title *janya-vyavahâra* (*janya* operation) and it begins as "Hereafter we shall give out the *janya* operations in calculations relating to measurement of areas." Mahâvîra's treatment of the subject has been explained fully by Bibhutibhusan Datta in a paper entitled: "On Mahâvîra's solution of rational triangles and quadrilaterals," *BCMS*, XX, 1928-9, pp. 267-294.

v2.c17.s18.44

¹ "Bîje dve trîṇi sakhe kṣetre janye tu saṁsthâpya"—*GSS*, vii. 92½.

v2.c17.s18.45

³ *Arithmetica*, Book III, 19; T. L. Heath, *Diophantus of Alexandria*, p. 167.

Right Triangles having a Given Side

v2.c17.s18.t1.1

In the *Śulba* we find an attempt to find rational right triangles having a given side, that is, rational solutions of x² + a² = z².

v2.c17.s18.t1.2

² "He dvitribîjakasya kṣetrabhujârdhena cânyamutthâpya"—*GSS*, vii. 111½.

v2.c17.s18.t1.3

In particular, we find mention of two such right triangles having a common side *a*, *viz.*, (a, 3a/4, 5a/4) and (a, 5a/12, 13a/12).⁴ The principle underlying these solutions will be easily detected to be that of the reduction of the sides of any rational right triangle in the ratio of the given side to its corresponding side. This principle of finding rational right triangles having a given side has been followed explicitly by Mahâvîra (850).¹

v2.c17.s18.t1.4

⁴ Datta, *Śulba*, p. 180.

v2.c17.s18.t1.5

¹ *Vide infra*, p. 213

v2.c17.s18.t1.6

It has been stated before that one rule of Brahmagupta² can be interpreted as giving rational solutions of x² + a² = z² as $a, ½((a²)/n-n), ½((a²)/n+n),$ where n is any rational number. In fact, he has used this solution in finding rational isosceles triangles having a given altitude.³ This solution has been expressly stated by Mahâvîra (850). He says:

v2.c17.s18.t1.7

³ *Vide infra*, p. 223

v2.c17.s18.t1.8

"The *saṅkramaṇa* between any optional divisor of the square of the given upright or the base and the (respective) quotient gives the diagonal and the base (or upright)."⁴

v2.c17.s18.t1.9

⁴ *GSS*, vii. 97½.

v2.c17.s18.t1.10

He has restated the solution thus:

v2.c17.s18.t1.11

"The *saṅkramaṇa* between any (rational) divisor of the upright and the quotient gives the elements; or any (rational) divisor of half the side and the quotient are the elements."⁵

v2.c17.s18.t1.12

⁵ *GSS*, vii. 95½.

v2.c17.s18.t1.13

The right triangles formed according to the first half of this rule are:⁶ $a, ½((a²)/(p²)-p²), ½((a²)/(p²)+p²),$ and those according to the second half are :¹ $(a²)/(4q²)-q², a, (a²)/(4q²)+q².$

v2.c17.s18.t1.14

² *Vide supra*, p. 206.

v2.c17.s18.t1.15

⁶ The "elements" here are ½(a/p+p), ½(a/p-p), where p is an optional number.

v2.c17.s18.t1.16

¹ The "elements" here are q, a/2q, where q is an optional number.

v2.c17.s18.t1.17

Bhâskara II gives two solutions one of which is the same as that of Brahmagupta. He says :

v2.c17.s18.t1.18

"The side is given : from that multiplied by twice an optional number and divided by the square of that optional number minus unity, is obtained the upright ; this again multiplied by the optional number and diminished by the given side becomes the hypotenuse. This triangle is a right-angled triangle.

v2.c17.s18.t1.19

"Or the side is given : its square divided by an optional number is put down at two places ; the optional number is subtracted (at one place) and added (at another) and then halved ; these results are the upright and the hypotenuse. Similarly from the given upright can be obtained the side and the hypotenuse."²

v2.c17.s18.t1.20

That is to say, the two solutions are $a, 2na/(n²-1), n(2na/(n²-1))-a,$ $and a, ½((a²)/n-n), ½((a²)/n+n).$

v2.c17.s18.t1.21

Bhâskara II illustrates this by finding four right triangles having a side equal to 12, *viz.*, (12, 35, 37), (12, 16, 20), (12, 9, 15) and (12, 5, 13).³

v2.c17.s18.t1.22

³ *L*, pp. 34f.

