135 MATH Assignment #4
Hand-In Problems
Fall 2007
Due: Thursday 11 October 2007, 8:20 a.m.
1. Which of the following linear Diophantine equations have solutions? In each case, explain briey why or why not. If there is a solution, determine the complete solution. (a) 28x + 91y = 40 (b) 2007x 897y = 15 2. (a) Determine all non-negative integer solutions to the linear Diophantine equation 133x + 315y = 98000. (b) Determine all non-negative integer solutions to the linear Diophantine equation 133x + 315y = 98000 with the additional property that x 640 2y. 3. Find the smallest positive integer x so that 141x leaves a remainder of 21 when divided by 31. 4. Suppose a, b, c Z. Prove that gcd(a, c) = gcd(b, c) = 1 if and only if gcd(ab, c) = 1. 5. Suppose a, b, n Z. Prove that if n 0, then gcd(an, bn) = n gcd(a, b). 6. Suppose that f (x) is a polynomial of degree n with coecients from Q and suppose that g(x) = x c for some c Q. (a) When f (x) is divided by g(x), we obtain a quotient q(x) and a remainder r(x). (See Assignment #3.) Explain why, in this case, r(x) is a constant r(x) = r Q. (b) Prove that r = f (c). 7. Consider the system of equations a + b = 2m2 b + c = 6m a+c = 2 Determine all real values of m for which a b c. (This problem is not directly related to the course material, but is included to keep your problem solving skills sharp.) Recommended Problems 1. Text, page 50, #42 2. Text, page 51, #44 3. Text, page 51, #48 4. Text, page 52, #75 5. Text, page 52, #79 ...continued
6. Let a, b and c be non-zero integers. Their greatest common divisor gcd(a, b, c) is the largest positive integer that divides all of them. (a) If d = gcd(a, b, c), prove that d is a common divisor of a and gcd(b, c). (b) If f is a common divisor of a and gcd(b, c), prove that f is a common divisor of a, b and c. (c) Prove that gcd(a, b, c) = gcd(a, gcd(b, c)).
MATH 135 Assignment #4 Solutions
Hand-In Problems 1. (a) By inspection, gcd(28, 91) = 7. (We could calculate this using the Euclidean Algorithm instead.) Since 7 | 40, there are no solutions. (b) Set z = y. We solve 2007x + 897z = 15 rst. We nd gcd(2007, 897) using the Extended Euclidean Algorithm: 1 0 2007 0 1 897 1 2 213 4 9 45 17 38 33 21 47 12 59 132 9 80 179 3 299 669 0
Fall 2007
2 4 4 1 2 1 3
So gcd(2007, 897) = 3 | 15 so there are solutions. Since 2007(80) + 897(179) = 3, then, multiplying both sides by 5, we get 2007(400) + 897(895) = 15. Therefore, (400, 895) is a particular solution to 2007x + 897z = 15. Therefore, the complete solution to 2007x + 897z = 15 is x = 400 + 897 n = 400 + 299n 3 2007 z = 895 3 n = 895 669n and so the complete solution to 2007x 897y = 15 is x = 400 + 299n y = z = 895 + 669n 2. (a) Using the Extended Euclidean Algorithm, 0 1 315 1 0 133 2 1 49 5 2 35 7 3 14 19 8 7 45 19 0 nZ nZ
2 2 1 2 2
so gcd(133, 315) = 7. Since 7 | 98000 (with quotient 14000), there are solutions. Since 133(19) + 315(8) = 7, then, multiplying both sides by 14000, we get 133(266000) + 315(112000) = 98000.
