Course notes (chap. 1 Number Theory, chap. 2 ... - McGill University

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EXAMPLE gcd(7,19) x=1,y=0,x’=0,y’=1 = gcd(19,7) x=0,y=1,x’=1,y’=0 = gcd(7,19 mod 7) = gcd(7,5) x=1,y=0,x’=-2,y’=1 = gcd(5,7 mod 5) = gcd(5,2) x=-2,y=1,x’=3,y’=-1 = gcd(2,5 mod 2) = gcd(2,1) x=3,y=-1,x’=-8,y’=3 = gcd(1,2 mod 1) = gcd(1,0) = 1 x=-8,y=3,x’=19,y’=-7
EXAMPLE gcd(7,19) = gcd(1,0) = 1 x=-8,y=3,x’=19,y’=-7 thus 1 = -8x7 + 3x19 and 7 -1 mod 19 = 11 = -8 mod 19 19 -1 mod 7 = 3 = 3 mod 7
Page 1 and 2: COMP-547A /B Cryptography and Data
Page 3 and 4: Euclid Leonhard Euler Greatest Comm
Page 7 and 8: 1.2.1 Fast modular exponentiation T
Page 9 and 10: 1.2.2 GCD calculations and multipli
Page 11: EXAMPLE gcd(7,19) = gcd(19,7) = gcd
Page 15 and 16: EXAMPLE 7x ≡ 5 (mod 9) 7 -1 ∙7x
Page 17 and 18: Finally, we know that a solution x
Page 19 and 20: 1.3.1 Chinese Remainder Theorem Qin
Page 21 and 22: 1.4 Quadratic Residues Quadratic re
Page 23 and 24: Properties ( 1 =+1 n) ( ) ab ( a )
Page 25 and 26: EXAMPLE Jacobi(527, 723) = -Jacobi(
Page 27 and 28: Pierre de Fermat 1.4.2 Fermat-Euler
Page 29 and 30: For prime numbers p ≡ 1(mod4)anda
Page 31 and 32: EXAMPLE p = 37, a = 16 z = 5 (5 +
Page 37 and 38: EXAMPLE n = 37*43 = 1591, a = 16
Page 39 and 40: Definition 1.7 (SQROOT) The square
Page 41 and 42: 1.5 Prime numbers If we want a rand
Page 43 and 44: EXAMPLE n = 37*43 = 1591, n-1 = 159
Page 45 and 46: EXAMPLE n = 1597, n-1 = 1596 = 399*
Page 47 and 48: Algorithm 1.6 ( Miller-Rabin prime(
Page 49 and 50: In August of 2002, Agrawal, Kayal,
Page 51 and 52: Remark: If n is a prime, then it is
Page 53 and 54: 3: Pick x ∈ R Z n ∗ 4: c i ←
Page 55 and 56: Examples F 2 =({0, 1}, ⊕, ∧). F
Page 57 and 58: Algorithm 2.1 ( Primitive(q) ) 1: L
Page 59 and 60: Factoring q − 1... The only effic
Page 61 and 62: EXAMPLE n = 10000000 S = [2710128,
Page 63 and 64:
EXAMPLE n = 10000000 S = [4721693,
Page 65 and 66:
We also have operations g(x) modh(x
Page 67 and 68:
F 2 x +1 x 9 + x 4 +1 x 2 + x +1 x
Page 69 and 70:
2.3 General Fields Let p be a prime
Page 71 and 72:
10011×01110 = 01001 since (x 4 +x+
Page 73 and 74:
EXTRA SLIDES
Page 75 and 76:
First, consider the case e =2. Sinc
Page 77 and 78:
Second, notice that the same exact
Page 79 and 80:
Suppose Alice wants to distribute a
Page 81:
Algorithm 2.6 ( Solve(x 1 ,x 2 ,...
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Course notes (chap. 1 Number Theory, chap. 2 ... - McGill University

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