M135F08A8S

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

This preview shows pages 1–3. Sign up to view the full content.

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, 0 1 34 1 0 23 - 1 1 11 1 3 - 2 1 2 - 34 23 0 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).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3. We split the congruence into two congruences x 13 + 7 x + 5 0 (mod 7) x 13 + 7 x + 5 0 (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 0 (mod 3) p 4 - q 4 0 (mod 5) by Proposition 3.64, since gcd(3 , 5) = 1. Since 5 is prime, 5 | p and 5 | q , then p 4 1 (mod 5) and q 4 1 (mod 5) by Fermat’s Little Theorem. Thus, p 4 - q 4 1 - 1 0 (mod 5). Since 3 is prime, 3 | p , and 3 | q , then p 2 1 (mod 3) (by Fermat’s Little Theorem) so p 4 = ( p 2 ) 2 1 2 1 (mod 3). Similarly, q 4 1 (mod 3), so p 4 - q 4 1 - 1 0 (mod 3). Therefore, p 4 q 4 (mod 15).
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern