MATH NUMBER THEORY SPRING Scientia Imperii Decus et Tutamen 1

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1 MATH NUMBER THEORY SPRING 2019 Scientia Imperii Decus et Tutamen 1 Robert R. Kallman University of North Texas Department of Mathematics 1155 Union Circle Denton, Texas office telephone: office fax: office robert.kallman@unt.edu January 4, Taken from the coat of arms of Imperial College London.

2 ******************************************************************************** CAVEAT EMPTOR! THIS INTRODUCTORY MATERIAL MAY BE CHANGED DY- NAICALLY THROUGHOUT THE SEMESTER! THE STUDENTS WILL BE NOTIFIED PROMPTLY OF ANY CHANGES. ******************************************************************************** SPRING 2019 COURSE: MATH , NUMBER THEORY. PREREQUISITES: MATH 2000 or MATH 3000 or CSCE CLASS MEETS: Monday, Wednesday, Friday 1:00 p.m. - 1:50 p.m., LANG 204. FINAL EXAM DATE AND TIME: The final is scheduled for Saturday, May 4, 2019 in LANG 204, 10:30 a.m. - 12:30 a.m. TEXT: Kenneth H. Rosen, Elementary Number Theory, Sixth Edition, Addison-Wesley, Boston, copyright 2011, ISBN , ISBN-10: INSTRUCTOR: Robert R. Kallman, 315 GAB [office], [office telephone], [fax], robert.kallman@unt.edu [ ] OFFICE HOURS: Monday, Wednesday, Friday, 8:00 a.m. - 9:30 a.m. & 11:00 a.m. - 12:30 pm ATTENDANCE POLICY: Mandatory. Specifically for TAMS students: if you are absent for any reason, you are required to file an absence report with the appropriate official in the TAMS Academic Office. ELECTRONIC DEVICES: No electronic devices of any sort are to be on let alone used during the class. Repeated flouting of this will result in a grade penalty. HOMEWORK: Homework will be assigned and some designated subset of it will be graded. The designated homework assigned on Monday, Wednesday and Friday of one week will be due at the beginning of class on the Friday of the following week. Homework sets will consist of 1-5 problems. Late homeworks will not be accepted under any circumstances. This applies even if the student is ill - give your homework to a classmate to hand in on time or convert your homework into a *.pdf file and to me with a time stamp earlier than the beginning of class. Each homework will be worth 1-5 points, which will be clearly 1

3 indicated. A subset of these homeworks will be graded, though of course the students will not know whihc ones beforehand. ACADEMIC INTEGRITY: There is no reason why students should not demonstrate complete academic integrity at all times, particularly on tests. Any transgression of this will result in a grade of zero on the test and a grade of F for the course. Consistent with this policy the instructor will retain xerox copies of a random sampling of all tests. GRADING POLICY: Grades will be based on the total number of points accrued from the assigned graded homeworks, from two in class one hour examinations (5 problems plus 1 bonus question), given circa in late February and late April, and from an in class 120 minute final (8 problems plus 1 bonus question). The number of points per test and final problem will normally be 10. There is no excuse for missing a test and no makeup tests will be given under any circumstances. A student missing a test will receive a grade of -1 on that test. If a student is unavoidably absent from a test and makes arrangements with the instructor well before the test date, then the grade assigned to the missing test will be prorated by the student s performance on the final examination minus 10 points. It is difficult a priori to determine the precise break points for the final grades. However, the golden rule in determining the final assigned grade is that if the number of points earned by person A is to the number of points earned by person B, then person A has a grade which is to the grade of person B. The only possible exception is that you must take the final examination and receive a passing grade on the final in order to get a grade greater than F. Students can obtain a real time good first approximation to their progress/status in this class by doing the following calculation. Let S = 0.4( A) + 0.6( C ), where A = number of B D homework points earned to date, B = maximum number of homework points possible to date, C = number of test and final points earned to date and D = maximum number of test points possible to date (not counting bonus points). Note that 0.0 S 1.0. Then the grade in this class at this instant is: A if S 0.9; B if 0.8 S < 0.9; C if 0.7 S < 0.8; D is 0.6 S < 0.7; F if S < 0.8. TOPICS: The topics to be covered can be found in most of Chapter , , , , , , , Some supplementary notes will be handed out. APPROXIMATE ITINERARY: The following is a first attempt, very rough approximation to what our schedule will be. This will perhaps be dynamically reconfigured as the semester progresses since it is of course impossible to make such a schedule with hard-andfast rules. 2

4 Test #1, Friday, February 22, 2019, covering , 6.8, 7.1 and 7.2. Test #2, Friday, April 12, 2019, covering , 7.8, 8.1, 8.2, 9.3, 9.5, Selected portions of Chapter 11, especially an independent study of Taylor s Formula with the integral form of the remainder. Final, Saturday, May 4, 2019, cumulative. ASK QUESTIONS in class so that we may all benefit. If you need help, it is your responsibility to seek me out. See me during my office hours. Empirical evidence suggests that there is a strong correlation between the amount of work done by the student and his/her final grade. STUDENTS WITH DISABILITIES: It is the responsibility of students with certified disabilities to provide the instructor with appropriate documentation from the Dean of Students Office. STUDENT BEHAVIOR IN THE CLASSROOM: The Powers That Be have strongly suggested that students be given the following statement: Student behavior that interferes with an instructor s ability to conduct a class or other students opportunity to learn is unacceptable and disruptive and will not be tolerated in any instructional forum at UNT. Students engaging in unacceptable behavior will be directed to leave the classroom and the instructor may refer the student to the Center for Student Rights and Responsibilities to consider whether the student s conduct violated the Code of Student Conduct. The university s expectations for student conduct apply to all instructional forums, including university and electronic classroom, labs, discussion groups, field trips, etc. The Code of Student Conduct can be found at In other words, cause trouble in the classroom and you will probably be cast into the Darkness and sent to the KGB. 3

5 How to Study for This Class Attend every class. Pay attention in class, take careful notes, and ask questions if needed. The evening of every class go over your classroom notes, list topics on which you have questions or need clarification, read the relevant section of the textbook, do the assigned homework to be graded, look over the additional homeworks to verify that you understand how to do them and make note of those additional homeworks that you do not understand to ask about them during the next class. It is important that you put a great deal of effort into the homework, both those to be turned in and those that are less formally assigned. One becomes adroit at any human activity - e.g., hitting a fast ball, throwing a slider, making foul shots or jump shots, or driving off a tee - only with a great deal of practice. The same applies to number theory. Do not waste your time memorizing endless formulas. This in fact is counterproductive. Instead, concentreate on general concepts involved in problem solving. All of the problems encountered in this class should be first approached by asking oneself what is a reasonable way to proceed. Then given the proper path or direction, you can then solve the problems by small, logical steps that inevitably lead one to the final solution. 4

