Grade 10

Indian, Islamic and Chinese mathematics

From Brahmagupta's cyclic-quadrilateral formula $S=\sqrt{(s-a)(s-b)(s-c)(s-d)}$ and his Pell equation $x^2-Ny^2=1$ , through al-Khwarizmi's algebra and Omar Khayyam's conic solutions of cubics, to Liu Hui's polygon-doubling $\pi$ and the Chinese remainder theorem.

IntuitionThree Civilizations, One River of Ideas

Between roughly the 3rd and the 13th century, while Europe kept relatively few written mathematical records, three civilizations pushed mathematics forward with remarkably concrete, hands-on methods: Indian astronomer-mathematicians such as Brahmagupta worked out exact formulas for cyclic quadrilaterals and integer solutions of $x^2-Ny^2=1$ ; scholars of the Islamic world such as al-Khwarizmi and Omar Khayyam turned ad-hoc tricks into a systematic algebra, solving equations by literally cutting and rearranging squares, or by intersecting curves; and Chinese mathematicians such as Liu Hui and the authors of the Sunzi Suanjing refined $\pi$ by relentlessly doubling polygons and solved calendar puzzles with what we now call the Chinese remainder theorem.

Inscribe a regular

n

-gon in a unit circle: adjacent vertices are separated by the central angle

\theta=360^\circ/n

. Drag

\theta

toward

30^\circ

(

n=12

) and imagine halving it again toward a

24

-gon — the gap between polygon and circle shrinks, which is exactly the geometric idea behind Liu Hui's method for approximating

\pi

SchoolBrahmagupta's Formula for Cyclic Quadrilaterals

Definition: Cyclic quadrilateral and semiperimeter

A quadrilateral with sides $a,b,c,d$ is cyclic if all four vertices lie on one circle; this forces opposite angles to be supplementary. Its semiperimeter is $s=\dfrac{a+b+c+d}{2}$ .

s=\dfrac{a+b+c+d}{2}

Brahmagupta's theorem (Brahmasphutasiddhanta, year 628) states that the area of such a quadrilateral depends only on its four side lengths, exactly as Heron's formula does for a triangle: $S=\sqrt{(s-a)(s-b)(s-c)(s-d)}$ .

S=\sqrt{(s-a)(s-b)(s-c)(s-d)}

Where does this come from? Draw diagonal $AC$ splitting the quadrilateral into triangles $ABC$ (sides $a,b$ ) and $ACD$ (sides $c,d$ ), with angles $B$ and $D$ opposite each other. Because opposite angles of a cyclic quadrilateral are supplementary, $B+D=180^\circ$ , hence $\cos B=-\cos D$ . The law of cosines in each triangle gives $AC^2=a^2+b^2-2ab\cos B$ and $AC^2=c^2+d^2-2cd\cos D$ ; substituting $\cos D=-\cos B$ into the second and equating both expressions for $AC^2$ lets you eliminate $\cos B$ in favor of the areas of the two triangles ( $\tfrac12 ab\sin B$ and $\tfrac12 cd\sin D$ , with $\sin B=\sin D$ ), and after algebraic simplification (completing squares and factoring using $s=\tfrac{a+b+c+d}2$ ) the four separate side lengths collapse into the single symmetric product above.

Liu Hui's polygon-doubling approximation of $\pi$ (unit circle, $r=1$ )
Sides $n$	Side length $l_n$	Estimate $n\,l_n/2$
6	1.0000	3.0000
12	0.5176	3.1058
24	0.2611	3.1326
48	0.1308	3.1394
96	0.0654	3.1410

UndergraduatePell's Equation and the Chakravala Method

In the same 628 treatise, Brahmagupta studied what is now called Pell's equation, $x^2-Ny^2=1$ , for a fixed non-square positive integer $N$ : find integers $x,y$ satisfying it. He discovered the bhavana (composition) identity, letting two solutions be combined into a new one, and centuries later Bhaskara II turned this into the chakravala (cyclic) algorithm: start from a rough solution to $x^2-Ny^2=k$ for some small $k$ , then repeatedly compose it with a suitably chosen trivial solution to shrink $|k|$ step by step until $k=\pm1$ , at which point a solution with $k=1$ is at hand or one more composition step produces it.

x^2-Ny^2=1

Nearly five centuries earlier, Omar Khayyam attacked a different problem: solving cubic equations, which cannot be done with straightedge and compass alone. In his Treatise on Demonstration of Problems of Algebra (c. 1070), he showed that every cubic can be solved by intersecting two conic sections.

