NUMBER THEORY (Module 1)

Home , Liouville function

[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

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 ϕ(n)...... 7 1.1.3 A relation connecting ϕ and µ...... 9 1.1.4 A product formula for ϕ(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 functions 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 associative property of Dirichlet multiplication and convolution. Generalized inversion formula is also included. The second section is devoted to averages 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 .

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1.1.1 The Möbiusfunction µ(n) Definition 1.1.2. The Mobius function µ is defined as:  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. d|n

Proof. We consider two cases. Case 1: Suppose n = 1. In this case X X µ(d) = µ(d) = µ(1) = 1. d|n d|1 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). d|n a1 a2 ak d|p1 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)

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Hence X µ(d) = µ(1) + [µ(p1) + µ(p2) + ··· + µ(pk)] + [µ(p1p2) + ··· + µ(pk−1pk)] + d|n

[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 ϕ(n) Deﬁnition 1.1.3. If n ≥ 1 the Euler totient function ϕ(n) is deﬁned as

ϕ(n) := #{a ∈ N : a ≤ n, (a, n) = 1}. Thus n 0 X ϕ(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 ϕ(n) = 1 = . (n, k) k=1 k=1 (k,n)=1 Theorem 1.1.2. If n ≥ 1 we have X ϕ(d) = n. d|n Proof. Let S = {1, 2, 3, . . . , n} We partition the set S into mutually disjoint subsets as follows: For each divisor d of n, let A(d) := {k :(k, n) = d, 1 ≤ k ≤ n}.

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Claim 1: If d1 and d2 are distinct divisors of n, then A(d1) ∩ A(d2) = ∅.

x ∈ A(d1) ∩ A(d2) ⇒ x ∈ A(d1) and x ∈ A(d2)

⇒ (x, n) = d1 and (x, n) = d2

⇒ d1 = d2, contradiction.

Claim 2: {1, 2, . . . , n} = ∪d|nA(d). Obviously

∪d|n A(d) ⊆ S. (1.1) Now

x ∈ S ⇒ (x, n) = d for some d, 1 ≤ d ≤ n ⇒ x ∈ A(d)

⇒ x ∈ ∪d|nA(d) for some d|n.

Thus

S ⊆ ∪d|nA(d). (1.2) From equations (1.1) and (1.2) it follows that

S = ∪d|nA(d).

Therefore X #(S) = #(A(d)). d|n This implies that X n = #(A(d)). (1.3) d|n Claim : #(A(d)) = ϕ(d). Now

k ∈ A(d) ⇔ (k, n) = d ⇔ (k/d, n/d) = 1, 0 < k/d ≤ n/d ⇔ (q, n/d) = 1, 0 < q ≤ n/d ⇔ q ∈ {k :(k, n/d) = 1, 0 < k ≤ n/d} | {z } B ⇔ q ∈ B

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This implies that there is a one to one correspondence between the elements of A(d) and B. Hence

#A(d) = #(B) = ϕ(n/d).

Hence equation (1.3) becomes: X n = ϕ(n/d) (1.4) d|n

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 ϕ(n) = ϕ(n/d). d|n d|n

Hence equation (1.4) can be written as X ϕ(d) = n. d|n

This completes the proof of the theorem.

1.1.3 A relation connecting ϕ and µ. Theorem 1.1.3. If n ≥ 1 we have X n ϕ(n) = µ(d) . d d|n

Proof. By the deﬁnition of ϕ, we have

n X 1 ϕ(n) = (n, k) k=1 n X X = µ(d) (by theorem 1.1.1) k=1 d|(n,k) n X X = µ(d) k=1 d|n&d|k  n  X X X =  µ(d) (1.5) d|n k=1 d|k

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Pn P Note that for a ﬁxed d, k=1 d|k µ(d) denotes the summation over all those k in the set {1, 2, . . . , n} which are multiples of d. Question: How many multiples of d are there in the set {1, 2, 3, . . . , n}? 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 {1, 2, . . . , n} there are n/d numbers which are multiples of d. These multiples of d are {d, 2d, . . . , (n/d)d} Hence equation (1.5) can be written as: X n ϕ(n) = µ(d) . d d|n This completes the proof of the theorem.

1.1.4 A product formula for ϕ(n) Theorem 1.1.4. For n ≥ 1 we have Y 1 ϕ(n) = n 1 − , p p|n where p is a prime divisor of n. Proof. Here we consider two cases. Case 1: Suppose n = 1. Note that when n = 1, ϕ(n) = ϕ(1) = 1. Also when n = 1, Y 1 Y 1 n 1 − = 1 − p p p|n p|1 Since there are no prime divide 1, we deﬁne Y 1 1 − = 1. p p|1

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Hence the theorem is true for n = 1. Case 2: Suppose n > 1. In this case n can be expressed in the form:

α1 α2 αr n = p1 p2 ··· pr , where pi’s are distinct primes and α1, α2, . . . , αr are integers ≥ 1. Consider

r Y 1 Y 1 1 − = 1 − p pi p|n i=1 1 1 1 = 1 − 1 − ··· 1 − p1 p2 pr 1 1 1 1 1 1 = 1 − + + ··· + + + + ··· + − p1 p2 pr p1p2 p1p3 pr−1pr 1 1 (−1)r + ··· + + ··· + p1p2p3 pr−2pr−1pr p1p2 ··· pr r r r X (−1) X (−1)2 X (−1)3 (−1)r = 1 + + + + ··· + pi pipj pipjpk p1p2 ··· pr i=1 i,j=1 i,j,k=1 i6=j i6=j6=k r r r X µ(pi) X µ(pipj) X µ(pipjpk) µ(p1p2 ··· pr) = 1 + + + + ··· + pi pipj pipjpk p1p2 ··· pr i=1 i,j=1 i,j,k=1 i6=j i6=j6=k X µ(d) = . d d|n

Hence

Y 1 X µ(d) n 1 − = n p d p|n d|n X n = µ(d) d d|n = ϕ(n) (by theorem 1.1.3 )

This completes the proof of the theorem.

Theorem 1.1.5. Euler’s totient has the following properties:

(a) ϕ(pα) = pα − pα−1, for prime p and α ≥ 1.

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d (b) ϕ(mn) = ϕ(m)ϕ(n) ϕ(d) , where d = (m, n) (c) ϕ(mn) = ϕ(m)ϕ(n) if (m, n) = 1. (d) a|b implies ϕ(a)|ϕ(b). (e) ϕ(n) is even for n ≥ 3. Moreover, if n has distinct r odd prime factors then 2r|ϕ(n). Proof. (a) We have Y 1 ϕ(n) = n 1 − (theorem 1.1.24) (1.6) p p|n Putting n = pα in equation (1.6), we obtain: Y 1 ϕ(pα)n 1 − (1.7) p p|pα 1 = pα 1 − (since p is the only prime divisor of pα) (1.8) p = pα − pα−1 (1.9) (b)We have Y 1 ϕ(n) = n 1 − (theorem 1.1.24) p p|n Therefore ϕ(n) Y 1 = 1 − . (1.10) n p p|n The above equation is true for all natural numbers n. Hence it is true for mn. Hence ϕ(mn) Y 1 = 1 − . (1.11) mn p p|mn Note that each prime divisor of mn is either a prime divisor of m or of n, and those prime which divide both m and n also divide d = (m, n). Hence ϕ(mn) Y 1 = 1 − mn p p|mn

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Q 1 Q 1 p|m 1 − p p|n 1 − p = Q 1 p|d 1 − p ϕ(m) ϕ(n) = m n (by theorem 1.1.24) ϕ(d) d d = ϕ(m)ϕ(n) . ϕ(d)

(c)By part (b) we have

d ϕ(mn) = ϕ(m)ϕ(n) , where d = (m, n). ϕ(d)

Putting d = 1 in the above equation, we obtain 1 ϕ(mn) = ϕ(m)ϕ(n) ϕ(1) = ϕ(m)ϕ(n) (since ϕ(1) = 1).

(d)Given that a|b.

a|b ⇒ b = ac, 1 ≤ c ≤ b.

Here we consider 3 cases. Case 1: Suppose b = 1. Since b = 1 and 1 ≤ c ≤ b, it follows that c = 1. This implies that a = 1. So ϕ(a) = ϕ(b) = 1. Hence ϕ(a)|ϕ(b). Case 2: Suppose b > 1, c = b. In this case a = 1. This implies that ϕ(a) = 1. Note that 1 divides every integer. In particular 1|ϕ(b). Hence ϕ(a)|ϕ(b). Case 3: b > 1, c < b. Now proof is by induction on b. Step 1: b = 2. Since 1 ≤ c < b and b = 2, it follows that c = 1. Hence a = 2. Therefore ϕ(a)|ϕ(b). Step 3: Assume that the result is true for all positive integers less than b( That is, whenever λ|µ, µ < b, then ϕ(λ)|ϕ(µ)). Step 2: Now to show that the result is true for b. Let d = (a, c). Then d|c. Since c < b, by induction hypothesis, we have

ϕ(c) ϕ(d)|ϕ(c) ⇒ ∈ . ϕ(d) N

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But now

ϕ(b) = ϕ(ac) d = ϕ(a)ϕ(c) ϕ(d) ϕ(c) = ϕ(a)d ϕ(d) Since ϕ(c)/ϕ(d) is a natural number, it follows that ϕ(a)|ϕ(b). (e)Let n ≥ 3. Then

n = 2km, for some odd numberm.

If m = 1, then n = 2k, where k ≥ 2(since n ≥ 3). Therefore

ϕ(n) = ϕ(2k) = 2k − 2k−1 = 2(2k−1 − 2k−2) = an even number.

Assume that m is an odd number greater than 1. Then m has an odd prime divisor say p. Now

k p|m ⇒ ϕ(p)|ϕ(n)(∵ n = 2 m) ⇒ ϕ(p)|ϕ(n) ⇒ (p − 1)|ϕ(n) ⇒ 2|ϕ(n)(p is an odd prime) ⇒ ϕ(n) is even.

Next assume that α1 α2 αk n = p1 p2 ··· pk , where p1, p2, . . . , pr ∈ {2, 3, 5, 7,...}. Then

Y 1 ϕ(n) = n 1 − p p|n α1 α2 αk 1 1 1 = p1 p2 ··· pk 1 − 1 − ··· 1 − p1 p2 pk α1−1 α2−1 αk−1 = p1 p2 ··· pk (p1 − 1)(p2 − 1) ··· (pn − 1) r Since each pi is odd it follows that each (pi − 1) is even. This implies 2 divides ϕ(n). This completes the proof.

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1.1.5 Dirichlet product of arithmetical functions Deﬁnition 1.1.4. Let f and g be arithmetical functions. Then the Dirichlet product of f and g is denoted by f ∗ g and is deﬁned as

X n (f ∗ g)(n) = f(d)g d d|n X = f(a)f(b), (1.12) ab=n where the sum on the right-hand side of (1.12) runs over all ordered pairs (a, b) of positive integers satisfying ab = n.

Deﬁnition 1.1.5. If α is a complex number, the power function N α is an arithmetical function is deﬁned as

α α N (n) = n , ∀n ∈ N.

Deﬁnition 1.1.6. The unit function u is deﬁned as

u(n) = 1 ∀n ∈ N.

Deﬁnition 1.1.7. The identity function I is deﬁned as ( 1 1 if n = 1, I(n) = = n 0 if n > 1.

Remark 1. Theorem 1.1.1 can be stated in the form ϕ = µ ∗ u.

We have X 1 µ(d) = n d|n Now X X n µ(d) = µ(d)u d d|n d|n = (µ ∗ u)(n)

Remark 2. Theorem 1.1.3 can be sated in the form µ ∗ N = ϕ.

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We have X n ϕ(n) = µ(d) d d|n X n = µ(d)N d d|n = (µ ∗ N)(n).

Theorem 1.1.6. Dirichlet multiplication is commutative and associative. That is, for any multiplicative functions f, g, h we have

f ∗ g = g ∗ f(Commutative law) f ∗ (g ∗ h) = (f ∗ g) ∗ h(Associative law)

Proof. Let f and g be two arithmetical functions. Then X n (f ∗ g)(n) = f(n)g d d|n X = f(a)g(b) ab=n X = g(b)f(a) ba=n = (g ∗ f)(n).

Hence the Dirichlet product is commutative. That is f ∗ g = g ∗ f. Next, we prove that Dirichlet multiplication is associative. Let k = g ∗ h and ` = f ∗ g. Then

(f ∗ (g ∗ h))(n) = (f ∗ k)(n) X = f(a)k(d) ad=n X = f(a)(g ∗ h)(d) ad=n X X = f(a) g(b)h(c) ad=n bc=d X = f(a)g(b)h(c). abc=n

((f ∗ g) ∗ h)(n) = (` ∗ h)(n)

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X = `(d)h(c) dc=n X = (f ∗ g)(d)f(c) dc=n X X = f(a)g(b)h(c) dc=n ab=d X = f(a)g(b)h(c). abc=n

Hence (f ∗ g) ∗ h = f ∗ (g ∗ h).

Theorem 1.1.7. For any arithmetical function f we have

f ∗ I = I ∗ f = f, where I is the identity function.

Proof. By the deﬁnition of Dirichlet product, we have X n (I ∗ f)(n) = I(d)f d d|n X n = I(1)f(n) + I(d)f d d|n d>1 X 1 n = f(n) + f d d d|n d>1 = f(n) + 0 (∵ I(1) = 1I(d) = 0 if d > 1.) = f(n).

Hence I ∗ f = f. Since Dirichlet product is commutative, we have f ∗ I = f.

