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Brahmagupta

Brahmagupta (ब्रह्मगुप्त), son of Jiṣṇugupta, author of the Brāhmasphuṭasiddhānta

A course in two strands: the astronomer of Bhillamāla who wrote a thousand verses at the age of thirty and a practical handbook at sixty-seven, and the mathematics inside them: arithmetic with zero and negative numbers, the quadratic rule, the composition law for the “square-nature” equation, cyclic quadrilaterals and second-order interpolation.

Born 598 (from his own statement); Bhillamāla is the usual attribution Died after 665; c. 668 is a conventional estimate Signature result \(x^2-92y^2=1\) solved by composition: \(1151^2-92\cdot120^2=1\) Level advanced undergraduate
Brahmagupta’s theorem. In a cyclic quadrilateral with perpendicular diagonals, the line from their crossing point perpendicular to one side bisects the opposite side.
Lesson 00

How to use this course

Brahmagupta wrote two books that survive, both in verse and both about astronomy. Two chapters of the first, on calculation and on algebra, contain some of the most important mathematics of the first millennium: the first full set of rules for computing with zero and negative numbers, a general rule for quadratic equations, a composition law that turns one solution of \(Nx^2+k=y^2\) into many, and exact formulas for the area and diagonals of a quadrilateral inscribed in a circle. His life is known only from a few lines in his own works and from later astronomers. The course keeps these two kinds of evidence apart.

Prerequisites for Part II
  • High-school algebra and trigonometry; the sine and cosine addition formulas.
  • Elementary number theory: divisibility, the Euclidean algorithm, congruences.
  • Euclidean geometry of the circle: inscribed angles, Ptolemy’s theorem is helpful.
  • The companion course on Āryabhaṭa, especially its lessons on the sine table and the pulverizer.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern notation, derivations and exercises with folded solutions. Verse references such as 12.21 or 18.44 give chapter and verse of the Brāhmasphuṭasiddhānta (abbreviated BSS) as they are usually cited in the modern literature (for example by Kim Plofker); editions differ by a verse or two. Brahmagupta stated rules in words, never in symbols. The formulas in this course are modern transcriptions.

Reference

Timeline

All dates CE. Only 598, 628 and 665 rest on Brahmagupta’s own words.

499Reference year of Āryabhaṭa’s Āryabhaṭīya.
598Born, by his statement that he wrote the Brāhmasphuṭasiddhānta at age 30 in Śaka year 550.
628Śaka 550: completes the Brāhmasphuṭasiddhānta, 24 chapters, under King Vyāghramukha of the Cāpa dynasty, ruling from Bhillamāla.
629Bhāskara I writes his commentary on the Āryabhaṭīya; the two contemporaries never mention each other.
665Composes the Khaṇḍakhādyaka, a handbook based on Āryabhaṭa’s midnight system, with a second part correcting it. He is 67.
c. 668Conventional date of death; no source fixes it.
754–775Reign of the caliph al-Manṣūr. Muḥammad ibn Ibrāhīm al-Fazārī renders Indian astronomy, largely of Brahmagupta’s school, into Arabic as the Sindhind.
9th c.Pṛthūdakasvāmin writes commentaries on both of his works.
c. 1030Al-Bīrūnī’s India discusses Brahmagupta at length.
1150Bhāskara II, in the Siddhāntaśiromaṇi and Bījagaṇita, completes the solution of the square-nature equation by the cakravāla method and calls Brahmagupta the crest-jewel of the circle of calculators.
1657Pierre de Fermat challenges European mathematicians to solve \(x^2-Ny^2=1\) in integers, with \(N=61\) among his test cases.
1768Joseph-Louis Lagrange proves that \(x^2-Ny^2=1\) has a non-trivial solution for every non-square \(N\).
1817Henry Thomas Colebrooke publishes English translations of BSS chapters 12 and 18, with Bhāskara II’s mathematical works.
1966R. S. Sharma and collaborators publish a four-volume edition of the BSS in New Delhi.
1970Bina Chatterjee’s edition and translation of the Khaṇḍakhādyaka with the commentary of Bhaṭṭotpala.
Part I · Formation
Lesson 01

Bhillamāla and the sources

Near the end of the Brāhmasphuṭasiddhānta, in chapter 24, Brahmagupta gives the facts that date him. He wrote the book, he says, when 550 years of the Śaka kings had elapsed, during the reign of King Vyāghramukha of the Cāpa (Cāvaḍa) dynasty, and he was thirty years old. Śaka 550 is 628 CE, so he was born in 598. He also names his father, Jiṣṇugupta, and later astronomers sometimes call him Jiṣṇusuta, “son of Jiṣṇu”.

Bhillamāla

The Cāpa kings ruled from Bhillamāla, modern Bhinmal in southern Rajasthan, then a capital of the Gurjara country. His ninth-century commentator Pṛthūdakasvāmin calls him Bhillamālakācārya, “the teacher from Bhillamāla”, and this is the main evidence that he lived and worked there. The epithet does not prove that he was born there, though most accounts assume it.

Ujjain

Many modern accounts, MacTutor among them, say that Brahmagupta headed the astronomical observatory at Ujjain, the city of Varāhamihira and the reference meridian of Indian astronomy. His own works do not say so, and no early source does either. Ujjain mattered to him as the prime meridian for all his calculations. The claim of an institutional post should be treated as later tradition.

Religion and outlook

The book’s title means “the corrected treatise of Brahmā”: Brahmagupta presented his work as a restoration of an older revealed astronomy, the Paitāmahasiddhānta of the creator god. This framing let him claim authority while changing parameters, and it explains the polemical tone toward rivals who, in his view, had strayed from tradition.

Exercise 1.1The Śaka era begins in 78 CE (Śaka year 0 begins in spring of 78). Check that Śaka 550 is 628 CE, and that a man aged 30 in that year was born in 597 or 598. Why do historians give 598?

Show solution
\(550+78=628\). A person aged 30 in 628 was born in 598 if the birthday had already passed, or in 597 if not. Since Brahmagupta states completed years of the era and of his age without months, the simplest reading \(628-30=598\) is the conventional one; a one-year uncertainty remains.
Lesson 02

The schools of astronomy

By 600 Indian astronomy was a set of competing systems. Each gave a long time cycle, the integer numbers of revolutions of the planets in it, and rules for corrections by epicycles. Varāhamihira’s Pañcasiddhāntikā in the sixth century summarized five older systems. Āryabhaṭa’s Āryabhaṭīya of 499 founded the school later called the Āryapakṣa, with a day counted from sunrise, and his lost second work founded the Ārdharātrika school, counting from midnight.

Brahmagupta’s book founded or renewed the Brāhmapakṣa, the school of Brahmā. It kept the traditional cosmic cycle of a kalpa of 4,320,000,000 years, and the traditional division of the great age into unequal quarters, which Āryabhaṭa had abandoned. Its parameters dominated astronomy in western and northern India for centuries, and Bhāskara II in the twelfth century still worked within it.

Why parameters mattered

A system’s revolution numbers determine the computed positions of the planets, and hence the dates of festivals, the timing of eclipses and the casting of horoscopes. Small differences in revolution numbers grow into visible errors over centuries. Disputes between schools were therefore practical, and they were also religious: a system’s claim to revealed origin was part of its authority.

Exercise 2.1In a kalpa of 4,320,000,000 years, suppose a planet makes \(n\) revolutions. Show that a change of one revolution per kalpa shifts its computed position by about \(0.3''\) per millennium. Why then did astronomers argue over revolution numbers that differed by thousands?

Show solution
One revolution is \(360^\circ=1{,}296{,}000''\). Over 1000 years the shift is \(1{,}296{,}000\times1000/4.32\times10^9=0.3''\). A difference of several thousand revolutions per kalpa gives errors of order \(1000\times0.3''\approx5'\) per millennium, and differences in the epoch positions add more. Such errors are visible in eclipse timings, which is what made the disputes testable.
Lesson 03

A thousand verses at thirty

The Brāhmasphuṭasiddhānta has 24 chapters, about 1008 verses in the āryā metre; some manuscripts add a 25th chapter of versified tables. The first ten chapters form a complete astronomical treatise: mean and true longitudes of the planets, the three problems of time, place and direction, lunar and solar eclipses, risings and settings, the Moon’s crescent, the Moon’s shadow, and conjunctions of planets with each other and with stars. The rest is a series of supplements:

  • Chapter 11, an examination of earlier treatises, largely a critique of Āryabhaṭa, Śrīṣeṇa, Viṣṇucandra and others.
  • Chapter 12, Gaṇita, calculation: arithmetic operations, fractions, the rule of three, series, plane figures including the cyclic quadrilateral, volumes, and shadows.
  • Chapters 13–17, further astronomical problems and refinements.
  • Chapter 18, Kuṭṭaka, the pulverizer, which is in fact a chapter of algebra: the pulverizer itself, rules for positive, negative and zero quantities, unknowns, equations, and the square-nature equation.
  • Chapters 19–24, on the gnomon, metres and prosody of numbers, the celestial sphere, instruments, measures, and a concluding chapter.

Style

Like the Āryabhaṭīya, the book states rules without proofs. Unlike it, it is long and often discursive, and it includes problems for the student. The algebra chapter ends with a famous boast attached to the equation \(92x^2+1=y^2\): whoever can solve it within a year is truly a mathematician. Lesson 07 shows how he did it.

Exercise 3.1Compare the Āryabhaṭīya (121 verses, 499) with the Brāhmasphuṭasiddhānta (about 1008 verses, 628). List three structural differences in how mathematics is placed within each book.

Show solution
(1) Āryabhaṭa puts mathematics in one compact section near the start, Brahmagupta in two long supplementary chapters (12 and 18) after the astronomy. (2) Brahmagupta separates calculation (gaṇita, procedures with known quantities) from algebra (kuṭṭaka, operations with unknowns), a division that later Indian authors kept as pāṭīgaṇita and bījagaṇita. (3) Brahmagupta treats negative numbers and zero explicitly as objects of computation, while Āryabhaṭa gives no such rules.
Part II · Mathematics and astronomy
Lesson 04

Zero, fortunes and debts

For the general reader. Zero as a placeholder in written numbers was in use in India before Brahmagupta. What he did was new: he treated zero, and negative quantities, as numbers that can be added, subtracted, multiplied and divided, and he wrote down the rules. He called positive numbers “fortunes” (dhana) and negative ones “debts” (ṛṇa). Most of his rules are exactly ours. One is not.

Goals
  • State the rules of BSS 18.30–35 in modern form.
  • See which rules agree with the axioms of a field and which do not.
  • Understand why division by zero cannot be given a value in a field.

The rules

Write \(a,b>0\) for fortunes and \(-a,-b\) for debts. Brahmagupta defines zero (śūnya) as the result of subtracting a number from itself, \(a-a=0\), and then states:

Rules (BSS 18.30–35, in modern form)
  1. Sums: \(a+b\) is a fortune, \((-a)+(-b)=-(a+b)\) is a debt; a fortune plus a debt is their difference, and zero if they are equal; \(a+0=a\), \(0+0=0\).
  2. Differences: \(0-a=-a\), \(0-(-a)=a\); subtracting a debt is adding a fortune.
  3. Products: fortune times fortune and debt times debt are fortunes; fortune times debt is a debt; anything times zero is zero.
  4. Quotients: the same sign rules as products.
  5. A fortune or a debt divided by zero is a quantity “having zero as divisor” (taccheda); zero divided by zero is zero.
  6. Squares of fortunes and of debts are fortunes; the square of zero is zero.

Rules 1–4 and 6 are those of the ring \(\mathbb Z\) and the field \(\mathbb Q\). Rule 5 is the exception. Brahmagupta leaves \(a/0\) as a formal expression, which is defensible, but sets \(0/0=0\), which is not. Later Indian authors tried other answers: Bhāskara II in the twelfth century treated \(a/0\) as an infinite quantity unchanged by adding anything to it.

Proposition

In a field \(F\), there is no element \(z\) with \(0\cdot z=1\). Consequently no value of \(a/0\) with \(a\ne0\) is consistent with the field axioms, and defining \(0/0\) to be any fixed element breaks the rule \((p/q)\cdot q=p\).

