# Intermediate Math Circles March 7, 2012 Linear Diophantine Equations II

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1 Intermediate Math Circles March 7, 2012 Linear Diophantine Equations II Last week: How to find one solution to a linear Diophantine equation This week: How to find all solutions to a linear Diophantine equation Recall: Linear Diophantine Equations are equations of the form ax + by = c, where a, b, c are given integers and we are solving for integers x and y. For example: 25x + 10y = 155 or 1053x + 481y = 13 Main question: If a, b, and c are integers, when does a solution exist and how can you find a solution to ax + by = c where x and y are integers? Sometimes trial-and-error works, but this is not always very efficient. And sometimes a solution does not even exist! Last week we saw: Important Fact: ax + by = c has a solution if and only if gcd(a, b) divides c. An integer d divides another integer e if and only if there is some integer q such that e = qb. The greatest common divisor of a and b is the largest integer that divides a and divides b. We write it as gcd(a, b). What s the best way to calculate gcd(a, b)? For small numbers a and b it s not too hard to find gcd by factoring a and b, but not as easy when trying to calculate gcd(104723, ) or gcd(3551, 4399) You can try to find all divisors, and see which ones divide both, but that could take all day! 1

2 Instead, we ll use the Euclidean Algorithm and the Important Fact: If a = qb + r, then gcd(a, b) = gcd(b, r). The Euclidean Algorithm for calculating gcd(a, b): Step 1: Arrange a and b so that a b. Step 2: Write a = qb + r, with 0 r < b. Step 3: If r = 0, then b divides a, so gcd(a, b) = b. STOP! If not then gcd(a, b) = gcd(b, r). Step 4: Go to Step 2 to calculate gcd(b, r). Since the numbers get smaller after each iteration, you will eventually get an answer. Example 1: Calculate gcd(2173, 2491). Solution: gcd(2173, 2491) = gcd(2491, 2173) 2491 = , so gcd(2491, 2173) = gcd(2173, 318) 2173 = , so gcd(2173, 318) = gcd(318, 265) 318 = , so gcd(318, 265) = gcd(265, 53) 265 = , so gcd(265, 53) = 53 We have gcd(2173, 2491) = 53. 2

3 An application of the Euclidean Algorithm: Solving linear Diophantine equations. In general, how do we solve ax + by = c for integers x and y? 1. Calculate gcd(a, b) by using the Euclidean Algorithm. 2. Does gcd(a, b) divide c? a) If NO, then there is no solution to the linear Diophantine equation. b) If YES then (i) Solve ax + by = gcd(a, b) by working backwards through your steps in the Euclidean Algorithm. c (ii) Multiply the x and y in the solution by gcd(a, b). Example 2: Find integers x and y such that 2173x y = 53. Solution: From the Euclidean Algorithm: 2491 = (1) 2173 = (2) 318 = (3) 265 = (4) So gcd(2173, 2491) = 53, which divides c = 53, so there is a solution to this linear Diophantine equation. We find the solution by working backwards through our steps from the Euclidean Algorithm: 53 = from (3) = 318 1( ) from (2) = = 7( ) from (1) = 7(2491) 8(2173) Therefore, a solution is x = 8, y = 7. (Check!) 3

4 Example 3: Find integers x and y such that 2173x y = 159. Solution: Since gcd(2173, 2491) = 53 and 53 divides 159, we know that there is a solution to this linear Diophantine equation. Our first step is to solve 2173x y = gcd(2173, 2491) = 53. We already know that Multiplying both sides by 3: And so Therefore, 2173( 8) (7) = 53 3(2173( 8) (7)) = 3(53) 2173(3( 8)) (3(7)) = ( 24) (21) = 159 So one solution is x = 24, y = 21. Example 4: Find an integer solution to 2173x y = 210. Solution: In Example 2 we saw that gcd(2173, 2491) = 53, and since 53 does not divide 210, there is no integer solution. 4

5 Exercise Set 1 1. Use the Euclidean Algorithm to calculate gcd(3689, 4182). 2. Find an integer solution to 3689x y = 17, or explain why one doesn t exist. 3. Find an integer solution to 3689x y = 102, or explain why one doesn t exist. 4. Find an integer solution to 3689x y = 100, or explain why one doesn t exist. Answers to Exercise Set 1: 1. gcd(3689, 4182) = (x, y) = ( 17, 15) is one (of many) solutions. 3. (x, y) = ( 102, 90) is one (of many) solutions. 4. There is no solution since gcd(3689, 4182) = 17, and 17 does not divide

