135 MATH Assignment 3 Solutions
Hand-In Problems
Exercise 2-11: 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.
Fall 2005
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 2-24:
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, 3953x + 1829y = r 1 0 1 -6 31 0 3953 1 1829 -2 295 13 59 -67 0 qi
2 6 5
Therefore 3953(-6) + 1829(13) = 59 = gcd(3953, 1829) so a(-6) + b(13) = gcd(a, b).
Exercise 2-36:
Find one integer solution, if possible, to the following Diophantine equation. 91x + 126y = 203 Solution: First, we use the Extended Euclidean Algorithm to find the gcd(91, 126). 126y + 91x = r 1 0 1 -2 3 -5 13 0 126 1 91 -1 35 3 21 -4 14 7 7 -18 0 qi
1 2 1 1 2
Thus gcd(91, 126) = 7 and 91(7) + 126(-5) = 7. Since 203 = 7(29), then 7 | 203, so the Diophantine equation 91x + 126y = 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 2-42: Find all the integer solutions of 169x - 65y = 91.
Solution: Set w = -y. We use the Extended Euclidean Algorithm to find a solution to the linear Diophantine equation 169x + 65w = 91, and at the end, set y = -w: 169x + 65w = r 1 0 1 -1 2 -5 0 169 1 65 -2 39 3 26 -5 13 13 0 qi
2 1 1 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 169x - 65y = 91 does have a solution. Therefore, a particular solution to 169x - 65y = 91 is x = 2 7 = 14, y = 5 7 = 35. This tells us that the complete integer solution to 169x - 65y = 91 is x = 14 + y = 35 -
-65 13 169 13
n = 14 - 5n n = 35 - 13n for all n Z.
Check: 169(14 - 5n) - 65(35 - 13n) = 2366 - 845n - 2275 + 845n = 91.
Exercise 2-44:
Find all the nonnegative integer solutions of 12x + 57y = 423. Solution: Either by inspection or by using the Euclidean Algorithm, we can see that gcd(12, 57) = 3 and 3 | 423, so integer solutions exist. To simplify the problem, we can divide through by the gcd to reduce the size of the numbers. This gives us the equation 4x + 19y = 141. (Note that gcd(4, 19) = 1.) By inspection 4(5)+19(-1) = 1. (Again, we could use the Euclidean Algorithm and reverse technique or the Extended Euclidean Algorithm if we couldn't find a solution just by looking.) Multiplying by 141 gives 4(705) + 19(-141) = 141, so a particular solution is x0 = 705 and y0 = -141. Thus, the complete solution is b x = x0 + n = 705 + 19n d for all n Z. a y = y0 - n = -141 - 4n d (Notice that we can think of a and b as 12 and 57 with d = gcd(12, 57) = 3 or we can think of a and b as 4 and 19 with d = gcd(4, 19) = 1: both give the same result.) Since we want non-negative solutions, then x = 705 + 19n 0 or n - 705 or n -37, since n is an integer. 19 Similarly, y = -141 - 4n 0 which yields that n -36. Therefore, n can take the values -37, -36, which give the non-negative solutions (x, y) = (2, 7), (21, 3). Check: 12(21) + 57(3) = 252 + 171 = 423 and 12(2) + 57(7) = 24 + 399 = 423.
Problem 2-74:
Show that gcd(ab, c) = gcd(b, c) if gcd(a, c) = 1. Is it true in general that gcd(ab, c) = gcd(a, c) gcd(b, c) ? Solution: Let d = gcd(b, c). We must show that d = gcd(ab, c). Since d = gcd(b, c), then d | b and d | c. Since d | b, then d | ab. Therefore, d is a common divisor of ab and c. We must show that d is the greatest of all common divisors of ab and c. By the Extended Euclidean Algorithm, since gcd(a, c) = 1, then there exist integers u and v such that au + cv = 1. 2
Hence abu + cbv = b (multiplying through by b). If e is a common divisor of ab and c then e | abu + cbv = b by Proposition 2.11.(ii), so e | b and e | c. Because e is also a common divisor of b and c then |e| d, since d = gcd(b, c). It follows that d = gcd(ab, c), since every common divisor of ab and c is at most d. Therefore, gcd(ab, c) = gcd(b, c). (Alternatively, we could say that there exist integers u, v such that au + cv = 1 and w, z such that bw + cz = d. Then (au + cv)(bw + cz) = 1(d) or ab(uw) + c(bvw + auz + cvz) = d so the equation (ab)x + cy = d has an integer solution, so by the GCD Characterization Theorem, d = gcd(ab, c).) In general it is not true that gcd(ab, c) = gcd(a, c) gcd(b, c). For example take a = b = c = 2. Then gcd(ab, c) = 2 = 2 2 = gcd(a, c) gcd(b, c) so the statement is not true in general.
Problem 2-76:
For what values of a and b does the Diophantine equation ax + by = c have an infinite number of positive solutions for x and y? Solution: If a = b = 0, then 0x + 0y = c has a solution if and only if c = 0. When c = 0, every pair of integers x and y is a solution, so there are infinitely many positive solutions. Suppose a and b are not both 0. In order to have any integer solutions, we must have gcd(a, b) | c. If a is nonzero, b = 0, and a and c have the same sign, then there is one positive solution for x, say x = x0 , but an infinite number of possible values for y, so any pair (x0 , y) with y a positive integer will be a positive solution to ax + 0y = c. Similarly, if b is nonzero, a = 0, and b and c have the same sign, then there will be an infinite number of positive solutions. If a and b are both nonzero, let d = gcd(a, b), and suppose there is at least one integer solution x = x0 , y = y0 . na nb , for all n Z. The general solution is x = x0 + , y = y0 - d d There are four cases: i) If a > 0 and b > 0, then for n sufficiently large and positive (say n > N , N a positive integer), the solutions will have opposite signs (ie. will not both be positive). If n is sufficiently large and negative (say n < M , M a negative integer), the solutions will have opposite signs. Therefore the only possible positive solutions are when M n N , so there are only finitely many possible values for n. Therefore there will only be finitely many positive solutions. ii) If a < 0 and b < 0 then, as in i), there will be only be finitely many positive solutions. iii) If a > 0 and b < 0, then for n is sufficient large and negative (say n < M , M a negative integer), both x and y will be positive. Therefore, all solutions with n < M will be positive solutions, so there are infinitely many positive solutions. iv) If a < 0 and b > 0, then for n sufficiently large, all solutions will be positive so we will have infinitely many positive solutions, as in iii). Hence there will be a infinite number of positive solutions if gcd(a, b) | c, a and b are both non-zero, and a and b have opposite signs, if gcd(a, b) | c, one of a and b is zero, and the non-zero one of a and b has the same sign as c, or if a = b = c = 0.
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Problem 2-83:
Let a, b, c be nonzero integers. Their greatest common divisor gcd(a, b, c) is the largest positive integer that divides all of them. Prove that gcd(a, b, c) = gcd(a, gcd(b, c)). Solution: Note that g = gcd(a, b, c) is non-negative. Let d = gcd(a, gcd(b, c)). By definition, d | a and d | gcd(b, c), and d 0. The latter implies that d | b and d | c since d | gcd(b, c) and gcd(b, c) | b, and d | gcd(b, c) and gcd(b, c) | c. Thus d is a common divisor of a, b and c, and so d g, since g = gcd(a, b, c). Now g | a, g | b, and g | c. Because g | b and g | c, Proposition 2.29.(iii) implies that g | gcd(b, c). Since g | a, then g is a common divisor of a and gcd(b, c), and so g d, since d = gcd(a, gcd(b, c)). Therefore, g d and d g. Hence gcd(a, b, c) = g = d = gcd(a, gcd(b, c)).
Recommended Problems
Exercise 2-29: Simplify
95 40 + . 646 391
Solution: First, we try to reduce each fraction. Using the Euclidean Algorithm we can find the gcd(95, 646) to simplify the first fraction. 646 = 6 95 + 76 95 = 1 76 + 19 76 = 4 19 + 0 Hence gcd(95, 646) = 19. So, we can simplify Next, taking a common denominator we get
95 646
into
5 34 .
(We can quickly see that
40 391
is in lowest terms.)
5 40 3315 + = 34 391 13294 To simplify the final answer we use again the Euclidean Algorithm to find the gcd(3315, 13294). 13294 = 4 3315 + 34 3315 = 17 + 34 17 34 = 2 17 + 0 Hence gcd(3315, 13294) = 17. Dividing by 17 we get 3315 195 = . 13294 782
Exercise 2-41: Find all the integer solutions of 243x + 405y = 123.
Solution: First, we use the Extended Euclidean Algorithm to find the gcd(243, 405): 243x + 405y = r 1 0 1 -1 2 -5 0 1 0 1 -1 3 243 405 243 162 81 0 4 qi
0 1 1 2
Since gcd(243, 405) = 81, and 81 does not divide 123 (by the Division Algorithm, 123 = 1 81 + 42), the Diophantine equation 243x + 405y = 123 has no integer solutions.
Exercise 2-45: Find all the non-negative integer solutions to the Diophantine equation 38x + 34y = 200.
