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Cryptography

This text provides a comprehensive overview of the concepts of the greatest common divisor (GCD), multiplicative inverses, and their applications in linear congruences within the context of cryptography. It explains the method of calculating GCD using the Euclidean algorithm and demonstrates how to find integer solutions to equations involving GCD. Through various examples, including solving modular equations, the materials illustrate the process of obtaining multiplicative inverses and discuss their utility in cryptographic applications.

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Cryptography

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  1. Cryptography Inverses and GCD Piotr Faliszewski

  2. gcd(a, 0) = a gcd(a, b) = gcd(b, a mod b) a = b*q + r Here: q = a / b r = a mod b (a – b*q) Key idea express the first argument in terms of the second GCD(a,b)

  3. Let a, n – two integers A number a-1 s.t. a*a-1= 1 (mod n) is called a multiplicative inverse of a Theorem if gcd(a,b) = d then there are integers x and y s.t.,ax + by = d Multiplicative Inverse

  4. Let a, n – two integers If gcd( a, n ) = 1 then there are integers x,y: ax + ny = 1 then, x is a-1 Note ax + ny = 1 (mod n) ax = 1 (mod n) Theorem if gcd(a,b) = d then there are integers x and y s.t.,ax + by = d Multiplicative Inverse

  5. gcd(a ,b), r0=a, r1 = b gcd( r0, r1 ) r0 = q1r1 + r2 r1 = q2r2 + r3 r2 = q3r3 + r4 ... rk-1 = qkrk + rk+1 rk = qk+1rk+1 +0 Idea: sequences (xi) and (yi) ri = axi + byi build as you go Computing x,y via GCD

  6. gcd(a ,b), r0=a, r1 = b gcd( r0, r1 ) r0 = q1r1 + r2 r1 = q2r2 + r3 r2 = q3r3 + r4 ... rk-1 = qkrk + rk+1 rk = qk+1rk+1 +0 x0 = 1, y0 = 0 x1 = 0, y1 = 1 x2 = x0 - q1x1, y2 = y0 - q1y1 x3 = x1 - q2x2, y3 = y1 - q2y2 x4 = x2 - q3x3, y4 = y2 - q3y3 ... rk+1 = axk+1 + byk+1 xj+1 = xj-1 – qjxj yj+1 = yj-1 – qjyj Computing x,y via GCD

  7. gcd(a ,b), r0=45, r1 = 20 gcd( 45, 20 ) r0 = q1 r1 + r2 45 = 2  20 + 5 r1 = q2  r2 + r3 20 = 4  5 + 0 x0 = 1, y0 = 0 x1 = 0, y1 = 1 x2 = x0 - q1x1, y2 = y0 - q1y1 x2 = 1 – 2  0, y2 = 0 – 2  1 x2 = 1, y2 = -2 r3 = 0  computation ended gcd(45, 20) = 5 = 451 – 220 Example: GCD(45, 20)

  8. gcd(a ,b), r0=19, r1 = 7 gcd( 19, 7 ) 19 = 2  7 + 5 7 = 1  5 + 2 5 = 2  2 + 1 2 = 2  1 + 0 x0 = 1, y0 = 0 x1 = 0, y1 = 1 x2 = 1 – 2  0 = 1 y2 = 0 – 2  1 = -2 x3 = 0 – 1  1 = -1 y3 = 1 – 1  (-2) = 3 x4 = 1 – 2  (-1) = 3 y4 = -2 – 2  (3) = -8 19  3 + 7 (-8) = 57 - 56 = 1 Example: GCD(19, 7)

  9. Problem: Solve 7x = 10 (mod 19) 11  7 = 1 (mod 19) Thus (11  7)x = 11  10 (mod 19) x = 110 (mod 19) x = 15 (mod 19) Getting the inverse via GCD we know that gcd(19, 7) = 1 7*(-8) + 19*3 = 1 -8 is the multiplicative inverse of 7 (mod 19) -8 = 19 - 8 = 11 (mod 19) Solving Linear Congruences

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