6 Miscellaneous Comments Re Integers in Everyday Life and Some Hard Conjectures and Exotica. Symbols. = for all or for every, = there exists, = in or an element of, / = not in or not an element of, = such that, = = implied by, = = implies, = if and only. Definition 1. Let N = {1, 2, 3,... } be the natural numbers, Z = the integers, and Q = the rationals. The Pythagorean Theorem and its proof leads to the diophantine equation a 2 + b 2 = c 2 and its Permat Conjecture generalization a n + b n = c n for n 3. A very, very difficult and advanced proof of Fermat s Conjecture was given in the last few decades. We will eventually show that every solution to the Pythagorean diophantine equation is of the form: Let u > v > 0, a = 2uv, b = u 2 v 2 and c = u 2 + v 2. It is simple algebra to check that given u > v > 0 the corresponding triple (a, b, c) satisfies a 2 + b 2 = c 2. On the other hand, is every Pythagorean triple of this form - we will eventually show that the answer is yes. This subject was intensively studied in many ancient cultures. Many ancient cultures also studied Pell s equation x 2 dy 2 = 1, for examle, Brahmagupta in the early 7th century. Of course this equadtion has the trivial solution x = 1, y = 0, but this is quite uninteresting. If case d is a square, say d = a 2, the equation is also uninteresting since it becomes x 2 = (ay) 2 + 1, which has no solutin other than the trivial solution. When d = 61 the study of Pell s equation points out the absurdities contained in a poem about the Battle of Hastings (Octobeer 14, 1066). More recently the study of Pell s equation has arisen in cryptograpyhy. The men of Harold stood well together, as their wont was, and formed sixty and one squares, with a like number of men in every square thereof, and woe to the hardy Norman who ventured to enter their redoubts; for a single blow of a Saxon war-hatchet would break his lance and cut through his coat of mail.... When Harold threw himself into the fray the Saxons were one mighty square of men, shouting the battle-cries, Ut! Olicrosse! Godemit e! Nontrivial solutions usually are gigantic. An example is provided by Archimedes s Cattle problem, which is equivalent to the Pell equation for d = In this case, x(d) and y(d), the smallest nontrivial solution, have and decimal digits, respectively. The Cattle Problem, which Archimedes set in epigrammatic form and sent to those interested in these matters in Alexandria, in the letter addressed to Eratosthenes of Cyrene. 5

7 If thou art diligent and wise, O stranger, compute the number of cattle of the Sun, who once upon a time grazed on the fields of the Thrinacian isle of Sicily, divided into four herds of different colours, one milk white, another a glossy black, the third yellow and the last dappled. In each herd were bulls, mighty in number according to these proportions: Understand, stranger, that the white bulls were equal to a half and a third of the black together with the whole of the yellow, while the black were equal to the fourth part of the dappled and a fifth, together with, once more, the whole of the yellow. Observe further that the remaining bulls, the dappled, were equal to a sixth part of the white and a seventh, together with all the yellow. These were the proportions of the cows: The white were precisely equal to the third part and a fourth of the whole herd of the black; while the black were equal to the fourth part once more of the dappled and with it a fifth part, when all, including the bulls went to pasture together. Now the dappled in four parts8 were equal in number to a fifth part and a sixth of the yellow herd. Finally the yellow were in number equal to a sixth part and a seventh of the white herd. If thou canst accurately tell, O stranger, the number of cattle of the Sun, giving separately the number of well-fed bulls and again the number of females according to each colour, thou wouldst not be called unskilled or ignorant of numbers, but not yet shall thou be numbered among the wise. But come, understand also all these conditions regarding the cows of the Sun. When the white bulls mingled their number with the black, they stood firm, equal in depth and breadth, and the plains of Thrinacia, stretching far in all ways, were filled with their multitude. Again, when the yellow and the dappled bulls were gathered into one herd they stood in such a manner that their number, beginning from one, grew slowly greater till it completed a triangular figure, there being no bulls of other colours in their midst nor none of them lacking. If thou art able, O stranger, to find out all these things and gather them together in your mind, giving all the relations, thou shalt depart crowned with glory and knowing that thou hast been adjudged perfect in this species of wisdom. Definition 2. Let 0 a Z and let b Z. Then b is said to be divisible by a or that a divides b or b is a multiple of a or a divides b or a goes into b if there is q Z such that b = qa. In symbols a b. If a 0 and a does not divide b, we write a b. Example 3. Let a Z. 2 6, 4 6, 3 4, 2 4, a 0 for every a 0, 1 a and 1 a for every a, a a and a a for every a 0. Definition 4. a Z is said to be even if 2 a, a is said to be odd otherwise, i.e., if 2 a. 6

8 A Mystery of the Universe. Consider the following iteration of N. n n/2 if n is even, n 3n + 1 if n is odd. A question raised to Collatz and Kakutani asks if applying this operation iteratively always leads to 1. For example It is easy to write programs to check this and it has been tested into the quintillions and so far no contradiction. Do this iteration sharting with 31 and you will see that something very remarkable and mysterious is going on here. It is theoretically possible that this question always has a positive answer but we may be unable to prove it. The 10 digit ISBN number trick to always get a number divisible by 11 if the ISBN number is correct. The easily proved algebraic identity (ac bd) 2 +(ad+bc) 2 = (a 2 +b 2 )(c 2 +d 2 ) for any four commuting variables a, b, c and d, familiar from complex numbers, shows that the product of any two positive integers, each a sum of two squares, is itself a sum of two squares. Not every positive integer is the sum of two squares, e.g., 3, and not every positive integer is the sum of three squares, e.g., 7. Remarkably, however, every positive integer is the sum of four squares, e.g, 19 = A very remarkable algebraic identity involving products of sums of four squares due to Euler plays a crucial role: (a a a a 2 4)(b b b b 2 4) = (a 1 b 1 a 2 b 2 a 3 b 3 a 4 b 4 ) 2 + (a 1 b 2 + a 2 b 1 + a 3 b 4 a 4 b 3 ) 2 + (a 1 b 3 a 2 b 4 + a 3 b 1 + a 4 b 2 ) 2 + (a 1 b 4 + a 2 b 3 a 3 b 2 + a 4 b 1 ) 2, communicated by Euler in a letter to Goldbach on April 12, The identity also follows from the fact that the norm of the product of two quaternions is the product of the norms. This identity proves that the product of two integers, each of which is the sum of four squares, is itself a sum of four squares. 7

9 Proofs by Induction. A Very Important Basic Observation. If = A N, then A has a smallest element. Give two intuitive proofs based on what we think the integers are. First Principle of Induction: Given a sequence of statements S 1, S 2,..., suppose that: (1) S 1 is true; (2) if S n is true, then S n+1 is true. Conclusion: all of the S n s are true. Give an inutitive proof and one based on the very important basic observation. Note that we can start with any integer. Suppose that (1) and (2) hold and some S k is false. Then by the Very Importaxnt Basic Observation we may assume that k is as small as possible. k > 1 since S 1 is true. Therefore k 1 1 and S k 1 is true since k is the smallest positive integer such that S k is false. But by (2) S k 1 true = S (k 1)+1 = S k is true, a contradiction. Therefore no S k is false, i.e., all of the S n are true. Note that both (1) and (2) must be carefully checked. For example, let S n be n = n 2. Then (1) is true, but (2) is not, for if n = n 2, then n + 1 = (n + 1) 2 = n 2 + 2n + 1 = 2n = 0 = n = 0, a contradiction. For another example, let S n be n = n + 1. Then (1) is false but (2) is true, for n = n + 1 = n + 1 = (n + 1) + 1. Example 5. S n is the statement n = n(n+1) 2. Recall Gauss s elementary proof. Example 6 (Bernoulli s Inequality). S n ia the statement that if x 1 then (1 + x) n 1 + nx. We have strict inequality if additionally x 0 and n 2. Example 7 Let S n be the statement (2n 1) = n 2. Example 8. Let S n be the statement n 2 = n(n+1)(2n+1) 6. 8

10 Example 9. Let S n be the statement n 3 = ( n) 2. Exercise 1 (5 points). For what integers n is 3 2n n+2 divisible by 7? If n = 1 then = 35 and At the (n + 1)-st stage, 3 2n n+3 = 9 3 2n n+2 = 7 3 2n+1 + 2(3 2n n+2 ), which is divisible by 7 by induction. Exercise 2 (4 points). For which integers n is 9 n 8n 1 divisible by 8? The summand 8n is always divisible by 8, so it suffices to show that 9 n 1 is always divisible by 8. If n = 1 then = 8. At the (n + 1)-st stage, 9 n+1 1 = 9 9 n 1 = 8 9 n + (9 n 1), which is divisible by 8 by induction. 9

11 Factorials, Binomial Coefficients, the Binomial Theorem. Definition 10. Define 0! = 1, 1! = 1, 2! = 2 1, 3! = = 3 2! and recursively n! = n (n 1)! for n 1. Lemma 11. If 1 k n then D k (x n ) = n!/(n k)!x n k. Definition 12. Let 0 k n. Then ( n k) = n!/k!(n k)!, a binomial coefficient, called n choose k. Corollary 13. Given n distinct objects, the number of subsets of size 0 k n is ( n k). Lemma ( 14. n ( k) = n ( n k). n ( 0) = n ( n) = 1 and n ) ( 1 = n n 1) = n. Lemma 15. If 1 k n then ( ( n k 1) + n ) ( k = n+1 ). k ( ) n + k 1 ( ) n k = = = = = n! (k 1)!(n (k 1))! + n! k!(n k)! n! (k 1)!(n k)! [ 1 n + 1 k + 1 k ] n! (k 1)!(n k)! n + 1 k(n + 1 k) (n + 1)! k!(n + 1 k)! ( ) n + 1 k This is also obvious by a simple combinatorial argument using a fixed distinguished element out of n + 1 distinct objects. 10

12 Theorem 16 (Binomial Theorem). If A and B are commuting variables and n 1, then (A + B) n = 0 l n ( n l) A l B n l. Induct. If n = 1 then ( 1 ) 0 l 1 l A l B 1 l = ( ) 1 0 A 0 B ( 1 1) A 1 B 1 1 = B + A = (A + B) 1. If the theorem is true for n, then (A + B) n+1 = (A + B)(A + B) n = (A + B) 0 l n ( n l ) A l B n l = ( ) n A l+1 B n l + ( ) n A l B n+1 l l l 0 l n 0 l n ( ) n = A l B n+1 l + ( ) n A l B n+1 l l 1 l 1 l n+1 0 l n = A n+1 + ( ) n A l B n+1 l + l 1 1 l n 1 l n = A n+1 + ( ) ( ) n n [ + ]A l B n+1 l + B n+1 l 1 l 1 l n = A n+1 + ( ) n + 1 A l B n+1 l + B n+1 l 1 l n ( ) n + 1 = A l B n+1 l. l 0 l n+1 ( ) n A l B n+1 l + B n+1 l 11

13 Divisibility Lemma 17. If a b, then a b, a b, a b and a b. If b = qa, then b = ( q)a, b = ( q)( a), b = q( a) and b = q a. Lemma 18. If a b and b c, then a c. b = q 1 a and c = q 2 b = c = (q 2 q 1 )a. Lemma 19. (1) ac bc = a b. (2) a b and c 0 = ac bc. (1) ac 0 = a 0 and c 0. bc = qac = b = qa = a b. (2) a 0 = ac 0. b = qa = bc = qac. Lemma 20. a b = a bc for all c. b = qa = bc = (qc)a. Lemma 21. a b and a c = a (b + c) and a (b c). b = q 1 a and c = q 2 a = b ± c = (q 1 ± q 2 )a. Lemma 22. a b and a c = a (bx + cy) for all x and y. 12

14 a bx and a cy = a (bx + cy). Theorem 23 (Dvision Algorithm). If a > 0 and b Z, then there is exactly one pair of numbers q and r such that b = qa + r,, 0 r < a (1) We first show that (1) has at least one solution. Among all the numbers of the form b ua there occur negative and positive ones, namely, for suffciently large positive u and negative u hasving sufficiently large absolute value, respectively. Suppose the smallest nonnegative number b ua occurs for u = q. Set r = b qa, then r = b qa 0 and r a = b (q +1)a < 0 = 0 r < a. To show uniqueness, suppose (1) holds and that u < q = u q 1 = b ua b (q 1)a = r + a a. If (1) holds and u > q = u q + 1 = b ua b (q + 1)a = r a < 0. The desired relations 0 b ua < athus holds only when u = q. Theorem 24. Let a, b naturals, neither 0. Of all the common multiples of a and b (there are such multiples and even positive ones, e.g., ab and 7ab), let m be the smallest positive one (the least common multiple or lcm) and let n Z be any one of them. Then m n, i.e., every common multiple is divisible by the least common multiple). m is often denoted [a, b] (not a closed bounded interval with endpoints a and b!). By Theorem 23 numbers q and r can be chosen so that n = qm + r, 0 r < m. From r = n qm = n 1m( q) and a n, a m, b n, and b m, it follows that a r and b r. Hence, by the definition of m, r cannot be > 0. Therefore r = 0, n = qm and m n. Proposition 25. If a 0 and b a, then b a, so that for every a 0 has only a finite number of divisors. a = qb and q 0 = q 1, a = q b b. 13

