# Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2)

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1 Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2) Problems from Prove that a b (mod n) if and only if a and b leave the same remainder when divided by n Proof Suppose a b (mod n) Then, by definition, we have a b = nk for some k Z Now by the Division Algorithm, a and b can be written uniquely in form (1) with 0 r, r < n But then a = nq + r b = nq + r (2) a = b + nk = (nq + r ) + nk = n(q + k) + r Comparing (??) and (??) we have a = nq + r, 0 r < n a = n(q + k) + r, 0 r < n By the uniqueness property of the division algorithm, we must therefore have r = r If a and b leave the same remainder when divided by n then we have a = nq + r b = nq + r Subtracting these two equations yields a b = n(q q ), so a b (modn) 212 If a Z, prove that a 2 is not congruent to 2 modulo 4 or to 3 modulo 4 By the Division Algorithm any a Z must have one of the following forms 4k 4k + 1 a = 4k + 2 4k + 3 This implies a 2 = 16k 2 = 4(4k 2 ) = 4q 16k 2 + 8k + 1 = 4(4k 2 + 2k) + 1 = 4r + 1 4(4k k + 4 = 4(4k 2 + 8k + 1) = 4s 16k k + 9 = 4(4k 2 + 6k + 2) + 1 = 4t + 1 1

2 2 So a 2 { 0 (mod 4) 1 (mod 4) 213 If a, b are integers such that a b (mod p) for every positive prime p, prove that a = b Since the set of prime numbers in Z is infinite, we can always find a prime number p larger than any given number In particular we can find a prime number p such that 0 a b < p Now by hypothesis, we have, for this prime p, a b = kp for some k Z (by the definition of congruence modulo p) Thus, p divides a b But 0 is the only non-negative number less than p that is also divisible by p Thus, a b = 0 or a = b 214 Which of the following congruences have solutions: (a) x 2 1 (mod 3) We need x 2 1 = 3k By the Division Algorithm, x must have one of three forms 3t 9t 2 1 x = 3t + 1 x 2 1 = 9t 2 + 6t 3t + 2 9t t + 3 Thus, if x has the form x = 3t + 1, then x 2 1 = 3(3t 2 + 2t) and so x 2 1 (mod 3) (b) x 2 2 (mod 7) We need x 2 2 = 3k By the Division Algorithm, x must have one of the seven forms 7k 49k 2 2 = 7 ( 7k 2) + 2 7k k k 1 = 7 ( 7k 2 + 2k 1 ) + 6 7k k k + 2 = 7 ( 7k 2 + 4k ) + 2 x = 7k + 3 x 2 1 = 49k k + 7 = 7 ( 7k 2 + 6k + 1 ) 7k k k + 14 = 7 ( 7k 2 + 8k + 2 ) 7k k k + 23 = 7 ( 7k k + 3 ) + 2 7k k = 7 ( 7k k + 4 ) + 6 Thus, if x has the form x = 7k + 3 or the form x = 7k + 4, then x 2 2 is an integer multiple of 7 and so x 2 2 (mod 7) (c) x 2 3 (mod 11) This is best handled by trial and error In order for x 2 3 (mod 11), we need x 2 3 = 11k for some choice of integers x and k For x = 0, 1, 2, 3, 4 there is no such k; but for x = 5 we have = 22 = 2 11,

3 3 so x = 5 is a solution x = 6 is also a solution since = 33 = If [a] = [b] in Z n, prove that GCD(a, n) = GCD(b, n) Since [a] = [b], a b (mod n) by Theorem 23 But then by the definition of congruence modulo n for some k Z But this implies a b = nk a = nk + b Now we apply Lemma 17 (if x, y, q, r Z and x = yq + r, then GCD(x, y) = GCD(y, r) taking x = a and y = n Thus, GCD(a, n) = GCD(n, b) 216 If GCD(a, n) = 1, prove that there is an integer b such that ab = 1 (mod n) Since GCD(a, n) = 1, we know by Theorem 13 that there exist integers u and v such that Hence If we now set b = u and k = v we have au + nv = 1 au 1 = nv ab 1 = nk which means that ab 1 (mod n) 217 Prove that if p 5 and p is prime then either [p] 6 = [1] 6 or [p] 6 = [5] 6 Let p be a prime 5 Then p is not divisible by 2 or 3 Now consider the a priori possible congruency classes of [p] 6 : viz, Z 6 = {[0] 6, [1] 6, [2] 6, [3] 6, [4] 6, [5] 6 } one by one [p] 6 cannot be [0] 6 since p is not divisible by 6 For p [0] 6 = p 0 (mod 6 ) = p 0 = k6 for some k Z = 6 p (contradiction!) Similarly, and p 2 = k6 = p = k6 2 = 2 (3k 1) = 2 p (contradiction!) p [3] 6 = p 3 = k 6 = p = 3 (2k 1) = 3 p (contradiction!) p [4] 6 = p 4 = k 6 = p = 2 (2k 2) = 2 p (contradiction!) The only possibilities left are [p] 6 = [1] 6 and [p] 6 = [5] 6