Proof. For any \(z\), \(0\cdot z=(0+0)z=0z+0z\), so \(0z=0\ne1\). If \(a/0=w\) means \(w\cdot0=a\), then \(a=0\). Setting \(0/0=0\) is consistent with \(w\cdot0=0\), but so is every other \(w\), so the definition picks one value arbitrarily and fails the cancellation rule \((x\cdot0)/0=x\).

Exercise 4.1Compute by Brahmagupta’s rules: \((-3)-(-5)\), \(0-(-4)\), \((-6)\times(-2)\), \((-12)/3\), \((-2)^2\). Name the rule used in each case.

Show solution
\((-3)-(-5)=-3+5=2\): subtracting a debt is adding a fortune, and a debt plus a larger fortune is a fortune. \(0-(-4)=4\). \((-6)(-2)=12\): debt times debt is a fortune. \((-12)/3=-4\): debt divided by fortune is a debt. \((-2)^2=4\): the square of a debt is a fortune.

Exercise 4.2Assume \(0/0=0\) and the usual rule \((x/y)\cdot y=x\) for all \(y\). Derive a contradiction. Then explain why the rule \(0\cdot a=0\) together with distributivity already forces \(0/0\) to be undefined if division is the inverse of multiplication.

Show solution
The rule \((x/y)\cdot y=x\) with \(x=0,y=0\) gives no contradiction. Use its companion \((x\cdot y)/y=x\), which says division undoes multiplication: with \(x=5,y=0\) it gives \((5\cdot0)/0=5\). But \(5\cdot0=0\) and \(0/0=0\), so the left side is 0, and \(5=0\). Division as the inverse of multiplication requires that \(w\cdot0=0\) determine \(w\) uniquely, and it does not, since every \(w\) satisfies it.

Exercise 4.3Brahmagupta also gives rules for square roots: a fortune has two roots, a fortune and a debt, while a debt has no root among his quantities. Verify that \(x^2=9\) has roots \(\pm3\) and explain how this rule feeds into the quadratic rule of Lesson 05.

Show solution
\(3^2=9\) and \((-3)^2=9\) by the sign rule for squares. When the quadratic rule takes \(\sqrt{4ac+b^2}\), the root could be taken with either sign, giving two candidate values of \(x\). Brahmagupta’s statement of the rule uses the positive root only; later authors, notably Bhāskara II, discuss both roots and reject a negative one when the problem’s context forbids it.
Lesson 05

Equations and the quadratic rule

For the general reader. Brahmagupta’s algebra uses words for unknowns: the first is yāvat-tāvat, “as much as so much”, and further unknowns are named after colours, kālaka (black), nīlaka (blue) and so on. With these he states rules to solve linear equations in one unknown and quadratic equations, as well as systems in several unknowns. His rule for the quadratic is the earliest clear general statement of the formula we learn at school.

Goals
  • State and derive the rule of BSS 18.44.
  • Solve a linear equation with unknowns on both sides by his rule.
  • Recognize the alternative form of the quadratic rule (BSS 18.45).

Linear equations (BSS 18.43)

For an equation with the unknown and a constant on each side, \(bx+c=dx+e\), Brahmagupta instructs: subtract the coefficient of the unknown on one side from that on the other, and the constant on the other side from this one; divide the difference of constants by the difference of coefficients. In symbols \(x=(e-c)/(b-d)\).

The quadratic

Rule (BSS 18.44, in modern form)

For \(ax^2+bx=c\) with \(a\ne0\): multiply the absolute number by four times the coefficient of the square, add the square of the middle coefficient, take the square root, subtract the middle coefficient, and divide by twice the coefficient of the square. That is,

\[ x=\frac{\sqrt{4ac+b^2}-b}{2a}. \]

Derivation. Multiply \(ax^2+bx=c\) by \(4a\): \(4a^2x^2+4abx=4ac\). Add \(b^2\): \((2ax+b)^2=4ac+b^2\). Take the root and solve for \(x\). This is completing the square, and the verse transcribes these steps.

The next verse (18.45) gives a second form, obtained by first dividing by \(a\) and halving the middle coefficient: \(x=\big(\sqrt{ac+(b/2)^2}-b/2\big)/a\). Both are standard in later Indian texts. The constant sits on the right, so with Brahmagupta’s rules for negatives the rule covers every case in which the discriminant \(4ac+b^2\) is a fortune.

Exercise 5.1Solve \(2x^2+3x=27\) by both forms of the rule.

Show solution
First form: \(4\cdot2\cdot27+9=225\), \(\sqrt{225}=15\), \((15-3)/4=3\). Second form: \(2\cdot27+(3/2)^2=54+2.25=56.25\), \(\sqrt{56.25}=7.5\), \((7.5-1.5)/2=3\). Check: \(18+9=27\).

Exercise 5.2Solve \(5x+7=2x+22\) by the rule of 18.43, and \(x^2-10x=-9\) by 18.44. For the second, what does the other root correspond to?

Show solution
Linear: \(x=(22-7)/(5-2)=5\). Quadratic with \(a=1,b=-10,c=-9\): \(4ac+b^2=-36+100=64\), root 8, \(x=(8-(-10))/2=9\). Check \(81-90=-9\). The other root, \(x=1\), comes from the negative square root: \((-8+10)/2=1\). The verse gives only the root with the positive square root.
Lesson 06

The pulverizer and residues

For the general reader. Āryabhaṭa gave a method, the kuṭṭaka or pulverizer, for finding whole numbers that leave given remainders on division by given numbers. Brahmagupta named his algebra chapter after it, restated it more clearly, and showed how astronomers could use it to run their calculations backwards: from an observed planet position to the number of days since the beginning of the cosmic cycle.

Goals
  • State the problem of simultaneous residues and its solvability condition.
  • Run the pulverizer on a two-modulus problem and chain it to more moduli.
  • See how residues of revolutions encode elapsed days.
Theorem (residues, modern form)

Let \(m_1,\dots,m_k\) be positive integers. The system \(N\equiv r_i \pmod{m_i}\) has a solution if and only if \(r_i\equiv r_j\pmod{\gcd(m_i,m_j)}\) for all \(i,j\). The solution is then unique modulo \(\operatorname{lcm}(m_1,\dots,m_k)\).

