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MATH_135_15

MATH_135_15 - Objectives Linear Congruences Polynomial...

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Objectives Linear Congruences Polynomial Congruences Preparing for RSA MATH 135 Faculty of Mathematics, University of Waterloo Lecture 15: Practice with Congruences Faculty of Mathematics, University of Waterloo MATH 135

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Objectives Linear Congruences Polynomial Congruences Preparing for RSA Content Content 1. Practice. Faculty of Mathematics, University of Waterloo MATH 135
Objectives Linear Congruences Polynomial Congruences Preparing for RSA Linear Congruences Solve 13 x 1 (mod 60) . Faculty of Mathematics, University of Waterloo MATH 135

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Objectives Linear Congruences Polynomial Congruences Preparing for RSA Linear Congruences Solve 13 x 1 (mod 60) . Solution Since gcd(13 , 60) = 1 and 1 | 1 we would expect to ﬁnd one congruence class as a solution to 13 x 1 (mod 60) . Faculty of Mathematics, University of Waterloo MATH 135
Objectives Linear Congruences Polynomial Congruences Preparing for RSA Linear Congruences Solve 13 x 1 (mod 60) . Solution Since gcd(13 , 60) = 1 and 1 | 1 we would expect to ﬁnd one congruence class as a solution to 13 x 1 (mod 60) . Now 13 x 1 (mod 60) is equivalent to the linear Diophantine equation 13 x + 60 y = 1 so we can use the EEA. Faculty of Mathematics, University of Waterloo MATH 135

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Objectives Linear Congruences Polynomial Congruences Preparing for RSA Linear Congruences Solve 13 x 1 (mod 60) . Solution Since gcd(13 , 60) = 1 and 1 | 1 we would expect to ﬁnd one congruence class as a solution to 13 x 1 (mod 60) . Now 13 x 1 (mod 60) is equivalent to the linear Diophantine equation 13 x + 60 y = 1 so we can use the EEA. x y r q 1 0 60 0 0 1 13 0 1 - 4 8 4 - 1 5 5 1 2 - 9 3 1 - 3 14 2 1 5 - 23 1 1 - 13 60 0 2 Faculty of Mathematics, University of Waterloo MATH 135
Objectives Linear Congruences Polynomial Congruences Preparing for RSA Linear Congruences Solve 13 x 1 (mod 60) . Solution Since gcd(13 , 60) = 1 and 1 | 1 we would expect to ﬁnd one congruence class as a solution to 13 x 1 (mod 60) . Now 13 x 1 (mod 60) is equivalent to the linear Diophantine equation 13 x + 60 y = 1 so we can use the EEA. x y r q 1 0 60 0 0 1 13 0 1 - 4 8 4 - 1 5 5 1 2 - 9 3 1 - 3 14 2 1 5 - 23 1 1 - 13 60 0 2 Thus 13( - 23) + 60(5) = 1 and so x ≡ - 23 37 (mod 60) is a Faculty of Mathematics, University of Waterloo MATH 135

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Objectives Linear Congruences Polynomial Congruences Preparing for RSA Polynomial Congruences No eﬃcient means to solve polynomial congruences. Faculty of Mathematics, University of Waterloo MATH 135
Objectives Linear Congruences Polynomial Congruences Preparing for RSA Polynomial Congruences No eﬃcient means to solve polynomial congruences. Solve x 2 1 (mod 8) by substitution. Faculty of Mathematics, University of Waterloo MATH 135

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Objectives Linear Congruences Polynomial Congruences Preparing for RSA Polynomial Congruences No eﬃcient means to solve polynomial congruences. Solve x 2 1 (mod 8) by substitution.
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MATH_135_15 - Objectives Linear Congruences Polynomial...

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