Concretely, take the cubic $x^3+a^2x=a^2b$ with $a,b>0$ . Intersect the parabola $x^2=ay$ with the circle $y^2=x(b-x)$ (the semicircle of diameter $b$ sitting on the positive $x$ -axis, passing through the origin). Substituting $y=x^2/a$ from the parabola into the circle's equation gives $\left(x^2/a\right)^2=x(b-x)$ , i.e. $x^4/a^2=bx-x^2$ ; dividing by $x\neq0$ and multiplying by $a^2$ recovers exactly $x^3+a^2x=a^2b$ . So the $x$ -coordinate of the (non-origin) intersection point is the root — a curve crossing replaces radicals.

Take $a=1,b=2$ : the cubic $x^3+x=2$ , the parabola $x^2=y$ , and the circle $y^2=x(2-x)$ . They meet (besides the origin) at $(1,1)$ : indeed $y^2=1$ and $x(2-x)=1\cdot1=1$ match, and $1^3+1=2$ checks the cubic directly. Khayyam's construction handles cubics whose roots are irrational just as well — the curves still cross, even when no clean number falls out.

UndergraduateLiu Hui's Polygon-Doubling Algorithm for $\pi$

Around 263, commenting on the classic Nine Chapters on the Mathematical Art, Liu Hui inscribed a regular hexagon in a circle of radius $r$ and repeatedly doubled the number of sides. If $l_n$ is the side length of the inscribed regular $n$ -gon, its apothem (distance from center to a side's midpoint) is $\sqrt{r^2-(l_n/2)^2}$ by the Pythagorean theorem, and the sagitta — the gap between that midpoint and the circle — is $r-\sqrt{r^2-(l_n/2)^2}$ . Connecting one original vertex to the new vertex sitting on the arc above that gap gives a right triangle, so the doubled side length obeys the recurrence $l_{2n}=\sqrt{\left(\dfrac{l_n}{2}\right)^2+\left(r-\sqrt{r^2-\left(\dfrac{l_n}{2}\right)^2}\right)^2}$ . Since the $n$ -gon's perimeter $n\,l_n$ approaches the circle's circumference $2\pi r$ as $n$ grows, $\pi\approx\dfrac{n\,l_n}{2r}$ gives better and better estimates.

l_{2n}=\sqrt{\left(\dfrac{l_n}{2}\right)^2+\left(r-\sqrt{r^2-\left(\dfrac{l_n}{2}\right)^2}\right)^2}

Starting from a hexagon ( $l_6=r$ , since a regular hexagon's side equals the radius) and doubling four times to a 96-gon, Liu Hui reached $\pi\approx3.14$ ; later, doubling all the way to a 3072-gon, he obtained the celebrated value $3.1416$ , accurate to four decimal digits — thirteen centuries before pocket calculators.

Theorem: Al-Khwarizmi's Completing-the-Square Construction

For $b>0$ and $c>0$ , the positive solution of $x^2+bx=c$ is $x=\sqrt{c+\left(\dfrac b2\right)^2}-\dfrac b2$ .

Why is it true?

Al-Khwarizmi (Baghdad, c. 820, in his book Al-Jabr wal-Muqabala — the origin of the word 'algebra') had no negative numbers or symbolic notation, so he solved quadratics by literally drawing squares and rectangles and physically completing a bigger square, a picture that makes the otherwise mysterious formula obvious.

Proof

Draw a square of side $x$ ; its area is $x^2$ , the first term of $x^2+bx=c$ . Attach to two adjacent sides of this square two thin rectangles, each of width $b/2$ and length $x$ , so their combined area is $2\cdot\tfrac b2\cdot x=bx$ , the second term. The resulting L-shaped figure (a "gnomon") has area $x^2+bx=c$ .

This gnomon is a big square of side $x+b/2$ with one small $\tfrac b2\times\tfrac b2$ corner square missing. To see this, notice the two rectangles plus the original square leave exactly a $(b/2)\times(b/2)$ gap at the outer corner where they meet. Fill that gap in: the whole figure — gnomon plus corner square — is now a genuine square of side $x+b/2$ , whose area is therefore $c+(b/2)^2$ (the gnomon's area $c$ plus the corner square's area $(b/2)^2$ , added exactly once).

So $(x+\tfrac b2)^2=c+\left(\tfrac b2\right)^2$ . Taking the positive square root of both sides (side lengths are positive) gives $x+\dfrac b2=\sqrt{c+\left(\dfrac b2\right)^2}$ , and isolating $x$ yields $x=\sqrt{c+\left(\dfrac b2\right)^2}-\dfrac b2$ .

Algebraic check: expanding $\left(\sqrt{c+(b/2)^2}-b/2\right)^2+b\left(\sqrt{c+(b/2)^2}-b/2\right)$ term by term reduces, after the $\pm b\sqrt{c+(b/2)^2}\,b/2$ cross terms cancel, exactly to $c$ — confirming the formula independently of the picture. Numerically, for $x^2+10x=39$ (so $b=10,c=39$ ): $x=\sqrt{39+25}-5=\sqrt{64}-5=8-5=3$ , and indeed $3^2+10\cdot3=9+30=39$ .