Solution: Since gcd(34, 38) = 2 (by inspection) and 2 | 200 a solution exists. Dividing the equation by 2 gives the equivalent equation 19x + 17y = 100. We use the Extended Euclidean Algorithm to find a particular solution to this Diophantine equation: 19x + 17y = r 1 0 1 -8 17 0 19 1 17 -1 2 9 1 -19 0 qi
1 8 2
We get 19(-8) + 17(9) = 1. Multiplying by 100 gives 19(-800) + 17(900) = 100. Therefore, a particular solution is x = -800 and y = 900. This tells us that the general solution is x = -800 + 17n y = 900 - 19n for all n Z.
Since we want non-negative solutions, we need x = -800 + 17n 0 or n 800 or n 48 since n is an integer. 17 Similarly, we need y = 900 - 19n 0 or n 900 or n 47, since n is an integer. 19 There are no integers less than or equal to 47 and greater than or equal to 48, so there are no non-negative solutions to the Diophantine equation 38x + 34y = 200.
Exercise 2-46:
Find all the nonnegative integer solutions of 11x - 12y = 13. Solution: Since gcd(11, 12) = 1, and 11(-1) - 12(-1) = 1, it follows that x0 = -13 and y0 = -13 is a particular solution for 11x - 12y = 13. Hence the general solution is x = -13 - 12n for all n Z. y = -13 - 11n
13 For non-negative solutions, we need -13 - 12n 0 (or n - 12 or n -2 since n is an integer) and -13 - 11n 0 13 (or n - 11 or n -2 since n is an integer). Therefore, we need n -2. Thus the complete set of non-negative solutions is
x = -13 - 12n y = -13 - 11n
for all integers n with n -2
Problem 2-75:
Show that the Diophantine equation ax2 + by 2 = c does not have any integer solutions unless gcd(a, b)|c. If gcd(a, b)|c, does the equation always have an integer solution? Solution: If a = b = 0 then the equation only has solutions if c = 0, and then any pair (x, y) with x, y Z will work. Now assume that a and b are not both 0, and suppose that the equation has a solution x0 and y0 . 2 Then ax2 + by0 = c. 0 Let d = gcd(a, b) = 0, so d | a and d | b. 2 By Proposition 2.11(ii), d | ax2 + by0 , so d | c. 0 The converse of this statement is not true, to see this let a = b = 1 and c = 3. Then gcd(1, 1) = 1, which does divide 3. However, if x2 + y 2 = 3, then x2 3 and y 2 3 (since x2 and y 2 are both at least 0), so |x| 1 and |y| 1. None of these integer pairs give solutions, so ax2 + by 2 = c does not have a solution in this case.
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Problem 2-77:
For what values of c does 8x + 5y = c have exactly one strictly positive solution? Solution: Let x0 , y0 be a positive solution to 8x + 5y = c. The general solution is given by x = x0 + 5n y = y0 - 8n
for all n Z .
If x0 , y0 is to be the only positive solution, then n = 0 is the only value of n to give a positive solution (since n = 0 gives (x, y) = (x0 , y0 )). Now, if n = 1, x0 + 5n = x0 + 5 will still be positive, so we must have y0 - 8 0 so y0 8, since n = 1 doesn't give a solution. Similarly, if n = -1, y0 - 8n = y0 + 8 will still be positive, so x0 - 5 0 so x0 5. So for there to be only one positive solution, we must have 1 x0 5 and 1 y0 8. Hence, c {8x0 + 5y0 | 1 x0 5, 1 y0 8}. This gives the 40 values, c = 13, 18, 21, 23, 26, 67, 70, 72, 75, 80 and all c in the range 28 c 65 except 30, 32, 35, 40, 53, 58, 61 and 63.
Problem 2-78:
An oil company has a contract to deliver 100000 liters of gasoline. Their tankers can carry 2400 liters and they can attach one trailer carrying 2200 liters to each tanker. All the tankers and trailers must be completely full on this contract, otherwise the gas would slosh around too much when going over some rough roads. Find the least number of tankers required to fulfill the contract. Each trailer, if used, must be pulled by a full tanker. Solution: Let the number of tankers by x, and the number of trailers be y, where y x (since each trailer must be pulled by a tanker). From the data given, we want to solve the equation 2400x + 2200y = 100 000. Dividing through by 200, we obtain 12x + 11y = 500. Clearly, gcd(12, 11) = 1, so this equation has an integer solution. Now 12(1)+11(-1) = 1, so 12(500)+11(-500) = 500, and the complete solution is x = 500 + 11n for all n Z. y = -500 - 12n We want x 0 and y 0 and x y. It is enough to look at y 0 and x y (since these together imply that x 0). So we want y = -500 - 12n 0 (or n - 500 or n -42 since n is an integer), and 500 + 11n -500 - 12n (or 12 23n -1000 or n - 1000 or n -43 since n is an integer). 23 Therefore, -43 n -42 and the two possible solutions are (x, y) = (27, 16), (38, 4). Therefore the least number of tankers required is 27.
