Name: There are 8 questions on 13 pages, including this cover. There are several blank pages at the end of your exam which you may as scrap paper or as additional space to continue an answer, if needed. You are permitted to use a calculator, but it must be an actual calculator, it cannot be a phone or tablet or device with other capabilities. You are not permitted to use any other electronic device other than a calculator during this exam. You are not permitted to look at notes or your book during this exam. Write your answers in the space provided. Ask for more scratch paper if needed. With the exception of the True/False and Short Answer questions, you must justify your claims and use complete sentences in proofs. Please ask questions if you are confused about anything. Problem 1 (15pts) Problem 2 (12pts) Problem 3 (12pts) Problem 4 (12pts) Score Problem 5 (12pts) Problem 6 (13pts) Problem 7 (12pts) Problem 8 (12pts) Score Page 1
1. For each statement below, CIRCLE true or false. You do not need to show your work. (a) (b) (c) (d) (e) True False True False True False True False True False (a) (21/35) = 1. (b) x 2 23 (mod 1000) has a solution. (c) If a bc but a does not divide b then a c. (d) There exists an integer x such that x 3 (mod 27) and x 4 (mod 100). (e) The well ordering principle says that any nonempy set of positive integers has a minimal element. Solution: (a) IGNORE. (b) False. Since 1000 = 2 3 5 3 this has a solution if and only if it has a solution moduli 2 3 and modulo 5 3. But x 2 3(mod 5) does not have a solution, and hence this has no solution. (c) False. 4 2 6 but 4 does not divide either 2 or 6. (d) True by the Chinese Reaminder Theorem, since gcd(27, 100) = 1. (e) True. Page 2
2. On this page only the answer will be graded. Circle your answer. (a) Compute the largest power of 48 that divides 50! Solution: Since 48 = 2 4 3 we must find the largest power of 2 and the largest power of 3 that divides 50!. For 2, the computation is [50/2] + [50/4] + [50/8] + [50/16] + [50/32] = 25 + 12 + 6 + 3 + 1 = 47. Since 47 = 11 4 + 3, the largest power of 2 4 that divides 50! is thus (2 4 ) 11. For 3 the computation is [50/3] + [50/9] + [50/27] = 16 + 5 + 1 = 22. Thus the largest power of 3 that divides 50! is 3 22. It follows that largest power of 48 that divides 50! is 48 11. (b) Find x such that 13x 2(mod 32). Solution: We first want to write 1 = 32a + 13b. For this we do the Euclidean algorithm and we get 32 = 2 13 + 6 and 13 = 2 6+1 which yields 1 = 5 13 2 32. Then we have 5 13x 65x x 5 2 10(mod 32). We can check: if x = 10 then we have 130 2 = 4 32. Thus x = 10 is one solution. (c) Compute 30 a=3 (a/31). Solution: Recall that 30 a=1 (a/31) = 0. Thus this is sum is ((1/31) + (2/31)) = (1 + (2/31)). But (2/31) = 1 since 31 7(mod 8). Thus this sum is (1 + 1) = 2. Page 3
3. On this page only the answer will be graded. Circle your answer. (a) Find an integer n where µ(n) = 1 and τ(n) = 4. Solution: primes. We can choose n = 6 or more generally any product of two distinct (b) What percent of the quadratic nonresidues of 31 are also primitive roots? Solution: There are (31 1)/2 = 15 quadratic non residues. There φ(φ(31)) = φ(30) = φ(2)φ(3)φ(5) = (1)(2)(4) = 8 primitive roots. Thus 8/15 (or 53%) of the quadratic non residues of 31 are also primitive roots. (c) Find a number 0 a 60 such that 2 1603 a (mod 60). Solution: φ(60) = φ(2 2 )φ(3)φ(5) = 2 2 4 = 16. So 2 1603 2 3 (mod 60). So a = 8. Page 4
4. (a) If 2 k 1 is prime, show that k must also be prime. Solution: Assume that k is not prime. Then we can factor k = rs with r, s > 1. Then 2 k 1 = (2 r 1)(2 r(s 1) + 2 r(s 2) + + 2 r + 1). Since r, s > 1 both factors are greater than 1. But this contradicts the fact that 2 k 1 is prime. Hence k must be prime. (b) Let a, b be positive integers and let S = {sa + tb s, t Z and sa + tb > 0}. Let d be the minimal element of S. Prove that d a. Solution: Using the Division Algorithm, we write a = qd + r where 0 r < d. We claim that r = 0. By assumption d = sa + tb for some s, t. Thus d r = (sa + tb) (a qd) = sa + tb a + q(sa + tb) = (s 1 + qs)a + (t + tq)b. Thus d r S. If r 0 then 0 < d r < d which contradicts the minimality of d. Thus r = 0 and thus d a. Page 5
5. (a) Give an example to show that a k b k (mod n) and k j(mod n) needs not imply that a j b j (mod n). Your answer should involve explicit values for a, b, k, j and n, and you must check that you have given a counterexample. Solution: Let a = 1, b = 2, n = 3, k = 0 and j = 3. Then a k b k (mod n) because 1 0 2 0 (mod 3). However a j b j (mod 3) because 1 3 2 3 (mod 3). (b) Let p > 3 be a prime integer such that p + 2 is also prime. Prove that 12 divides 2p + 2. Solution: Let p > 3 be a prime integer. By the Fundamental Theorem of Arithmetic, 12 (2p + 2) if and only if 4 (2p + 2) and 3 (2p + 2). (Alternately, since gcd(3, 4) = 1 it follows that 12 (2p + 2) 4 (2p + 2) and 3 (2p + 2).) First we prove that 3 (2p + 2). Since p > 3 and is prime, it cannot be divisible by 3. Thus p is congruent to either 1 or 2 modulo 3. However, if p 1( mod 3) then p + 2 0( mod 3) and thus 3 (p + 2). But p + 2 is also prime and greater than 3, so this would yield a contradiction. Thus p 2( mod 3) is the only possibility. Then 2p + 2 2 2 + 2 6 0( mod 3). So 3 (2p + 2). Next we prove that 4 (2p + 2). Since 2p + 2 = 2(p + 1), it suffices to show that p + 1 is even. But since p is prime and greater than 3, it is odd and there p + 1 is even. Thus 4 (2p + 2). Since 4 (2p + 2) and 3 (2p + 2) (and since gcd(3, 4) = 1)) we conclude that 12 (2p + 2). Page 6
6. (a) Let p, q be odd primes satisfying p q 1 (mod 4). It follows that p q = 4a for some a. Show that (a/p) = (a/q). Solution: Since 4 is square we have (4a/p) = (4/p)(a/p) = (a/p) since 4 is a square. Moreover, since 4a = p q this implies that (4a/p) = (p q/p) = ( q/p) = ( 1/p)(q/p). Since p 1(mod 4) we have ( 1/p) = 1. We conclude that (a/p) = (4a/p) = (q/p). Similarly: (4a/q) = (4/q)(a/q) = (a/q) and (4a/q) = (p q/q) = (p/q). But since p 1(mod 4) it follows that (p/q) = (q/p). And thus (a/p) = (a/q) as desired. (b) Let t(n) = τ(n) φ(n) and let T (n) = d n t(d). Prove that T (n) is a multiplicative function. You may use the fact that τ is multiplicative in your proof. Solution: We first observe that t is multiplicative, since if gcd(m, n) = 1 then t(mn) = τ(mn)φ(mn) = τ(m)φ(m)τ(n)φ(n) = t(m)t(n) where the middle equality is because τ and φ are multiplicative. Summing a multiplicative function over d n yields a new multiplicative function, and hence T is multiplicative as well. Page 7
7. Let f be a multiplicative function that is not identically zero. Let n = p k 1 1 p k 2 2 p kr Prove that µ(d)f(d) = (1 f(p 1 ))(1 f(p 2 )) (1 f(p r )). d n Solution: Let F = d n µ(d)f(d). Since µ, f are both multiplicative, so is their product, and thus so is F. We thus have that F (n) = r i=1 F (p k i i ) We now compute it explicitly for prime powers. We have: F (p k i i ) = µ(1)f(1) + µ(p i)f(p i ) + µ(p 2 i )f(p 2 i ) + + µ(p k i i )f(pk i i ) But µ(p s i ) = s for all s 2 and thus: = f(1) f(p i ) + 0 + + 0 = 1 f(p i) Since f(1) = 1 for any multiplicative function that is not identically zero (lemma from class, or you could quickly prove it.) It follows that r. F (n) = r r F (p k i i ) = (1 f(p i )). i=1 i=1 Page 8
8. Let p be an odd prime and let r be a primitive root for p. Let d = gcd(k, p 1). Prove that the values for which the congruence x k a (mod p) are solvable are precisely a = r d, r 2d,..., r (p 1). Solution: This problem had two typos: the first was a typo in the exponent. The seance is that a = 0 also solves the congruence, but I announced that you could ignore that solution, so it was not part of the grading. Since d = gcd(k, p 1) we can write k = k d and p 1 = md where gcd(k, p 1) = 1. ( :) Let a = r c and assume that x k a(mod p) has a solution. Then raising the lefthand side to the m power we get x km x k (p 1) 1(mod p) and hence a m 1(mod p). It follows that the order of a divides m. However, since a = r c and r is a primitive root, it follows that the order a is (p 1)/ gcd(c, p 1) and hence d must divide gcd(c, p 1) which implies that d divides c. ( :) Conversely, let a = r cd for some c = 1, 2,..., (p 1)/d. We must show that x k a(mod p) has a solution. Any such solution will be relatively prime to p and can thus be written as x = r y for some y. We thus want to solve the equation (r y ) k r yk r cd (mod p) Consider the congruence yk cd(mod p 1); this has a solution y since d = gcd(k, p 1) divides cd. For this value of y we then have p 1 divides yk cd and it follows that r yk r cd (mod p). Page 9