v2.c17.s18.t1.23

The *rationale* of the first solution has been given by Sûryadâsa (1538) thus : Starting with the rational right triangle n²-1, 2n, n²+1, he observes that if x, y, z be the corresponding sides of another right triangle, then $x/(n² - 1) = y/2n = z/(n² + 1) = k.(say).$

v2.c17.s18.t1.24

² *L*, p. 34.

v2.c17.s18.t1.25

Hence x = k(n² - 1), y = 2nk, z = k(n² + 1).

v2.c17.s18.t1.26

Therefore x + z = 2kn² = ny.

v2.c17.s18.t1.27

If now we have x = a, then $k = a/(n² - 1).$

v2.c17.s18.t1.28

Hence $y = 2na/(n² - 1),$ and $z = ny - a = n(2na/(n² - 1)) - a.$

v2.c17.s18.t1.29

The second rule has been demonstrated by Sûryadâsa, Gaṇeśa and Raṅganâtha thus:

v2.c17.s18.t1.30

Since x² + a² = z², we have a² = z² - x² = (z - x)(z + x).

v2.c17.s18.t1.31

Assume z - x = n, where n is any rational number; then $z + x = (a²)/n.$ $∴ z = ½((a²)/n + n), x = ½((a²)/n - n).$

v2.c17.s18.t1.32

Generalising the method of the *Âpastamba Śulba* the commentators obtained the solution¹ $a, ((m² + 2m)/(2m + 2))a, ((m² + 2m + 2)/(2m + 2))a.$

v2.c17.s18.t1.33

¹ Datta, *Śulba*, p. 16.

v2.c17.s18.t1.34

**Right Triangles having a Given Hypotenuse.** For finding all rational right triangles having a given hypotenuse (c), that is, for rational solutions of $x² + y² = c²,$ Mahâvîra gives three rules. The first rule is :

v2.c17.s18.t1.35

“The square-root of half the sum and difference of the diagonal and the square of an optional number are they (the elements).”¹

v2.c17.s18.t1.36

¹ *GSS*, vii. 95½.

v2.c17.s18.t1.37

In other words, the required solution will be obtained from the “elements” √((c + p²)/2) and √((c - p²)/2), where p is any rational number. Hence the solution is $p², √(c² - p⁴), c.$

v2.c17.s18.t1.38

The second rule is :

v2.c17.s18.t1.39

“Or the square-root of the difference of the squares of the diagonal and of an optional number, and that optional number are the upright and the base.”²

v2.c17.s18.t1.40

That is, the solution is $p, √(c² - p²), c.$

v2.c17.s18.t1.41

These solutions are defective in the sense that √(c² - p⁴) or √(c² - p²) might not be rational unless p is suitably chosen. Mahâvîra’s third rule is of greater importance. He says :

v2.c17.s18.t1.42

² *GSS*, vii. 97½.

v2.c17.s18.t1.43

“Each of the various figures (rectangles) that can be formed from the elements are put down ; by its diagonal is divided the given diagonal. The perpendicular, base and the diagonal (of this figure) multiplied by this quotient (give rise to the corresponding sides of the figure having the given hypotenuse).”³

v2.c17.s18.t1.44

³ *GSS*, vii. 122½.

v2.c17.s18.t1.45

Thus having obtained the general solution of the rational right triangle, *viz.*, m² - n², 2mn, m² + n², Mahâvîra reduces it in the ratio c/(m² + n²), so that all rational right triangles having a given hypotenuse c will be given by $((m²-n²)/(m²+n²))c, (2mn/(m²+n²))c, c.$

v2.c17.s18.t1.46

By way of illustration Mahâvîra finds four rectangles (39, 52), (25, 60), (33, 56) and (16, 63) having the same diagonal 65.¹

v2.c17.s18.t1.47

¹ *GSS*, vii. 123-124½.