Problem 2-84:
Prove that the Diophantine equation ax + by + cz = e has a solution if and only if gcd(a, b, c) | e. Solution: We follow the outline of the proof of Theorem 2.31 (i). Let d = gcd(a, b, c). (=) Suppose that there are integers x, y, z for which ax + by + cz = e. By definition of gcd, d | a, d | b, and d | c. Therefore d | ax + by by Proposition 2.11.ii, so d | ax + by and d | c. Thus, d | (ax + by) + cz by Proposition 2.11.ii again, i.e., d | e. (=) Suppose that d | e, so e = de1 for some e1 in Z. Since we are able to solve Linear Diophantine Equations in two variables, we try to use this to help solve such an 6
equation with three variables. Let f = gcd(b, c). By Problem 2-83 (see above), d = gcd(a, f ). By the Extended Euclidean Algorithm since d = gcd(a, f ), there exist integers x and y such that ax + f y = d By the Extended Euclidean Algorithm again since f = gcd(b, c), there exist integers w and z such that bw + cz = f. Substituting for f gives ax + bwy + czy = d. Multiplying each term by e1 , a(xe1 ) + b(wye1 ) + c(zye1 ) = de1 = e. Each product in parenthesis is an integer. So the given Diophantine equation does have an integer solution.
Problem 2-86:
Describe how to find all the solutions to the Diophantine equation ax + by + cz = e. Solution: In Problem 2-84, we found that there is a solution if and only if gcd(a, b, c)|e. Let's assume that there are solutions. We know that gcd(a, b, c) = gcd(a, gcd(b, c)). We know how to find the complete solution to a Linear Diophantine Equation with two variables, so let's try to use this. Rewrite the equation as ax + (by + cz) = e. The expression by + cz is divisible by gcd(b, c) and in fact can be equal to any possible multiple of gcd(b, c). So let by + cz = gcd(b, c)t. This means that we now have two equations ax + gcd(b, c)t by + cz = e = gcd(b, c)t.
Since gcd(a, gcd(b, c)) = gcd(a, b, c) | e, then both equations can be solved. The procedure for solving the three variable equation consists of finding the general solution of the first equation for x and t. Then given a fixed value of t, we can find the general solution of the second equation for y and z. If x = x0 , t = t0 is one solution to the equation ax + gcd(b, c)t = e (which we can find in theory by using the Extended Euclidean Algorithm) then the general solution is d1 x = x0 + n d for all n Z a t = t0 - n d where d = gcd(a, b, c) = gcd(a, gcd(b, c)) and d1 = gcd(b, c). Then the second equation becomes ad1 n d Since gcd(b, c) = d1 , then this equation has solutions (treating n as fixed and knowing that a, b, c, d1 , d are fixed). an We find a solution (y0 , z0 ) to by + cz = d1 (which we can do), and then multiply by t0 - to find a solution d y0 an z0 an ad1 y 0 t0 - , z 0 t0 - to by + cz = d1 t0 - n. The general solution is thus d d d y0 an c y = y 0 t0 - + m d d1 for all m Z z0 an b z = z0 t 0 - - m d d1 by + cz = d1 t = d1 t0 - 7
The general solution to the original equation is then x = y z d1 n d c y0 an + m = y 0 t0 - d d1 z0 an b b = z0 t 0 - - m- m d d1 d1 x0 +
for all n, m Z
Note that there are two independent integer parameters n and m. For example, if we wished to find the complete solution to the equation 5x + 4y + 6z = 27, we would split the equation into the equations 5x + 2t 4y + 6z The first equation has complete solution x = 5 + 2n t = 1 - 5n for all n Z = = 27 2t
We fix n and substitute this value for t into the second equation to obtain 4y + 6z = 2 - 10n Since 4(-1) + 6(1) = 2, then 4(-1 + 5n) + 6(1 - 5n) = 2 - 10n. So y = -1 + 5n, z = 1 - 5n is a particular solution to 4y + 6z = 2 - 10n. Thus, the complete solution to the equation 4y + 6z = 2 - 10n is y = (-1 + 5n) + 3m z = (1 - 5n) - 2m for all m Z
Therefore, the complete solution to the original equation 5x + 4y + 6z = 27 is x = 5 + 2n y = -1 + 5n + 3m for all m, n Z z = 1 - 5n - 2m
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