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
• Pells Equation in Brahmagupta
• Pells Equation in Bhâskara II
• Rational Triangles
• Biographical Notes Brahmagupta and Bhâskara

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5.1 The Euclidean Algorithm

• Major results known in ancient China and India
  (independently)
• Pythagorean theorem and triples
• Concept of p
• Euclidean algorithm (China, Han dynasty,200 BCE
  200 CE)
• Practical applications of Euclidean algorithm
• Chinese remainder theorem
• India solutions of linear Diophantine equations
  and Pells equation

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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 congruenciesx 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

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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
• 140 2 x (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 solution223 2 x 105 223
  210 23
• Question Why do we choose 140, 63 and 30 (or,
  more precisely 4, 3 and 2)?

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• 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 3140 2 x 70 2 x 2 (5 x 7) 4 (5 x 7)

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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!

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General method of finding inverses

• Qin Jiushao, 1247 - 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 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

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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 lt pi
• Then there exists integer n satisfying the system
  of k congruenciesn r1 mod p1n r2 mod
  p2.....................n rk mod pk

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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 ax by 1 where a a / gcd (a,b)
  and b b / gcd (a,b)
• Criterion for an integer solutionEquation ax
  by c has an integer solution if and only if gcd
  (a,b) divides c

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5.4 Pells 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 Pells equationBrahmagupta
  Brâhma-sphuta-siddhânta 628 CE

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Brahmaguptas method

• Pells 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

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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 x1x2Ny1y2 and
  y x1y2x2y1 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(x1x2Ny1y2, x1y2x2y1, 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)

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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 gt 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 thatx 1151, y 120 is a solution
  of x2 92y2 1

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5.5 Pells 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 Brahmaguptas approach

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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 (abm) / 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
  Brahmaguptas method

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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, 8m, 3(m2 61))
• Dividing by 3 we get ( (8m 61) / 3, (8m) / 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!

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5.6 Rational Triangles

• Definition A triangle is called rational if it
  has rational sides and rational area
• Equivalently rational sides and altitudes
• Brahmaguptas Theorem Parameterization of
  rational trianglesIf a, b, c are sides of a
  rational triangle then for some rational numbers
  u, v and w we havea u2 / v v, b u2 / w
  wc u2 / v v u2 / w w

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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
  with the altitude h 2usplitting side c into
  segments c1 (u2 v2 ) / v and c2 (u2 v2
  ) / v

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5.7 Biographical NotesBrahmagupta and Bhâskara
II

• Brahmagupta (598 (approx.) 665 CE)
• Brâhma-sphuta-siddhânta
• teacher from Bhillamâla (now Bhinmal, India)
• prominent in astronomy and mathematics
• Pells equation
• general solution of quadratic equation
• area of a cyclic quadrilateral (which generalizes
  Herons formula for the area of triangle)
• parameterization of rational triangles

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• Bhâskara II (1114 1185)
• greatest astronomer and mathematician in
  12th-century India
• head of the observatory at Ujjain
• Lilavati (work named after his daughter)