v2.c17.s18.t1.48

This method was later on rediscovered in Europe by Leonardo Fibonacci of Pisa (1202) and Vieta. It has been pointed out before that the origin of the method can be traced to the *Sulba*.

v2.c17.s18.t1.49

Bhâskara II (1150) says :

v2.c17.s18.t1.50

"From the given hypotenuse multiplied by an optional number and doubled and then divided by the square of the optional number added to unity, is obtained the upright ; this is again multiplied by the optional number ; the difference between that (product) and the given hypotenuse is the side.
"Or divide twice the hypotenuse by the square of an optional number added to unity. The hypotenuse minus the quotient is the upright and the quotient multiplied by the optional number is the side."²

v2.c17.s18.t1.51

Thus, according to the above, the sides of a right-angled triangle whose hypotenuse is c are : $2mc/(m²+1), m(2mc/(m²+1))-c, c;$ $or 2mc/(m²+1), c-2c/(m²+1), c.$

v2.c17.s18.t1.52

² *L*, pp. 35, 36.

v2.c17.s18.t1.53

By way of illustration Bhâskara II finds two right triangles (51, 68) and (40, 75) having the same hypotenuse 85.¹

v2.c17.s18.t1.54

¹ *L*, pp. 35f.

v2.c17.s18.t1.55

Sûryadâsa demonstrates the above substantially thus :

v2.c17.s18.t1.56

If (x, y, z) be the sides of the right triangle, we have $x/(m² - 1) = y/2m = z/(m² + 1) = k (say),$ where m is any rational integer. Then $x = k(m² - 1), y = 2mk, z = k(m² + 1).$ Therefore $x + z = 2km² = my.$

v2.c17.s18.t1.57

Since z is given to be equal to c, we have $k = c/(m² + 1).$ Hence $y = 2mc/(m² + 1),$ and $x = my - z = m(2mc/(m² + 1)) - c.$

v2.c17.s18.t1.58

**Problems Involving Areas and Sides.** Mahâvîra proposes to find rational rectangles (or squares) in which the area will be *numerically* (*saṁkhyayâ*) equal to any multiple or submultiple of a side, diagonal or perimeter, or of any linear combination of two or more of them. Expressed symbolically, the problem is to solve $.x² + y² = z², / mx + ny + pz = rxy; (1)$ m, n, p, r being any rational numbers (r ≠ 0). For the solution of this problem he gives the following rule :

v2.c17.s18.t1.59

"Divide the sides (or their sum) of any generated square or other figure as multiplied by their respective given multiples by the area of that figure taken into its given multiple. The sides of that figure multiplied by this quotient will be the sides of the (required) square or other figure."[^\mathcal]

v2.c17.s18.t1.60

That is to say, starting with any rational solution of x′ 2 + y′ 2 = z′ 2,2 we shall have to calculate the value of mx′ + ny′ + pz′ = Q, say.3 Then the required solution of (1) will be obtained by reducing the values of x′, y′, z′ in the ratio of Q/rx′ y′. Thus .x &= x^\prime Q/rx^\prime y^\prime = Q/ry^\prime, / y &= y^\prime Q/rx^\prime y^\prime = Q/rx^\prime, / z &= z^\prime Q/rx^\prime y^\prime.}\tag

v2.c17.s18.t1.61

Mahâvîra gives several illustrative examples some of which are very interesting:

v2.c17.s18.t1.62

"In a rectangle the area is (numerically) equal to the perimeter; in another rectangle the area is (numerically) equal to the diagonal. What are the sides (in each of these cases)?"²

v2.c17.s18.t1.63

Algebraically, we shall have to solve .x² + y² &= z², / 2(x + y) &= xy,}\tag{1.1} and .x² + y² &= z², / xy &= z.}\tag{1.2}

v2.c17.s18.t1.64

Starting with the solution s² - t², 2st, s² + t² of (2) and putting m = n = 2, p = 0, r = 1 in (4), we get the solution of (1.¹) as $(2(s² - t²) + 4st)/2st, (2(s² - t²) + 4st)/(s² - t²),$ $(2(s² - t²) + 4st)/(2st(s² - t²))(s² + t²).$

v2.c17.s18.t1.65

¹ *GSS*, vii. 112½.