UndergraduateThe Chinese Remainder Theorem

Theorem: Chinese Remainder Theorem (Sunzi, Qin Jiushao)

Let $m_1,m_2,\dots,m_k$ be pairwise coprime positive integers and let $a_1,\dots,a_k$ be any integers. Then the system of congruences $x\equiv a_i\pmod{m_i}$ for every $i$ has a solution $x$ , and this solution is unique modulo $M=m_1m_2\cdots m_k$ .

Why is it true?

First recorded around the 3rd–5th century in the Sunzi Suanjing ('the unknown number of things') and given a complete general algorithm by Qin Jiushao in 1247 (Da-yan rule), this theorem says that pairwise-coprime moduli carry independent information: knowing a number's remainder mod 3, mod 5, mod 7 pins it down uniquely mod 105, with no information lost or contradictory, because the moduli never 'overlap'.

Proof

Existence, step 1 (build the pieces): for each $i$ define $M_i=M/m_i$ , the product of all moduli except $m_i$ . Because the $m_j$ are pairwise coprime, every prime factor of $m_i$ is absent from every $m_j$ ( $j\neq i$ ), so $\gcd(M_i,m_i)=1$ . By Bezout's identity (extended Euclidean algorithm) there exists an integer $y_i$ — the modular inverse of $M_i$ mod $m_i$ — with $M_iy_i\equiv1\pmod{m_i}$ .

Existence, step 2 (assemble the solution): set $x=\sum_{i=1}^{k}a_iM_iy_i\bmod M$ . Fix any index $i$ and reduce this sum mod $m_i$ . For every $j\neq i$ , the factor $M_j=M/m_j$ contains $m_i$ as one of its factors (since $i\neq j$ ), so $m_i\mid M_j$ and the term $a_jM_jy_j\equiv0\pmod{m_i}$ . Only the $i$ -th term survives: $x\equiv a_iM_iy_i\equiv a_i\cdot1=a_i\pmod{m_i}$ (using $M_iy_i\equiv1\pmod{m_i}$ ). Since $i$ was arbitrary, $x$ satisfies every congruence $x\equiv a_i\pmod{m_i}$ at once.

Uniqueness mod $M$ : suppose $x'$ is another integer satisfying every $x\equiv a_i\pmod{m_i}$ . Then $x-x'\equiv0\pmod{m_i}$ for every $i$ , i.e. every $m_i$ divides $x-x'$ . Because the $m_i$ are pairwise coprime, their least common multiple equals their product $M$ , so $M\mid(x-x')$ as well — any common multiple of pairwise-coprime numbers must already be a multiple of their product. Hence $x\equiv x'\pmod M$ : the solution is unique modulo $M$ , exactly as claimed.

Toy verification: take $m_1=3,m_2=5$ with $a_1=2,a_2=3$ , i.e. the system $x\equiv2\pmod3,\ x\equiv3\pmod5$ . Here $M=15$ ; $M_1=5$ needs $5y_1\equiv1\pmod3$ , i.e. $2y_1\equiv1\pmod3$ , so $y_1=2$ ; $M_2=3$ needs $3y_2\equiv1\pmod5$ , so $y_2=2$ . Then $x=2\cdot5\cdot2+3\cdot3\cdot2=20+18=38\equiv8\pmod{15}$ , so $x=8$ . Check: $8=2\cdot3+2$ leaves remainder $2$ mod $3$ , and $8=1\cdot5+3$ leaves remainder $3$ mod $5$ — both congruences hold, confirming the construction.

UndergraduateReal-World Applications and Worked Examples

These are not museum pieces. Sunzi's problem was invented to synchronize calendars: ancient Chinese astronomers tracked several cycles at once (a 60-day sexagenary cycle, planetary periods, lunar months), and the remainder theorem let them pinpoint a date from a handful of remainders instead of counting days one by one — exactly the calendar-and-navigation use case for which it was built. The very same idea now powers modern cryptography: RSA decryption combines a huge modulus $n=pq$ with the Chinese remainder theorem (the 'CRT-RSA' optimization) to do the modular exponentiation separately mod $p$ and mod $q$ — each roughly half the bit-length — and then recombine, which is about four times faster than working mod $n$ directly.

Example: Brahmagupta's Formula on a Concrete Quadrilateral

A quadrilateral inscribed in a circle has consecutive sides $a=3,b=4,c=5,d=6$ . Find its area.

Solution

Compute the semiperimeter first: $s=\dfrac{3+4+5+6}{2}=\dfrac{18}{2}=9$ .

Now form each factor $s-a,s-b,s-c,s-d$ : $s-a=9-3=6$ , $s-b=9-4=5$ , $s-c=9-5=4$ , $s-d=9-6=3$ .