1.1.6 Dirichlet inverses and Mobius inversion formula Theorem 1.1.8. If f is an arithmetical function with f(1) 6= 0, then there is a unique inverse f −1 of f called Dirichlet inverse of f such that

f ∗ f −1 = I = f −1 ∗ f

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Moreover f −1 is given by

 1  f(1) if n = 1, f −1(n) = − 1 P f n f −1(d) if n > 1.  f(1) d|n d d

−1 1 −1 Take f (1) := f(1) . Then f (1) is unique. Also

X 1 1 (f ∗ f −1)(1) = f(d)f −1 = f(1)f −1(1) = f(1) = 1. d f(1) d|1 Hence the result is true for n = 1. Next assume that the result is true for all natural numbers less than n. That is f −1(r) is uniquely determined for all r < n, f ∗ f −1(r) = I for all r < n and 1 X r f −1(r) = − f f −1(d) ∀ r < n. f(1) d d|r d

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 

 1 X n −1  X n −1 = f(1) − f f (d) + f f (d)  f(1) d  d d|n d|n d

Theorem 1.1.9. Let

A := {f : f is arithmetical and f(1) 6= 0}

Then A is an abelian group under Dirichlet product. Proof.

Closure property: Let f, g ∈ A . Then f(1) 6= 0 and g(1) 6= 0. Also note that f ∗ g is an arithmetical function. Also

(f ∗ g)(1) = f(1)g(1) 6== 0.

Hence f ∗ g ∈ A . Associative property: By theorem 1.1.6 it follows that ∗ is associative.

Existence of identity: The existence of each element in A follows from theorem 1.1.8.

Existence of inverse: Note that I is the identity element (see theorem 1.1.7).

Commutative property: By theorem 1.1.6 it follows that ∗ is commutative.

Hence (A , ∗) is an abelian group. Remark 3. The functions µ and u are inverses of each other.

Proof. From theorem 1.1.1 we have X µ(d) = I(n). d|n

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The above equation can be written as X n µ(d)u = I(n). ( u(n) = 1 ∀n) d ∵ d|n That is (µ ∗ u)(n) = I(n) ∀n This completes the proof.

Theorem 1.1.10 (Mobius inversion formula). We have X X n f(n) = g(d) ⇔ g(n) = f(d)µ d d|n d|n That is f = g ∗ u ⇔ g = f ∗ µ. Proof. Note that u−1 = µ. Hence f = g ∗ u ⇔ f ∗ u−1 = (g ∗ u) ∗ u−1 ⇔ f ∗ u−1 = (g ∗ u) ∗ u−1 ⇔ f ∗ µ = g ∗ (u ∗ u−1) ⇔ f ∗ µ = g ∗ I = g. This completess the proof.

Corollary 1. If n ≥ 1 we have X n ϕ(n) = dµ . d d|n Proof. We have X ϕ(d) = n = N(n). d|n By Mobius inversion formula, ϕ(n) = (N ∗ µ)(n) X n = µ(d)N ( N(n) = n ∀n) d ∵ d|n X n = µ(d) . d d|n This completes the proof.

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1.1.7 The Mangoldt function Λ(n) Deﬁnition 1.1.8. For every integer n ≥ 1 the Mangoldt’s function Λ(n) is deﬁned as: ( log p if n = pm for some prime p and some m ≥ 1 Λ(n) = , 0 otherwise.

A short table values of Λ(n) is given below:

n 1 2 3 4 5 6 7 8 9 10 Λ(n) 0 log 2 log 3 log 2 log 5 0 log 7 log 2 log 3 0

Theorem 1.1.11. If n ≥ 1 we have X log n = Λ(n). d|n

Proof. Here we consider two cases: Case 1: Suppose n = 1. In this case X X Λ(d) = Λ(d) = Λ(1) = 0 d|n d|1 Hence the theorem is true for n = 1. Case 2: Suppose n > 1. In this case n can be expressed in the following form:

α1 α2 αk n = p1 p1 ··· p1 , where pi’s are distinct primes and αi ≥ 1. Taking logarithms we have

k X log n = αi log pi. (1.13) i=1 P Observe that non zero terms in d|n Λ(d) occcurs only when

2 α1 2 α2 2 αk d = p1, p1, . . . , pk ; p2, p2, . . . , p2 ; ... ; pk, pk, . . . , pk . Hence

X 2 α1 2 α2 Λ(d) = Λ(p1) + Λ(p1) + ··· + Λ(p1 ) + Λ(p2) + Λ(p2) + ··· + Λ(p2 ) + ··· d|n

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2 αk Λ(pk) + Λ(pk) + ··· + Λ(pk ) = (log p1 + log p1 + ··· + log p1) + (log p2 + log p2 + ··· + log p2) + ··· | {z } | {z } α1 times α2 times

(logk + log pk + ··· + log pk) | {z } αk times

= α1 log p1 + α2 log p2 + ··· + αk log pk k X = αi log pi. (1.14) i=1 From the light of equations (1.13) and (1.14), we have X log n = Λ(d). d|n This completes the proof.

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1.1.8 Multiplicative functions We have already proved that the set of all arithmetical function f with f(1) 6= 0(see theorem 1.1.9) forms an abelian group under Dirichlet multiplication. In this section we discuss an important subgroup of this group, called multiplicative functions. Deﬁnition 1.1.9. An arithmetical function f is called multiplicative if (a) f not identically zero,

(b) f(mn) = f(m)f(n), whenever (m, n) = 1. Deﬁnition 1.1.10. An arithmetical function f is called completely multiplicative if (a) f not identically zero,

(b) f(mn) = f(m)f(n), for all m, n ∈ N. Note :Consider the following sets:

A := {f : f is arithmetical and f(1) 6= 0}, M := {f ∈ A : f is multiplicative}, C := {f ∈ A : f is completely multiplicative}. Then we have M ⊂ C ⊆ A . Moreover M is a subgroup of (A , ∗). Unlike M , which is a group with respect to Dirichlet product, C is closed neither with respect to addition nor convolution. Theorem 1.1.13. If f is multiplicative then f(1) = 1. Proof. Since f is multiplicative, we have

f(mn) = f(m)f(n) whenever (m, n) = 1.

Note that for any n ∈ N, we have (n, 1) = 1. Hence by deﬁnition, f(n) = f(n · 1) = f(n)f(1)

This implies that f(n)[f(1) − 1] = 0. Since f is not identically zero, we have f(1) − 1 = 0. That is f(1) = 1.

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Theorem 1.1.14. Let f be an arithmetical function with f(1) = 1. Then

(a) f is multiplicative if and only if

α1 α2 αk α1 α2 αk f p1 p2 ··· pk = f (p1 ) f (p2 ) ··· f pk

for all primes pi and all integers αi ≥ 1. (b) If f is multiplicative, then f is completely multiplicative if and only if

f (pα) = f(p)α

for all primes p and all integers α ≥ 1.

