MATH 537 (Number Theory) FALL 2016 TENTATIVE SYLLABUS

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1 MATH 537 (Number Theory) FALL 2016 TENTATIVE SYLLABUS Class Meetings: MW 2:00-3:15 pm in Physics 144, September 7 to December 14 [Thanksgiving break November 23 27; final exam December 21] Instructor: Allen Bell Contact information: EMS E D2L: Go to then log in and find Number Theory (Math 537). (currently unavailable) Text and Topics Required text: An Introduction to Number Theory with Cryptography by James S. Kraft and Lawrence C. Washington; publisher Chapman and Hall/CRC. We will probably cover about half of the material in the book. I will cover some cryptographic content in class, but I will leave most to the interested student to read on his/her own. Optional supplemental texts: (1) Elementary Number Theory by Underwood Dudley, Dover Books. The second edition costs less than $10, and a copy of the first edition can be read or downloaded freely at (2) Elements of Number Theory by John Stillwell, Springer Undergraduate Texts in Mathematics. This introduction to number theory focuses on using ideas from abstract algebra. Topics will include factorization of integers, congruences, integer solutions of polynomial equations (Diophantine equations), sums of squares, some properties of (the set of) prime numbers, and the RSA cryptosystem. Other topics may include quadratic reciprocity, irrational numbers, Pell s equations, continued fractions, Bertrand s postulate, arithmetic functions, and Fermat s proof of his last theorem. Here is a sampling of questions we might consider. Let n be or one of your other favorite numbers. Is n prime? What is the prime factorization of n? What is the sum of the factors of n? Can we solve 14x + n y = 3 with x, y integers? Can we solve x 2 n (mod 11)? What is the smallest value of k such that 2 k 1 (mod n), if there is one? Can we write n = x 2 + y 2 for integers x, y ; what about n = x 2 + y 2 + z 2 or n = x 2 + y 2 + z 2 + w 2? Can we solve x 2 ny 2 = 1 for integers x, y? How many prime numbers p are there with p n? Office hours: My office hours will be 12:30-1:50 pm & 3:20-4:10 pm on Monday & Wednesday. I will generally be available on Tuesdays 1:00-1:50 pm, but you should check because there will be some department meetings during this time. You can also see me by appointment or any time you can find me in my office; do not hesitate to talk to me. All times are subject to change and to cancelation on some days due to other duties. Grades: Your grade will be based on examinations and homework, and possibly quizzes and presentations. The grading scale will be determined based on the class performance (i.e., there will be a curve). Here is an example of one weighting system for grades I have used in the past: homework counts for 20% of your grade, the two midterms 20% each, and the final exam 35%, with the remaining 5% for class participation and other activities.

2 Graduate students in the class will be required to do extra work. We will discuss this in class. Investment of time: A typical student should expect to spend 150 minutes per week in class and at least six hours per week studying and doing problems. The amount of time you need to spend outside of class may vary considerably from this estimate. When taking notes in class and when reading the text or any other material, try to work actively. Anticipate what the next step will be and attempt to come up with your own proofs and your own solutions. Other information: Links to UWM policies relevant to this class can be found at Note particularly the statements on academic misconduct and discriminatory conduct, as well as on students with disabilities. Please turn off and put away cell phones during class. If you have any special requirements or concerns regarding this course, please let me know as soon as possible. Friday, October 28 is the last day to drop the class with a W on your transcript. For other important dates, see dates-deadlines/. Homework: Homework will be assigned, collected, and graded regularly. There are many problems and exercises in the text, and it is vital that you work on a wide selection of them, including those that are not assigned to be turned in. It is impossible to really learn mathematics without doing problems! You are encouraged to review draft versions of your homework with me. We will discuss homework problems, graded and ungraded, in class whenever you have questions, and I encourage you to come to my office to talk about problems. You are free to discuss homework problems with other students, except that homework that is handed in for a grade must be your work. Please remember that if you don t do it yourself, you won t learn it. Exams: I anticipate that there will be two midterm exams during the semester and a comprehensive in-class written final exam. The first exam will be in October. The final exam will take place on Wednesday, December 21 from 12:30 to 2:30 p.m. You cannot take the final early. A make-up exam will not be given without a very good, documented reason that is acceptable to me. If you cannot come to an exam (for that very good reason), please let me know as far in advance as possible: you may call me, me, or leave a message at the Mathematics office, Sample: On the following pages is an exam, with solutions, from a previous offering of Math 537. This is meant to give you a general feeling for the class and its expectations. The exams in Fall 2016 will differ greatly from this exam, but there will always be an expectation that you justify your work and there will always be some proofs.