6 So we now know that ax + by = c has a solution (in the integers) if and only if gcd(a, b) divides c, and we know how to find a solution if there is one. How can we find more solutions? Let s go back to 2173x y = 53. In Example 2, we saw that one solution to this linear Diophantine equation is x = 8, y = 7. What is happening geometrically? This is the equation of a line through the point ( 8, 7): y (-8,7) 47 Slope of the line: ( a ) b ( Can we reduce? = a gcd(a,b) b gcd(a,b) ) x How do we find another lattice point? left 47, up 41 ( 55, 48) another? right 47, down 41 (39, 34) another? left 94, up 82 ( 102, 89) and so on... In general: Suppose x 0, y 0 is one solution to ax + by = c and let d = gcd(a, b). Then the full set of integer solutions is given by x = x 0 + n ( ) b ( a and y = y 0 n, where n is any integer. d d) 6

7 So for 2173x y = 53, since one solution is x = 8, y = 7, the complete solution is x = n = n 53 y = n = 7 41n where n is any integer What if we wanted, for example, x 150? Then we would need x = n 150 And so, 47n 158 Thus, n and since n is an integer, this implies that we must have n 4. What if we wanted, for example, x y? Then we would need x = n 7 41n = y And so, 88n 15 Thus, n and since n is an integer, this implies that we must have n 0. In general, to solve ax + by = c for integers x and y: 1. Calculate gcd(a, b) by using the Euclidean Algorithm. 2. Does gcd(a, b) divide c? a) If NO, then there is no solution to the linear Diophantine equation. b) If YES then (i) Solve ax + by = gcd(a, b) by working backwards through your steps in the Euclidean Algorithm. (Or, alternatively, find a solution by inspection if this is easy to do for the particular question). c (ii) Multiply the x and y in the solution by gcd(a, b). (iii) Use the formula ( ) to find all solutions to the Diophantine equation: b ( a x = x 0 + n and y = y 0 n d d) (iv) Use any restrictions in the problem (for example, x and y must be non negative) to restrict n and find the solutions you are interested in. 7

8 Example 5: A trucking company has to move 844 refrigerators. They have two types of trucks, one that carries 28 refrigerators and one that carries 34 refrigerators. The company only sends out full trucks, and the trucks return empty. List all possible ways to move all the refrigerators. Solution: Let x be the number of small trucks and y be the number of large trucks. Since a small truck can carry 28 refrigerators and a large truck can carry 34 refrigerators, we are interested in solving the linear Diophantine equation 28x + 34y = 844 First, let s use the Euclidean Algorithm to calculate gcd(28,34): 34 = (1) 28 = (2) 6 = (3) 4 = (4) So gcd(28, 34) = 2, which divides c = 844, so there is a solution to this linear Diophantine equation. We find a solution to 28x + 34y = 2 by working backwards through our steps from the Euclidean Algorithm: 2 = from (3) = 6 1(28 4 6) from (2) = = 5( ) 1 28 from (1) = 5(34) 6(28) Therefore, 28( 6) + 34(5) = 2. Let s use this to find a solution to 28x + 34y = 844. We already know that Multiplying both sides by 422: And so Therefore, 28( 6) + 34(5) = 2 422(28( 6) + 34(5)) = 422(2) 28(422 6) + 34(422 5) = ( 2532) + 34(2110) = 844 So a solution is x = 2532 and y = Are there more solutions? 8

9 Since x = 2532, y = 2110 is a particular solution, the complete solution is: x = n( 34) = n and y = 2110 n( 28 ) = n 2 2 Can x and y be negative? Since x and y represent trucks, we need x 0 and y 0: x 0 means that n 0, thus 17n 2532 and n Since n is an integer, we need n 149. y 0 means that n 0, thus 14n 2110 and n Since n is an integer, we need n 150. So 149 n 150. If n = 149, we obtain the solution x = (17) = 1 and y = (14) = 24 If n = 150, we obtain the solution x = (17) = 18 and y = (14) = 10 Therefore, there are two ways to move all the refrigerators. The trucking company could either send 1 small truck and 24 large trucks, or send 18 small trucks and 10 large trucks. 9

10 Exercise Set 2 1. Find all integer solutions to 3689x y = 17. Hint: See Exercise Set Find all integer solutions to 3689x y = 102. Hint: See Exercise Set a) Find all solutions to 12x + 57y = 3. b) Find all solutions to 12x + 57y = 423. c) Find all solutions to 12x + 57y = 423 where x 0 and y 0. Answers to Exercise Set 2: 1. x = n, y = n, where n is any integer. 2. x = n, y = n, where n is any integer. 3. a) x = n, y = 1 4n, where n is any integer. b) x = n, y = 141 4n, where n is any integer. c) (x, y) = (2, 7) or (x, y) = (21, 3) 10

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