Method. For two moduli, \(N=m_1x+r_1=m_2y+r_2\) gives \(m_1x-m_2y=r_2-r_1\). Run the Euclidean algorithm on \(m_1,m_2\), keep the quotients, and work back up to a solution, as in the Āryabhaṭa course. Having \(N\equiv N_{12}\pmod{\operatorname{lcm}(m_1,m_2)}\), combine it with the third congruence in the same way, and so on. Brahmagupta’s instructions follow this scheme and are clearer than Āryabhaṭa’s on the back-substitution.

Astronomy backwards

Suppose a planet makes \(R\) revolutions in \(D\) days of the cycle. After \(t\) days its mean position, in revolutions, is \(Rt/D\). An observer knows only the fractional part, measured as the residue \(Rt \bmod D\). Recovering \(t\) from the residue is a linear congruence \(Rt\equiv s\pmod D\). With several planets, one gets a system of such congruences, and the pulverizer gives the elapsed days. Chapter 18 poses problems of exactly this kind.

Exercise 6.1Find the smallest positive \(N\) with \(N\equiv 2\pmod 5\), \(N\equiv3\pmod 7\), \(N\equiv 4\pmod 9\).

Show solution
First two: \(N=5x+2=7y+3\), so \(5x-7y=1\). Euclid: \(7=1\cdot5+2\), \(5=2\cdot2+1\); back-substitute: \(1=5-2\cdot2=5-2(7-5)=3\cdot5-2\cdot7\), so \(x=3\), \(N=17\). All solutions \(N\equiv17\pmod{35}\). Now \(17+35s\equiv4\pmod9\): \(17\equiv8\), \(35\equiv8\), so \(8+8s\equiv4\), \(8s\equiv5\equiv-4\), \(-s\equiv-4\), \(s\equiv4\pmod 9\). \(N=17+140=157\). Check: \(157=31\cdot5+2=22\cdot7+3=17\cdot9+4\).

Exercise 6.2A toy planet makes \(R=7\) revolutions in a cycle of \(D=30\) days. Its observed mean position is \(\tfrac{11}{30}\) of a revolution past the starting point. How many days \(t\) (with \(0\le t<30\)) have elapsed since the epoch?

Show solution
Need \(7t\equiv11\pmod{30}\). Since \(\gcd(7,30)=1\), invert 7: \(30=4\cdot7+2\), \(7=3\cdot2+1\), so \(1=7-3\cdot2=7-3(30-4\cdot7)=13\cdot7-3\cdot30\). Hence \(7^{-1}\equiv13\), and \(t\equiv13\cdot11=143\equiv23\pmod{30}\). Check: \(7\cdot23=161=5\cdot30+11\). So 23 days have elapsed (plus any multiple of 30).
Lesson 07

Square-nature and composition

For the general reader. Which whole numbers \(x,y\) satisfy \(y^2=92x^2+1\)? Small trials give nothing; the smallest answer has \(y=1151\). Brahmagupta found a way to multiply solutions of related equations together, which lets one climb from an easy near-solution to a true one in two steps. Europeans call the equation Pell’s equation, after a misattribution by Leonhard Euler. In India it was varga-prakṛti, “square-nature”, and Brahmagupta’s law of composition, bhāvanā, was its key.

Goals
  • State and prove the composition identity.
  • Use it to solve \(x^2-92y^2=1\) and \(x^2-83y^2=1\) from near-solutions.
  • Understand the rules for additive constants \(\pm1,\pm2,\pm4\), and their limits.

Brahmagupta writes the equation as \(Nx^2+k=y^2\), with \(x\) the “lesser root” and \(y\) the “greater root”. Below we use the modern form \(X^2-NY^2=k\) (so \(X\) is his greater root) and call \((X,Y;k)\) a triple for \(N\).

Theorem (bhāvanā, BSS 18.64–65)

If \(X_1^2-NY_1^2=k_1\) and \(X_2^2-NY_2^2=k_2\), then

\[ (X_1X_2+NY_1Y_2)^2-N(X_1Y_2+X_2Y_1)^2=k_1k_2. \]

Proof. In \(\mathbb Z[\sqrt N]\), the norm \(\mathcal N(X+Y\sqrt N)=X^2-NY^2\) is multiplicative, because \(\mathcal N(\alpha)=\alpha\bar\alpha\). Multiply \((X_1+Y_1\sqrt N)(X_2+Y_2\sqrt N)=(X_1X_2+NY_1Y_2)+(X_1Y_2+X_2Y_1)\sqrt N\). Brahmagupta stated the identity as a rule, without this language; expanding both sides verifies it directly.

Two consequences

  • Many solutions from one. Composing a solution \((X,Y;1)\) with itself gives \((X^2+NY^2,\,2XY;\,1)\), and repetition gives infinitely many.
  • Reducing \(k\). Composing \((X,Y;k)\) with itself gives \((X^2+NY^2,\,2XY;\,k^2)\). Dividing both roots by \(k\) gives a rational solution with \(k=1\). When \(k=\pm2\) or \(\pm4\), Brahmagupta shows how to reach integers.
Rules for \(k=-1,\pm2,\pm4\)
  1. \(k=-1\): composing \((X,Y;-1)\) with itself gives the integer solution \((X^2+NY^2,\,2XY;\,1)\).
  2. \(k=\pm2\): composing with itself gives \((X^2+NY^2,\,2XY;\,4)\); since \(X^2-NY^2=\pm2\), the number \(X^2+NY^2=2NY^2\pm2\) is even, so \(\big(\tfrac{X^2+NY^2}{2},\,XY;\,1\big)\) is an integer solution.
  3. \(k=-4\): if \(X^2-NY^2=-4\), put \(r=\tfrac12(X^2+3)(X^2+1)\); then \(\big((X^2+2)(r-1),\ XYr;\ 1\big)\) is an integer solution. A similar rule handles \(k=4\).

The famous example

For \(N=92\): \(10^2-92\cdot1^2=8\), so \((10,1;8)\) is a triple. Composing it with itself gives \((100+92,\ 20;\ 64)=(192,20;64)\). Dividing the roots by 8: \((24,\tfrac52;1)\), a rational solution. Composing this with itself: \(\big(24^2+92\cdot\tfrac{25}{4},\ 2\cdot24\cdot\tfrac52;\ 1\big)=(576+575,\ 120;\ 1)=(1151,120;1)\). Check: \(1151^2=1{,}324{,}801=92\cdot14{,}400+1\).

What Brahmagupta lacked was a procedure guaranteed to find a starting triple with \(k\in\{\pm1,\pm2,\pm4\}\) for every \(N\). That came with the cakravāla (“cyclic”) method of Jayadeva and Bhāskara II, who used it to solve \(x^2-61y^2=1\) with \(x=1{,}766{,}319{,}049\), \(y=226{,}153{,}980\).

Exercise 7.1For \(N=83\), start from \(9^2-83\cdot1^2=-2\) and find an integer solution of \(X^2-83Y^2=1\). Then compose it with itself to get the next one.

Show solution
By the rule for \(k=\pm2\): \(X=(81+83)/2=82\), \(Y=9\). Check \(82^2=6724=83\cdot81+1\). Composing \((82,9;1)\) with itself: \((82^2+83\cdot81,\ 2\cdot82\cdot9)=(6724+6723,\ 1476)=(13447,1476)\). Check: \(13447^2-83\cdot1476^2=180{,}821{,}809-180{,}821{,}808=1\).

Exercise 7.2For \(N=13\), observe that \(3^2-13\cdot1^2=-4\). Use the rule for \(k=-4\) to find a solution of \(X^2-13Y^2=1\). Then find a different route, via \(18^2-13\cdot5^2=-1\).

Show solution
\(X=3,Y=1\): \(r=\tfrac12(12)(10)=60\), so \(\big((11)(59),\ 3\cdot1\cdot60\big)=(649,180)\). Check: \(649^2=421{,}201=13\cdot32{,}400+1\). Second route: compose \((18,5;-1)\) with itself: \((324+325,\ 180;\ 1)=(649,180;1)\), the same solution.

Exercise 7.3Show that composition is the same as multiplication of \(X+Y\sqrt N\) in \(\mathbb Z[\sqrt N]\), and deduce that the solutions of \(X^2-NY^2=1\) with \(X,Y>0\) are the powers of the smallest one. (You may assume the smallest solution exists.)

Show solution
The product formula is the composition law. Let \(\varepsilon=X_1+Y_1\sqrt N\) be the smallest unit greater than 1 with norm 1. If \(\eta>1\) has norm 1, choose \(n\) with \(\varepsilon^n\le\eta<\varepsilon^{n+1}\). Then \(\eta\varepsilon^{-n}\) has norm 1 and lies in \([1,\varepsilon)\). Its conjugate is its inverse, so if it exceeded 1 it would give a positive solution smaller than \(\varepsilon\). Hence \(\eta=\varepsilon^n\): every solution comes from repeated self-composition of the smallest.
Lesson 08

Cyclic quadrilaterals

For the general reader. The area of a triangle can be computed from its three sides alone. Brahmagupta gave the analogous rule for a quadrilateral whose four corners lie on a circle, and formulas for its diagonals in terms of the sides. He also showed how to build triangles and quadrilaterals whose sides, diagonals and areas are all whole numbers. These are among the most elegant results in the geometry of the first millennium.

Goals
  • State and prove the area formula of BSS 12.21.
  • Derive the diagonal formulas of BSS 12.28.
  • Prove the theorem on perpendicular diagonals (BSS 12.30–31) shown at the top of this page.
Theorem (area, BSS 12.21)

Let a quadrilateral with sides \(a,b,c,d\) be inscribed in a circle, and let \(s=\tfrac12(a+b+c+d)\). Its area is

\[ K=\sqrt{(s-a)(s-b)(s-c)(s-d)}. \]

Brahmagupta states this as the “exact” area, alongside a rough area (the product of half-sums of opposite sides). He does not say in the verse that the quadrilateral must be cyclic. Commentators and modern historians disagree about whether he assumed it; the formula is correct only in the cyclic case. Setting \(d=0\) gives Heron’s formula for a triangle.

Proof sketch. Let the diagonal \(p\) split the quadrilateral \(ABCD\) (with \(AB=a\), \(BC=b\), \(CD=c\), \(DA=d\)) into triangles with angle \(B\) between \(a,b\) and angle \(D\) between \(c,d\). Since the quadrilateral is cyclic, \(B+D=\pi\), so \(\sin D=\sin B\) and \(\cos D=-\cos B\). The law of cosines gives \(a^2+b^2-2ab\cos B=c^2+d^2+2cd\cos B\), so \(2\cos B\,(ab+cd)=a^2+b^2-c^2-d^2\). The area is \(K=\tfrac12(ab+cd)\sin B\). Then \(16K^2=4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2\), which factors as a difference of squares into \((2(ab+cd)-a^2-b^2+c^2+d^2)(2(ab+cd)+a^2+b^2-c^2-d^2)=((c+d)^2-(a-b)^2)((a+b)^2-(c-d)^2)\), and each factor splits again into the four terms \(16(s-a)(s-b)(s-c)(s-d)\).

Theorem (diagonals, BSS 12.28)

With the same notation, let \(p=AC\) and \(q=BD\). Then

\[ p^2=\frac{(ac+bd)(ad+bc)}{ab+cd},\qquad q^2=\frac{(ac+bd)(ab+cd)}{ad+bc},\qquad pq=ac+bd. \]

Derivation. From the proof above, \(\cos B=\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}\), and \(p^2=a^2+b^2-2ab\cos B\). Simplifying, \(p^2(ab+cd)=(a^2+b^2)cd+(c^2+d^2)ab=(ac+bd)(ad+bc)\). The formula for \(q\) follows by symmetry, and the product is Ptolemy’s theorem.

Theorem (Brahmagupta, BSS 12.30–31)

If a cyclic quadrilateral \(ABCD\) has perpendicular diagonals meeting at \(M\), then the line through \(M\) perpendicular to \(BC\) meets \(AD\) at its midpoint.

Proof. Let the perpendicular from \(M\) to \(BC\) meet \(BC\) at \(E\) and \(AD\) at \(F\). The angles \(\angle FAM=\angle DAC\) and \(\angle DBC=\angle MBC\) are inscribed angles on the same arc \(CD\), so they are equal. In the right triangle \(BMC\), \(\angle MBC=90^\circ-\angle BCM=\angle CME\), and \(\angle CME=\angle AMF\) (vertical angles). So \(\angle FAM=\angle AMF\), and triangle \(AFM\) is isosceles: \(FA=FM\). The same argument with \(D\) gives \(FD=FM\). Hence \(FA=FD\).

Rational figures

Brahmagupta also shows how to build triangles with rational sides and area by gluing two right triangles along a common leg, and cyclic quadrilaterals with rational sides and diagonals from two right triangles. For example, the right triangles 5–12–13 and 9–12–15 glued along the side 12 give the triangle with sides 13, 14, 15 and area 84.

Exercise 8.1A cyclic quadrilateral has sides \(AB=25\), \(BC=60\), \(CD=52\), \(DA=39\). Compute its area by Brahmagupta’s formula, both diagonals by the rule of 12.28, and verify with the right triangles 25–60–65 and 39–52–65.

Show solution
\(s=88\); \((s-a)(s-b)(s-c)(s-d)=63\cdot28\cdot36\cdot49=1764\cdot1764\), so \(K=1764\). Diagonals: \(ac+bd=1300+2340=3640\), \(ad+bc=975+3120=4095\), \(ab+cd=1500+2028=3528\). \(p^2=3640\cdot4095/3528=4225\), \(p=65\); \(q=3640/65=56\). Check: \(25^2+60^2=65^2\) and \(39^2+52^2=65^2\), so \(AC=65\) is a diameter and the area is \(\tfrac12(25\cdot60+39\cdot52)=750+1014=1764\).

Exercise 8.2Show that the formula of 12.21 fails for a non-cyclic quadrilateral by comparing a square of side 1 with a rhombus of side 1 and angle \(60^\circ\).

Show solution
Both have \(s=2\) and \((s-a)^4=1\), so the formula gives area 1 for both. The square (cyclic) has area 1, correct. The rhombus is not cyclic, since its opposite angles are \(60^\circ\) and \(60^\circ\), and has area \(\sin60^\circ\approx0.866\). The general formula, due to Bretschneider, subtracts \(abcd\cos^2\frac{B+D}{2}\), which vanishes exactly when \(B+D=\pi\).

Exercise 8.3Verify Brahmagupta’s theorem in coordinates for the circle \(x^2+y^2=25\), with \(A=(-4,3)\), \(C=(4,3)\), \(B=(0,5)\), \(D=(0,-5)\).

Show solution
\(AC\) is horizontal (\(y=3\)), \(BD\) vertical (\(x=0\)), so they are perpendicular and meet at \(M=(0,3)\). \(BC\) has direction \((4,-2)\), so the perpendicular through \(M\) has direction \((2,4)\), i.e. the line \(y=3+2x\). \(AD\) runs from \((-4,3)\) to \((0,-5)\), the line \(y=-5-2x\). Intersection: \(3+2x=-5-2x\), \(x=-2\), \(y=-1\). The midpoint of \(AD\) is \((-2,-1)\). Confirmed.
Lesson 09

Sines and interpolation

For the general reader. A table gives values at fixed steps; astronomers need values in between. The simplest method draws a straight line between neighbouring entries. Brahmagupta improved on it: he let the step size itself change smoothly across the interval, using the two neighbouring differences. The result is a second-order interpolation rule, a special case of a formula that Isaac Newton and James Stirling would write down a thousand years later.

Goals
  • Know Brahmagupta’s two sine tables.
  • State the interpolation rule of the Khaṇḍakhādyaka and show it agrees with Stirling’s formula to second order.
  • Measure its gain over linear interpolation.

Two sine tables

In the BSS (chapter 2) Brahmagupta gives 24 sines at steps of \(3^\circ45'\), like Āryabhaṭa, but with radius \(R=3270\). In the Khaṇḍakhādyaka he uses a coarse table with \(R=150\) and steps of \(15^\circ\) (\(900'\)). Its values and differences are:

Arc0°, 15°, 30°, 45°, 60°, 75°, 90°
\(150\sin\)0, 39, 75, 106, 130, 145, 150
Differences39, 36, 31, 24, 15, 5
Rule (Khaṇḍakhādyaka, second part; also in the earlier Dhyānagrahopadeśa)

Let \(f_r=f(x_r)\) be a tabulated value, \(h\) the step, \(D_r=f_r-f_{r-1}\) the difference already passed (gatakhaṇḍa) and \(D_{r+1}=f_{r+1}-f_r\) the difference to come (bhogyakhaṇḍa). For \(x=x_r+\theta h\) with \(0\le\theta\le1\), take as effective difference half their sum, corrected by \(\theta\) times half their difference, and multiply by \(\theta\):

\[ f(x_r+\theta h)\approx f_r+\theta\,\frac{D_r+D_{r+1}}{2}+\theta^2\,\frac{D_{r+1}-D_r}{2}. \]

The verse phrases the correction as added or subtracted according to whether the half-sum is less or greater than the coming difference. For a sine past its steepest point, \(D_{r+1}<D_r\), so the correction is subtracted. Stirling’s central-difference formula is

\[ f(x_r+\theta h)=f_r+\theta\,\frac{\Delta f_{r-1}+\Delta f_r}{2}+\frac{\theta^2}{2}\,\Delta^2 f_{r-1}+O(\Delta^3), \]

with \(\Delta f_{r-1}=D_r\), \(\Delta f_r=D_{r+1}\), \(\Delta^2f_{r-1}=D_{r+1}-D_r\). Brahmagupta’s rule is exactly its truncation after the second difference. It reproduces \(f_r\) at \(\theta=0\), \(f_{r+1}\) at \(\theta=1\) and \(f_{r-1}\) at \(\theta=-1\): it is the parabola through three consecutive table points.

