44.(ii) In this case we have that (12, 38) = 2 which does not divide 5 and so there are no solutions.

Transcription

1 Solutions to Assignment 3 5E More Properties of Congruence 40. We can factor 729 = so it is enough to show that (mod 7), (mod 3) and (mod 9) =(3 3 ) 576 = (27) 576 ( ) 576 (mod 7) (mod 7) =(3 3 ) 576 = (27) (mod 3) (mod 3) =(3 9 ) 92 ( ) 92 (mod 9) (mod 9) The last line follows from the following sequence of computations: 3 3 8(mod 9), so 3 4 5(mod 9), so 3 5 4(mod 9), so 3 6 7(mod 9), so 3 7 2(mod 9), so 3 8 6(mod 9) and finally 3 9 (mod 9). 42. By definition, 6a 0(mod 36) if and only if 36 divides 6a. By canceling from this equation 4 = (36, 6) we have that this happens if and only if 9 divides 4a. Since 9 and 4 are relatively prime, then this happens if and only if 9 divides a. The (positive) numbers satisfying this equation such that a apple 36 are then 9, 8, 27, and Let d =(r, m). It is clear that d r and d m so we have r 0 da r 0 db (mod m 0 d)wherer 0 = r d,m0 = m d 2 Z. Then by Proposition 6 we have that r 0 a r 0 b (mod m 0 ). Note that the definition of the gcd gives us that (r 0,m 0 ) =, so then by Proposition 7 we have that a b (mod m 0 ). 5F Linear Congruences and Bezout s Identity 44.(i) We first find that (2, 29) = which divides 5 and so by Proposition 8 we can find a solution. The first step of the EEA matrix gives us that ( 2) = 5 and so x=-2+29=27 is the smallest nonnegative solution. 44.(ii) In this case we have that (2, 38) = 2 which does not divide 5 and so there are no solutions. 44.(iii) This equation has a solution because (2, 47) =. Observe here that = and so x =4 5 = 20 is the smallest nonnegative solution to the system. 44.(iv) This equation does not have a solution because (2, 47) = (v) This equation has a solution because (2, 65) =. Note that 5 = ( nonnegative solution to the equation is x = 65 5 = 60. 5) and so the smallest 50. First we must find the inverse for 7 modulo 28. Note that the first step of the Euclidean algorithm gives us = , so the inverse is -3. Then multiplying both sides of our equation by -3 we get x 403 (mod 28) 33 (mod 28) Now (7, 28) =, so all solutions are give by x = k for all k 2 Z or equivalently x 33 (mod 28). 6C Arithmetic Modulo m

2 5. + mod mod mod mod (i) Note that (6, 0) = 2 and so there are 2 solutions (mod 0). One solution can be found by writing the equation as [6] 0 [x] 0 = [24] 0 which gives us that [4] 0 is a solution. Another solution can be found by writing the equation as [ 4] 0 [x] 0 = [4] 0 and this gives us that [ ] 0 = [9] 0 is the other solution. 8.(ii) In this case (5, 7) = and so there is only one solution given by writing the equation as [5] 7 [x] 7 = [0] 7 and thus [2] 7 is the unique solution. 8.(iii) This equation has only one solution and it can be seen from writing the equation as [4] 7 [x] 7 = [6] 7 which gives us the solution [4] 7. 8.(iv) The unique solution to this equation can be seen from writing it as [4] 7 [x] 7 = [36] 7 which gives us the solution [9] 7. 8.(v) The unique solution is [0] 9 which can be seen from first writing the equation as [ 6] 9 [x] 9 =[ 3] 9 and then noticing that [0] 9 is the multiplicative inverse of [2] 9. Another method that works is writing the equation as [ 6] 9 [x] 9 =[ 60] 9.

3 6D Complete Sets of Representatives (mod 3) 2 2(mod 3) 2 2 4(mod 3) 2 3 8(mod 3) 2 4 3(mod 3) 2 5 6(mod 3) (mod 3) 2 7 (mod 3) 2 8 9(mod 3) 2 9 5(mod 3) (mod 3) 2 7(mod 3) 2 2 (mod 3) 32. Note first that 2 is a primitive root modulo because 2 2(mod ) 2 2 4(mod ) 2 3 8(mod ) 2 4 5(mod ) (mod ) 2 6 9(mod ) 2 7 7(mod ) 2 8 3(mod ) 2 9 6(mod ) 2 0 (mod ) We use this primitive root to find other primitive roots. Note that 2 3 =8, 2 7 7(mod ), 2 9 6(mod ) are also primitive roots as we check concentrating on the exponents and using the fact that 2 0 (mod ). Note that as we exponentiate 2 3 the exponents are 2 3, 2 6, 2 9, 2 2 = (mod ), 2 5, 2 8, 2, 2 4, 2 7, 2 0. The idea is similar for the other exponents. The other exponents of 2 are not primitive roots since either 2 or 5 times the exponent will be divisible by 0. For example 5 (2 4 ) 5 =2 20 (mod ), and thus the powers of 5 cannot be a complete list of representatives of the nonzero elements mod. 37.(i) We will use the characterization for a complete set of representatives given in Proposition 4, part b). Suppose that ba i ba j (mod m), then by Proposition 7 on page 85 we have that since (b, m) = then a i a j (mod m). By Proposition 4 this means that i = j and so it follows that for every i, j with apple i<japple m, ba i 6 ba j (mod m) and thus the set of the ba i is a complete set of representatives mod m. 37.(ii) Suppose that (b, m) =d>. By Proposition 4 we have that there is some i such that a i m d (mod m) and some j such that a j 0(mod m). Since d> note that m d 6 0(mod m). Now ba i m b d 0 ba j (mod m) and so the set {ba,...,ba m } cannot be a complete set of representatives. 6E Units 46. The elements of Z/20Z that have inverses are the elements that are coprime with 20, and the inverses are