Proof. (a) First assume that f is multiplicative. Then f(mn) = f(m)f(n) whenever (m, n) = 1. We prove the the result by induction on k. Step 1:Assume that k = 1. In this case we have

α1 α1 f (p1 ) = f (p1 )

Hence the result is true for n = 1. Step 2:Assume that the result is true for all integers < k. That is

α1 α2 αi α1 α2 αi f (p1 p2 ··· pi ) = f (p1 ) f (p2 ) ··· f (pi ) ∀ i < k. Case 3:We will prove that the result is true for k. Now

α1 α2 αk α1 α2 αk−1 αk f p1 p2 ··· pk = f p1 p2 ··· pk−1 pk α1 α2 αk−1 αk = f p1 p2 ··· pk−1 f(pk )(∵ f is multiplicative) α1 α2 αk−1 αk = f (p1 ) f (p2 ) ··· f pk−1 f pk (by induction hypothesis) Hence the result is true for k. Hence by mathematical induction the result is true for all natural numbers. Conversely assume that

α1 α2 αk α1 α2 αk f p1 p2 ··· pk = f (p1 ) f (p2 ) ··· f pk ∀k ∈ N.

We will prove that f is multiplicative. Let m, n ∈ N such that (m, n) = 1. Let

α1 α2 αk m = p1 p2 ··· pk ,

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β1 β2 βs n = q1 q2 ··· qs where pi’s and qj’s are distinct primes and αi, βi ≥ 1 are integers. Then α1 α2 αk β1 β2 βs f(mn) = f p1 p2 ··· pk q1 q2 ··· qs α1 α2 αk β1 β2 βs = f (p1 ) f (p2 ) ··· f pk f q1 f q2 ··· f qk (by assumption) α1 α2 αk β1 β2 βs = f p1 p2 ··· pk f q1 q2 ··· qk = f(m)f(n)

(b) First assume that f is completely multiplicative. Then

f(mn) = f(m)f(n) ∀ m, n ∈ N

α α Claim : f (p ) = f(p) for all primes p and all integers αi ≥ 1. We will prove the theorem by induction on α. Step 1:Assume that α = 1. Note that f(p1) = f(p)1. Hence the result is true for n = 1. Step 2: Assume that the result is true for all integers < k. That is f pi = f(p)i ∀i < k. Step 3:We will prove that the result is true for k. Now

f (pα) = f pα−1p α−1 = f p f(p)(∵ f completely multiplicative) = [f(p)]α−1 f(p) = [f(p)]α (by assumption).

α α Hence by mathematical induction f (p ) = f(p) for all α ∈ N. Conversely assume that

f (pα) = f(p)α for all k.

We will prove that f is completely multiplicative. Let m, n ∈ N. Let (m, n) = d. Then m = ad and n = bd for some a, b ∈ N.Let

α1 α2 αk β1 β2 βs γ1 γ2 γt a = p1 p2 . . . pk , b = q1 q2 . . . qs , d = d1 d2 . . . dt Then

α1 α2 αk γ1 γ2 γt m = p1 p2 . . . pk d1 d2 . . . dt

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β1 β2 βs γ1 γ2 γt n = q1 q2 . . . qs d1 d2 . . . dt α1 α2 αk β1 β2 βs 2γ1 2γ2 2γt mn = p1 p2 . . . pk q1 q2 . . . qs d1 d2 . . . dt .

Now α1 α2 αk β1 β2 βs 2γ1 2γ2 2γt f(mn) = f p1 p2 . . . pk q1 q2 . . . qs d1 d2 . . . dt α1 α2 αk β1 β2 βs = f (p1 ) f (p2 ) ··· f pk f q1 f q2 ··· f qs

2γ1 2γ2 2γt f d1 f d2 ··· f dt (∵ f is multiplicative)

α1 α2 αk β1 β2 βs = f (p1) f (p2) ··· f (pk) f (q1) f (q2) ··· f (qs)

2γ1 2γ2 2γt f (d1) f (d2) ··· f (dt) (by assumption)

Again α1 α2 αk γ1 γ2 γt β1 β2 βs γ1 γ2 γt f(m)f(n) = f p1 p2 . . . pk d1 d2 . . . dt f q1 q2 . . . qs d1 d2 . . . dt

α1 α2 αk γ1 γ2 γt = f (p1 ) f (p2 ) ··· f pk f (d1 ) f (d2 ) ··· f (dt ) β1 β2 βs γ1 γ2 γt f q1 f q2 ··· f qs f (d1 ) f (d2 ) ··· f (dt )

α1 α2 αk β1 β2 βs = f (p1) f (p2) ··· f (pk) f (q1) f (q2) ··· f (qs)

2γ1 2γ2 2γt f (d1) f (d2) ··· f (dt)

Hence f(mn) = f(m)f(n) for all m, n ∈ N. That is f is completely multiplicative.

1.1.9 Examples of completely multiplicative functions

α α 1. The arithmetical function N (n) = n ∀ n ∈ N is completely multiplicative. For all m, n ∈ N, we have

N(mn) = (mn)α = N(m)N(m).

Hence N α is completely multiplicative. In particular N is completely multiplicative.

2. The unit function u(n) = 1 for all n ∈ N is completely multiplicative. For all m, n ∈ N, we have

u(mn) = 1 = 1.1 = u(m)u(n).

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3. The identity function is completely multiplicative. Here we consider several cases: Case 1: Suppose m = 1, n = 1. Then mn = 1. Therefore

I(m) = 1,I(n) = 1,I(mn) = 1

Hence I(mn) = I(m)I(n) Case 2: Suppose m = 1 and n > 1. In this case mn > 1. So we have

I(m) = 1,I(n) = 0,I(mn) = 0

Thus I(mn) = 0 = 1 × 0 = I(m)I(n). Case 3: Suppose n = 1, m > 1. In this case mn > 1. Moreover I(n) = 1, I(m) = 0 and I(mn) = 0. Hence I(mn) = 0 = 0 × 1 = I(m)I(n). Case 4:Suppose m, n > 1. In this case mn > 1. Moreover I(m) = I(n) = I(mn) = 0. Hence

I(mn) = 0 = 0 × 0 = I(m)I(n).

1.1.10 Examples of multiplicative functions 1. The Euler function ϕ is multiplicative but not completely multiplicative. Let m, n ∈ N such that (m, n) = 1. Then we have ϕ(mn) = ϕ(m)ϕ(n) (see theorem 1.20(c))

Observe that ϕ is not completely multiplicative. Let m = 4 and n = 2. Then (m, n) 6= 1. By the deﬁnition of ϕ we have

ϕ(8) = 4, ϕ(4) = 2, ϕ(2) = 1.

Hence

ϕ(mn) = ϕ(8) = 4 6= 2 × 1 = ϕ(4)ϕ(2) = ϕ(m)ϕ(n).

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2. The Mobius function is multiplicative but not completely multiplicative. Let m, n ∈ N such that (m, n) = 1. Case 1: If either m = 1 or n = 1, then obviously

µ(mn) = µ(m)µ(n).

