1. Chapter 4 Number Theory in Asia
The Euclidean Algorithm
The Chinese Remainder Theorem
Linear Diophantine Equations
Pell’s Equation in Brahmagupta
Pell’s Equation in Bhâskara II
Rational Triangles
Biographical Notes: Brahmagupta and Bhâskara

2. 5.1 The Euclidean Algorithm
Major results known in ancient China and India (independently)
Pythagorean theorem and triples
Concept of π
Euclidean algorithm (China, Han dynasty, 200 BCE – 200 CE)
Practical applications of Euclidean algorithm
Chinese remainder theorem
India: solutions of linear Diophantine equations and Pell’s equation

3. 5.2 The Chinese Remainder Theorem
Example: find a number that leaves remainder 2 on division by 3, rem. 3 on div. by 5, and rem. 2 on div. by 7
In terms of congruencies: x ≡ 2 mod 3, x ≡ 3 mod 5, x ≡ 2 mod 7
Solution: x = 23
“General method” - Mathematical Manual by Sun Zi (late 3rd century CE):
If we count by threes and there is a remainder 2, put down 140
If we count by fives and there is a remainder 3, put down 63
If we count by seven and there is a remainder 2, put down 30
Add them to obtain 233 and subtract 210 to get the answer

4. Explanation
140 = 4 x (5 x 7) leaves remainder 2 on division by 3 and remainder 0 on division by 5 and 7
63 = 3 x (3 x 7) leaves remainder 3 on division by 5 and remainder 0 on division by 3 and 7
30 = 2 x (3 x 5) leaves remainder 2 on division by 3 and remainder 0 on division by 5 and 7
Therefore their sum 233 leaves remainders 2, 5, and 2 on division by 3, 5, and 7, respectively
Subtract integral multiple of 3 x 5 x 7 = 105 to obtain the smallest solution: 233 – 2 x 105 = 233 – 210 = 23
Question: Why do we choose 140, 63 and 30 (or, more precisely 4, 3 and 2)?

5. Sun Zi:
If we count by threes and there is a remainder 1, put down 70
If we count by fives and there is a remainder 1, put down 21
If we count by seven and there is a remainder 1, put down 15
We have:
70 = 2 x (5 x 7) – smallest multiple of 5 and 7 leaving remainder 1 on division by 3
Multiply it by 2 to get remainder 2 on division by 3: 140 = 2 x 70 = 2 x 2 (5 x 7) = 4 (5 x 7)

6. Inverses modulo p
Definition b is called inverse of a modulo p if ab ≡ 1 mod p
Examples:
2 is inverse of 3 modulo 5
3 is inverse of 3 modulo 8
2 is inverse of 35 modulo 3
Does inverse of a modulo p exist ?
Example: does inverse of 4 modulo 6 exist?
If we had 4b ≡ 1 mod 6 then 4b – 1 were divisible by 6 and therefore 1 were divisible by 2 = gcd (4,6), which is impossible!

7. General method of finding inverses
Qin Jiushao, used Euclidean algorithm to find inverses
Suppose ax = 1 mod p where x is unknown
Then ax – 1 = py ↔ ax - py = 1
This equation has solutions if and only if gcd (a,p) = 1 and in this case solutions can be found from Euclidean algorithm!
In particular, if p is prime then inverse always exist

8. Chinese remainder theorem
p1, p2, … pk - relatively prime integers, i.e. gcd (pi, pj) =1 for all i ≠ j
remainders: r1, r2, …, rk such that 0 ≤ri < pi
Then there exists integer n satisfying the system of k congruencies: n ≡ r1 mod p1 n ≡ r2 mod p n ≡ rk mod pk

9. 5.3 Linear Diophantine Equations
ax + by = c
Euclidean algorithm:
China: between 3rd century CE and (Qin Jiushao)
India: Âryabhata (499 CE)
Bhaskara I (India, 522) reduced problem to finding a’x + b’y = 1 where a’ = a / gcd (a,b) and b’ = b / gcd (a,b)
Criterion for an integer solution: Equation ax + by = c has an integer solution if and only if gcd (a,b) divides c

10. 5.4 Pell’s Equation in Brahmagupta
China: development of algebra and approximate methods, integer solutions for linear equations, but not integer solutions for nonlinear equations
India: less progress in algebra but success in finding solutions of Pell’s equation: Brahmagupta “Brâhma-sphuta-siddhânta” 628 CE

11. Brahmagupta’s method Pell’s equation: x2 – Ny2 = 1
Method is based on Brahmagupta's discovery of identity:
Note: if we let N = -1 we obtain identity discovered by Diophantus:

12. Composition of triples
Consider equations (1) x2 – Ny2 = k1 and (2) x2 – Ny2 = k2
x1, y1 – sol. of (1), x2, y2 – sol. of (2)
Then the identity implies that x =x1x2+Ny1y2 and y =x1y2+x2y1 is a solution of x2 – Ny2 = k1 k2
We therefore define composition of triples (x1, y1, k1 ) and (x2, y2, k2 ) equal to the triple (x1x2+Ny1y2, x1y2+x2y1, k1 k2 )
Thus if k1= k2 =1 one can obtain arbitrary large solutions of x2 – Ny2 = 1 (e.g. start from some obvious solution and compose it with itself)

13. Moreover…
It turns out that we can obtain solutions of x2 – Ny2 = 1 composing solutions of x2 – Ny2 = k1 and x2 – Ny2 = k2 even when k1, k2 > 1
Indeed: composing (x1, y1, k1 ) with itself gives integer solution of x2 – Ny2 = (k1 ) 2 and hence rational solution of x2 – Ny2 = 1
Example (Brahmagupta: “a person solving this problem within a year is a mathematician”) x2 – 92y2 = 1
Consider “auxiliary” equation x2 – 92y2 = 8
It has obvious solution (10, 1, 8)
Composing it with itself we get (192, 20, 64) which is a solution of x2 – 92y2 = 82
Dividing both sides by 82 we obtain (24, 5/2, 1) which is a rational solution of x2 – 92y2 = 1
Composing it with itself we get (1151, 120, 1) which means that x = 1151, y = 120 is a solution of x2 – 92y2 = 1

14. 5.5 Pell’s Equation in Bhâskara II
Brahmagupta:
invented composition of triples
proved that if x2 – Ny2 = k has an integer solution for k = 1, 2 ,4 then x2 – Ny2 = 1 has integer solution
Bhâskara:
first general method for solving the Pell equation (“Bîjaganita” 1150 CE)
method is based on Brahmagupta’s approach

15. Method of Bhâskara
Goal: find a non-trivial integer solution of x2 – Ny2 = 1
Let a and b are relatively prime such that a2 – Nb2 = k
Consider trivial “equation” m2 – N x 12 = m2 – N
Compose triples (a, b, k) and (m, 1, m2-N)
We get (am + Nb, a + bm, k (m2 – N) )
Dividing by k we get: ((am + Nb) / k, (a + bm) / k, (m2 – N) ) / k
Choose m so that (a+bm) / k = b1 is an integer AND so that m2 – N is as small as possible
It turns out that (am + Nb) / k = a1 and (m2-N) / k = k1 are integers
Now we have (a1)2 – N (b1)2 = k1
Repeat the same procedure to obtain k2 and so on
The goal is to get ki = 1, 2 or 4 and use Brahmagupta’s method

16. Example
Consider x2 – 61y2 = 1
Equation 82 – 61 x 12 = 3 gives (a, b, k) = (8, 1, 3)
Composing (8, 1, 3) with (m,1, m2 – 61) we get (8m + 61, 8+m, 3(m2 – 61))
Dividing by 3 we get ( (8m + 61) / 3, (8+m) / 3, m2 – 6)
Letting m = 7 we get (39, 5, - 4)
(Brahmagupta) Dividing by 2 (since 4 = 22) we get (39/2, 5/2, -1)
Composing it with itself we get (1523 / 2, 195 /2, 1)
Composing it with (39/2, 5/2, -1) we get (29718, 3805, -1)
Composing it with itself we get ( , , 1) which is a solution of x2 – 61y2 = 1 !
In fact, it is the minimal nontrivial solution!

17. 5.6 Rational Triangles
Definition A triangle is called rational if it has rational sides and rational area
Equivalently: rational sides and altitudes
Brahmagupta’s Theorem: Parameterization of rational triangles If a, b, c are sides of a rational triangle then for some rational numbers u, v and w we have: a = u2 / v + v, b = u2 / w + w c = u2 / v – v + u2 / w – w

18. Stronger Form
Any rational triangle is of the form a = (u2 + v2) / v, b = (u2 + w2) / w c = (u2 – v2 ) / v + (u2 – w2 ) / w for some rational numbers u, v, w with the altitude h = 2u splitting side c into segments c1 = (u2 – v2 ) / v and c2 = (u2 – v2 ) / v

19. 5.7 Biographical Notes: Brahmagupta and Bhâskara II
Brahmagupta (598 – (approx.) 665 CE)
“Brâhma-sphuta-siddhânta”
teacher from Bhillamâla (now Bhinmal, India)
prominent in astronomy and mathematics
Pell’s equation
general solution of quadratic equation
area of a cyclic quadrilateral (which generalizes Heron’s formula for the area of triangle)
parameterization of rational triangles

20. Bhâskara II (1114 – 1185)
greatest astronomer and mathematician in 12th-century India
head of the observatory at Ujjain
“Līlāvatī” (work named after his daughter)