4 [] 20 = [] 20 [3] 20 = [7] 20 [7] 20 = [3] 20 [9] 20 = [9] 20 [] 20 = [] 20 [3] 20 = [7] 20 [7] 20 = [3] 20 [9] 20 = [9] (i) The elements of Z/pZ that are units are the elements that are coprime with p and thus there are p such elements. A complete list of these elements is {, 2,...,p } since there are p many of them, they are all distinct mod p, and they are all coprime to p. 49.(ii) The elements of Z/0Z that are coprime to 0 are, 3, 7, 9 which are the units modulo 0. Moreover, 7 3(mod 0) and 9 (mod 0). 49.(ii) The elements of Z/24Z that are coprime to 24 are, 5, 7,, 3, 7, 9, 23 and so these numbers give a complete set of units modulo We already know that {, 2,...,p } is a complete set of units modulo p so it su ces to show that the p p two sets are the same, modulo p. Note that p (mod p), 2 p 2(mod p),..., 2 p 2 = p+ 2 (mod p) and so the two sets are indeed equal modulo p.

5 Solutions to Assignment 4 7A 5. We know that elements in Z/mZ are either units or zero divisors. We also know the units in Z/mZ are all of the form [a] where(a, m) =. So the units are: Z/6Z : [], [5] Z/7Z : [], [2], [3], [4], [5], [6] Z/8Z : [], [3], [5], [7] 7C 23. Obviously, x = 0 is a solution of the equation ax =0 Since a is a zero divisor, there is an nonzero element y so that ay = 0. But then x = 0 and x = y are two di erent solutions of the equation. 30.As in the first question, the zero divisors of Z/8Z are the elements [a] with(a, 8) 6=. complementary zero divisors are: [2] : [9] [3] : [6], [2] [4] : [9] [6] : [3], [6], [9], [2], [5] [8] : [9] [9] : [2], [4], [6], [8], [0], [2], [4], [6] [0] : [9] [2] : [3], [6], [9], [2], [5] [4] : [9] [5] : [6], [2] [6] : [9] The 32. We need to solve ax (mod365) for a = 53, 73, 93, 3. First note that (73, 365) = 73 so [73] is not a unit, hence it has no multiplicative inverse. For the other cases, we just need to solve the above equation as in previous homeworks: [53] = [62] [93] = [57] [3] = [42] 35. (i) The elements are {0+0i, +0i, 2+0i, 0+i, +i, 2+i, 0+2i, +2i, 2+2i}. (ii) =, 2 = 2, i =2i, (2i) = i, (+i) =2+i, (2+i) =+i, (2+2i) =+2i, ( + 2i) =2+2i. For instance, ( + 2i)(2 + 2i) =2+4i +2i +4i 2 =2+4i 2 +6i =2 4+6i 2+0i (iii) ( + i) 2 =2i, (2i) 2 = 4 2, 2 2. Therefore ( + i) 8 =. And hence the order of ( + i) must be a factor of 8, i.e., 2 or 4. However, we have already seen from the above computation that ( + i) 2, ( + i) 4 are not equal to. Hence the order of ( + i) must be 8.

6 36.The four elements are {0,,i,+i}. Obviously 2 ==i 2 so both and i have themselves as inverses. However, ( + i)( + i) =2i 0. It is a zero divisor, which has no inverses. 7D 40. We need to check f g satisfies the first three conditions in page 40. (i) f g(a + b) =f(g(a + b)) = f(g(a)+g(b)) = f(g(a)) + f(g(b)) = f g(a)+f g(b). (ii) f g(a b) =f(g(a b)) = f(g(a) g(b)) = f(g(a)) f(g(b)) = f g(a) f g(b). (iii) f g() = f(g()) = f() =. The other conditions can be deduced from the first three. 42. Again, check the first three conditions in page 40: a + b 0 a 0 b 0 (i)f(a + b) = = + = f(a)+f(b) 0 a + b 0 a 0 b ab 0 a 0 b 0 (ii)f(ab) = = = f(a)f(b) 0 ab 0 a 0 b 0 (iii) f() = = Id 0 2, the identity matrix. Note that the identity matrix is the multiplicative identity in M 2 (R), hence we are done.