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# Chapter 4 Number Theory in Asia - PowerPoint PPT Presentation

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 Bh â skara II Rational Triangles Biographical Notes: Brahmagupta and Bh â skara. 5.1 The Euclidean Algorithm.

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Chapter 4 Number Theory in Asia

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### Chapter 4Number 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

### 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

### 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

### Explanation

• 140 = 4x (5 x 7) leaves remainder 2 on division by 3 and remainder 0 on division by 5 and 7

• 63 = 3x (3 x 7) leaves remainder 3 on division by 5 and remainder 0 on division by 3 and 7

• 140 = 2x (3 x 5) leaves remainder 2 on division by 3 and remainder 0 on division by 5 and 7

• Therefore their sum 223 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:223 – 2 x 105 = 223 – 210 = 23

• Question: Why do we choose 140, 63 and 30 (or, more precisely 4, 3 and 2)?

• 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)

### Inverses modulo p

• Definitionb is called inverse of a modulo p if ab ≡ 1modp

• 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 ≡ 1mod6 then 4b – 1 were divisible by 6 and therefore 1 were divisible by 2 = gcd (4,6), which is impossible!

### General method of finding inverses

• Qin Jiushao, 1247 - used Euclidean algorithm to find inverses

• Suppose ax = 1mod p where x is unknown

• Then ax – 1 = py↔ ax - py = 1

• This equation has solutions if and only ifgcd (a,p) = 1 and in this case solutions can be found from Euclidean algorithm!

• In particular, if p is prime then inverse always exist

### 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 p1n ≡ r2 mod p2.....................n ≡ rk mod pk

### 5.3 Linear Diophantine Equations

• ax + by = c

• Euclidean algorithm:

• China: between 3rd century CE and 1247 (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

### 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

### 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:

### 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 thatx =x1x2+Ny1y2and y =x1y2+x2y1is 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 ofx2 – Ny2 = 1 (e.g. start from some obvious solution and compose it with itself)

### 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 solutionofx2 – 92y2 = 82

• Dividing both sides by 82 we obtain (24, 5/2, 1) which is a rational solutionof x2 – 92y2 = 1

• Composing it with itself we get (1151, 120, 1) which means thatx = 1151, y = 120 is a solution of x2 – 92y2 = 1

### 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

### 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)

• 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

### 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 (1766319049, 226153980, 1)which is a solution of x2 – 61y2 = 1 !

• In fact, it is the minimal nontrivial solution!

### 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 trianglesIf a, b, care sides of a rational triangle then for some rational numbers u, v and w we have:a = u2 / v + v, b = u2 / w + wc = u2 / v – v + u2 / w – w

### Stronger Form

• Any rational triangle is of the forma = (u2 + v2) / v, b = (u2 + w2) / wc = (u2 – v2 ) / v + (u2 –w2 ) / wfor some rational numbers u, v, w withthealtitude h = 2usplittingside c intosegmentsc1 = (u2 – v2 ) / v and c2 = (u2 – v2 ) / v

### 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

• Bhâskara II (1114 – 1185)

• greatest astronomer and mathematician in 12th-century India

• head of the observatory at Ujjain

• “Līlāvatī” (work named after his daughter)