Therefore, (266000, 112000) is a particular solution to 133x + 315y = 98000. The complete solution is thus x = 266000 + 315 n = 266000 + 45n 7 y = 112000 133 n = 112000 19n 7 For non-negative solutions,
5 x 0 266000 + 45n 0 n 266000 = 5911 45 n 5911 since n Z 45 y 0 112000 19n 0 n 112000 = 5894 14 n 5895 since n Z 19 19
nZ
Combining these inequalities, the non-negative solutions are when 5911 n 5895, so x = 266000 + 315 n = 266000 + 45n 7 with 5911 n 5895 y = 112000 133 n = 112000 19n 7 (b) We use the solutions from (a) and add the condition that x 640 2y. Using the results for x and y in terms of n from (a), we have 266000 + 45n 640 2(112000 19n) 266000 + 45n 640 + 224000 + 38n 7n 41360 n 41360 = 5908 4 7 7 Since n Z, then n 5909. Since x, y 0, we also have to combine this with the restriction 5911 n 5895 from (a), to get 5911 n 5909. Therefore, n = 5911, 5910, 5909. Using the formulae for x and y, these give (x, y) = (5, 309), (50, 290), (95, 271), respectively. 3. We want 141x = 21 + 31q for some q Z. We solve 141x + 31y = 21 (setting y = q). Using the Extended Euclidean Algorithm, 1 0 141 0 1 31 1 4 17 1 5 14 2 9 3 9 41 2 11 50 1 3 141 0
4 1 1 4 1 2
Since 141(11) + 31(50) = 1, then, multiplying both sides by 21, 141(231) + 31(1050) = 21. Therefore, (231, 1050) is a particular solution to 141x + 31y = 21. Thus, the complete solution to 141x + 31y = 21 is x = 231 + 31 n = 231 + 31n 1 y = 1050 141 n = 1050 141n 1 nZ
So the complete solution to 141x = 21 + 31q is x = 231 + 31n q = y = 1050 + 141n For x > 0, 231 + 31n > 0 or n > 231 = 7 14 . 31 31 Since n Z, n 7. The smallest value of x occurs when n = 7, so x = 14. 4. = Suppose that gcd(ab, c) = 1. Then, there exist integers x and y so that abx + cy = 1 by Proposition 2.27(i). Thus, a(bx) + cy = 1, so gcd(a, c) = 1 by Proposition 2.27(i), since there exist integers u and v with au + cv = 1. Similarly, b(ax) + cy = 1, so gcd(b, c) = 1. Therefore, if gcd(ab, c) = 1, then gcd(a, c) = gcd(b, c) = 1. = Suppose that gcd(a, c) = gcd(b, c) = 1. Then there exist integers x, y, u, v so that ax + cy = 1 and bu + cv = 1. Combining these equations, (ax + cy)(bu + cv) = 1 1 abux + acvx + bcuy + c2 vy = 1 ab(ux) + c(avx + buy + cvy) = 1 Since a, b, c, u, v, x, y Z, then ux Z and avx + buy + cvy Z. Since ab(ux) + c(avx + buy + cvy) = 1, then gcd(ab, c) = 1 by Proposition 2.27(i). Therefore, if gcd(a, c) = gcd(b, c) = 1, then gcd(ab, c) = 1. Thus, gcd(a, c) = gcd(b, c) = 1 if and only if gcd(ab, c) = 1. 5. If n = 0, L.S.= gcd(0, 0) = 0 and R.S.= 0, so the result is true. Suppose now that n > Let 0. d = gcd(a, b). Since d | a and d | b, then a = dq and b = dQ for some q, Q Z. Thus, an = ndq and bn = ndQ, so nd | an and nd | bn, so nd is a common divisor of an and bn. Since d = gcd(a, b), then there exist x, y Z such that ax + by = d. Then anx + bny = nd. By Proposition 2.24, gcd(an, bn) = nd, since nd is a common divisor of an and bn. Therefore, if n 0, gcd(an, bn) = n gcd(a, b). 6. (a) From the given information, f (x) = q(x)g(x) + r(x). From the Division Algorithm, r(x) = 0 or the degree of r(x) is less than the degree of g(x) (which is 1). Thus, r(x) = 0 or the degree of r(x) is 0. In either case, r(x) is a constant, say r(x) = r. nZ
(b) We know that f (x) = q(x)g(x) + r(x) = q(x)(x c) + r. Substituting x = c gives f (c) = q(c)(c c) + r = q(c)(0) + r = r, as required. 7. Adding the three equations, we obtain 2a + 2b + 2c = 2m2 + 6m + 2. Dividing by 2, we obtain a + b + c = m2 + 3m + 1. Thus, a = m2 + 3m + 1 (b + c) = m2 + 3m + 1 6m = m2 3m + 1. Similarly, b = m2 + 3m + 1 (a + c) = m2 + 3m + 1 2 = m2 + 3m 1. Also, c = m2 + 3m + 1 (a + b) = m2 + 3m + 1 2m2 = m2 + 3m + 1. 1 For a b, we have m2 3m + 1 m2 + 3m 1 or 2 6m or m 3 . 2 2 2 2 For b c, we have m + 3m 1 m + 3m + 1 or 2m 2 or m 1 so 1 m 1. 1 Since we want a b c, then m 3 and 1 m 1, so 1 m 1. 3 Recommended Problems 1. Set z = y. We solve 169x + 65z = 91 using the Extended Euclidean Algorithm: 1 0 169 0 1 65 1 2 39 1 3 26 2 5 13 5 13 0
2 1 1 2
Therefore, gcd(169, 65) = 13 | 91, so there are solutions. Since 169(2) + 65(5) = 13, then, multiplying both sides by 7, we get 169(14) + 65(35) = 91. Therefore, (14, 35) is a particular solution to 169x + 65z = 91. Therefore, the complete solution to 169x + 65z = 91 is x = 14 + 65 n = 14 + 5n 13 z = 35 169 n = 35 13n 13 and so the complete solution to 169x 65y = 91 is x = 14 + 5n z = y = 35 + 13n 2. Using the Extended Euclidean Algorithm, 0 1 57 1 0 12 4 19 5 1 3 19 40 nZ nZ
4 1 3
so gcd(12, 57) = 3 | 423, so there are solutions. Since 12(5) + 57(1) = 3, then, multiplying both sides by 141, we get 12(705) + 57(141) = 423.