v2.c17.s18.t1.66

² *GSS*, vii. 115½.

v2.c17.s18.t1.67

And putting m = n = 0, p = r = 1, in (4), we have the solution of (1.2) as $(s² + t²)/2st, (s² + t²)/(s² - t²), ((s² + t²)²)/(2st(s² - t²)).$

v2.c17.s18.t1.68

Bhâskara II solves a problem similar to the second one above:

v2.c17.s18.t1.69

Find a right triangle whose area equals the hypotenuse.¹

v2.c17.s18.t1.70

¹ *BBi*, p. 56.

v2.c17.s18.t1.71

He starts with the rational right triangle (3x, 4x, 5x); then by the condition, area = hypotenuse, finds the value x = 5/6. So that a right triangle of the required type is (5/2, 10/3, 25/6). He then observes: "In like manner, by virtue of various assumptions, other right triangles can also be found."² The general solution in this case is $(s² + t²)/st, (2(s² + t²))/(s² - t²), ((s² + t²)²)/(st(s² - t²)).$

v2.c17.s18.t1.72

² "Evamiṣṭavaśâdanye'pi"—*BBi*, p. 56.

v2.c17.s18.t1.73

Another example of Mahâvîra runs as follows:

v2.c17.s18.t1.74

"(Find) a rectangle of which twice the diagonal, thrice the base, four times the upright and twice the perimeter are together equal to the area (numerically)."³

v2.c17.s18.t1.75

³ *GSS*, vii. 117½.

Problems Involving Sides but not Areas

v2.c17.s18.t2.1

Mahâvîra also obtained right triangles whose sides multiplied by arbitrary rational numbers have a given sum. Algebraically, the problems require the solution of $.x² + y² = z², / rx + sy + tz = A;$ where r, s, t, A are known rational numbers. His method of solution is the same as that described above. Starting with the general solution of $x'² + y'² = z'²$ we are asked to calculate the value $rx' + sy' + tz' = P, say.$ Then, says Mahâvîra, the required solution is $x = x'A/P, y = y'A/P, z = z'A/P.$

v2.c17.s18.t2.2

One illustrative problem given by Mahâvîra is :

v2.c17.s18.t2.3

"The perimeter of a rectangle is unity. Tell me quickly, after calculating, what are its base and upright."¹

v2.c17.s18.t2.4

¹ *GSS*, vii. 118½.

v2.c17.s18.t2.5

Starting with the rectangle m² - n², 2mn, m² + n², we have in this case P = 2(m² - n² + 2mn). Hence all rectangles having the same perimeter unity will be given by $(m² - n²)/(2(m² - n² + 2mn)), mn/(m² - n² + 2mn),$ m, n being any rational numbers.

v2.c17.s18.t2.6

The isoperimetric right triangles will be given by $((m-n)/2m)p, np/(m+n), (m²+n²)/(2m(m+n))p ;$ where p is the given perimeter.

v2.c17.s18.t2.7

Another example is :

v2.c17.s18.t2.8

"(Find) a rectangle in which twice the diagonal, thrice the base, four times the upright and the perimeter together equal unity."²

v2.c17.s18.t2.9

² *GSS*, vii. 119½.

v2.c17.s18.t2.10

**Pairs of Rectangles.** Mahâvîra found "pairs of rectangles such that

v2.c17.s18.t2.11

(*i*) their perimeters are equal but the area of one is double that of the other, or
(*ii*) their areas are equal but the perimeter of one is double that of the other, or
(*iii*) the perimeter of one is double that of the other and the area of the latter is double that of the former."

v2.c17.s18.t2.12

These are particular cases of the following general problem contemplated in his rule:

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To find (*x*, *y*) and (*u*, *v*) representing the base and upright respectively of two rectangles which are related, such that $.2m(x + y) &= 2n(u + v), / pxy &= quv; (A)$ where *m*, *n*, *p*, *q* are known integers.