Exercise 9.1Compute \(150\sin57^\circ\) from the table by Brahmagupta’s rule and by linear interpolation. Compare with the true value 125.80.

Show solution
\(x_r=45^\circ\), \(f_r=106\), \(D_r=31\), \(D_{r+1}=24\), \(\theta=12/15=0.8\). Brahmagupta: \(106+0.8\cdot27.5+0.64\cdot(-3.5)=106+22-2.24=125.76\). Linear: \(106+0.8\cdot24=125.2\). The second-order rule has error 0.04, the linear one 0.6.

Exercise 9.2Show that the rule is the unique quadratic through \((x_{r-1},f_{r-1})\), \((x_r,f_r)\), \((x_{r+1},f_{r+1})\), and use it for \(150\sin20^\circ\) (true value 51.30).

Show solution
At \(\theta=-1\): \(f_r-\tfrac{D_r+D_{r+1}}2+\tfrac{D_{r+1}-D_r}2=f_r-D_r=f_{r-1}\). At \(\theta=1\): \(f_r+D_{r+1}\). A quadratic is fixed by three points. For \(20^\circ\): \(x_r=15^\circ\), \(f_r=39\), \(D_r=39\), \(D_{r+1}=36\), \(\theta=1/3\): \(39+12.5+\tfrac19(-1.5)=51.33\). Linear interpolation gives 51.00.
Lesson 10

The astronomy of the Brāhmapakṣa

For the general reader. Brahmagupta was an astronomer first. His books compute the positions of the Sun, Moon and planets, predict eclipses, and explain the Moon’s phases. He held that the Earth is a sphere at rest at the centre of the cosmos, rejected Āryabhaṭa’s rotating Earth, and explained why things fall in terms that al-Bīrūnī later quoted with admiration.

Goals
  • Compare his year lengths with Āryabhaṭa’s and the modern sidereal year.
  • Derive the illuminated fraction of the Moon from the elongation.
  • Understand the arguments in the dispute over the Earth’s rotation.

The year

The BSS gives a sidereal year of 365 days 6 hours 5 minutes 19 seconds; the Khaṇḍakhādyaka, following the Ārdharātrika system, gives 365 days 6 hours 12 minutes 36 seconds. The modern sidereal year is 365 days 6 hours 9 minutes 10 seconds.

The Moon (BSS chapter 7)

Brahmagupta argues that the Moon is nearer than the Sun, and that the illuminated part seen from the Earth depends on the angular separation, the elongation, between them. In modern terms, if the Sun is far away and the elongation is \(E\), the illuminated fraction of the disc is

\[ k=\frac{1-\cos E}{2}, \]

which is the versine of \(E\) divided by the diameter. Indian rules express this through the versine (utkramajyā), \(R(1-\cos E)\).

A static Earth, and why things fall

Brahmagupta rejected the rotation of the Earth that Āryabhaṭa had proposed. According to al-Bīrūnī’s India, he explained falling bodies by the nature of the Earth: it is the nature of the Earth to attract and hold things, as it is the nature of water to flow. Modern claims that he formulated a theory of gravitation read too much into this; it is a statement about natural places, close in spirit to Aristotle.

Exercise 10.1Convert both of Brahmagupta’s year lengths and Āryabhaṭa’s 365.25868 days into days, and give the error of each in minutes against 365.25636 days.

Show solution
BSS: \(6^{\text h}5^{\text m}19^{\text s}=21{,}919\) s \(=0.25369\) d, so 365.25369 d, error \(-0.00267\) d \(\approx-3.8\) min. Khaṇḍakhādyaka: \(6^{\text h}12^{\text m}36^{\text s}=22{,}356\) s \(=0.25875\) d, error \(+0.00239\) d \(\approx+3.4\) min. Āryabhaṭa: \(+0.00232\) d \(\approx+3.3\) min. The Khaṇḍakhādyaka value differs from Āryabhaṭa’s by about 6 seconds, showing its dependence on the midnight system.

Exercise 10.2Derive \(k=(1-\cos E)/2\) for a distant Sun, and find the elongation at which the Moon is exactly one quarter illuminated.

Show solution
Look at the Moon’s disc from the Earth. The terminator is a great circle perpendicular to the Sun direction; seen at angle \(\pi-E\) from that direction, it projects to a half-ellipse with semi-minor axis \(r\cos(\pi-E)=-r\cos E\). The lit area is half the disc plus or minus the half-ellipse: \(\tfrac{\pi r^2}{2}(1-\cos E)\). Divide by \(\pi r^2\). For \(k=\tfrac14\): \(\cos E=\tfrac12\), \(E=60^\circ\).
Part III · Life and legacy
Lesson 11

The handbook of 665 and the critic

In 665, thirty-seven years after the BSS, Brahmagupta wrote the Khaṇḍakhādyaka, a karaṇa or practical handbook for calendar makers. Its title is the name of a sweet, a hint at its intended ease of use. The first part reproduces, in convenient form, the astronomy of Āryabhaṭa’s midnight system. The second part, often called the Uttarakhaṇḍakhādyaka, corrects that system with Brahmagupta’s own refinements, including the interpolation rule of Lesson 09. The handbook was widely used and commented on, notably by Pṛthūdakasvāmin and Bhaṭṭotpala.

The critic of Āryabhaṭa

Chapter 11 of the BSS and passages throughout the book attack Āryabhaṭa: his equal division of the great age, his rotating Earth, his eclipse rules and many parameters. The tone is sharp. Yet the Khaṇḍakhādyaka is built on Āryabhaṭa’s midnight system. Historians explain the contrast in several ways: the handbook was meant for practitioners who already used that system, Brahmagupta’s polemic targeted the claim to authority more than the arithmetic, and the rival schools borrowed from each other while disputing in public.

Death

Nothing is known about Brahmagupta’s life after 665. The death date of about 668 found in many reference works has no documented basis.

Exercise 11.1A karaṇa uses a recent epoch instead of the start of the kalpa. Explain, with the formula for mean position, why this makes calculation easier, and what the cost is.