Case 2: m is square free and n is not square free. Then clearly mn is not square free. Then by the deﬁnition of µ, we have µ(mn) = 0 and µ(n) = 0. Hence

µ(mn) = 0 = µ(m)µ(n).

Case 3: m is not square free and n is square free. This case is exactly similar to case 2. Case 4: Both m and n are square free. Since m and n are relatively prime, mn is square free. Let

m = p1p2 ··· pm,

where pi’s are distinct primes. Similarly, let

q = q1q2 . . . qn,

where qi’s are distinct primes. Then

µ(m) = (−1)m, µ(n) = (−1)n and µ(mn) = (−1)m+n.

Hence µ(mn) = (−1)m+n = (−1)m(−1)n = µ(m)µ(n).

Hence µ is completely multiplicative. Next we will show that µ is not completely multiplicative. Let m = 2 and n = 2. Then µ(4) = 0 and µ(2) = −1. Note that µ(4) 6= µ(2)µ(2). Hence µ is not completely multiplicative.

1.1.11 Multiplicative functions and Dirichlet multiplication

Theorem 1.1.15. If f and g are multiplicative, so is their Dirichlet product f ∗ g.

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Proof. Since f and g are multiplicative functions, we have f(1) = 1 and g(1) = 1. Now X (f ∗ g)(1) = f(d)g(1/d) = f(1)g(1) = 1 d|1

Hence (f ∗ g)(1) 6= 0. Let m, n ∈ N such that (m, n) = 1. Note that

d|mn ⇔ d = ab, where a|m, b|m and(a, b) = 1.

This completes the proof.

Remark 4. The Dirichlet product of two completely multiplicative functions need not be completely multiplicative.

Proof. Observe that the arithmetical functions u and N are completely multiplicative. We will prove that the function f = u ∗ N is not completely multiplicative. To prove that f is not completely multiplicative it suﬃces to prove that for any prime number p, we have f (pα) 6= f(p)α. Now

f (pα) = (u ∗ N)(pα) X = u(d)N(pα/d) d|pα

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X pα = d d|pα pα pα pα pα = + + + ··· + 1 p p2 pα = pα + pα−1 + ··· + 1.

Again

f(p) = (u ∗ N)(p) X = u(d)N(p/d) = u(1)N(p) = p d|1

Therefore f(p)α = pα 6= pα + pα−1 + ··· + 1 = f (pα) . Hence f is not completely multiplicative.

Theorem 1.1.16. Let f, g be arithmetic functions. If both g and f ∗ g are multiplicative, then f is also multiplicative.

Proof. Since g and f ∗g are multiplicative, we have g(1) = 1 and (f ∗g)(1) = 1. Now X 1 1 = (f ∗ g)(1) = f(d)g = f(1)g(1) = f(1)(1) = f(1). d d|1

Hence f 6= 0. Claim :f is multiplicative.That is, f(mn) = f(m)f(n) whenever (m, n) = 1. We prove the claim by induction on mn. Step 1: Assume that mn = 1. In this case m = n = 1. Note that

f(mn) = f(1) = 1 = f(m)f(n).

Hence the result is true for mn = 1. Step 2: Assume that the result is true for all products ab less than mn. Step 3: We prove the claim for mn. Note that X mn (f ∗ g)(mn) = f(d)g d d|mn

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X mn = f(ab)g ab a|m b|n X mn = f(mn)g(1) + f(ab)g ( (a, b) = 1) ab ∵ a|m b|n ab

1.1.12 The inverse of a completely multiplicative function Theorem 1.1.18. Let f be a multiplicative function. Then f is multiplicative if and only if f −1(n) = µ(n)f(n) for all n ≥ 1.

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Proof. Assume that f is completely multiplicative. Let g(n) = µ(n)f(n). We want to show that f −1(n) = g(n). Now

Hence g ∗ f = I = f ∗ g. This implies that g = f −1. That is

f −1(n) = µ(n)f(n) for alln ≥ 1.

Conversely assume that f be multiplicative and f −1(n) = µ(n)f(n). We want to show that f is completely multiplicative. To prove that f is completely multiplicative it suﬃces to show that

f(pα) = f(p)α for all α.

Note that

f −1 ∗ f = I −1 ∴ (f ∗ f)(n) = I(n)

This implies that X n f −1(d)f = I(n) d d|n

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That is, X n µ(d)f(d)f = I(n)( f −1(n) = µ(n)f(n)) d ∵ d|n

Putting n = pα in the above equation we obtain:

X pα µ(d)f(d)f = I(pα) d d|pα

That is

µ(1)f(1)f(pα) + µ(p)f(p)f(pα−1) + ··· + µ(pα)f(pα)f(pα−1) = 0.

That is,

α α−1 3 α f(p )−f(p)f(p )+0+···+0 = 0 (∵ f(1) = 1, µ(p ) = ··· = 0,I(p ) = 0) This implies that f(pα) = f(p)f(pα−1) That is

f(pα) = f(p)f(p)f(pα−2) = [f(p)]2f(pα−2) = [f(p)]3f(p)f(pα−3) . . α α = f(p) f(1) = f(p) (∵ f(1) = 1). Hence f is completely multiplicative. Remark 5. The Dirichlet inverse of the Euler totient function is given by X ϕ−1(n) = µ(d)d. d|n

Proof. Since ϕ = µ ∗ N we have

ϕ−1 = µ−1 ∗ N −1

We have already proved that N is completely multiplicative. Therefore by previous theorem, N −1(n) = µ(n)N(n).

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Hence ϕ−1 = µ−1 ∗ N −1 = µ−1 ∗ µN = u ∗ µN Therefore

ϕ−1(n) = (u ∗ µN)(n) X n = µN(d)u d d|n X = (µN)(d) d|n X X = µ(d)N(d) = µ(d)d. d|n d|n

This completes the proof.

Theorem 1.1.19. If f is multiplicative we have X Y µ(d)f(d) = (1 − f(p)). d|n p|n

Proof. Let X g(n) = f(d)µ(d) d|n X = (fµ)(d) d|n X n = (fµ)(d)u d d|n = (fµ ∗ u)(n)

Hence g = fµ ∗ u. Since f and µ are multiplicative, so is fµ. Also observe that u is multiplicative. This implies that fµ ∗ u is multiplicative. Now

g(pα) = (fµ ∗ u)(pα) X pα = (fµ)(d)u d d|pα

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X = (fµ)(d) d|pα X = f(d)µ(d) d|pα = f(1)µ(1) + f(p)µ(p) + f(p2)µ(p2) + ··· + f(pα)µ(pα) 2 3 = 1 − f(p)(∵ f(1) = 1, µ(p ) = µ(p ) = ··· = 0)

Let

α1 α2 αk n = p1 p2 ··· pk , where pi’s are distinct primes and α ≥ 1.. Since g is multiplicative we have

α1 α2 αk g(n) = g(p1 )g(p2 ) ··· g(pk ) = (1 − f(p1)(1 − f(p2)) ··· (1 − f(pk)) Y = (1 − f(p)) p|n

This completes the proof.

