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Fall2004 M2

# Fall2004 M2 - Faculty of Mathematics University of Waterloo...

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Unformatted text preview: Faculty of Mathematics University of Waterloo MATH 135 MIDTERM EXAM #2 Fall 2004 Monday 15 November 2004 19:00 — 20:15 Solutions [8) 1. Solve the system of congruence equations 4 (mod 11) 50 (mod 109) (E :13 Solution: By the Chinese Remainder Theorem, we know that there will be a unique solution modulo 11 X 109 = 1199 since gcd(11,109) : 1. From the second congruence, a: = 50 + 109:]; for some 3/ E Z. Substituting into the ﬁrst congruence, we obtain 50 + 1093/ E 4 (mod 11) 1093; E —~46 (mod 11) —y E —2 (mod 11) y E 2 (mod 11) Therfore, y z 2 + 112 for some 2 E Z. Substituting back into :17, we get a: = 50 + 109(2 + 112) = 268 + 11992. Therefore, an E 268 (mod 1199) is the unique solution mod 1199. [7] 2. Find the complete solution to 29933 E 117 (mod 949). Solution: We convert this congruence to the Linear Diophantine Equation 299117 + 9491/ = 117. Next7 we solve this Linear Diophantine Equation using the Extended Euclidean Algorithm: 299:1: +9491: ; r ODD-40103 Therefore, gcd(299, 949) = 13. Since 13I117 (with quotient 9), then the Linear Diophantine Equation has solutions, so the linear congruence has solutions. From the EEA, we see that 299(«19) + 949(6) : 13. MATH 135, Midterm #2 Solutions Page 2 of 5 Multiplying through by 9, we obtain 299(-17 1) + 949(54) : 117. Therefore, at = —171 is a partigcliiglar solution to the linear congruence. Since gcd(299, 949) = 13 and 1—3 : 73, then we can write the complete solution as m E ——171 (mod 73) which can be rewritten as a: E 48 (mod 73) or we can write the complete solution as ICE—171+73’n (mod 949) forn:0,1,2,...,12 which can be rewritten as x2778+73n (mod949) forn:0,1,2,...,12 or x E 48,121, 194, 267, 340, 413,486, 559,632, 705, 778, 851, 924 (mod 949) 3. Deﬁne f : Z51 —-> Z51 by f([a:]) : [28][:I:] + [17]. [3] (a) Find the (multiplicative) inverse of [28] in Z51. Solution: To ﬁnd the inverse of [28] in Z51, we solve the equation [28] [11:] = [1] in Z51, which is equivalent to solving the linear congruence 28m .=—_ 1 (mod 51). Since gcd(28, 51) = 1, this equation has a solution, and the solution is unique mod 51. By the Extended Euclidean Algorithm (or otherwise), the solution to this equation is an E 31 (mod 51). Therefore, in Z51, [28]_1 2 [31]. [4] (b) Let [y] : [28][x] + [17] for some [11:] E Zsl. Find positive integers a and b so that [x] : [a][y] + [b]. Solution: In (a), we determined that [28]”1 = [31] in Z51. So, we multiply both sides of the equation [3/] : [28][:c] + [17] by [31] to obtain [31llyl I [31] [281196] + [31lll7] [311M = [\$144527] (Sincel31ll281 =[1D BUM-[527] = [37] l3lllyl+[—527] I [37] [31][y]+[34] : [:13] (since -—527E34 (mod 51)) Therefore, a = 31 and b = 34 are two positive integers that satisfy the required. MATH 135, Midterm #2 Solutions Page 3 of 5 [2] (c) Prove that the function f is bijective. Solution: In (b), we showed that f is invertible by ﬁnding that f“1([y]) : [31] [y] + [34]. Therefore, since f is invertible, it is bijective. 4. In Sarah’s RSA scheme, 19 2 11, q = 7 and e 2 13. [5] (a) Calculate Sarah’s private key (d, n). Solution.- From the set—up of RSA, n : pq : 77 and ¢(n) = (p -— 1)(q —— 1) : 60. Also, d satisﬁes the congruence ed E 1 (mod ¢(n)). So to ﬁnd d, we must solve 13d .=_ 1 (mod 60). Since gcd(13, 60) = 1, this has a unique solution mod 60. By the Extended Euclidean Algorithm (or otherwise), we can determine that the solution is d E 37 (mod 60). Since d is an integer satisfying 1 S d < ¢(n), then d : 37. Therefore, Sarah’s private key is (37,77). [4] (b) Using Sarah’s public key (6, n), encrypt the message M 2 61. Solution: To encrypt M, we ﬁnd an integer C with 1 S C < n and C E M6 (mod n) Thus, here we ﬁnd an integer C with 1 S C’ < 77 and C E 6113 (mod 77). We use square and multiply to calculate C’. First, we write 13 = (1101)2 = 8 + 4 + 1, so 6113 : 618 - 614 - 611. Calculating, 612: (611)25 6125 3721; 25 (mod 77) 614: (612)22 2525 6252 9 (mod 77) 613: (614)25 925 815 4 (mod77) So 6113 ; 618-614-611 4- 9 - 61 (mod 77) 2196 (mod 77) E 40 (mod 77) Therefore, the encrypted message is C = 40. MATH 135, Midterm #2 Solutions Page 4 of 5 [10] 5. A sequence {mu} is deﬁned by \$1 2 —3, x2 2 23, m3 2 «45 and xn 2 7xn—2 "‘ 6xn—3 for all n 2 4. Prove that as" 2 2(—3)" + 2” + 1 for all n E 1?. Solution: We prove this result by strong induction on n. Base Cases For n 2 1, m1 —3 and 2(—3)1 + 21 + 1 2 —6 + 2 + 1 2 —3, so the result is true. For n 2 2, x2 23 and 2(—-3)2 + 22 + 1 2 18 + 4 + 1 2 23, so the result is true. For n 2 3, £63 2 —45 and 2(——3)3 + 23 + 1 2 —54 + 8 + 1 2 ~45, so the result is true. H H! Induction Hypothesis Assume that the result is true for n 2 1, 2, 3, . . . , k, for some k E P, k 2 3. (That is, assume that :5” 2 2(—3)” + 2n + 1 for n 2 1, 2,3, . . . , k, for some 19 E If”, k 2 3. ) Induction Conclusion Consider n 2 k + 1. Then xk+1 2 7xk_1 —= 6xk12 (since 16 + 1 Z 4) 2 7(2(—3)k‘1 + 2"“1 + 1) - 6(2(——3)k_2 + 2"—2 + 1) (by Induction Hypothesis) 2 14(4))“1 — 12(——3)k—2 + 7- 2k—1 — 6- 2H + 7 — 6 = i4(—3)k—1 + 4(—3)"~1 + 7- 2k—1 — 3 . 2’6-1 +1 = 18(—3)k"1 + 4 - 2"—1 + 1 : 2(_3)k+1 + 2k+1 + 1 Therefore, the result holds for n 2 k + 1. Therefore, 20,, 2 2(—3)" + 2" + 1 for all n E P by the Principle of Strong Induction. [5] 6. (a) Let p be a prime number. Prove that the congruence equation x2 E p (mod p3) has no solutions. Solution: Assume that there is an integer a: for which :32 E p (mod 193). By the deﬁnition of congruence, 2:2 2 p + kp?’ for some ’9 E Z. Since 10 divides into the right side, then p divides into the left side, ie. p | 3:2. Since p19? 2 m - :12, then 1) | :1: or p | a; (since 19 is a prime number) so 1) | 3:. Thus, 53 2 Qp for some Q E Z. Substituting back into our equation, we get Q2 2 2 p + 19193. Dividing by p, we get Q21) 2 1 + 19122 or Q21) —— kpz 2 1 or 19(6)2 — lop) 2 1. Since 19 divides the left side, then 1) divides the right side, ie. p l 1. But since p is a prime number, this is a contradiction. Therefore, the congruence equation x2 E p (mod p3) has no solutions. MATH 135, Midterm #2 Solutions Page 5 of 5 [2] (b) Find a. positive integer a > 1 and an integer a: so that 9:2 E a (mod a3). Solution: Ifx=2 anda=4, then 22 E4 (mod 64) ie. x2 Ea (moda3). ...
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