ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

PRINCIPLE OF COMPOSITION 249 Principle of Composition The above results have been technically known amongst Indian algebraists as Bhāvanā (demonstrated or proved, hence theorem or lemma). The word bhāvanā also means "composition or combination" in algebra. Bhāvanā may be of two types : Samāsa Bhāvanā (or addition Lemma, or additive composition) and Antara Bhāvanā (or subtraction Lemma or subtractive composition). Whenever, again, the bhāvanā is made with two equal sets of roots and interpolators, it is technically named as Tulya Bhāvanā (or composition of equals), and when with two unequal sets of values then it is known as Atulya Bhāvanā (or composition of unequals). Proof of Brahmagupta's Lemmas It is significant to be indicated that Brahmagupta's Lemmas were rediscovered by Euler in 1764 and by Lagrange in 1768, and a considerable importance was attached to them. Kṛṣṇa, (1580 A.D.) the commentator on the Bījagaṇita of Bhāskara II gives the following proof of Brahmagupta's Lemmas : Let (α,β) and (α',β') be the two solutions of the equation nx² + k = y². we have Nα² + k = β² Nα'² + k' = β'² Multiplying the first equation by β'², we get Nα²β'² + kβ'² = β²β'² Now, substituting the value of factor β²' of the interpolator from the second equation, we get Nα² β'² + k (Nα'² + k') = β²β'² or N(α²β'² + Nkα'² + kk' = β²β'² Again, substituting the value of k from the first equation in the second term of the left-hand side expression, we have Nα²β'² + Nα'²(β² - Nα²) + kk' = β²β'² or N(α²β'² + α'²β²) + kk' = β²β'² + N²α²α'² Adding ±2Nαβα'β' to both sides, we get N(αβ' ± α'β)² + kk' = (ββ' ± Nαα')²

250 BRAHMAGUPTA AS AN ALGEBRAIST Brahmagupta's Corollary also follows at once from the above by putting α'=α, β'=β and k'=k. N (2αβ)²+k²=(β²±Nα²)² Thus the roots are x=2αβ and y=β²±Nα² which is the Corollary. It would be seen that modern historians of mathematics are incorrect when they say that Fermat (1657) was the first to state that the equation Nx²+1=y², where N is a non-square integer has an unlimited number of solutions in integers. For this assertion, history takes us to the early Seventh Century A.D. when Brahmagupta wrote his classical treatise, the Brāhmasphu- tasiddhānta, and gave the well known two Lemmas and the Corollary to the first Lemma. Second Lemma of Brahmagupta In the Brāhmasphuṭa siddhānta, we find another important Lemma by Brahmagupta stated as follows : On dividing the two roots (of a square- Nature) by the square-root of its additive or subtractive, the roots for interpolator unity (will be found).¹ This Lemma when expressed in the modern language of algebra would mean that if x=α, y=β be a solution of the equation. Nx²+k²=y² then x=α/k, y=β/k is a solution of the equation Nx²+1=y². This rule, at another place, has been re-enunciated as follows : If the interpolator is that divided by a square then the roots will be those multiplied by its square- root.²

1. प्रक्षेपशोधक हृते मूले प्रक्षेपके रूपे । —BrSpSi. XVIII. 65
2. वर्गाच्छिन्ने क्षेपे तत्पदगुणिते तदा मूले । —BrSpSi. XVIII. 70