Corollary 2. The Dirichlet inverse of ϕ is given by

Y ϕ−1 = (1 − f(p)). p|n

Proof. From remark 5, we have

X ϕ−1(n) = µ(d)d d|n X = µ(d)N(d) d|n Y = (1 − N(p)) (by theorem 1.1.19) p|n Y = (1 − p) p|n

This completes the proof of the theorem.

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1.1.13 Liouville’s function λ(n) Definition 1.1.11. The Liouville’s function λ is defined as: ( 1 if n = 1, λ(n) = α1+α2+···+αk α1 α2 αk (−1) if n = p1 p2 ··· pk , where pi’s distinct primes and αi ≥ 1 are integers. Remark 6. The Liouville’s function is a completely multiplicative function. Proof. From the definition of λ we have

λ(pα) = (−1)α = [λ(p)]α.

This implies that λ is completely multiplicative.

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= λ(1) + λ(p) + ··· + λ(pα) = 1 + (−1) + (−1)2 + (−1)3 + ··· + (−1)α ( 1 if α is even, = 0 if α is odd.

Let α1 α2 αk n = p1 p2 ··· pk , where pi’ s are distinct primes and i ≥ 1 are integers. Therefore

α1 α2 αk g(n) = g(p1 )g(p2 ) ··· g(pk ) ( 1 if each α is even, = i 0 if at least one αi is odd.

Assume that each αi is even. Then αi = 2βi for βi,i = 1, 2, . . . , k. This implies that

2β1 2β2 2βk n = p1 p2 ··· pk 2 2 2 β1 β2 βk = p1 p2 ··· pk , = a square

Hence ( X 1 if n is a square, g(n) = λ(d) = 0 otherwise. d|n Note that λ is completely multiplicative. Therefore

λ−1(n) = λ(n)µ(n) (1.15)

Claim : λ(n)µ(n) = |µ(n)| for all n. We have  1 if n = 1,  k µ(n) = (−1) if n = p1p2 ··· pk, for distinct primes, 0 otherwise.  1 if n = 1,  α1+α2+···+αk α1 α2 αk λ(n) = (−1) if n = p1 p2 ··· pk , 0 otherwise.

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Here we consider 3 cases: Case 1:Suppose n = 1. In this case λ(n)µ(n) = 1 = |µ(1)|. Case 2:Suppose n = p1p2 ··· pk, where pi’s are distinct primes. Then µ(n) = (−1)k, λ(n) = (−1)k. Therefore λ(n)µ(n) = (−1)k(−1)k = 1 = |µ(n)|.

α1 α2 αk Case 3: Suppose n = p1 p2 ··· pk , where pi’s are distinct primes and αi ≥ 1. In this case µ(n) = 0, λ(n) = (−1)α1+α2+···+αn Therefore λ(n)µ(n) = 0 = |µ(n)|. Hence λ−1(n) = µ(n)λ(n) = |µ(n)| for all n.

1.1.14 The divisor function σα(n) Deﬁnition 1.1.12. For any real or complex α and any integer n ≥ 1 the divisor function is deﬁned by

X α σα(n) := d , d|n = sum of the αth powers of the divisors of n

Remark 7. σα(n) is multiplicative. Proof. Note that N α and u are multiplicative functions. Now

X α σα(n) = d d|n X n = N α(d)u d d|n = (N α ∗ u)(n).

α Hence σα is the Dirichlet product of two multiplicative functions N and u. Thus σα is multiplicative.

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Remark 8. If α = 0, then the divisor function σα can be written as:

X 0 σ0(n) = d d|n X = 1 d|n = Number of the divisors of n.

The function σ0 is usually denoted by d(n).

Remark 9. If α = 1, then the divisor function σα can be written as: X σ1(n) = d d|n = sum of the divisors of n.

The function σ1 is denoted by σ(n). Theorem 1.1.21. For n ≥ 1, we have X n σ−1(n) = dαµ(d)µ . α d d|n Proof. We have

X α σα(n) = d d|n X n = N α(d)u d d|n = (N α ∗ u)(n)

α Hence σα = N ∗ u. Therefore

−1 −1 −1 −1 σα = (Nα) ∗ u = (Nα) ∗ µ. (1.16) Since N α is completely multiplicative, we have

(N α)−1(n) = N α(n)µ(n) = (N αµ)(n).

Inserting the value of (N α)−1 in equation (1.16) we obtain

−1 α σα (n) = (N µ) ∗ µ(n)

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X n = (N αµ)(d)µ d d|n X n = N α(d)µ(d)µ d d|n X n = dαµ(d)µ . d d|n

α1 α2 αk Theorem 1.1.22. If n > 1 and n = p1 p2 ··· pk , where pi’s distinct primes and αi ≥ 1., then

k Y d(n) = (αi + 1). i=1 Proof. Since d is a multiplicative function we have

α1 α2 αk α1 α2 αk d p1 p2 ··· pk = d (p1 ) d (p2 ) ··· d pk (1.17)

αi αi To compute d (pi ) we note that the divisors of pi are

2 αi 1, pi, pi , . . . , pi

αi Hence the number of divisors of pi are (αi + 1). Therefore,

αi d (pi ) = (αi + 1) Hence k Y d(n) = (αi + 1) i=1

α1 α2 αk Theorem 1.1.23. If n > 1 and n = p1 p2 ··· pk , where pi’s distinct primes and αi ≥ 1., then

k α +1 Y p i − 1 σ(n) = i pαi − 1 i=1 i Proof. Since σ is a multiplicative function we have

α1 α2 αk α1 α2 αk σ p1 p2 ··· pk = σ (p1 ) σ (p2 ) ··· σ pk (1.18)

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αi αi To compute σ (pi ) we note that the divisors of pi are

2 αi 1, pi, pi , . . . , pi Hence (αi+1) αi 2 αi pi − 1 σ (pi ) = 1 + pi + pi + ... + pi = . pi − 1 Hence k α +1 Y p i − 1 σ(n) = i pαi − 1 i=1 i

1.1.15 Generalised convolution Deﬁnition 1.1.13. Let F be a real or complex valued function deﬁned on the positive real axis (0, ∞) such that F (x) = 0 for 0 < x < 1. Let α be an arithmetical function. Then the sum X x α(n)F n n≤x is called the generalized convolution of G and α and is denoted by α ◦ F . Thus X x (α ◦ F )(x) := α(n)F . n n≤x Theorem 1.1.24 (Associative property relating α and ∗). For any arithmetical functions α and β, we have

α ◦ (β ◦ F ) = (α ∗ β) ◦ F.

