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NUMBER THEORY (Module 1) NUMBER THEORY (Module 1) [MTH1C05] STUDY MATERIAL I SEMESTER CORE COURSE M.Sc. Mathematics (2019 Admission onwards) UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION CALICUT UNIVERSITY- P.O MALAPPURAM- 673635, KERALA 190555 MTH1C05-NUMBER THEORY (MODULE-1) SCHOOL OF DISTANCE EDUCATION UNIVERSITY OF CALICUT STUDY MATERIAL FIRST SEMESTER M.Sc. Mathematics (2019 ADMISSION ONWARDS) CORE COURSE: MTH1C05-NUMBER THEORY (Module-1) Prepared by: Dr. Anil Kumar V Professor Department of Mathematics University of Calicut Scrutinized By: Dr. Preethi Kuttipulackal Associate Professor & Head Department of Mathematics University of Calicut MTH1C05-NUMBER THEORY (MODULE-1) Anil Kumar V. Number Theory c 2020 University of Calicut All rights reserved. This work may be distributed and/or modified under the conditions of the University of Calicut. The Current Maintainer of this work is School of Distance Education, University of Calicut. First edition: July 2020 Author Anil Kumar V. Professor. Department of Mathematics University of Calicut ISBN The University of Calicut Press, Malappuram 2 School of Distance Education, University of Calicut Contents 1 Arithmetical Functions and its Averages 5 1.1 Arithmetical functions . .5 1.1.1 The M¨obiusfunction µ(n)................6 1.1.2 Euler totient function '(n)...............7 1.1.3 A relation connecting ' and µ..............9 1.1.4 A product formula for '(n)............... 10 1.1.5 Dirichlet product of arithmetical functions . 15 1.1.6 Dirichlet inverses and Mobius inversion formula . 17 1.1.7 The Mangoldt function Λ(n).............. 21 1.1.8 Multiplicative functions . 23 1.1.9 Examples of completely multiplicative functions . 26 1.1.10 Examples of multiplicative functions . 27 1.1.11 Multiplicative functions and Dirichlet multiplication . 28 1.1.12 The inverse of a completely multiplicative function . 31 1.1.13 Liouville's function λ(n)................. 36 1.1.14 The divisor function σα(n)............... 38 1.1.15 Generalised convolution . 41 1.1.16 Derivatives of arithmetical functions . 43 1.1.17 The Selberg identity . 45 1.1.18 Exercises . 46 1.2 Averages of arithmetical functions . 47 1.2.1 The big oh notation. Asymptotic equality of functions 48 1.2.2 Partial sums of a Dirichlet product . 56 1.2.3 Applications to µ(n) and Λ(n)............. 58 1.2.4 Another identity for the partial sums of a Dirichlet product . 64 1.2.5 Exercises . 66 3 Anil Kumar V. Number Theory Bibliography 69 1.A Syllabus . 70 1.A Notations . 71 4 School of Distance Education, University of Calicut Module 1 Arithmetical Functions and its Averages This module consists of two sections. Let N be the set of natural numbers. Then f : N ! C is called a sequence. In number theory such functions are called arithmetical functions. In the first section introduces several arithmetical functions. The Dirichlet product of two arithmetical func- tions are included. Moreover we will prove that the set of all arithmetical functions which do not vanish at 1 form an abelian group under Dirichlet multiplication. Furthermore, M¨obiusinversion is included. This section also includes generalized convolutions and gives relation between asso- ciative property of Dirichlet multiplication and convolution. Generalized inversion formula is also included. The second section is devoted to av- erages of Arithmetical functions. The well known Euler's summation formula is derived. Several asymptotic formulas are also derived. 1.1 Arithmetical functions Definition 1.1.1. A complex valued function defined on the set of natural numbers is called an arithmetical function or a number theoretic function. That is, a function f : N ! C is called an arithmetical function . 