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
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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
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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. 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
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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 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
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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. 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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This note was uploaded on 04/02/2012 for the course MATH 135 taught by Professor Andrewchilds during the Winter '08 term at Waterloo.

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MATH_135_15 - Objectives Linear Congruences Polynomial...

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