Proof. For x > 0 we have X x (α ◦ (β ◦ F ))(x) = α(n)(β ◦ F ) n n≤x x X X = α(n) β(m)F n m n≤x x m≤ n X X x = α(n) β(m)F mn n≤x mn≤x X x = α(n)β(m)F mn mn≤x

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X k x = α(n)β F , k = mn n k k≤x   X X k x = α(n)β F  n  k k≤x n|k X x = (α ∗ β)(k)F k k≤x = ((α ∗ β) ◦ F )(x).

Therefore α ◦ (β ◦ F ) = (α ∗ β) ◦ F . This completes the proof.

Theorem 1.1.25 (Generalized inversion formula). Let α has Dirichlet inverse α−1. Then X x X x G(x) = α(n)F ⇔ F (x) = α−1(n)G . n n n≤x n≤x That is G = α ◦ F ⇔ F = α−1 ◦ G. Proof. First assume that G = α ◦ F . Then

α−1 ◦ G = α−1 ◦ (α ◦ F ) = (α−1 ∗ α) ◦ F (by theorem 1.1.24) = I ◦ F = F.

Conversely assume that F = α−1 ◦ G. Then

α ◦ F = α ◦ (α−1 ◦ G) = (α ∗ α−1) ◦ G(by theorem 1.1.24) = I ◦ G = G.

This completes the proof.

Theorem 1.1.26 (Generalized Mobius inversion formula). Let α be completely multiplicative. Then X x X x G(x) = α(n)F ⇔ F (x) = α(n)µ(n)G n n n≤x n≤x That is G = α ◦ F ⇔ F = α−1 ◦ G = (αµ) ◦ G.

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Proof. First assume that G = α ◦ F . Then

α−1 ◦ G = α−1 ◦ (α ◦ F ) = (α−1 ∗ α) ◦ F = I ◦ F = F.

That is

−1 F = α ◦ G = (αµ) ◦ G (∵ α is completely multiplicative) Conversely assume that F = α−1 ◦ G. Then

α ◦ F = α ◦ (α−1 ◦ G) = (α ∗ α−1) ◦ G = I ◦ G = G.

This completes the proof.

1.1.16 Derivatives of arithmetical functions Deﬁnition 1.1.14. Let f be an arithmetical function. Then the derivative of f is deﬁned as f 0(n) := f(n) log n, n ≥ 1.

Example 1. The derivative of the function I is 0.

By deﬁnition, the derivative of I is

I0(n) = I(n) log(n) 1 = log n n ( 1 × log 1 if n = 1 = 0 × log n if n > 1. = 0.

Remark 10. We have Λ ∗ u = u0.

Proof. By the deﬁnition of derivative, we have

u0(n) = u(n) log n = log n (u(n) = 1)

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X = Λ(d) (by theorem 1.1.11) d|n X n = Λ(d)u d d|n = (Λ ∗ u)(n).

This completes the proof.

Theorem 1.1.27. If f and g are arithmetical functions we have:

(a) (f + g)0 = f 0 + g0.

(b) (f ∗ g)0 = f 0 ∗ g + f ∗ g0.

Proof. (a) By the deﬁnition of derivative, we have

(f + g)0(n) = (f + g)(n) log n = (f(n) + g(n)) log n = f(n) log n + g(n) log n = f 0(n) + g0(n).

(b)Note that

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(c) Note that I0 = 0. This implies that (f ∗ f −1)0 = 0. Then by part (b) we have 0 = f 0 ∗ f −1 + f ∗ (f −1)0 This implies that f ∗ (f −1)0 = −(f 0 ∗ f −1) This implies that

f −1 ∗ f ∗ (f −1)0 = −f −1 ∗ (f 0 ∗ f −1)

That is, (f −1 ∗ f) ∗ (f −1)0 = −f 0 ∗ (f −1 ∗ f −1) That is I ∗ (f −1)0 = −f 0 ∗ (f ∗ f)−1 That is (f −1)0 = −f 0 ∗ (f ∗ f)−1. This completes the proof.

1.1.17 The Selberg identity Theorem 1.1.28. For n ≥ 1 we have X n X Λ(n) log n + Λ(d)Λ = µ(d) log2(n/d). d d|n d|n

Proof. By Remark 10 we have:

Λ ∗ u = u0. (1.19)

Diﬀerentiation of this equation gives:

Λ0 ∗ u + Λ ∗ u0 = u00.

That is

Λ0 ∗ u + Λ ∗ (Λ ∗ u) = u00.

Now multiply both sides by µ = u−1 to obtain:

(Λ0 ∗ u) ∗ u−1 + (Λ ∗ (Λ ∗ u) ∗ u−1) = u00 ∗ µ.

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That is

Λ0 + Λ ∗ Λ = u00 ∗ µ

This implies that

(Λ0 + Λ ∗ Λ)(n) = (u00 ∗ µ)(n)

This completes the proof.

1.1.18 Exercises 1. Find all integers n such that

(a) ϕ(n) = n/2, (b) ϕ(n) = ϕ(2n) (c) ϕ(n) = 12.

2. For each of the following statements either give a proof or exhibit a counter example.

(a) If (m, n) = 1 the (ϕ(m), ϕ(n)) = 1, (b) If n is composite, then (n, ϕ(n)) > 1, (c) If the same prime divide m and n, then nϕ(m) = mϕ(n).

3. Prove that n X µ2(d) = ϕ(n) ϕ(d) d|n

4. Prove that ϕ(n) > 6 for all n with at most 8 distinct prime factors.

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P 2 5. Prove that d2|n µ(d) = µ (n). √ √ 6. Let f(n) = [ n] − n − 1. Prove that f is multiplicative but not completely multiplicative. 7. Assume that f is multiplicative. Prove that (a) f −1(n) = µ(n)f(n) for every square free n. (b) f −1(p2) = f(p)2 − f(p2) for every prime p. (a) Assume that f is multiplicative. Prove that f is completely multiplicative if, and only if, f −1(pα) = 0 for all primes p and all integers a ≥ 2. (b) Prove that the following statement or exhibit a counter example. P If f is multiplicative, then F (n) = d|n f(d) is multiplicative.

1.2 Averages of arithmetical functions

An arithmetical function is defined to be a function f(n), defined for n ∈ N, which maps to a complex number such that f : N → C. Examples of arithmetic functions include: the number of primes less than a given number n, the number of divisors of n, etc. While the behavior of values of arithmetic functions are hard to predict, it is easier to analyze the behavior of the averages of arithmetic functions which we define as f(1) + f(2) + ··· + f(n) lim = L n→∞ n where L the average value of f(n). 1. Let r(n) be the number of representations of n as a sum of two squares n = x2 + y2 Then average number of representations of a natural number as a sum of two squares is π. That is, r(1) + r(2) + ··· + r(n) lim = π n→∞ n 2. Let f(n) be the number of decompositions of a natural number n into a sum of one or more consecutive prime numbers. For example, f(395) = 2 because 395 = 127 + 131 + 137 = 71 + 73 + 79 + 83 + 89.