3 MATH 537 Final Exam December 18, 2002 Be sure to show your work and give clear, complete explanations in order to obtain full credit. If you re not sure about something, ask me. There are 100 points on this exam, plus two bonus questions; students had two hours. The highest score was 85, the average was 57.6, and the median was 62. On this test, σ(n) is the sum of the positive divisors of n and φ(n) is the Euler phi function of n. 1. (10) Find the order of 13 mod 17. Then use this information to find the multiplicative inverse of 13 mod (15) For what positive integers n, k is φ(kn) = kφ(n)? 3. (16) Prove the following constructive version of the Chinese Remainder Theorem. Suppose that m 1,..., m k are pairwise relatively prime positive integers and that a 1,..., a k are any integers. For each i, define M i to be the product of all m j except m i. (For example, M 1 = m 2 m k.) Set x = k i=1 a i M φ(m i) i. Show that x a i (mod m i ) for all i = 1,..., k. 4. (15) Recall that an affine cipher is one where we encode the letter x (where x = 0,..., 25 ) as ax + b mod 26 for some a, b with gcd(a, 26) = 1. Suppose that we guess (correctly!) that the letter E (04) has been encoded as W (22) and that T (19) has been encoded as B. What are the values of a, b? 5. (14) Let p be a prime number larger than 5. (a) Show that p divides infinitely many numbers of the form (b) [A little trickier] What about numbers of the form ? 6. (12) (a) Prove that if k > 1, then σ(kn) > kσ(n). (b) Prove that no multiple and no divisor of a perfect number is perfect, except the number itself. [Don t assume the perfect number is even.] Continues on next page.

4 7. (18) Recall that an integer n > 1 that is not prime is called a Carmichael number if a n a (mod n) for all integers a. In class, we stated that n is a Carmichael number if and only if n = p 1 p k for distinct primes p 1,..., p k, k 2, with the property that (p i 1) (n 1) for all i. We never proved that every Carmichael number has this form. We will do that in this problem. Thus we assume n is a Carmichael number and n = k i=1 p a i i for some positive integers a 1,..., a k and distinct primes p 1,..., p k. Fill in the details in the parts below. (a) If n is even, explain why n cannot be a Carmichael number. Hint: Consider a = 1. From now on, assume n is an odd Carmichael number. (b) Let r i be a primitive root mod p a i i for each i. By the Chinese Remainder Theorem, there is an integer x such that x r i (mod p a i i ) for each i. Explain why such a number x is relatively prime to n. (c) Explain why x n 1 1 (mod n). (d) Using parts (b) & (c), explain why p a i 1 i (p i 1) (n 1). (e) Using part (d), conclude that a i = 1 and (p i 1) (n 1) for each i. Bonus Question A. Prove that the equation x 2 3y 2 = 2z 2 has no solutions for positive integers x, y, z. Bonus Question B. Show that if p is a prime number, ( ) p (mod ). p

5 MATH 537 Final Exam Solutions December 18, 2002 These are sample solutions. Remember, there may be other ways to solve some problems (mod 17) and ( 4) 2 1 (mod 17), so 13 4 is congruent to 1 mod 17, but 13 2 is not. Thus ord = 4. Since 13 4 = , it follows that 13 3 is the multiplicative inverse of 13 mod 17. Note (mod 17). 2. First, note that if all prime factors of k occur in the prime factorization of n, we have φ(kn) = kφ(n), since only the first power of a prime p contributes a p 1 to the phi-function: higher powers contribute p s. Now write k = rs where gcd(r, n) = 1 and the primes factors of s are those of k that occur in n. Then gcd(r, sn) = 1, so we have φ(kn) = φ(rsn) = φ(r)φ(sn) = φ(r)sn. This equals kn = rsn only if φ(r) = r, that is, r = 1. Thus φ(kn) = kφ(n) iff the prime factors of k all occur in the prime factorization of n. 3. Note that for any j, gcd(m j, m j ) = 1. Thus M φ(m j) j 1 (mod m j ) by Euler s Theorem. If i j, on the other hand, we have m j M i. Thus M φ(m i) i 0 (mod m j ) for these i. It follows that x a j 1 = a j (mod m j ) for each j. 4. The information we re given yields two congruences: a4 + b 22 (mod 26) and a19 + b 1 (mod 26). Subtracting these yields 15a 21 (mod 26), which has solution a = 9. Substituting this back into either of the original congruences yields b = (a) These are numbers of the form 10 n 1, where n is the number of digits. We wish to show such a number is congruent to 1 mod p for infinitely many n, that is 10 n 1 (mod p). Since gcd(10, p) = 1, we know 10 has some order m mod p. Now take n = km for any positive integer k and we get a value of n that works. (b) This time the number is n 1 i=0 10 i = (10 n 1)/(10 1). Since gcd(9, p) = 1, such a number is divisible by p if and only if 10 n 1 (mod p). This happens infinitely often, just as in (a). 6. (a) Let d 1,..., d r be the divisors of n. Then kd 1,..., kd r are divisors of kn. Furthermore, this list does not include 1. Therefore, σ(kn) 1 + kd kd r = 1 + kσ(n) > kσ(n). (b) If m = kn, then σ(m) > kσ(n) = 2m, so m is not perfect. If n = km, then 2km = 2n = σ(n) > kσ(m), so m is not perfect. 7. (a) If n is even, ( 1) n = 1, and this is not congruent to 1 unless n = 2. But 2 is prime, so n cannot be 2. (b) If gcd(x, n) > 1, then there must be a prime p i dividing both x and n. But x r i (mod p i ), and gcd(r i, p i ) = 1. Thus this is impossible.

6 (c) We have x x n 1 x 1 (mod n). Since gcd(x, n) = 1, we can cancel x and get x n 1 1 (mod n). (d) By part (c), we get x n 1 1 (mod p a i i ). Since x is congruent to r i, it has the same order, which must divide n 1. Thus φ(p a i i ) n 1, as desired. (e) Clearly gcd(p i, n 1) = 1, so the result in (d) can only hold if a i 1 = 0, i.e., a i = 1. Putting this together with part (d), we re done. Bonus A Assume the equation has a solution. If there is a solution with all of x, y, z even, we can divide each of them by 2. If we keep doing this, we ll eventually get a solution where at least one of them is odd. Furthermore, if x, y are both even, x 2 3y 2 is divisible by 4, and this would force z to be even. Thus there must be a solution with at least one of x, y odd. On the other hand, if exactly one of x, y is odd, we would have x 2 3y 2 odd, and hence it could not equal 2z 2. Thus there Must be a solution with both x, y odd. Now consider the congruence x 2 3y 2 2z 2 (mod 8). Since x, y are odd, x 2, y 2 are both congruent to 1, so we have 2 2z 2 (mod 8). This is impossible, since mod 8, z 2 is one of 0, 1, 4. Thus there is no solution after all. Bonus B This is true when p = 2, so assume p is odd. Let x = ( 1)( 2) ( p+1). Then x ( 1) p (p 1) (mod ). Since p 1 is even, x (p 1)! (mod ). Now ( ) ( ) p = px/(p 1)!, so we have (p 1)! p p(p 1)! (mod ). Since (p 1)! is relatively prime to p, we can cancel it to get ( ) p p (mod ). Question: numerical evidence suggests that ( ) p p (mod p 3 ) for an odd prime p. Is this true?

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