The first exam will be on Monday, June 8, 202. The syllabus will be sections. and.2 in Lax, and the number theory handout found on the class web site, plus the handout on the method of successive squaring in computing high powers of an integer modulo n. In the following lists, pages and results on the number theory handout will be referred by preceding the number with H. Thus, Page H.23 refers to page 23 of the handout, while Proposition H..2.3 refers to Proposition.2.3 of the handout. Following are some of the concepts and results you should know: Know the Well-ordering principle: Any set of positive integers which has at least one element contains a smallest element. Know the Induction Principle and how to use it to do proofs by induction. The cardinality of X, denoted X, is the number of elements of X. Some formulas for the cardinality of combinations of sets X and Y :. X Y X + Y X Y. 2. X Y X Y. 3. P(X) 2 X where P(X) denotes the power set of X, that is, P(X) is the set of all subsets of X. 4. {all functions f : X Y } Y X. The number of ways to choose r elements (without replacement) from an n-element set is ( ) n n! r r!(n r)!. Know the Binomial Theorem: If x and y are any numbers, then (x + y) n n x n k y k, k0 where ( ) n k n! k!(n k)!. Know the Division Algorithm. Know the definition of a divides b for integers a and b (notation: a b). Know the definition of the greatest common divisor of the integers a and b (notation: (a, b)). Know the Euclidean Algorithm and how to use it to compute the greatest common divisor of integers a and b. Know how to use elementary row operations to codify the calculations needed for the Euclidean algorithm into a sequence of matrix operations as illustrated on Pages H. and H.2. Know the definition of relatively prime integers. Be sure to know Theorem H.2.3 (Page H.6) which relates the relative primeness of a and b to various divisibility conclusions.
Know the definition of least common multiple of integers a and b (notation: [a, b]). Know the relationship between the greatest common divisor, least common multiple, and the product of integers a and b: ab (a, b)[a, b]. (Page H.22.) Know the definition of prime number. Know Euclid s Lemma: If p is a prime, a and b are integers, and p ab, then p a or p b (Corollary 2.9, Page 4). Know the Fundamental Theorem of Arithmetic (Theorem 2.2, Page 4 or Theorem H..2.7, Page H.8), and how to use it to compute the greatest common divisor and least common multiple of two integers a and b (Proposition 2.22, Page 5 or Proposition H..2.0, Page H.2). Know what it means for an integer a to be congruent modulo n to another integer b (notation a b mod n). Know the definition of congruence class of a modulo n (notation [a] n ). Know the definition of the number system Z n, and how to do arithmetic in Z n : [a] n + [b] n [a + b] n [a] n [b] n [ab] n Know the definition of [a] n is invertible in Z n, (Definition H..4.4, Page H.38) and know the criterion of invertibility of [a] n : An element [a] n Z n is invertible (or has a multiplicative inverse) if and only if (a, n), that is, if and only if a and n are relatively prime. Moreover, if r and s are integers such that ar + ns, then [a] n [r] n. (Proposition H..4.5, Page H.38.) Know how to use the Euclidean algorithm to compute [a] n, when the inverse exists. Know what it means for a nonzero congruence class [a] n Z n to be a zero divisor: there is [b] n [0] n with [a] n [b] n [0] n. (Definition H..4.3, Page H.38) Every nonzero element of Z n is either invertible or a zero divisor. (Proposition H..4.5, Page H.38.) Z n is the set of invertible elements of Z n. Z n is closed under multiplication, i.e., the product of any two elements of Z n is in Z n. If p is a prime, then Z p Z p {[0] n }. (Corollary H..4.6, Page 39). Know what it means to solve a linear congruence ax b mod n. In particular, know that the solutions form congruence classes modulo n, so that they can be identified with elements of Z n. Know the criterion for solvability of a linear congruence, namely Theorem H..3.5, Page H.28. 2
Know how to find all of the solutions to the linear congruence ax b mod n using the algorithm extracted from the proof of Theorem H..3.5, as described in class:. Calculate d (a, n). 2. Test whether d divides b. (a) If d does not divide b, then there is no solution. (b) If d divides b then there are d solutions mod n. 3. To find the solutions in case (b), `divide the congruence through by d to get (a/d)x (b/d) mod (n/d). Notice that since a/d and n/d have greatest common divisor, this congruence will have a unique solution. 4. Calculate the inverse [e] n/d of [a/d] n/d, either by inspection or use the Euclidean algorithm. 5. Multiply the divided congruence by [e] n/d to get and calculate a solution c for x. [x] n/d [e] n/d [b/d] n/d 6. The solutions to the original congruence ax b mod n will the d congruence classes mod n: [c] n, [c + (n/d)] n, [c + 2(n/d)] n,..., [c + (d )(n/d)] n. Know the Chinese Remainder Theorem (Theorem H..3.6, Page H.3), and how to solve simultaneous congruences. Know Fermat s theorem: if p is prime and p does not divide a then a p mod p. (Corollary H..4.2, Page H.42). The Euler phi-function at n, denoted φ(n), is defined to be the number of positive integers less than or equal to n which are relatively prime to n. (Definition H..4.7, Page H.40) It is also true that φ(n) Z n. (Page H.40). Know how to compute φ(n) from the two rules (Exercises 7 and 29, Pages H.44 45, and discussed in class):. If p is a prime k is any positive integer, then 2. If n and m are relatively prime, then φ(p k ) p k p k. φ(nm) φ(n)φ(m). Know how to to compute φ(n) from the prime factorization of n using the two rules listed above (Proposition H..4.8, Page H.4): If n p k pkr r, then ) ) φ(n) (p k pk ) (p k r pk r ) n ( p ( pr. 3
Know Euler s Theorem: If n 2 and a is relatively prime to n, then a φ(n) mod n. Review Exercises Be sure that you know how to do all assigned homework exercises. The following are a few supplemental exercises similar to those already assigned as homework. These exercises are listed randomly. That is, there is no attempt to give the exercises in the order of presentation of material in the text. [ ] [ 2 2. Let A. Using mathematical induction, show that A 0 2 n n n2 n ] 0 2 n for all positive integers n P. Solution. For n P, let P (n) be the statement [ A n 2 n n2 n ] 0 2 n. [ ] [ 2 2 Since A 2 A ] 0 2 0 2, it follows that P () is true, which establishes the base step for induction. Now [ assume that P (k) is true for some k P. That is, assume that k P and A k 2 k k2 k ] 0 2 k. Then A k+ A [ k A 2 k k2 k ] [ ] 2 0 2 k 0 2 [ 2 k 2 2 k + k2 k ] 2 0 2 k 2 [ 2 k+ (k + )2 k ] 0 2 k+. [ 2 n n2 The last matrix is of the form n ] 0 2 n for n k +. Hence P (k + ) is true whenever P (k) is true, which verifies the induction step, and by the principle of mathematical induction, P (n) is true for all n P, i.e., the proposed formula for A n is valid for all n P. 2. Using mathematical induction, verify that for all positive integers n P. 2 + 2 3 + + n (n + ) n n + 4
Solution. For n P, let P (n) be the statement: For n, P () is the statement: which is clearly a true statement. verified. 2 + 2 3 + + n (n + ) Now assume that P (k) is true. This means that n n +. 2 2, Thus P () is true, and the base step for induction is 2 + 2 3 + + k (k + ) k k +. Consider the left hand side of the statement P (n) for n k +. That is: ( ) 2 + 2 3 + + k (k + ) + (k + ) (k + 2). Because of the assumption that P (k) is true, ( ) can be written as 2 + 2 3 + + k (k + ) + (k + ) (k + 2) k k + + (k + ) (k + 2) k(k + 2) (k + )(k + 2) + k(k + 2) + (k + )(k + 2) (k + )(k + 2) (k + ) 2 (k + )(k + 2) k + k + 2. This is the statement that P (k + ) 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. 3. (a) If S and T are finite sets, write the formula relating S T, S, T, and S T. It is not necessary to verify the formula, just write it down. Solution. S T S + T S T. (b) The following is a portion of a report submitted by a marketing analysis employee: Number of consumers interviewed: 00. Number of consumers using brand X: 78. Number of consumers using brand Y : 7. Number of consumers using both brands: 48. Explain why this report cannot be correct. (Hint: Part (a) may be useful.) Solution. Let S {consumers interviewed who use brand X}, and T {consumers interviewed who use brand Y }. Then from the data given, S 78, T 7, S T 48. The formula from part (a) then gives S T S + T S T 78 + 7 48 0. But the total number of consumers interviewed is 00, so S T 00. This contradiction shows that some error must have been made. 5
4. Let X and Y be sets with cardinalities X 4 and Y 3. The following are 5 sets constructed from these two given sets: A X Y B P(X) C P(Y ) D {Functions f : X Y } E {Functions f : Y X} Recall that P(X) denotes the power set of X, that is, the set of all subsets of X. List these 5 sets in order according to their cardinality, starting with the set with the smallest number of elements and ending with the set with the largest number of elements. Solution. A X Y 4 3 2; B 2 X 2 4 6; C 2 Y 2 3 8; D Y X 3 4 8; and E X Y 4 3 64. Therefore, in increasing order of cardinality, they are: C A B E D 5. Find the remainder when b is divided by a if: (a) a 6, b 25 Solution. 25 6 4 +, so the remainder is. (b) a 6, b 25 Solution. 25 ( 6) 5 + 4, so the remainder is 4. 6. This problem involves arithmetic modulo 6. All answers should only involve expressions of the form [a] 6, with a an integer and 0 a < 6. (a) Compute [4] 6 + [5] 6. Solution. [4] 6 + [5] 6 [4] 6 + [ ] 6 [3] 6. (b) Compute [4] 6 [5] 6. Solution. [4] 6 [5] 6 [4] 6 [ ] 6 [ 4] 6 [2] 6. (c) Compute [5] 6. Solution. [5] 6 [ ] 6 [ ] 6 [5] 6. (d) List the invertible elements of Z 6. Solution. [] 6, [3] 6, [5] 6, [7] 6, [9] 6, [] 6, [3] 6, [5] 6. (e) List the zero divisors in Z 6. Solution. [2] 6, [4] 6, [6] 6, [8] 6, [0] 6, [2] 6, [4] 6. 7. (a) Show that the square of an arbitrary integer x is congruent to either 0 or modulo 4. 6
Solution. An arbitrary integer is congruent to either 0,, 2, or 3 modulo 4, that is [x] 4 [0] 4, [x] 4 [] 4, [x] 4 [2] 4, or [x] 4 [3] 4. Since [0] 2 4 [0] 4, [] 2 4 [] 4, [2] 2 4 [4] 4 [0] 4, and [3] 2 4 [9] 4 [] 4, it follows that [x 2 ] 4 is either [0] 4 or [] 4. That is, x 2 is congruent to either 0 or modulo 4. (b) Let x and y Z and assume that x 2 + y 2 r mod 4 for a non-negative integer r which is chosen as small as possible. What are the possible values of r? Why? (Hint: Part (a) will be useful.) Solution. From Part (a), [x 2 ] 4 and [y 2 ] 4 can be either [0] 4 or [] 4. Hence, the possibilities for [x 2 + y 2 ] 4 [x 2 ] 4 + [y 2 ] 4 are as follows: [x 2 ] 4 [y 2 ] 4 [x 2 + y 2 ] 4 [0] 4 [0] 4 [0] 4 [0] 4 [] 4 [] 4 [] 4 [0] 4 [] 4 [] 4 [] 4 [2] 4 Hence, the possibilites for r are 0,, 2. In particular, note that no integer that is congruent to 3 modulo 4 can be written as the sum of two squares of integers. (c) Show that not every positive integer is the sum of two squares of integers. Solution. This was answered in the previous part. (d) List the first 4 positive integers that are not sums of two squares. (Note that 2 + 0 2 is a sum of two squares.) Solution. From Part (b) we know that 3, 7,, and 5 are not sums of two squares since they are congruent to 3 modulo 4. However, there may be additional numbers that are not sums of two squares. The first few squares are 0 0 2, 2, 4 2 2, and 9 3 2. So we ask if there are any numbers smaller than 5 that cannot be written as a sum of two of these squares: 2 + 0 2 ; 2 + 2 ; 4 2 2 + 0 2 ; 6 cannot be written as a sum of two of the integers 0,, and 4; 8 2 2 + 2 2 ; 9 3 3 + 0 2 ; 0 3 2 + 2. Therefore, 3, 6, 7, are the four smallest integers that cannot be written as a sum of two squares. 8. Let a, b, c, and d be positive integers. Determine if each of the following statements is True or False. If False, provide a counterexample. (a) If a c and b c, then ab c. Solution. False. 2 2 and 2 2 but 2 2 2. (b) If (a, b) and (c, d), then (ac, bd). Solution. False. (2, 3) and (3, 2), but (2 3, 3 2) 6. (c) If there exist integers r and s such that ra + sb d, then d (a, b). Solution. False. 4 3 5 2 2 but (3, 2). 7
(d) [a, b] > (a, b). Solution. False. If a is any positive integer, then [a, a] a (a, a). (e) For every positive integer n, the number system Z n always contains nonzero zero divisors. Solution. False. If n is a prime, then every nonzero element of Z n is invertible, and hence not a zero divisor. (f) Every nonempty set of positive integers contains a largest element. Solution. False. The set of all positive integers does not have a largest element. 9. Express 24 and 02 a products of primes and use this information to calculate (2, 02) and [2, 02]. Solution. 24 8 3 2 3 3 and 02 2 3 7. Then (2, 02) 2 3 6 and [2, 02] 2 2 3 7 204. 0. How many elements does Z 8 contain? List them. Solution. G 8 {[] 8, [3] 8, [5] 8, [7] 8 } so G 8 has 4 elements.. Determine all x Z that solve the linear congruence 6x 9 mod 5. Solution. If we divide the congruence by 3, we conclude that 6x 9 mod 5 is equivalent to the congruence 2x 3 mod 5. But [2] 5 [3] 5 and [3 2 ] 5 [4] 5, so multiplying this last congruence by [3] 5 gives a congruence x 4 mod 5. Hence the solutions of the original congruence are all integers of the form x 4 + 5k for k Z. 2. (a) Find the greatest common divisor d (803, 54) of 803 and 54, using the Euclidean Algorithm. Solution. Use the matrix method illustrated on Page 4. [ ] [ ] [ ] 0 803 5 33 4 5 33 0 54 5 0 54 4 2 22 [ ] [ ] 5 26 2 5 26 4 2 22 4 73 0 Hence (803, 54). (b) Write d (803, 54) in the form d s 803 + t 54. Solution. From the top row of the last matrix in Part (a), it follows that 5 803 26 54. 8
(c) Find the least common multiple [803, 54]. Solution. [803, 54] 803 54 (803, 54) 23662 242. 3. Solve the system of congruences: x 5 (mod 25) x 23 (mod 32). Solution. Use the Euclidean algorithm to write 9 25 7 32. Then x 9 25 23 7 32 5 4055 modulo 25 32 800. Reducing modulo 800 gives x 55 modulo 800. Thus all solutions of the simultaneous congruences are x 55 + 800k for k Z. 4. Solve the system of congruences: 5x 4 (mod 7) 3x 2 (mod 3). Solution. First note that [5] 7 [7] 7 since 5 7 35 + 2 7 and [7] 7 [4] 7 [98] 7 [3] 7, so the congruence 5x 4 (mod 7) can be multiplied by [7] 7 to get an equivalent congruence x 3 (mod 7). Similarly, the second congruence can be multiplied by [3] 3 [9] 7 to give an equivalent congruence x 5 (mod 3). Hence, we need to solve the equivalent system: x 3 (mod 7) x 5 (mod 3). Use the Euclidean algorithm to write 4 3 3 7. Then the solution of the system of congruences is x 3 4 3 5 3 7 42 modulo 3 7 22. Hence all of the solutions are of the form x 200 + 22k for some k Z. 5. Give an example of integers a, b, m, n such that the system of congruences has no solution. Solution. One example is 6. (a) Compute 4 26 mod 27. x a (mod m) x b (mod n) x (mod 2) x 0 (mod 2) 9
Solution. Since 27 is prime and 27 does not divide 4, Fermat s theorem applies to give 4 26 mod 27. (b) Compute 4 63 mod 27. Solution. Since 4 4 2 8 256 2 mod 27, it follows that 4 7 4 4 4 3 2 64 28 mod 27. Hence (c) Compute 5 97 mod 27. 4 63 (4 7 ) 9 mod 27. Solution. First write the exponent 97 in binary form as a sum of powers of 2: 97 64 + 32 +. Then compute the binary powers of 5 by successive squaring: 5 5 2 25 5 4 (5 2 ) 2 625 7 0 mod 27 5 8 (5 4 ) 2 ( 0) 2 00 27 mod 27 5 6 (5 8 ) 2 ( 27) 2 94 33 mod 27 5 32 (5 6 ) 2 ( 33) 2 089 73 mod 27 5 64 (5 32 ) 2 73 2 22 5 mod 27. Then 5 97 5 64+32+ 5 64 5 32 5 ( 5) 73 5 825 80 mod 27. 7. Compute the Euler phi function φ(n) for each of the following natural numbers n. (a) n 22 Solution. n 22 3 7 so φ(22) 2 6 92. (b) n 625 Solution. n 625 5 3 7 2 so φ(625) (25 25) (49 7) 4200. (c) n 34 Solution. n 34 3 so φ(34) 0 30 300. (d) n 6860 Solution. n 6860 2 2 5 7 3 so φ(6860) (2 2 2) 4 (7 3 7) 5376. 0