All of the following exercises are based on the material in the handout on integers found on the class website. 1. Find d = gcd(475, 385) and express it as a linear combination of 475 and 385. That is write d = 475s + 385t for some integers s and t. Solution. The greatest common divisor d is computed by applying the Euclidean algorithm: 475 = 1 385 + 90 385 = 4 90 + 25 90 = 3 25 + 15 25 = 1 15 + 10 15 = 1 10 + 5 10 = 2 5 + 0. Hence d = gcd(475, 385) = 5. To write d = 5 as a linear combination, d = 475s + 385t, reverse the above Euclidean algorithm computation: 5 = 15 10 Thus 5 = 30 475 37 385. Alternate Solution: = 15 (25 15) = 2 15 25 = 2(90 3 25) 25 = 2 90 7 25 = 2 90 7(385 4 90) = 30 90 7 385 = 30(475 385) 7 385 = 30 475 37 385. Use the matrix method illustrated on Page 11. The notation k between two matrices means that the second matrix is obtained from the first by subtracting k times the second row from the first, while, if the k is on top (as in k ), then k times the first row is subtracted from the second. [ ] 1 0 475 1 1 90 4 1 1 90 0 1 385 1 0 1 385 4 5 25 13 16 15 1 13 16 15 3 4 5 25 17 21 10 30 37 5 2 30 37 5 1 17 21 10 77 95 0 Hence d = (475, 385) = 5 = 30 475 37 385. 2. (a) Calculate d = gcd(4307, 1121) and express it as a linear combination of 4307 and 1121. Math 4023 1
Solution. The simplest method is to use the Euclidean algorithm. The following is the matrix format of this algorithm: [ ] 1 0 4307 1 3 944 1 1 3 944 0 1 1121 3 0 1 1121 1 4 177 6 23 59 3 6 23 59 5 1 4 177 19 73 10 Hence d = (4307, 1121) = 59 = 6 4307 23 1121. (b) Calculate m = lcm[4307, 1121]. Hint: You may want to refer to the formula on page 22 of the handout. Solution. By the formula on page 22 of the handout, we have (a, b)[a, b] = ab for any positive integers a and b. Hence, using part (a): [4307, 1121] = 4307 1121 (4307, 1121) = 4828147 = 81833. 59 3. Which integers can be expressed in the form 12m + 20n, where m and n are integers? Solution. We proved in class that if a and b are natural numbers with at least one not zero, then {ax + by : x, y Z} = az + bz = dz, where d = (a, b). This is essentially the same argument that is used to prove Theorem 1.1.4 and Theorem 1.1.6 in the handout. Thus, since (12, 20) = 4 it follows that {12m + 20n : m, n Z} = (12, 20)Z = 4Z. Thus the integers that can be expressed in the form 12m + 20n are all of the multiples of 4. 4. Give a proof by induction to show that 5 2n 1 is divisible by 24, for all positive integers n. Solution. Let P (n) be the statement 5 2n 1 is divisible by 24. We will prove that P (n) is true for all positive integers n by induction. The statement P (1) is the statement 5 2 1 is divisible by 24. Since 5 2 1 = 24, this is certainly a true statement. For the induction step, assume that the statement P (k) is true. That is, assume that 5 2k 1 is divisible by 24. Using this, we must show that it follows that P (k + 1) is also Math 4023 2
true. That is, we must show that 5 2(k+1) 1 is divisible by 24, provided that 5 2k 1 is divisible by 24. Assuming that 5 2k 1 = 24m for some m P, we conclude that 5 (2(k+1) 1 = 5 2k+2 1 = 5 2k 5 2 1 = 5 2k 5 2 (5 2 24) (since 1 = 5 2 24) = 5 2 (5 2k 1) + 24 = 5 2 (24m) + 24 (by the induction hypothesis) = 24(5 2 m + 1). Hence 24 divides 5 2(k+1) 1 provided 24 divides 5 2k 1. This is the statement that P (k + 1) is true, provided P (k) is true, and by the principle of induction, we conclude that P (n) is a true statement for all n P. 5. Determine if each of the following statements is true or false. (a) 40 13 mod 9 Solution. True since 40 13 = 3 9. (b) 29 1 mod 7 Solution. False since 29 1 = 30 = ( 5) 7+5, so 29 1 is not divisible by 7. (c) 29 6 mod 7 Solution. True since 29 6 = 35 = ( 5) 7. (d) 132 0 mod 11 Solution. True since 132 = 11 12. 6. Find all the solutions (when there are any) of the following linear congruences: (a) 8x 6 mod 14 Solution. gcd(8, 14) = 2 so there are 2 distinct solutions modulo 14. Divide the congruence by 2 to get 4x 3 mod 7. Since [4] 1 7 = [2] 7, it follows that x 6 mod 7. Hence, x 6 mod 14 or x 13 mod 14. (b) 66x 100 mod 121 Solution. Since gcd(66, 121) = 11 and 11 100, there are no solutions to this congruence. (c) 21x 14 mod 91 Math 4023 3
Solution. gcd(21, 91) = 7 and 7 14 so there are 91/7 = 13 distinct solutions mod 91. Dividing by 7 gives an equivalent congruence 3x 2 mod 13, which has the solution x 5 mod 13. Thus, the solutions mod 91 are x r mod 91, where r {5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89}. (d) 21x 14 mod 89 Solution. Since gcd(21, 89) = 1 there is a unique solution mod 89. Applying the Euclidean algorithm gives 17 21 4 89 = 1, so that the inverse of 21 modulo 89 is 17. Multiplying the congruence by 17 gives x 17 21x 17 14 238 60 mod 89. 7. Find the inverse of 13 modulo 35 and use it to solve the equation [13] 35 x = [9] 35 in Z 35. Solution. From the Euclidean algorithm, 3 35 8 13 = 1. Hence, [13] 1 35 = [ 8] 35 = [27] 35. Multiplying the equation [13] 35 x = [9] 35 by [13] 1 35 = [ 8] 35 gives x = [ 8] 35 [9] 35 = [ 72] 35 = [ 72 + 105] 35 = [33] 35. 8. Find the multiplicative inverses of the given elements (if possible). (a) [91] 2565 in Z 2565 Solution. Use the Euclidean algorithm to write 451 91 16 2565 = 1. Hence, [91] 1 2565 = [451] 2565. (b) [423] 6087 in Z 6087 Solution. Use the Euclidean algorithm to check that (423, 6087) = 3 = 1 so there is not multiplicative inverse of 423 modulo 6087. 9. Solve the following system of linear congruences: x 4 (mod 24) x 7 (mod 11) Solution. By inspection or using the Euclidean algorithm, find the equation ( 5)24 + 11 11 = 1. Then the solutions to the simultaneous congruence are given by x 4 (11 11) + 7(( 5)24) = 356 (mod 264). Math 4023 4
10. In Z 18 find all units and all zero divisors. Solution. The units consist of all [a] 18 such that (a, 18) = 1, while the zero divisors consist of all [a] 18 such that (a, 18) > 1. Thus the units are and the zero divisors are 11. Find {[1] 18, [5] 18, [7] 18, [11] 18, [13] 18, [17] 18 } {[0] 18, [2] 18, [3] 18, [4] 18, [6] 18, [8] 18, [9] 18, [10] 18, [12] 18, [14] 18, [15] 18, [16] 18 }. (a) 6 76 mod 13, Solution. 6 12 1 mod 13 by Euler s theorem, so write 76 = 6 12 + 4 to get 6 76 = 6 6 12+4 = (6 12 ) 6 6 4 6 4 (6 2 ) 2 100 9 mod 13. (b) 7 1001 mod 11, Solution. By Euler, 7 10 1 mod 11. Since 1001 = 10 100 + 1 it follows that 7 1001 7 1 = 7 mod 11. (c) 2 25 mod 21, Solution. By Euler, φ(21) = 8 so 2 8 1 mod 21. Since 25 = 8 3 + 1, we have 2 25 2 1 = 2 mod 21. (d) 7 66 mod 120. Solution. Since φ(120) = 32, Euler gives 7 32 1 mod 120. Since 66 = 2 32 + 2 it follows that 7 66 7 2 = 49 mod 120. 12. Calculate φ(32), φ(33), φ(120), and φ(384). (φ(n) is the Euler φ function evaluated at n.) Solution. 32 = 2 5 so 4φ(32) = 2 5 2 4 = 16; 33 is prime so φ(33) = 33 1 32; φ(120) = φ(2 3 )φ(3)φ(5) = (2 3 2 2 ) 2 4 = 32; 384 = 2 7 3 so φ(284) = φ(2 7 )φ(3) = (2 7 2 6 )(3 1) = 128. Math 4023 5