15 Theorem 26. Let a and b not both be 0 and let d be the greatest common divisor of a and b, the gcd of a and b and often denoted (a, b) (not an open interval withn endpoints a and b!). (1) if f is any common divisor of a and b, then f d; (2) if a > 0 and b > 0 and m is the smallest positive common multiple of a and b, then md = ab. In particular, then if a > 0 and b > 0 and d = 1, then m = ab. d exists and is > 0 since at least one of the numbers a, b is 0 and hence has only finitely many divisors and 1 is certainly a common divisor of both. Note that (1) says that every common divisor goes into the greatest common divisor. Case I, a > 0 and b > 0. Since ab is a common multiple of a and b, then m ab by Theorem 24 and g = ab ab N. Note m = N, a = m N and b = m N and therefore g m g g b g a is a common divisor of a and b. We will prove that if f a and f b then f g = f g = g = (a, b) = d and d = ab. If fact, if f a and f b, then a a b and b b a, ab thus is a common multiple of a m f f f and b. Hence, m ab, ab ab ab f, so the quotient f g f ab = g is an integer and therefore f g. f g Case II. Suppose the assumptions a > 0 and b > 0 are not satisfied but a and b are both still not 0. Then (1) follows from Case I since a has the same divisors as a and b. In fact, d is the greatest common divisor not only of a and b but of a and b as well. Case III. If one of the two numbers is 0, say a = 0, so that b 0, then obviously d = b and from f 0 and f b, it follows that f d. Example 27. (4, 6) = 2, (0, 3) = 3, ( 4, 6) = 2, and (1, 0) = 1. Lemma 28. For any a and b, not both 0, then (a, b) = (b, a) and (a, b) = (± a, ± b ). The definition of (a, b) is obviously symmetrical in a and b. Definition 29. If (a, b) = 1, that is, if 1 is the only positive common divisor of a and b, then a and b are called relatively prime. We also say that a is relatively prime to b. 1 and 1 are then the only common divisors of a and b. 14

16 Example 30. (6, 35) = 1 since 6 has only 1, 2, 3 and 6 as its only positive divisors and none of the numbers 2, 3 or 6 goes into 35. (a, 0) = 1 for a = 1 and for a = 1 but for no otheer a. Theorem 31. If (a, b) = d, then ( a d, b d ) = 1. If f > 0, f a, f b d d then fd a and fd b = fd d = f 1 and therefore f = 1. Corollary 32. Every fraction (i.e., rational number) may be written in lowest terms. Specifically, if r = a/b Q, where a, b Z and b 0, then we may assume that (a, b) = 1. If (a, b) = d > 0, then ( a d, b d ) = 1 and a/b = a d / b d. Theorem 33. If c > 0, c a, c b and ( a, b ) = 1, then c = (a, b). c c Since a/c and b/c do not both vanish, a and b are not both 0. If we set (a, b) = d, then c d, so that d/c is an integer. From (d/c)(a/d) = a/c and (d/c)(b/d) = b/c, it follows that (d/c) (a/c) and (d/c) (b/c) and therefore since ( a, b) = 1, d > 0, c > 0, d = 1 and c = d. c c c Theorem 34 (Euclidean Algorithm). If a > 0 and b Z, run the algorsithm: b = q 1 a + r 2,, 0 < r 2 < a a = q 2 r 2 + r 3, 0 < r 3 < r 2 r 2 = q 3 r 3 + r 4, 0 < r 4 < r 3 r n 2 = q n 1 r n 1 + r n, 0 < r n < r n 1 r n 1 = q n r n Then r n = (a, b) = gcd of a and b and (a, b) may be expressed as ax + by for certain x and y by back solving these equations.. 15

17 Example 35. Find (48257, 11739): = = = = = = = = 2 1 Therefore (48257, 11739) = 1. Back solving: 1 = ( 1) 2 1 = ( 1)(8 + ( 2) 3) = ( 1) = ( 1) 8 + 3(11 + ( 1) 8) = ( 4) 8 1 = ( 4)(30 + ( 2) 11) = ( 4) = ( 4) ( ( 43) 30) = ( 477) 30 1 = ( 477)( ( 9) 1301) = ( 477) = ( 477) ( ( 4) 11739) = ( 17693) Corollary 36. If a, b Z, not both 0, and m > 0, then (am, bm) = (a, b)m. We may assume that a > 0. Run the Euclidean Algorithm: b = q 1 a + r 2,, 0 < r 2 < a a = q 2 r 2 + r 3, 0 < r 3 < r 2 r 2 = q 3 r 3 + r 4, 0 < r 4 < r 3 r n 2 = q n 1 r n 1 + r n, 0 < r n < r n 1 r n 1 = q n r n and then multiply these equations by m to obtain: bm = q 1 am + r 2 m,, 0 < r 2 m < am am = q 2 r 2 m + r 3 m, 0 < r 3 m < r 2 m 16

18 r 2 m = q 3 r 3 m + r 4 m, 0 < r 4 m < r 3 m r n 2 m = q n 1 r n 1 m + r n m, 0 < r n m < r n 1 m r n 1 m = q n r n m and conclude that (am, bm) = r n m = (a, b)m. Corollary 37. If a, b Z, not both 0, d > 0, d a and d b, then d (a, b) and (a/d, b/d) = (a, b)/d. a/d, b/d Z and (a, b) = ((a/d)d, (b/d)d) = (a/d, b/d)d. Corollary 38. If (a, b) = d, then ( a d, b d ) = 1. The Least Remainder Algorithm. This may take fewer steps than the Euclidean Algorithm. For example: 253 = = = = = 4 1 vs. 253 = = = = 4 1. Leopold Kronecker ( ) proved that the Least Remainder Algorithm never takes more steps than any other Euclidean Algorithm (a quite difficult theorem). The following four homeworks are due at the beginning of class on Friday, February 8, Please take these homeworks seriously. Exercise 3 (3 points). Let a, b Z, not both 0. Then (a, b) is the smallest positive value taken over all expressions of the form ax + by, where x, y Z. Suppose that r > 0 and that there exist integers x and y such that r = ax + by. Then (a, b) a and (a, b) b = (a, b) r = (a, b) r. Of course there do exist intgers x and y such that (a, b) = ax + by by backsolving in the Euclidean Algorithm. 17