SECOND LEMMA OF BRAHMAGUPTA 251 This rule may be expressed in terms of symbols as follows. Suppose the Varga-prakṛti (Square-nature) to be Nx² ± p²d = y², so that its interpolator (kṣepa) p²d is exactly divisible by the square p². Then, putting therein u = x/p, v = y/p, we derive the equation Nu² ± d = v² whose interpolator is equal to that of the original Square-nature divided by p². It is clear that the roots of the original equation are p times those of the derived equation. Rational Solution Indian algebraists have usually suggested the following method to obtain a first solution of Nx² + 1 = y² : Take an arbitrary small rational number, α, such that its square multiplied by the guṇaka N and increased or diminished by a suitably chosen rational number k will be an exact square. In other words, we shall have to obtain empirically a rela- tion of the form Nα² ± k = β² where α, k, and β are rational numbers. Let us call this relation as the Auxiliary Equation. Then by Brahmagupta's Coro- llary, we get from it the relation N(2αβ)² + k² = (β² + Nα²)², or N(2α β / k)² + 1 = ((β² + Nα²) / k)² Hence, one rational solution of the equation Nx² + 1 = y² is given by x = 2αβ / k , y = (β² + Nα²) / k Work on the rational solution of the Square-nature has been also done by Śrīpati. In fact, his solution, given in 1039 A.D. is of historical significance. He derives the rational solution without the aid of the "auxiliary equation." He gives the follo- wing rule :

252 BRAHMAGUPTA AS AN ALGEBRAIST Unity is the lesser root. Its square multiplied by the prakṛti is increased or decreased by the prakṛti com- bined with an (optional) number whose square-root will be the greater root. From them will be obtained two roots by the Principle of Composition¹ Thus if m² be the rational number optionally chosen, one shall have the identity : N.1²+(m²—N)=m², or N.1²—(N—m²)=m² Then by applying Brahmagupta's Corollary we get N(2m)²+(m² ∼ N)²=(m²+N)² ∴ N (2m / (m² ∼ N))² + 1 = ((m² + N) / (m² ∼ N))² Hence x = 2m / (m² ∼ N), y = (m² + N) / (m² ∼ N) where m is any rational number, is a solution of the equation Nx² + 1 = y². This rational solution of the varga-prakṛti which was used by Śrīpati in 1039 A.D. was rediscovered in Europe by Broun- cker in 1657. We shall close this discussion by taking an illustration from Bhāskara II : Problem : Tell me, O mathematician, what is that square which multiplied by 8 becomes, together with unity, a square; and what square multiplied by 11 and increased by unity, becomes a square. This means that we have to solve the equations : 8x² + 1 = y² ......(i) 11x² + 1 = y² ......(ii) In the second example, let us assume 1 as the lesser root. Following the method of Śrīpati, let us multiply its square by the prakṛti (here in eq. ii, prakṛti is 11), then let us subtract 2 (an optional number) and then extracting the square-roots we

1. Śrīpati, Siddhānta-śekhara XIV. 33

RATIONAL SOLUTION 253 get the greater root as 3. Hence the statement for the com- position is m=11 l=1 g=3 i=-2 l=1 g=3 i=-2 Here m=multiplier (guṇaka or prakṛti), l=lesser root (kaniṣṭha-mūla), g=greater root (jyeṣṭha-mūla) and i=interpola- tor (kṣepa). Here we have set down successively the lesser root, greater root and interpolator, and below them again set down the same (See Brahmagupta's Lemmas described by Bhāskara II). Now proceeding as before we obtain the roots for the additive 4 : l=6, g=20, (for) i=4. Then by the rule : "If the interpolator (of a varga-prakṛti or Square-nature) divided by the square of an optional number be the interpolator (of another Square-nature), then the two roots (of the former) divided by that optional number will be the roots (of the other). Or, if the interpolator be multiplied, their roots should be multiplied."¹ are found the roots for the additive unity l=3, g=10 (for) i=1. Whence by the Principle of Composition of Equals, we get the lesser and greater roots : l=60, g=199 (for) i= 1. In this way an infinite number of roots can be deduced. Alternative method:—Bhāskara II has given another method for finding the two roots for the additive unity : Or divide twice an optional number by the difference between the square of that optional number and the prakṛti. This (quotient) will be the lesser root (of a Square-nature) when unity is the additive. From that (follows) the greater root.²

1. इष्टवर्गहृतः क्षेपः क्षेपः स्यादिष्टभाजिते । मूले ते स्तोऽथवा क्षेपः क्षुण्णः क्षुण्णे तदा पदे ॥ Bījagaṇita II. 5.
2. Siddhānta-śekhara, XIV. 32.

254 BRAHMAGUPTA AS AN ALGEBRAIST Let us solve the first example 8x²＋1＝y². We assume the optional number to be 3. Its square is 9; the prakṛti of multiplier is 8, their difference is 9–8＝1. Dividing by this twice the optional number (2×3, i.e. 6), namely 6, we get the lesser root for the addi- tive unity as 6. Whence proceeding as before, we get the greater to be 17. Thus here x＝6 and y＝17. Let us use this method for the equation 11x²＋1＝y². Let the optional number be 3. Its square is 9: multiplier or prakṛti is 11; the difference is 11–9＝2; dividing by this twice the optional number (2×3), namely 6, we get 6/2＝3, which is the lesser root. Consequently the greater root would be 10. Thus for this equation x＝3 and y＝10. Solution in Positive Integers The Indian algebraists usually aimed at obtaining solutions of the varga-prakṛti or Square-nature in positive integers or abhinna. The tentative methods of Brahmagupta and Śrīpati always did not furnish solutions in positive integers. These auth- ors, however, discovered that if the interpolator of auxiliary equa- tion in the tentative method be ±1, ±2 or ±4, an integral solu- tion of the equation Nx²＋1＝y² can always be found. Thus Śrīpati says : If 1, 2 or 4 be the additive or subtractive (of the auxi- liary equation), the lesser and greater roots will be integral (abhinna)¹. (i) If k＝±1, then the auxiliary equation will be Nα²±1＝β where α and β are intergers. Then by Brahmagupta' Corollary we get x＝2αβ and y＝β²＋Nα² as the required first solution in positive integers of the equation Nx²＋1＝y²

1. इष्टवर्गं प्रकृत्योर्यद्विवरं तेन वा भजेत् । द्विघ्नमिष्टं कनिष्ठं तत् पदं स्यादेक संयुतौ । ततो ज्येष्ठमिहानन्त्यं भावनाभिस्तथेष्टतः ।। Bījagaṇita, Varga-Prakṛti, 5-6

Here is the complete line-by-line transcription of the page into clean Markdown, preserving all text, mathematical expressions (without LaTeX math mode), and Devanagari Sanskrit:

SOLUTION IN POSITIVE INTEGERS 255 (ii) Let k=±2; then the auxiliary equation is Nα²±2=β² By Brahmagupta's Corollary, we have N(2αβ)²+4=(β²+Nα²)² or N (αβ)²+1=( (β²+Nα²)/2 )² Hence the required first solution is x=αβ, y=½(β²+Nα²) Since Na²=β² ∓ 2, we have ½ (β²+Nα²)=β² ∓ 1=a whole number. (iii) Now suppose k=+4: so that Nα²+4=β² With an auxiliary equation like this, the first integral solution of the equation Nx²+1=y² is x=½αβ y=½(β²-2); if α is even; or x=½

256 BRAHMAGUPTA AS AN ALGEBRAIST Substituting the value of N in the right-hand side expres- sion from (i), we have N.( αβ / 2 )² + 1 = ( (β²-2) / 2 )² (iii) Composing (ii) and (iii), N { α/2 (β²-1) }² + 1 = { β/2 (β²-3) }² Hence x= ½ αβ, y= ½ (β²--2); and x= ½α(β²-1), y= ½β(β²-3); are solutions of Nx²+1=y². If β be even, the first values of (x,y) are integral. If β be odd, the second values are integral. (iv) Finally, suppose k=-4; the auxiliary equation is Nα²-4 = β² Then the required first solution in positive integers of Nx²+1=y² is x= ½αβ(β²+3) (β²+1) y=(β²+2) { ½(β²+3) (β²+1)-1 }. Brahmagupta says : In the case of 4 as subtractive, the square of the second is increased by three and by unity; half the product of these sums and that as diminished by unity (are obtained). The latter multiplied by the first sum less unity is the (required) second root; the former multi- plied by the product of the (old) roots will be the first root corresponding to the (new) second root.¹ The rationale of this solution, as given by Datta and Singh is as follows : Nα²-4=β² (i) N(α/2)²-1=(β/2)² Hence by Brahmagupta's Corollary, we get N ( αβ / 2 )² + 1 = ( β²/4 + N α²/4 )²

1. चतुरूनेऽन्यपद कृती त्र्येकयुते वधदलं पृथग्व्येकम् । व्येकाद्वाहतमन्त्यं पदवध गुणमाद्यमान्त्यपदम् ॥ BrSpSi. XVIII. 68

CAKRAVĀLA OR CYCLIC METHOD 257 = {½(β² + 2)}² (ii) Again applying the Corollary, we get N {½αβ(β² + 2)}² + 1 = {½(β⁴ + 4β² + 2)}² (iii) Now by the Lemma we obtain from (ii) and (iii) N {½αβ(β² + 3) (β² + 1)}² + 1 = [(β² + 2){½(β² + 3) (β² + 1) − 1}]² Hence x = ½αβ(β² + 3) (β² + 1), y = (β² + 2){½(β² + 3) (β² + 1) − 1} is a solution of Nx² + 1 = y² This can be proved without difficulty that these values of x and y are integral. Since if β is even, β² + 2 is also even. And hence the above values of x and y are integral. On the other hand, if β is odd, β² is also odd; under these conditions β² + 1 and β² + 3 are even. In this also, therefore, the above values must be integral. Putting p = αβ, q = β² + 2, we can write the above solution in the form x = ½p (q² − 1). y = ½q (q² − 3). This was the form in which the solution was found by Euler. Cakravāla or Cyclic Method We have shown in the preceding articles that the most fundamental step in Brahmagupta's method for the general solution in positive integers of the equation Nx² + 1 = y² where N is a non-square integer, is to form an auxiliary equation of the kind Na² + k = b² where a and b are positive integers and k = ±1, ±2 or ±4. From this auxiliary equation, by the Principle of Composition, applied repeatedly whenever necessary, one can derive, as we have alrea- dy shown above, one positive integral solution of the original Varga-prakṛti or Square-Nature. And thence again, by means of the same principle, an infinite number of other solutions in integers can be obtained. How to form an auxiliary equation of

258 BRAHMAGUPTA AS AN ALGEBRAIST this type was a problem, write Datta and Singh, which could not be solved completely nor satisfactorily by Brahmagupta. In fact, Brahmagupta had to depend on trial. Success in this direc- tion was, however, remarkably attained by Bhāskara II. He evol- ved a simple and elegant method which assisted in deriving an auxiliary equation having the required interpolator ±1, ±2, or ± 4, simultaneously with its two integral roots, from another auxi- liary equation empirically formed with any simple integral value of the interpolator, positive or negative. This method has been technically known as Cakravāla or the cyclic method. This is so called because it proceeds as in a circle, the same set of opera- tions being applied again and again in a continuous round. For the details of this method, our reader is requested to consult the Algebra of Bhāskara II and the narrative on this method as given by Datta and Singh under the title "Cyclic Method" in their History of Hindu Mathematics: Algebra, 1962 Edition, pp. 161-72. Solution of Indeterminate Quadratic Equation It is remarkable to see that Brahmagupta was the first algebraist in the history of mathematics to find a general solu- tion of the indeterminate quadratic equation Nx²±c=y² in positive integers. We have the following verse in the Brāhmasphuṭasiddhānta in this connection: From two roots (of a Square-nature or varga-prakṛti) with any given additive or subtractive, by making (combination) with the roots for the additive unity other first and second roots (of the equation having) the given additive or subtractive (can be found).¹ Let us take the following two equations: a₁k=an+b; and b₁k=bn+Na From them we get : by eliminating n a₁b−ab₁=1

1. रूप प्रक्षेपपदे पृथगिष्टक्षेपशोध्यमूलाभ्याम् । कृत्वाऽऽन्त्याथपदे ये प्रक्षेपे तेने ॥ —BrSpSi. XVIII. 66

SOLUTION OF INDETERMINATE 259 Hence b₁ = (a₁b - 1) / a = a whole number. Now n² - N = ((a₁k - b)² - Na²) / a² = (a₁²k² - 2bka₁ + k) / a² = k(a₁²k - 2ba₁ + 1) / a² Therefore (k / a²)(a₁²k - 2ba₁ + 1) is a whole number. Since a, k have no common factor, it follows that (a₁²k - 2ba₁ + 1) / a² = (n² - N) / k = k₁ = an integer. Also k₁ = (n² - N) / k = (a₁²k - 2ba₁ + 1) / a² = (a₁²(b² - Na²) - 2ba₁ + 1) / a² = ((a₁b - 1) / a)² - Na₁². Thus having known a single solution in positive integers of the equation Nx² ± c = y², says, Brahmagupta, an infinite number of other integral solutions can be obtained by making use of the integral solutions of Nx² + 1 = y². If (p, q) be a solution of the former equation found empirically and if (α, β) be an integral solution of the latter, then by the principle of Composition x = pβ ± qα; y = qβ ± Npα will be a solution of the former. Repeating the operations, we can easily deduce as many solutions as we like. FORM Mn²x² ± c = y² : In this connection, Brahmagupta says : If the remainder is that divided by a square, the first root is that divided by its root¹. This seems to mean that if we have the equation Mn²x² ± c = y² (i) such that the multiplier (i.e. the coefficient of x²) is divisible

1. वर्गेच्छिन्ने गुणके प्रथमं तन्मूल भाजितं भवति । —BrSpSi. XVIII.70

260 BRAHMAGUPTA AS AN ALGEBRAIST by n², then we are justified in saying that if we put nx=u, the equation (i) becomes Mu²±c=y² (ii), and clearly the first root of (i) is equal to the first root of (ii) divided by n. The corresponding second root will be the same for both the equations. FORM a²x²±c=y² : We find Brahmagupta giving the following rule in this connection : This is a solution of a particular form of a varga-prakṛti or Square-nature. If the multiplier be a square, the interpolator divided by an optional number and then increased and decreased by it, is halved. The former (of these results) is the second root; and the other divided by the square-root of the multiplier is the first root.¹ Thus the solutions of the equation a²x²±c=y² are : x = 1/(2a) (±c/m - m) y = 1/2 (±c/m + m) where m is an arbitrary number. Bhāskara II and Nārāyaṇa have also given the same solutions as proposed by Brahmagupta. Rational Geometrical Figures In the days of the Taittirīya Saṁhitā and the Śatapatha Brāhmaṇa, Indian mathematicians got familiarity with the solution of such equations x²+y²=z² and the results were arrived geometrically on the basis of the law of rectangle as propounded by Baudhāyana in the Śulba Sūtras and which goes by his name. The reader is referred to the Chapter on Baudhāyana, the first Geometer in the author's "Founders of Sciences in Ancient India". Baudhāyana (c. 800 B.C.) gave a

1. वर्गे गुणके क्षेपः केनचिदुद्धृतयुतोनितो दलितः । प्रथमोऽन्यन्मूलमन्त्यो गुणकारपदोद्धृतः प्रथमः ॥ —BrSpSi. XVIII. 69

, is there a space or a matra? Let's compare त in कोटिः or तद्दलं: त has a left loop. In आयतचतुरस्रस्य: आ य त च तु र स्र स्य. No, it's आयतचतुरस्रस्य. Wait, let me re-examine: आयताचतुरस्रस्य? Wait, look at त: the right side of त goes down, then there is a च. It's आयतचतुरस्रस्य. Wait, what about चेष्टस्येष्टाधिका? Let's look at चे ष्टस्येष्टाधिका: चे has a space after it? Wait, look at the footnote line 1: 1. इष्टस्य भुजस्य कृतेर्भक्तो नेष्टेन तद्दलं कोटिः । Line 2: आयतचतुरस्रस्य चे ष्टस्येष्टाधिका कर्णः ॥ —BrSpSi. XII. 35 Wait, look at चे ष्टस्येष्टाधिका: There is चे, then a space, then ष्टस्येष्टाधिका. Wait, could it be चेष्टस्येष्टाधिका with just a broken space? Yes, in De

262 BRAHMAGUPTA AS AN ALGEBRAIST 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.¹ Thus if m,n be two integers such that m is not equal to n, the sides of all rational isosceles triangles with integral sides are given by m² + n², m² + n², 2(m² - n²) and the altitude of the triangle is 2mn. This method was also followed by Mahāvīra and other Indian mathematicians. In fact, their solutions are based on the juxtaposition of two rational right triangles, equal so that they have a common leg. It is remarkably a powerful device, for every rational triangle or quadrilateral may be formed by the juxtaposition of two or four rational right triangles. Isosceles Triangles with a Given Altitude Here we have a rule given by Brahmagupta for finding out all rational isosceles triangles possessing the same altitude : The (given) altitude is the producer (karaṇī). 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 Thus if m be any rational number then for a given definite altitude a, the sides of the rational isosceles triangles are ½ (a²/m + m) each and the base is (a² - m²)/m . We shall illustrate it by an example taken from the commentary of Pṛthūdaka Svāmī The given altitude is 8; let us take any rational number m = 4 then the two equal sides of the isosceles are given by ½ ((8² + 4²)/4) = 10 each and the base is (8² - 4²)/4 = 12. Thus the three sides of the

1. कृति युतिर सदृशराश्योर्बाहुर्घातो द्विसंगुणो लम्बः । कृत्यन्तरमसदृशयोर्द्विगुणं द्विसमत्रिभुज भूमिः ॥ —BrSpSi. XII. 33
2. करणी लम्बस्तत्कृतिरिष्टहतेष्टोन संयुताऽल्पा भूः । अधिको द्विहृतो बाहुः संक्षेप्यो यद्वधो वर्गः ॥ —BrSpSi. XII. 37.

ISOSCELES TRIANGLES 263 rational isosceles triangle with altitude 8 are (10,10, 12). Rational Scalene Triangles: Brahmagupta lays down the following rule in the case of rational scalene triangle : 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.¹ In other words, if m, p, q are any rational numbers, then the sides of a rational scalene triangle are : ½ ( m²/p + p ), ½ ( m²/q + q ), ½ ( m²/p - p ) + ½ ( m²/q - q ) Here the altitude (m), area and segments of the base of this triangle are all rational. Thus putting m=12, p=6, and q=8 in Brahmagupta's gene- ral equation, Pṛthūdaka Svāmī derives a scalene triangle with sides (13,15) and (14) altitude (12), area (84 and the segments of the base (5) which are all integral numbers. ½ ( m²/p + p ) = ½ ( 12²/6 + 6 ) = 15; ½ ( m²/q + q ) = ½ ( 12²/8 + 8 ) = 13 B' A C' A' B' C' B H C H B C Fig. 19 Fig. 20

1. इष्टद्वयेन भक्तो द्विधेष्ट वर्ग फलेष्टयोगार्धे । विषमत्रिभुजस्य भुजाविष्टोनफलार्धयोगो भूः ॥ —BrSpSi. XII, 34.

264 BRAHMAGUPTA AS AN ALGEBRAIST Thus the two sides of the rational scalene triangle are 15 and 13. The base is . ½ ( 12²/6 - 6 ) + ½ ( 12²/8 - 8 ) = 9 + 5 = 14 The altitude is m = 12; area is equal to (base × altitude) / 2 = (14 × 12) / 2 = and the segments are √(13² - 12²) = 5 and √(15² - 12²) = 9. Thus they are all integers. Rational Isosceles Trapeziums Brahmagupta has given us a method of obtaining such isosceles trapeziums whose sides, diagonals, altitude, segments and area are all rational numbers. His rule is as follows : The diagonals of the rec- tangle (generated) are the flank sides of an isosceles trapezium; the square of its side is divided by an optional number and then lessened by that optional number and divided by two; (the result) increased by the upright is the base and lessened by it is the face.¹ Fig. 21 Here in the figure, we have the isosceles trapezium ABCD of which C D is the base and A B is known as the fase. Accord- ing to Brahmagupta's rule, we have ( p being the optional number). CD = ½ ( (4m²n² - p) / p ) + ( m² - n² ) (base) AB = ½ [ 4m²n²/p - p ] - [ m² - n² ] (face) DH = ( m² - n² ) (upright)

1. आयतकर्णौ बाहू भुजकृतिरिष्टेन भाजितेष्टोना । द्विहृता कोट्यधिका भूर्मुखमूना दिसमचतुरस्रे ॥ —BrSpSi. XII. 36

RATIONAL TRAPEZIUMS 265 AD = BC = m²+n² (the sides of the trapezium) HC = base-upright = ½ [ 4m²n²/p - p ] (segment) AC = BD = [ 4m²n²/p + p ] (diagonal) AH = 2 mn (altitude) ABCD = mn [ 4m²n²/p - p ] (area) By chosing the values of m n and p suitably, the values of all the dimensions of the isosceles trapezium can be made integral. Pṛthūdaka Svāmī starts with the rectangle (5, 12, 13) and suitably takes p as 6; then he calculates out the dimensions of the trapezium : flank sides (AD and BC) = 13, base =14, and base = 4, altitude (AH) = 12, segments of base (DH and HC) = 5, and 9, diagonals (AC and BD) = 15, area ABCD = 108. All these values are integers. In this example, the rectangle chosen is (5, 12, 13) which is AA' DH, where AD = m² + n² = 13 and DH = m² - n² = 5 whence by adding the two we have 2m² = 18 This gives the value of m = 3, and hence n = 2. Pṛthū- daka Svāmī has taken the value of p = 6 by choice. Putting these values of m, n and p, the values for the dimensions of the isosceles trapezium follow from the expressions given by Brahma- gupta. CD = ½ ( (4.3².2² / 6) - 6 ) + ( 3²-2² ) = 9+5 = 14 (base) Face = 9-5=4 Sides AD = BC = 3²+2² = 13 and so on for the other dimensions. Rational Trapeziums With Three Equal Sides This problem is very much the same as one of the rational isosceles tpapezium with the only difference that in this case one of the parallel sides is also equal to the slant sides. We

266 BRAHMAGUPTA AS AN ALGEBRAIST have the following solution of this problem from Brahma- gupta : The square of the diagonal (of a generated rectangle) gives three equal sides; the fourth (is obtained) by subtracting the square of the upright from thrice the square of the side (of that rectangle). If greater, it is the base; if less, it is the face.¹ As before, the rectangle generated from m, n is given by (m²–n², 2mn, m²+n²), that is these are the three sides of the right triangle, which correspond to the two sides and the diagonal of the rectangle generated by them. Let us suppose, we have a trapezium ABCD whose sides AB, BC and AD are equal, then AB = BC = AD = (m²+n²)² CD = 3(2mn)² – (m²–n²)² = 14 m²n² – m⁴ – n⁴ or CD = 3(m²–n²)² – (2mn)² = 3m⁴+3n⁴ – 10 m²n². Pṛthudaka Svāmī has taken an illustration, where m=2, n=1 and he deduces two rational trapeziums with three equal sides (25, 25, 25, 39) and (25, 25, 25, 11). The segment (CH), altitude (AH), diagonals (AC, BD) and area of this trapezium are also rational, and given by : CH (segment) = 6 m²n²—m⁴ – n⁴ AH (altitude) = 4 mn (m²—n²) AC = BD (diagonals) = 4 mn (m²+n²) ABCD (area) = 32m³n³ (m²—n²). Rational Inscribed Quadrilaterals We find in the Brāhmasphuṭasiddhānta a remarkable proposition formulated by Brahmagupta : To find all quadrilaterals which will be inscribable within circles and whose sides, diagonals, perpendi- culars, segments (of sides and diagonals by perpendi- culars from vertices as also of diagonals by their intersection), areas, and also the diameters of the

1. कर्णकृतित्रिसम भुजास्त्रयश्चतुर्थो विशोष्य कोटि कृतिम् । बाहुकृतेस्त्रिगुणाया यद्यधिको भूर्मुखं हीनः ॥ —BrSpSi. XII. 37

RATIONAL INSCRIBED QUADRILATERALS 267 circumscribed circles will be expressible in integers. Such quadrilaterals we shall call as Brahmagupta Quadrilaterals. The solution of this formidable problem has been given by Brahmagupta as follows : The upright and bases of two right-angled triangles being reciprocally multiplied by the diagonals of the other will give the sides of a quadrilateral of unequal sides : ( of these ) the greatest is the base, the least is the face, and the other two sides are the two flanks.¹ Taking Brahmagupta's integral solution, the sides of the two right triangles of reference are given by : (1) m²-n², 2 mn, m²+n²; (ii) p²-q², 2 pq, p²+q²; where m, n, p, q are integers. Then the sides of the Brah- magupta Quadrilateral are. (m²-n²)(p²+q²), (p²-q²)(m²+n²), 2mn(p²+q²), 2pq(m²+n²) (Arrangement A) Pṛthūdaka Svāmī has- illustrated the rational inscribed quadrilateral by taking an example of the right angle triangles. (i) (3,4,5) (m²-n²=3, m²+n²=5, whence m=2, n=1) (ii) (5,12,13)(p²-q²=5, p²+q²=13, whence p=3, q=2) Fig. 22 Substituting these values in the above equations, we get the sides of the quardilateral as ( 39, 25, 52 and 60).²

1. जात्यद्वय कोटिभुजाः परकर्णगुणा भुजाश्चतुर्विषमे । अधिको भूर्मुखंहीनो बाहुद्वितयं भुजावन्यौ ॥ —BrSpSi. XII. 38 2 The diagonals of this quadrilateral are given by Bhāskara II as 56 (=3.12+4.5) and 63 (=4.12+3.5). (Cont. on page 268)

270 BRAHMAGUPTA AS AN ALGEBRAIST Put in other words, this means that one has to solve the following equations : (i) 5x—25 = y² (ii) 10x—100 = y² (iii) 83x—7635 = y² Pṛthūdaka Svāmī, the commentator on the Brāhmasphuṭa- siddhānta proceeds to solve these equations as follows : (1.1) Suppose y = 10; then x = 125. Or put y = 5; then x = 10. (2.1) Suppose y = 10; then x = 20. (3.1) Assume y = 1; then x = 92. He then remarks that by virtue of the multiplicity of suppositions there will be an infinitude of solutions in every case, But no method has been given either by Brahmagupta or his commentator to obtain the general solution. Double Equations of the First Degree Perhaps we have the earliest reference of the simultaneous indeterminate quadratic equations of the type x ± a = u² x ± b = v² in the Bhakaśālī Manuscript (Folio 59, recto). Brahmagupta gives the solution of such simultaneous inde- terminate quadratic equations of a general case as follows : The difference of the two numbers by the addition or subtration of which another number becomes a sq- uare, is divided by an optional number and then incre- ased or decreased by it. The square of half the result diminished or increased by the greater or smaller (of the given number) is the number (required).¹ Expressed in the language of algebra, shall have : = ½ { ½ ( (a - b)/m ± m ) }² ∓ a

1. याभ्यां कृतिरधिको नस्तदन्तरं हृत युतो न मिष्टेन । तद्दल कृतिरधिकोऽधिकयो रविको न यो राशिः ॥ —BrSpSi. XVIII. 74