5 Anil Kumar V. Number Theory 1.1.1 The M¨obiusfunction µ(n) Definition 1.1.2. The Mobius function µ is defined as: 8 1 if n = 1; <> µ(n) = (−1)k if n is a product k distinct primes; :>0 otherwise: From the definition of µ it clear that µ(n) = 0 if and only if n has a square factor > 1.Obviously, µ(1) = 1; µ(2) = −1; µ(3) = −1; µ(4) = 0; µ(5) = −1; µ(6) = 1; µ(8) = 0 and so on. Theorem 1.1.1. If n ≥ 1 we have ( X 1 1 if n = 1; µ(d) = = n 0 if n > 1: djn Proof. We consider two cases. Case 1: Suppose n = 1. In this case X X µ(d) = µ(d) = µ(1) = 1: djn dj1 Case 2: Suppose n > 1. Then by fundamental theorem of arithmetic, n can be expressed in the following form: a1 a2 ak n = p1 p2 ··· pk ; where pi's are distinct primes and α1; α2; : : : ; αk are integers greater than or equal to 1. Now consider X X µ(d) = µ(d): djn a1 a2 ak djp1 p2 ···pk Note that µ(d) = 0 if and only if d has a square factor. Therefore µ(d) 6= 0 if and only if d = 1; p1; p2; : : : ; pk; p1p2; : : : ; pk−1pk; p1p2p3; : : : pk−2pk−1pk; ··· ; p1p2 ··· pk : | {z } | {z } k | {zk } | {zk } k (1) (2) (3) (k) 6 School of Distance Education, University of Calicut Anil Kumar V. Number Theory Hence X µ(d) = µ(1) + [µ(p1) + µ(p2) + ··· + µ(pk)] + [µ(p1p2) + ··· + µ(pk−1pk)] + djn [µ(p1p2p3) + ··· + µ(pk−2pk−1pk)] + [µ(p1p2 ··· pk)] = 1 + [(−1) + (−1) + ··· + (−1)] + (−1)2 + (−1)2 + ··· + (−1)2 + | {z } k | {zk } (1) (2) (−1)3 + (−1)3 + ··· + (−1)3 + ··· + (−1)k | {z } | {zk } k (3) (k) k k k k = 1 + (−1) + (−1)2 + (−1)3 + ··· + (−1)k 1 2 3 k = (1 − 1)n = 0: This completes the proof. 1.1.2 Euler totient function '(n) Definition 1.1.3. If n ≥ 1 the Euler totient function '(n) is defined as '(n) := #fa 2 N : a ≤ n; (a; n) = 1g: Thus n 0 X '(n) = 1; k=1 where dash denotes the sum is taken over those k which are relatively prime to n. That is n n X X 1 '(n) = 1 = : (n; k) k=1 k=1 (k;n)=1 Theorem 1.1.2. If n ≥ 1 we have X '(d) = n: djn Proof. Let S = f1; 2; 3; : : : ; ng We partition the set S into mutually disjoint subsets as follows: For each divisor d of n, let A(d) := fk :(k; n) = d; 1 ≤ k ≤ ng: 7 School of Distance Education, University of Calicut Anil Kumar V. Number Theory Claim 1: If d1 and d2 are distinct divisors of n, then A(d1) \ A(d2) = ;: x 2 A(d1) \ A(d2) ) x 2 A(d1) and x 2 A(d2) ) (x; n) = d1 and (x; n) = d2 ) d1 = d2; contradiction: Claim 2: f1; 2; : : : ; ng = [djnA(d). Obviously [djn A(d) ⊆ S: (1.1) Now x 2 S ) (x; n) = d for some d; 1 ≤ d ≤ n ) x 2 A(d) ) x 2 [djnA(d) for some djn: Thus S ⊆ [djnA(d): (1.2) From equations (1.1) and (1.2) it follows that S = [djnA(d): Therefore X #(S) = #(A(d)): djn This implies that X n = #(A(d)): (1.3) djn Claim : #(A(d)) = '(d): Now k 2 A(d) , (k; n) = d , (k=d; n=d) = 1; 0 < k=d ≤ n=d , (q; n=d) = 1; 0 < q ≤ n=d , q 2 fk :(k; n=d) = 1; 0 < k ≤ n=dg | {z } B , q 2 B 8 School of Distance Education, University of Calicut Anil Kumar V. Number Theory This implies that there is a one to one correspondence between the elements of A(d) and B. Hence #A(d) = #(B) = '(n=d): Hence equation (1.3) becomes: X n = '(n=d) (1.4) djn Observe that if d is a divisor of n, then n=d is also a divisor of n. The divisors d and n=d are called conjugate divisors of n. Hence X X '(n) = '(n=d): djn djn Hence equation (1.4) can be written as X '(d) = n: djn This completes the proof of the theorem. 1.1.3 A relation connecting ' and µ. Theorem 1.1.3. If n ≥ 1 we have X n '(n) = µ(d) : d djn Proof. By the definition of ', we have n X 1 '(n) = (n; k) k=1 n X X = µ(d) (by theorem 1.1.1) k=1 dj(n;k) n X X = µ(d) k=1 djn&djk 2 n 3 X X X = 4 µ(d)5 (1.5) djn k=1 djk 9 School of Distance Education, University of Calicut Anil Kumar V. Number Theory Pn P Note that for a fixed d, k=1 djk µ(d) denotes the summation over all those k in the set f1; 2; : : : ; ng which are multiples of d. Question: How many multiples of d are there in the set f1; 2; 3; : : : ; ng? Suppose 1 ≤ k ≤ n and k = qd. Now 1 ≤ k ≤ n , 0 < k ≤ n k n , 0 < ≤ d d k n , 1 ≤ ≤ d d n , 1 ≤ q ≤ d This implies that in the set f1; 2; : : : ; ng there are n=d numbers which are multiples of d. These multiples of d are fd; 2d; : : : ; (n=d)dg Hence equation (1.5) can be written as: X n '(n) = µ(d) : d djn This completes the proof of the theorem. 1.1.4 A product formula for '(n) Theorem 1.1.4. For n ≥ 1 we have Y 1 '(n) = n 1 − ; p pjn where p is a prime divisor of n. Proof. Here we consider two cases. Case 1: Suppose n = 1. Note that when n = 1, '(n) = '(1) = 1. Also when n = 1, Y 1 Y 1 n 1 − = 1 − p p pjn pj1 Since there are no prime divide 1, we define Y 1 1 − = 1: p pj1 10 School of Distance Education, University of Calicut Anil Kumar V.
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