This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 135 Fall 2005 Assignment 3 Solutions HandIn Problems Exercise 211: Prove that gcd( ad,bd ) =  d  gcd( a,b ) . Solution: If d = 0, then both sides of the equation are equal to 0, so the result is true. Suppose d > 0 and let e = gcd( a,b ). Then the right side of the equation equals de . We must show that gcd( ad,bd ) = de . Since e  a , then a = qe for some integer q . This tells us that ad = qde and so de  ad by definition. Similarly, since e  b , then de  bd . Therefore, de is a common divisor of ad and bd . Since e = gcd( a,b ), by the Extended Euclidean Algorithm, there exist integers x and y so that ax + by = e . Multiplying through by d , we obtain ( ad ) x + ( bd ) y = de . But de is a common divisor of ad and bd , so by the GCD Characterization Theorem, de = gcd( ad,bd ). If d < 0, then set D = d =  d  . Then gcd( ad,bd ) = gcd( ad, bd ) = gcd( aD,bD ) = D gcd( a,b ) =  d  gcd( a,b ) from the result for positive D above. This concludes the proof, as we have proven the result for d = 0, d > 0 and d < 0. Exercise 224: If a = 3953 and b = 1829, write gcd( a,b ) in the form ax + by where x,y Z . Solution: Using the Extended Euclidean Algorithm, 3953 x +1829 y = r q i 1 3953 1 1829 1 2 295 2 6 13 59 6 31 67 5 Therefore 3953( 6) + 1829(13) = 59 = gcd(3953 , 1829) so a ( 6) + b (13) = gcd( a,b ). Exercise 236: Find one integer solution, if possible, to the following Diophantine equation. 91 x + 126 y = 203 Solution: First, we use the Extended Euclidean Algorithm to find the gcd(91 , 126). 126 y +91 x = r q i 1 126 1 91 1 1 35 1 2 3 21 2 3 4 14 1 5 7 7 1 13 18 2 Thus gcd(91 , 126) = 7 and 91(7) + 126( 5) = 7. Since 203 = 7(29), then 7  203, so the Diophantine equation 91 x + 126 y = 203 has a solution. Since 91(7) + 126( 5) = 7, then 91(29)(7) + 126(29)( 5) = 203, or 91(203) + 126( 145) = 203. Therefore, a solution is x = 203 and y = 145. Check: 91 203 126 145 = 18473 18270 = 203. 1 Exercise 242: Find all the integer solutions of 169 x 65 y = 91. Solution: Set w = y . We use the Extended Euclidean Algorithm to find a solution to the linear Diophantine equation 169 x + 65 w = 91, and at the end, set y = w : 169 x +65 w = r q i 1 169 1 65 1 2 39 2 1 3 26 1 2 5 13 1 5 13 2 Therefore, gcd(169 , 65) = 13 and 169(2) + 65( 5) = 13. Since gcd(169 , 65) = 13, then gcd(169 , 65) = 13. Also, we have 169(2) 65(5) = 13. Notice that 91 = 13 7, so the Diophantine equation 169 x 65 y = 91 does have a solution. Therefore, a particular solution to 169 x 65 y = 91 is x = 2 7 = 14, y = 5 7 = 35. This tells us that the complete integer solution to 169 x 65 y = 91 is x = 14 + ( 65 13 ) n = 14 5 n y = 35 ( 169 13 ) n = 35 13 n for all n Z ....
View Full
Document
 Spring '08
 Vanderburgh

Click to edit the document details