M135F08A8S

M135F08A8S - MATH 135 Fall 2008 Assignment #8 Solutions...

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Unformatted text preview: MATH 135 Fall 2008 Assignment #8 Solutions Hand-In Problems 1. From the first congruence, 2 x 13 (mod 59) implies 2 x 72 (mod 59). Since gcd(2 , 59) = 1, we can divide out the common factor of 2 to obtain x 36 (mod 59). Thus, x = 36 + 59 y for some y Z . Substituting into the second congruence, 5(36 + 59 y ) 42 (mod 34) 180 + 295 y 42 (mod 34) 295 y - 138 (mod 34) 23 y 32 (mod 34) This is equivalent to the Diophantine equation 23 y + 34 z = 32. Using the Extended Euclidean Algorithm, 1 34 1 23- 1 1 11 1 3- 2 1 2- 34 23 11 Therefore, 23(3) + 34(- 2) = 1 so 23(96) + 34(- 64) = 32. Thus, since gcd(23 , 32) = 1, the solution to 23 y 32 (mod 34) is unique, so is y 96 28 (mod 34). So y = 28 + 34 w for some w Z . Thus, x = 36 + 59(28 + 34 w ) = 1688 + 2006 w Therefore, x 1688 (mod 2006). 2. We can rewrite the first and third congruences as x - 2 (mod 5) x - 2 (mod 19) Since gcd(5 , 19) = 1, then x - 2 (mod 95) by Proposition 3.64. Thus x =- 2 + 95 y for some y Z . Substituting into the second congruence,- 2 + 95 y 7 (mod 11) 7 y 9 (mod 11) 7 y 42 (mod 11) y 6 (mod 11) (since gcd(7 , 11) = 1) . Therefore, y = 6 + 11 z for some z Z . Thus, x =- 2 + 95(6 + 11 z ) = 568 + 1045 z , so x 568 (mod 1045). 3. We split the congruence into two congruences x 13 + 7 x + 5 (mod 7) x 13 + 7 x + 5 (mod 13) by using Proposition 3.64. Consider the first congruence. Since 7 is prime, then x 7 x (mod 7) for all x Z . Therefore, x 13 = x 6 x 7 x 6 x x 7 x (mod 7). So the congruence becomes x + 0 x + 5 0 (mod 7) or x - 5 2 (mod 7). Consider the second congruence. Since 13 is prime, then x 13 x (mod 13) for all x Z . So the congruence becomes x + 7 x + 5 0 (mod 13) or 8 x - 5 8 (mod 13). Since gcd(8 , 13) = 1, we can divide out the common factor of 8 to get x 1 (mod 13). Therefore, the original congruence is equivalent to the system x 2 (mod 7) x 1 (mod 13) From the second congruence, x = 1 + 13 y for some y Z . Substituting into the first of this pair, we get 1 + 13 y 2 (mod 7)- y 1 (mod 7) y - 1 (mod 7) Thus, y =- 1 + 7 z for some z Z . Therefore, x = 1 + 13(- 1 + 7 z ) =- 12 + 91 z , so x - 12 79 (mod 91). 4. p 4 q 4 (mod 15) is equivalent to p 4- q 4 0 (mod 15) which is equivalent to p 4- q 4 (mod 3) p 4- q 4 (mod 5) by Proposition 3.64, since gcd(3 , 5) = 1. Since 5 is prime, 5 6 | p and 5 6 | q , then p 4 1 (mod 5) and q 4 1 (mod 5) by Fermats Little Theorem. Thus, p 4- q 4 1- 1 0 (mod 5). Since 3 is prime, 3 6 | p , and 3 6 | q , then p 2 1 (mod 3) (by Fermats Little Theorem) so p 4 = ( p 2 ) 2 1 2 1 (mod 3)....
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M135F08A8S - MATH 135 Fall 2008 Assignment #8 Solutions...

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