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His rule for the solution of this general problem is :

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"Divide the greater multiples of the area and the perimeter by the (respective) smaller ones. The square of the product of these ratios multiplied by an optional number is the upright of one rectangle. That diminished by unity will be its base, when the areas are equal. Otherwise, multiply the bigger ratio of the areas by that optional number and subtract unity; three times the upright diminished by this (difference) will be the base. The upright and base of the other rectangle should be obtained from its area and perimeter (thus determined) with the help of the rule, 'From the square of half the perimeter, etc.,' described before."¹

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¹ *GSS*, vii. 131½-133. The reference in the concluding line is to rule vii. 129½.

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In other words, to solve (A), assume $y = s (ratio of perimeters)(ratio of areas)², (1)$ and x = y - 1, if p = q, \hfill (2) or x = 3 [y - s (ratio of areas) - 1], if p ≠ q, \hfill (2') where s is an arbitrary number, and the ratios are to be so presented as always to remain greater than or equal to unity.

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Let m n, q p. Then we shall have to assume .y &= s (m² q²)/(n² p²), / x &= 3 (s (m² q²)/(n² p²) - s q/p + 1).} (3)

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Substituting these values in (A), we get $u + v = m/n (4 s (m² q²)/(n² p²) - 3 s q/p + 3), (4)$ $u v = 3 s (m² q)/(n² p) (s (m² q²)/(n² p²) - s q/p + 1).$

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Then $(u - v)² = (m²)/(n²) (4 s (m² q²)/(n² p²) - (9 s q)/(2 p) + 3)² + (3 s q)/(4 p) ((s q)/p - 4).$

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Now, if the arbitrary multiplier s be chosen such that $(s q)/p = 4, (5)$ we have $u - v = m/n (4 s (m² q²)/(n² p²) - (9 s q)/(2 p) + 3). (6)$

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From (4) and (6) we get .u &= m/n (4 s (m² q²)/(n² p²) - (15 s q)/(4 p) + 3), / v &= (3 s m q)/(4 n p).} (7)

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Substituting the value of s from (5) in (3) and (7) we have finally the solution of (A), when m n, q p, as

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y &= 4(m²q)/(n²p), & v &= 3m/n, \\ x &= 3(4(m²q)/(n²p) - 3), & u &= 4m/n(4(m²q)/(n²p) - 3). \end{aligned}} (I)

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Mahâvîra has observed that “when the areas are equal” we are to assume¹ y &= s(m²)/(n²), / x &= s(m²)/(n²) - 1.

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¹ Bibhutibhusan Datta has shown that this restriction is *not necessary*. In fact, starting with the assumption .y &= s(m²q²)/(n²p²), / x &= s(m²q²)/(n²p²) - 1;} m \geqslant n, q \geqslant p, and proceeding in the same way as above, he has obtained another solution of (A) in the form .y &= 2(m²q)/(n²p), & v &= m/n, / x &= 2(m²q)/(n²p) - 1, & u &= m/q(4(m²q)/(n²p) - 2).} (II) Datta finds two general solutions of (A), *viz.* .y &= (rm²q²)/(n²p²) + t, / x &= (rm²q²)/(tn²p²)((rm²q²)/(n²p²) - rq/p + t), / v &= rmq/np, / u &= m/nt((rm²q²)/(n²p²) + t)((rm²q²)/(n²p²) - rq/p + t);} (III)

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**Isosceles Triangles with Integral sides.** Brahmagupta says :

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“The sum of the squares of two unequal numbers is the side ; their product multiplied by two is the altitude, and twice the difference of the squares of those two unequal numbers is the base of an isosceles triangle.”

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*BrSpSi*, xii. 33.

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Mahâvîra gives the following rule for obtaining an isosceles triangle from a single generated rectangle:

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“In the isosceles triangle (required), the two diagonals (of a generated rectangle) are the two sides, twice its side is the base, the upright is the altitude, and the area (of the generated rectangle) is the area.”

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A rectangle generated from the numbers *m* and *n* has its sides equal to m² - n² and 2mn and its diagonal equal to m² + n². *Cf.* pp. 208-9.

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*GSS*, vii. 108½.

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Thus if *m*, *n* be two integers such that *m* ≠ *n*, the sides of all rational isosceles triangles with integral sides are :

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$(i) m² + n², m² + n², 2(m² - n²) ;$ or (ii) m² + n², m² + n², 4mn.$

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and

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.y &= (rm²q²)/(n²p²) - t, & v &= m/n(rq/p - t), / x &= (rm²q²)/(n²p²)(rq/tp - 1), & u &= rmq/np((rm²q²)/(tn²p²) - 1)} (IV)

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where m n, q p and *r*, *t* are any two integers.

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See Datta, “On Mahâvîra’s solution of rational triangles and quadrilaterals,” *BCMS*, XX, 1928-9, pp. 267-294; particularly p. 285.

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The altitude of the former is 2mn and of the latter m² - n² and the area in either case is the same, that is, 2mn(m² - n²).

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**Juxtaposition of Right Triangles.** It will be noticed that the device employed by Brahmagupta and Mahâvîra to find the above solutions is to juxtapose two rational right triangles—equal in this case—so as to have a common leg. It is indeed a very powerful device. For, every rational triangle or quadrilateral may be formed by the juxtaposition of two or four rational right triangles. So, in order to construct such rational figures, it suffices to know only the complete solution of x² + y² = z² in integers. The beginning of this principle is found as early as the *Baudhâyana Śulba*¹ (800 B.C.) wherein is described the formation of a kind of brick, called *ubhayî* (born of two), by the juxtaposition of the eighths of two suitable rectangular bricks of the same breadth (and thickness) but of different lengths.

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¹ *BŚl*, iii. 122; Compare Datta, *Śulba*, p. 45, where necessary figures are given.

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**Isosceles Triangles with a Given Altitude.** Brahmagupta gives a rule to find all rational isosceles triangles having the same altitude. He says :

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“The (given) altitude is the producer (*karaṇî*). Its square divided by an optional number is increased and diminished by that optional number. The smaller is the base and half the greater is the side.”²

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That is to say, the sides and bases of rational isosceles triangles having the same altitude a are respectively, $½((a²)/m + m), ½((a²)/m + m) and ((a² - m)/m),$ where m is any rational number.

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² *BrSpSi*, xviii. 37.

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In particular, let the given altitude be 8. Then taking m = 4 Pṛthûdakasvâmî (860) obtains the rational isosceles triangle (10, 10, 12).

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**Pairs of Rational Isosceles Triangles.** Mahâvîra gives the following rule for finding two isosceles triangles whose perimeters, as also their areas, are related in given proportions :

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“Multiply the square of the ratio-numbers of the perimeters by the ratio-numbers of the areas mutually and then divide the larger product by the smaller. Multiply the quotient by 6 and 2 (severally) and then diminish the smaller by unity : again (find severally) the difference between the results, and twice the smaller one : these are the two sets of elements for the figures to be generated. From them the sides, etc., can be obtained in the way described before.”¹

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¹ *GSS*, vii. 137.

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If (s₁, s₂) and (Δ₁, Δ₂) denote the perimeters and areas of two rational isosceles triangles, such that $s₁ : s₂ = m : n, Δ₁ : Δ₂ = p : q, (1)$ where the ratio-numbers m, n, p, q are known integers, then the triangles will be obtained, says Mahâvîra, from the rectangles *generated from* $(6(n²p)/(m²q), 2(n²p)/(m²q) - 1) and (4(n²p)/(m²q) + 1, 4(n²p)/(m²q) - 2),$ where n²p > m²q, when the dimensions of the first are multiplied by m and those of the second by n.

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The dimensions of the isosceles triangle formed from the first set of *bîja* are : $side = m (6(n²p)/(m²q))² + (2(n²p)/(m²q) - 1)²,$

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$base = 24m (n²p)/(m²q)(2 (n²p)/(m²q) - 1),$ $altitude = m(6 (n²p)/(m²q))² - (2 (n²p)/(m²q) - 1)²;$ and from the second set $side = n(4 (n²p)/(m²q) + 1)² + (4 (n²p)/(m²q) - 2)²,$ $base = 4n(4 (n²p)/(m²q) + 1)(4 (n²p)/(m²q) - 2),$ $altitude = n(4 (n²p)/(m²q) + 1)² - (4 (n²p)/(m²q) - 2)².$

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It can be easily verified that the perimeters and areas of the isosceles triangles thus obtained satisfy the conditions (1).

