MATH 3400.001 NUMBER THEORY SPRING 2019 Scientia Imperii Decus et Tutamen 1 Robert R. Kallman University of North Texas Department of Mathematics 1155 Union Circle 311430 Denton, Texas 76203-5017 office telephone: 940-565-3329 office fax: 940-565-4805 office e-mail: robert.kallman@unt.edu January 4, 2019 1 Taken from the coat of arms of Imperial College London.
******************************************************************************** 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 3400.001, NUMBER THEORY. PREREQUISITES: MATH 2000 or MATH 3000 or CSCE 2100. 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-13 978-0-321-50031-1, ISBN-10: 0-321-50031-8. INSTRUCTOR: Robert R. Kallman, 315 GAB [office], 940-565-3329 [office telephone], 940-565-4805 [fax], robert.kallman@unt.edu [e-mail] 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 e-mail 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
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 3.1-3.7, 4.1-4.6, 5.1-5.5, 6.1-6.3, 7.1-7.4, 8.1-8.5, 9.1-9.6, 11.1-11.5. 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
Test #1, Friday, February 22, 2019, covering 6.1-6.6, 6.8, 7.1 and 7.2. Test #2, Friday, April 12, 2019, covering 7.3-7.5, 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 www.unt.edu/csrr In other words, cause trouble in the classroom and you will probably be cast into the Darkness and sent to the KGB. 3
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
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 = 410 286423 278424. In this case, x(d) and y(d), the smallest nontrivial solution, have 103273 and 103265 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
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
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 17 52 26 13 40 20 10 5 16 8 4 2 1. 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 = 4 2 + 1 2 + 1 2 + 1 2. A very remarkable algebraic identity involving products of sums of four squares due to Euler plays a crucial role: (a 2 1 + a 2 2 + a 2 3 + a 2 4)(b 2 1 + b 2 2 + b 2 3 + 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, 1749. 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
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 1 + 2 + + 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 1 + 3 + 5 + + (2n 1) = n 2. Example 8. Let S n be the statement 1 2 + 2 2 + 3 2 + + n 2 = n(n+1)(2n+1) 6. 8
Example 9. Let S n be the statement 1 3 + 2 3 + 3 3 + + n 3 = (1 + 2 + + n) 2. Exercise 1 (5 points). This one is tricky. For what integers n is 3 2n+1 + 2 n+2 divisible by 7? Exercise 2 (4 points). For which integers n is 9 n 8n 1 divisible by 8? 9
Factorials, Binomial Coefficients, the Binomial Theorem. Definition 10. Define 0! = 1, 1! = 1, 2! = 2 1, 3! = 3 2 1 = 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
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 0 + ( 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
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
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
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
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 loest 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
Example 35. Find (48257, 11739): 48257 = 4 11739 + 1301 11739 = 9 1301 + 30 1301 = 43 30 + 11 30 = 2 11 + 8 11 = 1 8 + 3 8 = 2 3 + 2 3 = 1 2 + 1 2 = 2 1 Therefore (48257, 11739) = 1. Back solving: 1 = 1 3 + ( 1) 2 1 = 1 3 + ( 1)(8 + ( 2) 3) = ( 1) 8 + 3 3 1 = ( 1) 8 + 3(11 + ( 1) 8) = 3 11 + ( 4) 8 1 = 3 11 + ( 4)(30 + ( 2) 11) = ( 4) 30 + 11 11 1 = ( 4) 30 + 11(1301 + ( 43) 30) = 11 1301 + ( 477) 30 1 = 11 1301 + ( 477)(11739 + ( 9) 1301) = ( 477) 11739 + 4304 1301 1 = ( 477) 11739 + 4304(48257 + ( 4) 11739) = 4304 48257 + ( 17693) 11739 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
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 = 2 122 + 9 122 = 13 9 + 5 9 = 1 5 + 4 5 = 1 4 + 1 4 = 4 1 vs. 253 = 2 122 + 9 122 = 14 9 4 9 = 2 4 + 1 4 = 4 1. Leopold Kronecker (1823-1891) proved that the Least Remainder Algorithm never takes more steps than any other Euclidean Algorithm (a quite difficult theorem). The following three homeworks are due at the beginning of class on Friday, February 8, 2019. 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. Exercise 4 (4 points). Find (525, 231) and then find x, y such that (525, 231) = 525x + 231y. 17
Exercise 5 (4 points). Find (6188, 4709) and then find x, y such that (6188, 4709) = 6188x + 4709y. 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. 18