19 Exercise 4 (4 points). Find (525, 231) and then find x, y such that (525, 231) = 525x + 231y. 525 = = = = 2 21 Therefore (525, 231) = 21 Back solving: 21 = ( 1) = ( 1)(231 + ( 3) 63) = ( 1) = ( 1) (525 + ( 2) 231) = ( 9) 231 Exercise 5 (4 points). Find (6188, 4709) and then find x, y such that (6188, 4709) = 6188x y = = = = = = 2 17 Therefore (6188, 4709) = 17 Back solving: 17 = ( 3) = ( 3)(272 + ( 2) 119) = ( 3) = ( 3) ( ( 5) 272) = ( 38) = ( 38)( ( 3) 1479) = ( 38) = ( 38) ( ( 1) 4709) = ( 159)

20 Exercise 6 (7 points). Find (81719, 52003, 33649, 30107). This is the gcd of these four numbers. It should be clear that one can compute this by computing the gcd of the first two, then the gcd of this first gcd and the third number, etc. Let d 1 = (81719, 52003), d 2 = (d 1, 33649), d 3 = (d 2, 30107), and d = (81719, 52003, 33649, 30107). Then d = d 3. To see this note that d 3 is a common divisor of all four numbers and therefore d 3 d. On the other hand d d 3 = d d = = = = Therefore d 1 = = = = = Therefore d 2 = = = = = 2 23 Therefore d = d 3 = 23 Theorem 39. If c ab and b and c are relatively prime, then c a. (b, c) = 1 = (ab, ac) = a. c ab and c ac = c a. Corollary 40. If (a, c) = 1 and (b, c) = 1, then (ab, c) = 1. 19

21 If (ab, c) = d > 1, then d ab, (b, c) = 1 = (b, d) = 1 = d a = (a, c) d > 1, a contradiction. Proof #2: ax + cy = 1 and br + cs = 1 = ab(xr) + c(axs + bry + csy) = 1 = (ab, c) = 1. Corollary 41. If each of the integers a 1, a 2, a 3,..., a n is relatively prime to m, then their product will be relatively prime to m. We may assume that each a l > 0. Induct. If n = 1 then the corollary is trivial. So assume true for n and let us prove it true for n + 1. (a 1 a 2 a n, m) = 1 by induction and (a n+1, m) = 1. The previous theorem now implies (a 1 a 2 a n a n+1, m) = 1. Corollary 42. If each of the integers a 1, a 2, a 3,..., a m is relatively prime to each of the integers b 1, b 2, b 3,..., b n, then (a 1 a 2 a m, b 1 b 2 b n ) = 1. (a 1 a 2 a m, b l ) = 1 by the previous corollary. Now conclude the proof of this corollary by another appeal to the previous corollary. Corollary 43. If (a, b) = 1, then (a m, b n ) = 1. Corollary 44. If a, n N, then n a is either an integer or an irrational number. Suppose r, s N with (r, s) = 1 and a = ( r s )n = r n = as n = s r n = (r n, s) = s. But then (r n, s) = 1 by the previous corollary which implies s = 1 and a = r n. Corollary 45. If a, b and n N and a n b n, then a b. b n /a n N and n b n /a n = b/a Q = b/a N. 20

22 Theorem 46. If a, b N with (a, b) = 1 and ab = c n, an exact power, for some c N, then both a and b are exact nth powers. By hypothesis we have ab = c n. Let (a, c) = α, a = γα and c = βα. Therefore (γα, βα) = α = (γ, β) = 1. ab = c n = γαb = β n α n = γb = β n α n 1. (γ, β n ) = 1 and β n γb = β n b and we can set b = β n d = γβ n d = β n α n 1 = γd = α n 1. (α, d) (a, b) = 1 = (α, d) = 1. Therefore α n 1 γd and (α n 1, d) = 1 = α n 1 γ = γ = α n 1 ϵ. Therefore γd = α n 1 = α n 1 ϵd = α n 1 = ϵd = 1 = ϵ = d = 1 since ϵ, d N. Finally a = γα = α n and b = dβ n = β n and c = βα. 21

23 Pythagorean Triples - the solutions (x, y, z) of the Diophantine equation x 2 + y 2 = z 2 in positive integers. The ancients attributed to Pythagorous the observation that ( n2 1, n, n2 +1) is always a 2 2 Pythagorean triple. Lemma 47. Let M N. Then (x, y, z) is a Pythagorean triple if and only if (M z, M y, M z) is a Pythagorean triple. We therefore look only for primitive solutions, i.e., those where (x, y, z) = 1. Lemma 48. If (x, y, z) is a Pythagorean triple and M = (x, y), then M z. Similar statements hold for (x, z) and (y, z). Therefore a Pythagorean triple (x, y, z) is a primitive Pythagorean triple if and only if (x, y) = (x, z) = (y, z) = 1. In particular, at most one of x, y, z is even. Note that if, for example, M = (x, y), then M 2 z 2 = M z = M (x, y, z). So (x, y, z) = 1 (x, y) = 1 and (x, z) = 1 and (y, z) = 1 for primitive solutions. In particular, at most one of x, y, z is even. Lemma 49. If (x, y, z) is a primitive Pythagorean triple, then one of x, y must be odd and the other even. If x and y are both even, then (x, y) 2 = (x, y, z) is not primitive. Suppose that both x and y are odd. Then, for example, x = 1+2a = x 2 = 1+4a+4a 2 = 1+4a(a+1) = 1+8b since one of a or a + 1 is even. Therefore x 2 + y 2 = 2 + 8c = z 2 = 2 z 2 = z is even = 4 z 2. This is a contradiction since c. Therefore one of x, y must be even and the other odd. Theorem 50. All primitive solutions to the Pythagorean equation x 2 + y 2 = z 2 are precisely of the form (x, y, z) = (r 2 s 2, 2rs, r 2 + s 2 ), where r, s N, (r, s) = 1, of different parity, and r > s. Suppose that (x, y, z) is a primitive Pythagorean triple with y even. Then x 2 + y 2 = z 2 = z 2 x 2 = y 2 = z+x z x = (y/2) 2. So there exists r > 0, s > 0 such that z+x = r

24 and z x 2 = s 2 = z = r 2 + s 2, x = r 2 s 2 and y = 2rs. Here r and s are positive integers, necessarily relatively prime, of different parity, and r > s. Conversely, suppose that r and s are positive integers, relatively prime, of different parity, and r > s. Then (x, y, z) = (r 2 s 2, 2rs, r 2 + s 2 ) is a Pythagorean triple since (r 2 s 2 ) 2 + (2rs) 2 + (r 2 + s 2 ) 2. Here x, y and z are positive integers and it remains to show that they do not have nontrivial common divisors. Let d = (x, y, z) = d z+x = 2r 2, d z x 2 = 2s 2 and d y = 2rs. But (r 2, s 2, rs) = 1 since (r 2, s 2 ) = 1. Hence, (2r 2, 2s 2, 2rs) = 2 = d 2, i.e., d = 1 or d = 2. But d 2 since x and z are odd. Hence, d = 1. Culture: One can use the result about Pythagorean triples in a clever way to show that there are no triples of positive integers (x, y, z) such that x 4 + y 4 = z 2. From this it is easy to show that there are no triples of positive integers which satisfy x 4 + y 4 = z 4, thus proving Fermat s Conjecture in this very special case and therefore also in all cases of the form x 4n +y 4n = z 4n. We may find time to do this later. The following four exercises are due on Friday, February 15, Exercise 7 (2 points). If a, b N and (a, b) = 1, then (a + b, a b) = 1 or 2. Suppose d > 0, d a + b and d a b. Then d 2a = (a + b) + (a b) and d 2b = (a + b) (a b). But (a, b) = 1 = (2a, 2b) = 2 by Corollary 36. Hence, d 2 and d = 1 or d = 2. Note that (3, 1) = 1 and (3 + 1, 3 1) = (4, 2) = 2, (3, 2) = 1 and (3 + 2, 3 2) = (5, 1) = 1, so both possibilities do indeed occur. Exercise 8 (2 points). Show that the fraction a+a b+b is expressed in lowest terms if ab a b = ±1. Suppose that d > 0, d (a+a ) and d (b+b ). Then d (a+a )b a (b +b) = ab a b = ±1 = d = 1 = (a + a, b + b ) = 1. Exercise 9 (5 points). If α, β, γ, δ, a, b N and αδ βγ = ±1, then the greatest common divisor of m = αa + βb and n = γa + δb is the same as that of a and b. 23

25 [ α β γ δ ] [ ] a = b [ ] m n = [ ] [ ] [ ] δ β m a = ± γ α n b or, without using matrices, directly solve the two simultaneous equations to express a and b as linear combinations of m and n. So if d > 0, d a and d b = d m and d n, and conversely, so (a, b) = (m, n). Exercise 10 (7 points). The sum of two positive integers is 5, 432 and their least common multiple is 223, 020. Find the two numbers and their greatest common divisor. Use what we have learned and a modest amount of computation to find the answer - do not use prime factorization since we do not know this yet. Suppose a, b N with a + b = 5432 and [a, b] = Let d = (a, b), we have d = [a, b]d = ab = a(5432 a), or a a + d = 0. a N, using the quadratic formula, d must be an integer. This necessitates that 0 < d < /( ) < 34. This square root is an integer only for d = 28 for 1 d 34, in which case the square root is 2128, a = 5432 ± 2128/2 = 3780 or Let a be one of these numbers and let b be the other. 24

26 Primes Definition 51. If p N p is said to be a prime number if p > 1 and if a N, a p = a = 1 or a = p. If n > 1 is not a prime number, then n is said to be a compositive number. Thus n is composite if and only if there exists a N such that a n and 1 < a < n. If so, then an a is called a proper divisor of n. Therefore p N is a prime if and only if it does not have proper divisors. Example 52. The Sieve of Eratosthenes. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, etc. Lemma 53. If p, q are primes and p q, then p = q. This is a trivial consequence of the definitions. Lemma 54. Arbitrarily long blocks of consecutive composite integers exist. Suppose that 2, 3,, p are the primes up to p and q = p = q. then the consecutive p 1 numbers q + 2, q + 3, q + 4,, q + p all all composite. Are there infinitely many prime pairs p, p + 2, e.g., 3, 5 and 5, 7 and 11, 13 and 17, 19 and 29, 31 etc.? This is terra incognita. Lemma 55. n N is compositive if and only if n has a proper divisor n. Suppose n = ab, where a is a proper divisor of n. Then b is also a proper divisor of n. If a < n and b < n, then n = ab < ( n) 2 = n, a contradiction. Lemma 56. If a > 1 and d is the smallest positive divisor of a greater than 1, then d is a prime. 25

27 If r d is a proper divisor, then 1 < r < d and r a, then r a and r is a proper divisor of a smaller than d. Contradiction. Hence, d is a prime. Corollary 57. n N is compositive if and only if n has a prime divisor n. We only need to prove =. n has a proper divisor a n by Lemma 55. If d is the smallest positive divisor of a greater than 1, then d is a prime by Lemma and d a n. Lemma 58. If a N and p is a prime, then either (a, p) = 1 or (a, p) = p. (a, p) p and therefore (a, p) = 1 or (a, p) = p. Lemma 59. If a, b N and p is a prime, then p ab and (a, p) = 1 = p b.. p ab and a and p are relatively primes implies p b by Theorem 39. Lemma 60. If a, b and p are primes, then p ab = p = a or p = b. If (a, p) = 1, then p b by Lemma 59 and therefore p = b by Lemma 53. Otherwise p = (a, p) = a by Lemma 58. Corollary 61. If a l N, p is a prime and p a 1 a 2 a n, then p a i for some i. In particular, if a i itself is a prime, then p = a i. (p, a i ) = 1 or p by Lemma 58. If (p, a i ) = 1 for every i, then p a 1 a 2 a n by Corollary 41. Therefore (p, a i ) > 1 for some i, (p, a i ) = p by Lemma 58 and therfore p a i. If a i is a prime, then p = (p, a i ) = a i. 26

28 Theorem 62. If n > 1 then n can be expressed as a product of primes (with possible repetitions). 2 is a prime. If n is a prime then we are done. If n is not a prime then n has proper divisors and the smallest positive proper divisor p n is a prime. n = pb, where b N. Now use the Second Principle of Induction. Theorem 63 (Fundamental Theorem of Arithmetic). If a > 1 then a = p r 1 1 p r 2 2 p rm m, where the p l s are distinct primes, p l < p l+1 and each r l > 0. This representation, the so-called standard representation, of a is unique. a is a product of primes and may be put into the desired form using commutativity and associativity. Suppose that a = q s 1 1 q s 2 2 qn sn is another such representation of a. Then each p i is some q j and conversely. Therefore m = n and, since the sets are arranged in increasing order, each p i = q i. If r i > s i, then cancel p s i i = q s i i from both sides and then get a contradiction since p i would then divide the left hand side but not the right hand side. Theorem 64 (Euclid). The number of primes is infinite. If p 1, p 2,..., p n are the only primes then n = p 1 p 2... p n + 1 is not a product of primes since p i n for all i. Proposition 65. Arrange the primes in increasing order. If p n is the nth prime, then p n+1 < p n n + 1 Use Euclid s argument. Let q = p n + 1. Then p n+1 < p n n + 1. Proposition 66. Arrange the primes in increasing order. Then p n < 2 2n. True for n = 1. Euclid s argument shows that p n+1 p 1 p 2 p n + 1 < n + 1 < 2 2n+1. 27