4 4 Problems from Write out the addition and multiplication tables for Z 4 Addition in Z 4 [0] [1] [2] [3] [0] [0] [1] [2] [3] [1] [1] [2] [3] [0] [2] [2] [3] [0] [1] [3] [3] [0] [1] [2] Multiplication in Z 4 [0] [1] [2] [3] [0] [0] [0] [0] [0] [1] [0] [1] [2] [3] [2] [0] [1] [0] [2] [3] [0] [3] [2] [1] 222 Prove or disprove: If ab = 0 in Z n, then a = 0 or b = 0 Disproof by Counter-Example Consider multiplication in Z 4 as given in the previous problem One has [2] [2] = [0], but [2] [0] in Z Prove that if p is prime then the only solutions of x 2 + x = 0 in Z p are 0 and p 1 Let us revert to the original explicit notation for elements of Z p We want to prove (3) ([x] [x]) [x] = [0] (inz p ) [x] = [0] or[p 1] Now, by the definition of addition and multiplication in Z p statement (??) is equivalent to [x(x 1)] = [0] [x] = [0] or[p 1] Now if the congruence class in Z p of x 2 + x is the same as that of 0, then the difference between x 2 + x and 0 must be divisible by p Hence, p divides x 2 + x 0 = x 2 + x Now x 2 + x = x(x + 1) Since p is prime, and p divides x(x + 1), p must divide either x or x + 1 (by Corollary 19) If p divides x, then qp = x = x 0 so x is in the same congruence class as 0; ie, [x] = [0] If p does not divide x, then it must divide x + 1; so x + 1 = q p [x] = [ 1] = [p 1] 224 Find all [a]in Z 5 for which the equation ax = 1 has a solution Let us write down the multiplication table for Z 5

5 5 So we have [0] [1] [2] [3] [4] [0] [0] [0] [0] [0] [0] [1] [0] [1] [2] [3] [4] [2] [0] [2] [4] [1] [3] [3] [0] [3] [1] [4] [2] [4] [0] [4] [3] [2] [1] [1] [1] = 1 [2] [3] = 1 [4] [4] = 1 so if a = [1], [2], [3], or [4]then ax = 1 has a solution in Z Prove that there is no ordering of Z n such that (i) ifa b, andb c, thena c; (ii) ifa b, thena + c b + c for everyc Z n By an ordering on Z n we mean a rule that tells you whether or not pairs of elements of Z n In addition to the conditions given above, we must assume that the ordering is complete in the sense that if a b then either a b or b a So assume we have such a relation on Z n Since [0]and [1]are distinct congugacy classes in Z n, we must then have either [0] [1] or [1] [0] Assume [0] [1] Then by property (ii) we must have [0] + [c] [1] + [c], [c] Z n Since [0] + [c] = [c] and [1] + [c] = [c + 1], we then have Thus, [c] [c + 1], [c] Z n (4) [0] [1] [2] [n 1] [n] [n + 1] Applying Property (i) recursively, [1] [2] and[2] [3] [1] [3] [1] [3] and[3] [4] [1] [4] [1] [4] and[4] [5] [1] [5] etc, we can conclude that [1] [n] But [n] = [0] in Z n So [1] [0] But this contradicts our assumption that [0] [1] Hence no such ordering exists The case when [1] [0] is treated similiarly Problems from If n is composite, prove that there exists a, b Z n such that a [0] and b [0] but ab = [0] Assume n to be positive (otherwise, we have to define Z n for n < 0; which can be done, but with no particular gain) If n is composite then n has a factorization n = pq

6 6 with 1 < p q < n In view of the inequality above n does not divide p nor does n divide q, so [p] [0] and [q] [0] However, [p][q] = [pq] = [n] = [0] Setting a = [p] and b = [q] we arrive at the desired conclusion 232 Let p be prime and assume that a 0 in Z p Prove that for any b Z p, the equation ax = b has a solution By Theorem 28, the equation ax = 1 always has a solution in Z p, for every a [0] if p is prime Multiplying both sides by b Z p, yields bax = b Setting x = bx we see that every b Z p has a factorization b = ax for every [a] [0] in Z p 233 Let a [0] in Z n Prove that ax = [0] has a nonzero solution in Z n if and only if ax = [1] has no solution Suppose a [0], b [0] and that ab = [0] We aim to show that ax = [1] has no solution We will use a proof by contradiction Suppose c is a solution of ax = [1] Then b = b 1 = b(ac) = (ab)c = [0] c = 0 But this contradicts our original hypothesis that b is a nonzero solution of ax = [0] Hence, there can be no solution of ax = [1] Suppose a [0] and ax = [1] has no solution We aim to show that ax = [0] has a nonzero solution in Z n Let z be the integer, lying between 1 and n 1 representing the congruence class of a Z n ; ie, [z] = a We first note that, by Corollary 29, GCD(z, n) = 1 if and only if ax = [1] has a solution in Z n Since the latter is not so, GCD(z, n) 1 and so z and n must share a common divisor greater than 1, call it t We thus have z = rt, n = st By construction 1 s < n, and so the congruence class of s is not equal to [0] But a[s] = [z][s] = [rt][s] = [r][st] = [r][0] = [0] Hence, [s] is a nonzero solution of ax = [0] in Z n 234 Solve the following equations (a) 12x = 2 in Z 19

7 7 The fastest approach to this problem might be trial and error Simply compute the multiples 0 12, 1 12,, and figure out which of these products have remainder 2 when divided by 19 Then we d have k 12 = q or 2 k 12 (mod 19) and so [k] 19 hence the solution of X will be [k] 19 Such a trial and error procedure reveals 192 = (10)(19) + 2 [16] 19 x = [16] 19 Next, we give a more systematic approach which is also applicable for large integers (where the trial and error procedure because tedious if not impractical) Apply the Euclidean Algorithm to the pair (19, 12) 19 = (1) (12) = (1) (7) = (1) (5) = (2) (2) = (1) (1) + 0 The point here is not to figure out the GCD of 19 and 12 (which is obviously 1 since 19 is prime), but to obtain a useful arrangemet of substitutions what will allows us to express 1 as an integer linear combination of 19 and 12 That is to find numbers u and v so that 1 = u (19) + v (12) The utility of this equation well become clear once we get a suitable choice of v and u We re-write the sequence of Euclidean Algorithm equations so the remainders are isolated on the left hand side 1 = 5 (2) (2) (a) 2 = 7 (1) (5) = 7 5 (b) 5 = 12 (1) (7) = 12 7 (c) 7 = 19 (1) (12) = (d) Now the idea is to use back substitution to eliminate all the intermediary remainders: substituting the right hand side of (d) for the number 7 in (c) yields (e) 5 = 12 (19 12) = (2)(12) 19 We ve now expressed 7 and 5 in the form 12u + 19v Substituting the right hand sides of (d) and (e) into (b) yields (f) 2 = (19 12) ((2) (12) 19) = (2)(19) (3)(12) Finally, we substitute the right hands sides of (e) and (f) into (a) to get 1 = ((2) (12) 19) 2 ((2) (19) (3) (12)) = ( 5) (19) + (8) (12) The last equality just being a check on our calculation We now have (12) (8) (5) (19) = 1 Taking congruence classes of both sides modulo 19 we get or since [19] 19 = 0, [8] 19 [5] 19 [19] 19 = [1] 19 [8] 19 = [1] 19

8 8 Now simply multiply both sides by [2] 19, to obtain [2] 19 [8] 19 [1] 19 or using [2] 19 [8] 19 = [16] 19, and [2] 19 [1] 19 [16] 19 Thus, is the solution to X = [16] 19 X (b) 7x = 2 in Z 24 Either method used in part (a) will produce (7) (14) = 98 = (4) (24) + 2 [7] 24 [14] 24 = [2] 24 x = [14] 24 (c) 31x = 1 in Z 50 Either method used in part (a) will produce (31)(20) = 651 = (7)(50) + 1 [31] 50 [20] 50 = [1] 50 x = [20] 50 (d) 34x = 1 in Z 97 Here only the second method of part (a) is actually practical We ll do the calculation explicitly First we apply the Euclidean algorithm to the pair (97, 34) 97 = (2) (34) = (1) (29) = (5) (5) = (1) (4) + 1 Now we back-substitute to express 1 as an integer linear combination of 97 and 34 We have and so or 29 = = 5 (1) (4) (a) 4 = 29 (5) (5) (b) 5 = 34 (1) (29) (c) 29 = 97 (2) (34) (d) 5 = = 34 ( ) = = = ( ) 5 ( ) = = 5 4 = ( ) ( ) = = 1

9 9 So so and so [20] 97 is the solution of 1 (20) (34) (mod 97) [34] 97 [20] 97 = [1] 97 [34] 97 X = [1] 97

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