Apply Brahmagupta's formula: $S=\sqrt{(s-a)(s-b)(s-c)(s-d)}=\sqrt{6\cdot5\cdot4\cdot3}=\sqrt{360}$ .

Simplify the radical: $360=36\cdot10$ , so $S=\sqrt{36\cdot10}=6\sqrt{10}$ , approximately $18.97$ square units — a single formula that needed no knowledge of the individual angles or diagonals.

Example: Sunzi's Original Remainder Problem

The Sunzi Suanjing (3rd–5th century) poses: find the smallest positive $x$ with $x\equiv2\pmod3,\ x\equiv3\pmod5,\ x\equiv2\pmod7$

Solution

The moduli $3,5,7$ are pairwise coprime, so the Chinese remainder theorem applies with $M=3\cdot5\cdot7=105$ .

Compute each $M_i$ and its inverse $y_i$ : $M_1=105/3=35$ , and $35y_1\equiv1\pmod3$ means $2y_1\equiv1\pmod3$ (since $35\equiv2$ ), so $y_1=2$ ( $2\cdot2=4\equiv1\pmod3$ ). Next $M_2=105/5=21$ , and $21y_2\equiv1\pmod5$ means $1\cdot y_2\equiv1\pmod5$ (since $21\equiv1$ ), so $y_2=1$ . Finally $M_3=105/7=15$ , and $15y_3\equiv1\pmod7$ means $1\cdot y_3\equiv1\pmod7$ (since $15\equiv1$ ), so $y_3=1$ .

Assemble: $x=a_1M_1y_1+a_2M_2y_2+a_3M_3y_3=2\cdot35\cdot2+3\cdot21\cdot1+2\cdot15\cdot1=140+63+30=233$ .

Reduce mod $M=105$ : $233=2\cdot105+23$ , so $233\equiv23\pmod{105}$ , giving $x=23$ .

Check all three congruences: $23=7\cdot3+2$ (remainder $2$ mod $3$ ), $23=4\cdot5+3$ (remainder $3$ mod $5$ ), $23=3\cdot7+2$ (remainder $2$ mod $7$ ) — all match, and since $23<105$ it is the smallest positive solution.

Common mistake. Two classic slips. First, in completing the square, students often add

b^2

or add

(b/2)^2

twice — but the geometric picture shows the missing corner square has area

(b/2)^2

and it is added exactly once to close the gnomon into a square; adding it again (or using

b^2

) breaks the equality with

c

. Second, the Chinese remainder theorem's guarantee of a unique solution mod

M=m_1\cdots m_k

requires the moduli to be pairwise coprime, not merely

\gcd(m_1,\dots,m_k)=1

overall — for instance

m_1=4,m_2=6

are not pairwise coprime (

\gcd=2

) even though no single modulus divides both simultaneously; the construction

x=\sum a_iM_iy_i

silently breaks down here because

M_i

and

m_i

are no longer coprime, so no inverse

y_i

exists, and a solution may fail to exist at all for some choices of

a_i

Historical note

Around 820, at the House of Wisdom in Baghdad, al-Khwarizmi wrote Al-Jabr wal-Muqabala, systematizing the solution of quadratics by the completing-the-square construction described above; the Latinized name of the book (and of the author) gave us the words 'algebra' and 'algorithm'. He was building on — and transmitting westward — ideas already circulating two centuries earlier in Brahmagupta's 628 Brahmasphutasiddhanta from India, and on an even older Chinese tradition: Liu Hui had refined the polygon-doubling computation of $\pi$ in the 3rd century, and Omar Khayyam would extend the geometric spirit of these methods to cubic equations two and a half centuries after al-Khwarizmi, around 1070. Read together, the chain India → Islamic world → (later) Europe, running alongside an independent and equally rich Chinese tradition, shows mathematics as a genuinely global relay race, not a single-culture invention.

Muhammad ibn Musa al-Khwarizmi Brahmagupta Liu Hui Omar Khayyam

A cyclic quadrilateral has sides $3,4,5,6$ . Using Brahmagupta's formula $S=\sqrt{(s-a)(s-b)(s-c)(s-d)}$ , what is its area?

Which equation, studied by Brahmagupta in 628 and later solved algorithmically by the chakravala method, is now called Pell's equation?

In his Treatise on Demonstration of Problems of Algebra, Omar Khayyam solved cubic equations geometrically by intersecting which two curves?

Sunzi's remainder problem was invented for calendar and astronomical calculations; the same Chinese remainder theorem speeds up modern RSA decryption mainly by...

References

Victor J. Katz (2009). A History of Mathematics: An Introduction
Oliver Knill (2012). A Multivariable Chinese Remainder Theorem · arXiv:1206.5114