Therefore, (705, 141) is a particular solution to 12x + 57y = 423. Therefore, the complete solution to 12x + 57y = 423 is x = 705 + 57 n = 705 + 19n 3 y = 141 12 n = 141 4n 3 For non-negative solutions,
2 x 0 705 + 19n 0 n 705 = 37 19 n 37 since n Z 19
nZ
y 0 141 4n 0 n 141 = 35 1 n 36 since n Z 4 4 Combining these inequalities, we obtain 37 n 36. Therefore, the non-negative solutions occur when n = 36 and 37, so (x, y) = (21, 3), (2, 7). 3. The problem is asking whether or not the linear Diophantine equation 11x + 17y = 120 has positive solutions. By inspection, gcd(11, 17) = 1 and 11(3) + 17(2) = 1 so 11(360) + 17(240) = 120. (The Extended Euclidean Algorithm could be used instead to nd a particular solution.) Thus, the complete solution to 11x + 17y = 120 is x = 360 + 17n y = 240 11n For positive solutions, x > 0 360 + 17n > 0 n > y > 0 240 11n > 0 n <
3 360 = 21 17 17 240 9 = 21 11 n 11
nZ
n 22 since n Z 21 since n Z
Thus two inequalities do not yield an admissible value for n, so there are no positive solutions to the linear Diophantine equation. Thus, 120 cannot be written as the sum of a positive multiple of 11 and a positive multiple of 17. 4. We want to show If ax2 + by 2 = c has a solution, then gcd(a, b) | c. Suppose ax2 + by 2 = c has a solution x = X and y = Y (that is, aX 2 + bY 2 = c) and d = gcd(a, b). Then d | a and d | b, so d | aX 2 + bY 2 by Proposition 2.11(ii). Since ax2 + by 2 = c, then d | c. If gcd(a, b) | c, there are not necessarily solutions. For example, x2 + y 2 = 3 has no integer solutions since x2 , y 2 0 and the only possible values for x2 and y 2 less then 3 are 0 and 1. We cannot combine 0 and 1 to get 3. Thus, x2 + y 2 = 3 has no integer solutions, so if gcd(a, b) | c the equation ax2 + by 2 = c does not necessarily have integer solutions. 5. Let x be the number of small trucks being used and y the number of large trucks being used. We want to solve 28x + 34y = 844, where x, y are non-negative integers.
Using the Extended Euclidean Algorithm, 0 1 34 1 0 28 1 16 5 4 4 6 52 17 14 0
1 4 1 2
so gcd(28, 34) = 2 | 844, so there are solutions. Since 28(6) + 34(5) = 2, then, multiplying both sides by 422, we get 28(2532) + 34(2110) = 844. Therefore, (2532, 2110) is a particular solution to 28x + 34y = 844. Therefore, the complete solution is x = 2532 + 34 n = 2532 + 17n 2 y = 2110 28 n = 2110 14n 2 For non-negative solutions, x 0 2532 + 17n 0 n y 0 2110 14n 0 n
2532 16 = 148 17 17 2110 = 150 10 n 14 14
nZ
n 149 since n Z 150 since n Z
Combining these inequalities, 149 n 150, so (x, y) = (1, 24) or (18, 10), so the company must use 1 small truck and 24 large trucks, or 18 small trucks and 10 large trucks. 6. (a) Since d = gcd(a, b, c). Then d | a, d | b and d | c. Since d | a, then d | a. Since d | b and d | c, then d | gcd(b, c) by Proposition 2.29. Therefore, d is a common divisor of a and gcd(b, c). (b) Suppose f is a common divisor of a and gcd(b, c). Since f | gcd(b, c) and gcd(b, c) | b, then f | b by Proposition 2.11 (i). Similarly, f | c. Thus, f | a, f | b, f | c (that is, f is a common divisor of a, b and c). (c) If f is a common divisor of a and gcd(b, c), then f | a, f | b, f | c by (b). Since d = gcd(a, b, c), then f d, because f is a common divisor of a, b and c and d is the greatest common divisor of a, b and c. Since d is a common divisor of a and gcd(b, c) by part (a) and f d for every common divisor f of a and gcd(b, c), then d = gcd(a, gcd(b, c)).