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Then average number of decompositions of a natural number into a sum of one or more consecutive prime numbers is log 2. That is

f(1) + f(2) + ··· + f(n) lim = log 2 n→∞ n

3. Consider the divisor function d(n) deﬁned by

X d(n) = 1. d|n

Note that d(p) = 2 for all primes numbers and d(n) is large when n has a large number of divisors. Thus the values of d(n) ﬂuctuate as n increases. In this case it is diﬃcult to determine their behavior as n increases. We can prove that

d(1) + d(2) + ··· + d(n) = 1 n log n

In chapter paper, we will examine averages of several diﬀerent arithmetic functions.

1.2.1 The big oh notation. Asymptotic equality of functions Deﬁnition 1.2.1. Let f and g be real valued functions such that g(x) > 0 for all x ≥ a. We write f(x) = O(x)(and say that f(x) is big oh of g(x)) if there exists a constant M > 0 such that

|f(x)| ≤ Mg(x) for all x ≥ a.

• f(x) = O(g(x)) means that |f(x)/g(x)| is bounded for all x ≥ a.

• The equation of the form

f(x) = h(x) + O(g(x))

means that f(x) − h(x) = O(g(x))

R x R x • If f(t) = O(g(t)) for some t ≥ a, then a f(t) dt = O( a g(t) dt)

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Proof. Note that f(t) = O(g(t)) means that there exists a constant M > 0 such that |f(t)| ≤ Mg(t), for t ≥ a This implies that Z x Z x |f(t)|dt ≤ M g(t)dt, for x ≥ a a a That is, Z x Z x

f(t)dt ≤ M g(x), for x ≥ a a a That is Z x Z x f(t) dt = O g(t) dt . a a

Example 2. Note that

2 2 2 n x x = O(x ), x + 10x + 7 = O(x ), x = O(e ) for all n ∈ N. Theorem 1.2.1.

1. If f1 is O(g1) and f2 is O(g2), then (f1 + f2) is O(max{g1, g2}).

2. If f1 and f2 are both O(g), then (f1 + f2) is O(g).

3. If f1 is O(g1) and f2 is O(g2), then (f1f2) is O(g1g2).

4. If f1 is O(f2) and f2 is O(f3), then f1 is O(f3). 5. If f is O(g), then (af) is O(g) for any constant Definition 1.2.2. We say f(x) is asymptotic to g(x) as x → ∞ if f(x) lim = 1 x→∞ g(x) and write f(x) ∼ g(x), x → ∞. Here we need the following: Definition 1.2.3. Let f be an arithmetical function and x is any positive P real number. Then n≤x f(n) is defined as:

[x] X X f(n) = f(n) n≤x n=1

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Deﬁnition 1.2.4. Let f be an arithmetical function. Then the average of f is deﬁned as 1 X f ∼(n) := f(n) n n≤x

Theorem 1.2.2. (Euler summation formula) If f has a continuous derivative f 0 on the interval [y, x], where 0 < y < x, then

X Z x Z x f(n) = f(t)dt+ (t − [t]) f 0(t)dt+f(x) ([x] − x) − f(y) ([y] − y) . y

Equivalently,

X Z x Z x f(n) = f(t)dt + {t}f 0(t)dt + f(y){y} − f(x){x}, y

Let m = [y] and k = [x]. If n, n − 1 ∈ [y, x], then:

k Z k X Z n [t]f 0(t) dt = [t]f 0(t) dt m n=m+1 n−1 k X Z n = (n − 1)f 0(t) m+1 n−1 k X = (n − 1)[f(n) − f(n − 1)] m+1 k X = ([nf(n) − (n − 1)f(n − 1)] − f(n)) m+1 k k X X = ([nf(n) − (n − 1)f(n − 1)]) − f(n) m+1 m+1 X = kf(k) − mf(m) − f(n) y

X Z k f(n) = − [t]f 0(t) dt + kf(k) − mf(m) y

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Z x Z y Z x = − [t]f 0(t)dt − [t]f 0(t) + [t]f 0(t)dt + kf(k) − mf(m) y m k Z x Z k Z x Z k Z y Z x ∵ = + = − + y y k m m k Z x Z y Z x = − [t]f 0(t)dt − mf 0(t) + kf 0(t)dt + kf(k) − mf(m) y m k Z x = − [t]f 0(t)dt − m[f(y) − f(m)] + k[f(x) − f(k)] + kf(k) − mf(m) y Z x = − [t]f 0(t)dt − mf(y) + kf(x) y Z x = − [t]f 0(t)dt − [y]f(y) + [x]f(x) y Z x = − [t]f 0(t)dt + (y − [y]) f(y) − (x − [x]) f(x) + xf(x) − yf(y) y Z x = − [t]f 0(t)dt + {y}f(y) − {x}f(x) + xf(x) − yf(y) y Z x Z x Z x = − [t]f 0(t)dt + {y}f(y) − {x}f(x) + tf 0(t)dt + f(t)dt y y y Z x Z x = (t − [t])f 0(t)dt + {y}f(y) − {x}f(x) + f(t)dt y y Z x Z x = {t}f 0(t)dt + f(t)dt + {y}f(y) − {x}f(x). y y

Z x Z x 0 x tf (t)dt =tf(t)]y − f(t)dt y y Z x =xf(x) − yf(y) − f(t)dt. y

Definition 1.2.5. The Euler’s constant is defined by the equation 1 1 1 C = lim 1 + + + ··· + − log n . n→∞ 2 3 n Definition 1.2.6. The Riemann zeta function is defined by the equation

(P∞ 1 n=1 ns if s > 1 ζ(s) = P 1 x1−s limx→∞ n≤x ns − 1−s if 0 < s < 1.

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Theorem 1.2.3. If x ≥ 1 we have P 1 1 (a) n≤x n = log x + C + O x , where C is Eulers constant.

P 1 x1−s −s (b) n≤x ns = 1−s + ζ(s) + O(x ) if s > 0, s 6= 1. P 1 1−s (c) n>x ns = O(x ) if s > 1. P s xα+1 s (d) n≤x n = α+1 + O(x ) if α ≥ 0. Proof. (a)By Euler’s Summation formula, we have X Z x Z x f(n) = f(t)dt + (t − [t]) f 0(t)dt + f(x) ([x] − x) − f(y) ([y] − y) . y

Putting y = 1 and f(t) = 1/t in the above formula, we obtain: X 1 Z x 1 Z x = dt − (1 − [t])(−1/t2)dt + (1/x)([x] − x) n t 1

X 1 Z x t − [t] 1 = log x − dt + 1 + O n t2 x 1≤n≤x 1

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Z ∞ t − [t] Z ∞ t − [t] 1 = log x − 2 dt − 2 dt + 1 + O 1 t x t x Z ∞ t − [t] Z ∞ t − [t] 1 = log x − 2 dt + 2 dt + 1 + O (1.22) 1 t x t x Note that Z ∞ t − [t] Z ