Number Theory Math 420 Silverman Exam #1 February 27, 2018

Transcription

1 Name: Number Theory Math 420 Silverman Exam #1 February 27, 2018 INSTRUCTIONS Read Carefully Time: 50 minutes There are 5 problems. Write your name neatly at the top of this page. Write your final answer in the answer box, if one is provided. Show all your work. Partial credit will be given for substantial progress towards the solution. No credit will be given for answers with no explanation. Problem Value Points Total 50

2 Math 420 Solutions for Exam #1 Page 1 Problem 1. (10 points) For this problem, you can use the fact that (a) Find a number b satisfying is a prime number. b (mod 48611) and 0 b < (b) Find a number x between 0 and that is a solution to the congruence x (mod 48611). b = x = (c) Find a number z between 0 and that is a solution to the congruence z (mod 48611). z = Solution. For all three parts we use Fermat s Little Theorem, which says that if p a, then a p 1 1 (mod p). For the prime that we re using, it says a (mod 48611). (a) Here the exponent is p + 2, so we compute = (mod 48611). It remains to compute 11 3, which can be done in two steps using paper and pen: 11 2 = 121, and then 11 3 = = (b) For this part, the exponent is p, so x x x x (mod 48611) for every x that is not a multiple of p. We want this to be congruent to 37, so we must take x = 37. (c) For this part, the exponent is p 2. So if we multiply both sides by z, we ll get an exponent of p 1 and can use Fermat s theorem. Thus z (mod 48611) congruence to be solved, z z (mod 48611) multiply both sides by z, 1 2z (mod 48611) use Fermat s little theorem. So we want to divide 2z 1, so we can take z = =

3 Math 420 Solutions for Exam #1 Page 2 Problem 2. (10 points) Here is the Euclidean algorithm computation showing that gcd(373, 231) = = = = = = = = = Use this computation to find integers x and y that satisfy the equation 373x + 231y = 1. Be sure to explain your reasoning. x = y = Solution. Let a = 373 and b = 231. We have a = 1 b = a b, b = b = a b = a + 2b, 142 = a b = a + 2b = 2a 3b 89 = a + 2b = 2a 3b = 3a + 5b 53 = a 3b = 3a + 5b = 5a 8b 36 = a + 5b = 2(5a 8b) = 13a + 21b 17 = a 8b = 8( 13a + 21b) = 109a 176b So x = 109 and y = 176 are solutions to 373x + 231y = 1. Problem 3. (10 points) For this problem, p is a large prime number, say p > 100. Also, when we talk about distinct solutions to a congruence, we mean that two solutions a and b are considered to be the same if they are congruent modulo p. (a) Give an upper bound for the maximum possible number of distinct solutions to the congruence x 23 3x x 0 (mod p)? Also give a lower bound for the minimum number of distinct solutions. Justify both of your answers. There are at most solutions and at least solutions.

4 Math 420 Solutions for Exam #1 Page 3 (b) Exactly how many distinct solutions are there to the congruence x p (mod p). Be sure to justify your answer. There are solutions. Solution. If f(x) is a polynomial of degree d having integer coefficients, then we proved in class: ( ) f(x) 0 (mod p) has at most d solutions. We use this fact several times. (a) Using ( ), we know that there are at most 23 solutions. On the other hand, it is clear that x = 0 is a solution, so there is at least 1 solution. I ll give full credit for that answer, but you might notice that x = 1 and x = 1 are also solutions, so in fact, since p 3, there are at least 3 solutions. For those who are interested, I did a computer search and found that for p = , there are 15 solutions, and that s the most for any prime smaller than one million. Using more advanced methods from number theory, one can prove that there exist primes for which the congruence has 23 solutions. As for the minimum, there are lots of primes for which the congruence has exactly 3 solutions, including p = 107 and p = 113. (b) The given congruence has at most p 1 solutions from ( ). But Fermat s Little Theorem tells us that every integer 1, 2, 3,..., p 1 is a solution, so there are at least p 1 solutions. Hence there are exactly p 1 solutions. Problem 4. (10 points) Fix an integer k 2. For this problem, we live in K-World, where the only numbers that exist are multiples of k. In other words, the only numbers that exist are the numbers in the set K = {..., 3k, 2k, k, 0, k, 2k, 3k,...}. As usual, if a and b are in K, then we say: b K-divides a if there is some c K so that a = bc. a is a K-prime if a has no K-divisors.

5 Math 420 Solutions for Exam #1 Page 4 For this problem, be sure to provide justification for your answers. If you don t see how to do the problem in general, try doing it for k = 7 and you will receive partial credit. (a) What is the smallest positive element of K that is a K-prime? Smallest Positive K-Prime = (b) What is the smallest positive element of K that is not a K-prime? Smallest Positive Non-K-Prime = (c) For this part, you may assume that k 7. Write down a positive K-number that can be factored as a product of K-primes in two truly different ways, and give the two factorizations. Solution. (a) The number k is the smallest positive number in K, and I claim that k is a K-prime. To see this, suppose that we try to write k = bc with b and c in K. Then b and c are multiples of k, so bc k 2, so bc cannot equal k. Hence k is the smallest positive K-prime. (b) Next suppose that a > 0 and the a is not a K-prime. This means that a = bc, where b, c K, so b and c are multiples of k, say b = kd and c = ke. Then a = bc = k 2 de, so certainly a k 2. On the other hand, k 2 is not a K-prime, since k 2 = k k is a product of two numbers in K. Hence k 2 is the smallest non-prime in K-World. (c) The numbers k, 2k, 3k, and 6k are K-primes, since the smallest K- non-prime is k 2, and for this part we have been told that k 7, so 6k is smaller than the smallest K-non-prime. Then 6k 2 has two different factorizations into K-primes, 6k 2 = (2k) (3k) = k (6k). Problem 5. (10 points) Suppose that gcd(a, b) = 1, and also suppose that a divides bc. Prove that a divides c. Solution. From class we know that we can find integers x and y that satisfy ax + by = 1, since gcd(a, b) = 1. Multiply both sides by c, we get acx + bcy = c. We are given that a bc, which means that bc = ad for some integer d. Substituting ad for bc gives acx + ady = c.

6 Math 420 Solutions for Exam #1 Page 5 Thus c = a(cx + dy), so a c. An alternative way to do this problem is to use unique factorization. Thus factor a, b, c into a product of primes. The assumption that gcd(a, b) = 1 means that a and b have no prime factors in common, while the assumption that a bc implies that every prime power dividing a appears in the factorization of bc. Since none of those primes can appear in b, they must all appear in c.