29 Definition 67. If x 0, +, let π(x) be the number of primes x. Lemma 68. π(x) + as x +. Use Theorem 64. Proposition 69. π(x) log(log(x)) for x 2. Suppose that n 4 and e en 1 < x e en. Then e n 1 > 2 n (not true for n = 3) = e en 1 > 2 2n = π(x) π(e en 1 ) π(2 2n ) n log(log(x)) for x > e e3. It is obvious that the inequality holds for 2 x e e3. Theorem 70. There are infinitely many primes of the form 4n + 3. Let q = p 1. Then q is of the form 4n + 3 and is not divisible by any of the primes up to p. q cannot be a product of primes of the form 4n + 1 only, since a productd of two numbers of this form is of this form. Therefore it is divisible by a prime of the form 4n + 3 for some prime greater than p. Lemma 71. The harmonic series diverges, i.e., l 1 1/l = +. Definition 72. If x 0, + let π(x) be the number of primes x. then π(x) + as x +. Lemma 73. Let 0 < x 1/2. Then ln( 1 1 x ) 2x. Let f(x) = 2x + ln(1 x). Then f (x) = 1 2x > 0 for 0 < x < 1/2, so f is strictly 1 x increasing on 0, 1/2. But f(0) = 0 = f(x) 0 = 2x ln(1 x) = 2x ln( 1 ) 1 x on 0, 1/2. 28

30 Theorem 74. The series p, summed over all the primes in ascending order, diverges. 1 p Use Theorem 63 to conclude that p ξ p = p ξ (1+ 1 p + 1 p 2 + ) = a 1 1, where the a symbol signifies that the sum is taken over all those numbers a that are not divisible by any p > ξ. These include all natural numbers ξ. Therefore 1 p ξ a [ξ] = a p 1 p ξ + as ξ + = 1 1 p ξ log( 1 ) + = p ξ + by p p p Lemma 73. Recall that the pair p, p + 2 is called a pair of two primes if both p and p + 2 are primes, e.g., 3&5; 5&7; 11&13; 17&19; 29&31, etc. It is a complete mystery of the univese if there are an infinite number of two primes. However, even if there are an infinite number of two primes, they must be rather sparse. This last assertion is somewhat justified by an amazing theorem of Viggo Brun, viz., even if there are an infinite number of twin primes, the sum of their reciprocals does indeed converge (c.f. Theorem 74). Brun created an ingenious and very nontrivial extrapolation of the Sieve of Eratosthenes to prove this. Proposition 75. n! = p n p l 1 [ n p l ] = p p l 1 [ n p l ]. The number of multiples of p that are n is [ n p ]. The number of multiples of p2 that are n is [ n p 2 ]. The number of multiples of p 3 that are n is [ n p 3 ]. Etc. Lemma 76. Let {p l } l 1 be the primes arranged in increasing order. Then l 5 = p l 1 l + 2. If l = 5, then p 5 1 = p 4 = Now induct. If the lemma is true for l 5 then p (l+1) 1 = p l p l l > (l + 1) + 2. Lemma 77. Let {p l } l 1 be the primes arranged in increasing order. Let n 10 and choose the smallest l such that n l + 1 < p l. Then 1 < l < n and n l l. 29

31 Note that the set of l such that n l + 1 < p l is nonempty since l = n 1 is in this set. Note that 1 is not in this set since n = n 10 > 2 = p 1. So the smallest valid l satisfies 1 < l < n. Since l is minimal it must be true that n (l 1) + 1 p l 1 or n l p l 1 2 l by Lemma 76 if l 5. Now if n 10, the smallest value of l such that n l + 1 < p l, or n < p l + l 1 must be 5 because if it were 4 then n < p = = 10. Thus, n 10 = n l l. Theorem 78 (Bonse [1907]). Let {p l } l 1 be the primes arranged in increasing order. Then n 4 = p 2 n+1 < p 1 p 2 p n. The first eleven primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. Then: p = 3 2 = 9 > 2 = p 1 p = 5 2 = 25 > 6 = 2 3 p = 7 2 = 49 > 30 = p = 11 2 = 121 < 210 = p = 13 2 = 169 < 2, 310 = p = 17 2 = 289 < 39, 270 = p = 19 2 = 361 < 746, 130 = p = 23 2 = 529 < 17, 160, 990 = p = 23 2 = 529 < 17, 160, 990 = p = 29 2 = 841 < 497, 668, 710 = p = 31 2 = 961 < 15, 427, 730, 010 = Thus direct calculations shows that Theorem 78 is true if 4 n 11. Divide the primes p 1, p 2,..., p n into two camps, p 1, p 2,..., p l 1 and p l, p l+1,..., p n, where the choice of l is limited only by the requirement that 1 < l < n. Consider the system of p l numbers p 1 p 2 p l p 1 p 2 p l p 1 p 2 p l p 1 p 2 p l 1 p l 1 None of these numbers is divisible by p 1, p 2,..., p l 1 and at most one can be divisible by a given prime out of the series p l, p l+1,..., p n. For suppose that two of the numbers 30

32 p 1 p 2 p l 1 j 1 and p 1 p 2 p l 1 j 1 (j < j ) are divisible by the same prime p α, where l α n. Then their difference p 1 p 2 p l 1 (j j) are divisible by p α, which is impossible since all of the factors p 1, p 2,..., p l 1, j j are all less than p l p α. Suppose now that l is chosen so that n l + 1 < p l. It is always possible to satisfy this condition, e.g., take l = n 1. Then the number of primes p l, p l+1,..., p n, which is n l+1, is less than the numbers integers in the listing. Since two numbers of this system cannot be divisible by the same prime p α (l α n), there must be in the system at least one number nondivisible by any of the primes p l, p l+1,..., p n and naturally also by any of the primes p 1, p 2,..., p l 1. Let this number be N = p 1 p 2 p l 1 j 1. Now every prime divisior of N, being different from p 1, p 2,..., p n, must be p n+1. Moreover, j p l and consequently p n+1 N p 1 p 2 p l 1 p l 1 and hence p n+1 < p 1 p 2 p l 1 p l provided n l + 1 < p l. Let l be the smallest number satisfying this inequality. Then n 10 = n l l by Lemma 77. Comparing the products p 1 p 2 p l and p l+1 p n, we see that the second consists of at least as many factors as the first, and besides p l+1 > p 1, p l+2 > p 2,..., p 2i > p i. Consequently p 1 p 2 p l < p l+1 p l+2 p n and (p 1 p 2 p l ) 2 < p 1 p 2 p n. Since, on the other hand, p n+1 < p 1 p 2 p l we shall have p 2 n+1 < p1p 2 p n for n 10. Corollary 79. The numbers 2, 3, 4, 6, 8, 12, 18, 24, 30 are the only positive integers n with the property numbers less than n and relatively prime to n are either 1 or primes. Check directly that this is true for numbers up to 49. Suppose that n 50 has the desired property. Choose k 4 so that p 2 k n < p2 k+1. Note that actually p2 k < n, for if p 2 k = n, then p2 k 49 = (22, n) = (2 2, p 2 k ) = 1, a contradiction. So p2 k < n < p2 k+1. Therefore if 1 l k and (p l, n) = 1, then (p 2 l, n) = 1 and p2 l p2 k < n, a contradiction. Therefore (p l, n) = p l, p l n and so p 1 p 2 p k n. But p 2 k+1 < p 1p 2 p k n < p 2 k+1 by Theorem 78, a contradiction. 31

33 TEST #1 This is a take home test. Each problem is valued at 10 points. Do the first five problems and the sixth and/or the seventh and/or the eighth and/or ninth if you can. Therefore it is possible to get 90 out of 50 points. Problem 1. Use the Fundamental Theorem of Arithmetic to prove: if (a, b) = 1 and ab = c n, then there exists α and β such that c = αβ, a = α n and b = β n. Let c = p po(p) be the prime factorization of c. Then c n = p pno(p) is the prime factorization of c n. Let α = p a po(p) and β = p b po(p). Then a = α n and b = β n. Problem 2. Show that (a + b, a b, ab) = 1 if (a, b) = 1. Proof #1: Exercise 7 shows that (a, b) = 1 = (a + b, a b) = 1 or 2. So if d = (a + b, a b, ab) then d = 1 or d = 2. a and b cannot be both even since (a, b) = 1. If one of a and b is even and the other odd, then a + b and a b are both odd and d = 2 = d (a + b) and d (a b). So both a and b must be odd. But then ab is odd, 2 ab and d ab, a contradiction. Hence, d = 1. Proof #2: Actually, more is true: (a, b) = 1 = (a+b, ab) = 1. If p is a prime and p (a+b, ab) = p ab = p a or p b. p b = p a = (a + b) b = p (a, b), a contradiction. Similarly, p a = p b = (a + b) a = p (a, b), again a contradiction. Therefore (a + b, ab) = 1. Problem 3. Find g = (b, c), where b = and c = and then determine x and y such that bx + cy = g = = = = =

34 = = = = = = = = = = = = = = = = = = 29 1 Therefore g = ( , ) = 1. Back solving: 1 = ( 5) 29 1 = ( 5)(175 + ( 1)146) = ( 5) = ( 5) (321 + ( 1) 175) = ( 11) = ( 11)(496 + ( 1) 321) = ( 11) = ( 11) (817 + ( 1)496) = ( 28) = ( 28)( ( 1) 817) = ( 28) = ( 28) ( ( 8) 1313) = ( 388) = ( 388)( ( 6) 11321) = ( 388) = ( 388) ( ( 2) 69239) = ( 5134) = ( 5134)( ( 1)149799) = ( 5134) = ( 5134) ( ( 2) ) = ( 20148) = ( 20148)( ( 1) ) = ( 20148) = ( 20148) ( ( 1) ) = ( 47803) = ( 47803)( ( 1) ) = ( 47803)

35 1 = ( 47803) ( ( 3) ) = ( ) = ( )( ( 7) ) = ( ) = ( ) ( ( 1) ) = ( ) ) 1 = ( )( ( 8) ) = ( ) = ( ) ( ( 1) ) = ( ) = ( )( ( 1) ) = ( ) = ( ) ( ( 1) ) = ( ) = ( )( ( 1) ) = ( ) = ( ) ( ( 1) ) = ( ) Problem 4. Suppose that a and b are positive integers that satisfy (a, b) = [a, b]. Then a = b. Suppose that (a, b) = [a, b]. Then (a, b) 2 = (a, b)[a, b] = ab = 1 = (a/(a, b)) (b/(a, b)). But a/(a, b) N and b/(a, b) N = a/(a, b) = b/(a, b) = 1 = a = (a, b) = b. Problem 5. Prove that if a, b, c, d N, then if a b and c d are in lowest terms and a b + c d N = b = d. Suppose that (a, b) = 1, (c, d) = 1 and a + c = e N. Then ad + bc = bde = ad = b d b(de c) = b ad. Since (a, b) = 1 = b d = b d. A similar argument proves d b = b = d. Problem 6 (Bonus Question). Prove that n is composite for all n > 1. Check directly that n = (n 2 + 2n + 2)(n 2 2n + 2) and each factor is > 1. 34

36 Problem 7 (Bonus Question). Prove that n 4 + n is composite if n > 1. Check directly that n 4 + n = (n 2 + n + 1)(n 2 n + 1) and each factor is > 1. Problem 8 (Bonus Question). Prove that if a, b N and (a, b) = 1, then (a + b, a 2 ab + b 2 ) = 1 or 3. If a = 3 and b = 1, then (a, b) = (3, 1) = 1 and (a + b, a 2 ab + b 2 ) = (4, 7) = 1. If a = 5 and b = 1, then (a, b) = (5, 1) = 1 and (a + b, a 2 ab + b 2 ) = (6, 21) = 3, so both cases can occur. Suppose p is a prime and p (a + b, a 2 ab + b 2 ) = p (a + b) 2 (a 2 ab + b 2 ) = 3ab. If p 3 = p = 3 and we are done. Otherwise p ab = p a or p b. If p a = p b = (a + b) a = p (a, b), a contradiction. If p b = p a = (a + b) b = p (a, b), again a contradiction. Problem 9 (Bonus Question). Prove that if a, b N and (a, b) = 1, then (a 2 ab + b 2, a 4 a 3 b + a 2 b 2 ab 3 + b 4 ) = 1. We use prime factorization in this proof. Suppose that p is a prime and p (a 2 ab+b 2, a 4 a 3 b+a 2 b 2 ab 3 +b 4 ) = p (a 2 ab+b 2 ) and p (a 2 (a 2 ab+b 2 ) b 3 (a b)) = p (a b)b 3. If p b 3 = p b = p a 2 = (a 2 ab + b 2 ) + b(a b) = p a = p (a, b), a contradiction. Otherwise, p (a b) = p 2 (a 2 2ab + b 2 ) = p ab = (a 2 ab + b 2 ) (a 2 2ab + b 2 ). Therefore either p a or p b. As before, p b = p a. Similarly, p a = p b. In either case, p (a, b), again a contradiction. Therefore no such prime p exists and the gcd in question is 1. 35