Chapter 4 number theory in asia
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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 4 number theory in asia

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

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

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

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)?


Chapter 4 number theory in asia

  • 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

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

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

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

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

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

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

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

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 bh skara ii

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 bh skara

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

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

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

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 bh skara ii

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


Chapter 4 number theory in asia

  • Bhâskara II (1114 – 1185)

    • greatest astronomer and mathematician in 12th-century India

    • head of the observatory at Ujjain

    • “Līlāvatī” (work named after his daughter)


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