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In particular, putting m = n = p = q = 1, we have two isosceles triangles of sides, bases and altitudes (29, 40, 21) and (37, 24, 35) which have equal perimeters (98) and equal areas (420). This particular case was treated by Frans van Schooten the Younger (1657), J. H. Rahn (1697) and others.¹

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¹ Dickson, *Numbers*, II, p. 201.

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It is evident that multiplying the above values by m⁴q² we get pairs of isosceles triangles whose dimensions are integral.

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**Rational Scalene Triangles.** Brahmagupta says:

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“The square of an optional number is divided twice by two arbitrary numbers; the moieties of the sums of the quotients and (respective) optional numbers are the sides of a scalene triangle; the sum of the moieties of the differences is the base.”²

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² *BrSpSi*, xii. 34.

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That is to say, the sides of a rational scalene triangle are $½((m²)/p+p),½((m²)/q+q),½((m²)/p-p)+½((m²)/q-q)$ where m, p, q are any rational numbers. The altitude (m), area and segments of the base of this triangle are all rational.

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Mahāvîra gives the rule :

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“Half the base of a derived rectangle is divided by any optional number. With this divisor and the quotient is obtained another rectangle. The sum of the uprights (of these two rectangles) will be the base of the scalene triangle, the two diagonals, its sides and the base (of either rectangle) its altitude.”

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If m, n be any two rational numbers, the rational rectangle (AB′ BH)

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Fig. 3

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Fig. 4

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formed from them is $m²-n², 2mn, m²+n².$

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If p, q be any two rational factors of mn, that is, if mn=pq, the second rectangle (AC′ CH) is $p²-q², 2pq, p²+q².$

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¹ *GSS*, vii. 110½.

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Now, juxtaposing these two rectangles so that they do not overlap (Fig. 3), the sides of the rational scalene triangle are obtained as $p² + q², m² + n², (p² - q²) + (m² - n²),$ where mn = pq. Evidently the two rectangles can be juxtaposed so as to overlap (Fig. 4). So the general solution will be $p² + q², m² + n², (p² - q²) ± (m² - n²).$ The altitude of the rational scalene triangle thus obtained is 2mn or 2pq, its area pq(p² - q²) ± mn(m² - n²) and the segments of the base are p² - q² and m² - n².

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In particular, putting m = 12, p = 6, q = 8 in Brahmagupta's general solution, Pṛthûdakasvâmî derives a scalene triangle of which the sides (13, 15), base (14), altitude (12), area (84) and the segments of the base (5, 9) are all integral numbers.

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In order to get the above solutions of the rational scalene triangle the method employed was, it will be noticed, the juxtaposition of two rational right triangles so as to have a common leg. In Europe, it is found to have been employed first by Bachet (1621). The credit for the discovery of this method of finding rational scalene triangles should rightly go to Brahmagupta (628), but not to Bachet as is supposed by Dickson.¹

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¹ Dickson, *Numbers*, II, p. 192.

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**Triangles having a Given Area.** Mahâvîra proposes to find all triangles having the same given area A. His rules are :

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“Divide the square of four times the given area by three; The quotient is the square of the square of a side of the equilateral triangle.”\∗2

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\∗2 *GSS*, vii. 154½.

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"Divide the given area by an optional number; the square-root of the sum of the squares of the quotient and the optional number is a side of the isosceles triangle formed. Twice the optional number is its base and the area divided by the optional number is the altitude."¹

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¹ *GSS*, vii. 156½.

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"The cube of the square-root of the sum of eight times the given area and the square of an optional number is divided by the product of the optional number and that square-root; the quotient is diminished by half the optional number which is the base (of the required triangle). The *saṅkramaṇa* between this remainder and the quotient of the square of the optional number divided by twice that square-root will give the two sides."²

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The last rule has been re-stated differently.³

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