Show solution
Mean position after \(t\) days is \(\lambda_0+\omega t\), where \(\omega=2\pi R/D\). From the kalpa epoch, \(t\) is about \(1.97\times10^{12}\) days and \(R\) is enormous, so the products are large. A recent epoch makes \(t\) small and lets \(\omega\) be given as a convenient fraction, with \(\lambda_0\) tabulated. The cost is that small errors in \(\omega\) or \(\lambda_0\) are not constrained by the requirement of an integer number of revolutions per cycle, and the handbook loses accuracy far from its epoch.

DiscussionBrahmagupta attacked Āryabhaṭa in 628 and built on him in 665. Compare with a modern case in which a scientist publicly rejects a rival programme but adopts its tools. What role did the claim of revealed authority play in his polemic?

Lesson 12

Baghdad, Bhāskara II and Fermat

Baghdad

In the reign of the caliph al-Manṣūr (754–775), an Indian astronomer arrived at the Abbasid court with a Sanskrit astronomical text, which Muḥammad ibn Ibrāhīm al-Fazārī rendered into Arabic. The result, the Zīj al-Sindhind, transmitted parameters and methods largely of the Brāhmapakṣa, and Arabic astronomers also knew a work derived from the Khaṇḍakhādyaka under the name Arkand. Muḥammad ibn Mūsā al-Khwārizmī’s astronomical tables drew on this material. Around 1030 al-Bīrūnī, who read Sanskrit, discussed Brahmagupta repeatedly in his India, both praising and criticizing him.

Bhāskara II

The greatest later Indian astronomer of the Brāhmapakṣa, Bhāskara II, called Brahmagupta gaṇakacakracūḍāmaṇi, the crest-jewel of the circle of calculators. His Bījagaṇita of 1150 completed the theory of the square-nature equation with the cakravāla method and took a new position on division by zero.

Europe

In 1657 Pierre de Fermat challenged European mathematicians to solve \(x^2-Ny^2=1\) in integers, choosing \(N=61\) as a test; William Brouncker and John Wallis found a method. Euler, following a misreading of Wallis, named the equation after John Pell. Lagrange proved in 1768 that a non-trivial solution exists for every non-square \(N\), using continued fractions. Europeans learned of Brahmagupta’s algebra from Colebrooke’s translation of 1817. In modern algebraic number theory his composition is multiplication of norms, and its generalization to binary quadratic forms is Gauss’s composition of forms.

Assessment

  • Arithmetic: the first systematic rules for zero and negative numbers as quantities.
  • Algebra: a general quadratic rule, symbolic names for several unknowns, and the composition law that made the square-nature equation tractable.
  • Geometry: exact area and diagonals of cyclic quadrilaterals, rational triangles and quadrilaterals.
  • Numerical analysis: second-order interpolation.
  • Limits: \(0/0=0\); an area rule stated without its cyclic hypothesis; no proofs in the text.

Exercise 12.1Fermat’s test case: the smallest solution of \(x^2-61y^2=1\) is \(x=1{,}766{,}319{,}049\), \(y=226{,}153{,}980\). Bhāskara II started from \(8^2-61\cdot1^2=3\). Compose \((8,1;3)\) with itself and divide by 3 to get a rational triple. Why does Brahmagupta’s method alone not finish the job?

Show solution
Composition gives \((64+61,\ 16;\ 9)=(125,16;9)\); dividing by 3, \((125/3,\ 16/3;\ 1)\), which is rational but not integral. Brahmagupta’s integer rules apply only for \(k\in\{\pm1,\pm2,\pm4\}\), and \(k=3\) is not among them. Bhāskara II’s cakravāla repeatedly composes with \((m,1;m^2-61)\) and divides, choosing \(m\) so that the result stays integral and \(|k|\) small, until it reaches \(k=-4\) via \((39,5;-4)\); Brahmagupta’s rule for \(k=-4\) then finishes.

Final projectChoose one of: the rules for zero, the quadratic rule, the bhāvanā, or the cyclic quadrilateral. Write two pages for a strong undergraduate that paraphrase the verse, reconstruct the mathematics in modern notation with a proof, give a worked example, and explain what later mathematicians (Bhāskara II, Fermat, Euler, Lagrange or Gauss) added.

Reference

Connections

People linked to Brahmagupta across this collection. A name without a link has no page here yet. See Brahmagupta in the network of all 337 lives →

Influenced by

  • Varahamihira

Influenced

Colleagues and collaborators

  • Bhaskara I

Rivals

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Texts and translations

  1. H. T. Colebrooke, Algebra, with Arithmetic and Mensuration, from the Sanscrit of Brahmegupta and Bháscara, London, 1817 (BSS chapters 12 and 18).
  2. R. S. Sharma (ed.), Brāhma-Sphuṭa Siddhānta, 4 vols., Indian Institute of Astronomical and Sanskrit Research, New Delhi, 1966.
  3. B. Chatterjee, The Khaṇḍakhādyaka of Brahmagupta, with the Commentary of Bhaṭṭotpala, 2 vols., New Delhi, 1970.

Biography and history

  1. Brahmagupta, MacTutor History of Mathematics.
  2. Brahmagupta and Brāhmasphuṭasiddhānta, Wikipedia.
  3. Brahmagupta, Biographical Encyclopedia of Astronomers, Springer, 2007.
  4. K. Plofker, Mathematics in India, Princeton University Press, 2009, ch. 5.
  5. D. Pingree, Brahmagupta, Dictionary of Scientific Biography, vol. 2, 1970.

Mathematics

  1. A. K. Dutta, The bhāvanā in mathematics, Bhāvanā 1 (2017).
  2. Pell’s equation, MacTutor history topic.
  3. Pell’s equation (history section) and Brahmagupta’s interpolation formula, Wikipedia.
  4. N. Handa and P. Taneja, Methods of interpolation in Indian astronomy, Indian Journal of History of Science 49 (2014) 251–259.
  5. Brahmagupta theorem and Brahmagupta’s formula, Wikipedia.
  6. A. Weil, Number Theory: An Approach through History from Hammurapi to Legendre, Birkhäuser, 1984 (on the square-nature equation).