mod m y x x y 1 (mod m ) gcd( x , m ) = 1 3 mod 5 2 3 2 6 - - PowerPoint PPT Presentation

mod m y x x y 1 mod m gcd x m 1 3 mod 5 2 3 2 6 1 mod 5
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mod m y x x y 1 (mod m ) gcd( x , m ) = 1 3 mod 5 2 3 2 6 - - PowerPoint PPT Presentation

mod m y x x y 1 (mod m ) gcd( x , m ) = 1 3 mod 5 2 3 2 6 1 (mod 5) 10 mod 12 10 x 1 (mod 12) mod m x , y a ax + my = 1 1 ax y mod m ax 1 (mod m ) x a 3 mod 5 3 x + 5 y = 1 1 3(2) + 5(1) = 1 3


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y x mod m x ⋅ y ≡ 1 (mod m) gcd(x, m) = 1 3 mod 5 2 3 ⋅ 2 ≡ 6 ≡ 1 (mod 5) 10 mod 12 10x ≡ 1 (mod 12)

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a mod m x, y ax + my = 1 ax 1 y mod m ax ≡ 1 (mod m) x a 3 mod 5 3x + 5y = 1 3(2) + 5(−1) = 1 ⇒ 3 ≡ 2 (mod 5) x, y

−1

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10 mod 12 10x + 12y = 1 10 12 5x + 6y = x, y a mod m gcd(a, m) = 1 2 1

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ax + my = 1 a, m 1

# Assumes a > b def gcd(a, b): if b == 0: return a return gcd(b, a % b)

gcd(26, 15) = gcd(15, 11) = gcd(11, 4) = gcd(4, 3) = gcd(3, 1) = gcd(1, 0) = 1 x, y

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(1) gcd(26, 15) ⇒ 26 = 1 ⋅ 15 + 11 ⇒ 11 = 26 − 1 ⋅ 15 (2) gcd(15, 11) ⇒ 15 = 1 ⋅ 11 + 4 ⇒ 4 = 15 − 1 ⋅ 11 (3) gcd(11, 4) ⇒ 11 = 2 ⋅ 4 + 3 ⇒ 3 = 11 − 2 ⋅ 4 (4) gcd(4, 3) ⇒ 4 = 1 ⋅ 3 + 1 ⇒ 1 = 4 − 1 ⋅ 3 gcd(a, b) = 1 gcd(some number, 1) (3) (4) (2) (1)

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1 = 4 − 1 ⋅ 3 = 4 − 1 ⋅ (11 − 2 ⋅ 4) = 3 ⋅ 4 − 11 = 3 ⋅ (15 − 1 ⋅ 11) − 11 = 3 ⋅ 15 − 4 ⋅ 11 = 3 ⋅ 15 − 4 ⋅ (26 − 1 ⋅ 15) = 7 ⋅ 15 − 4 ⋅ 26 15 ≡ 7 (mod 26) −4 ≡ 3 ≡ 26 (mod 7)

−1 −1

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9 (mod 14)

−1

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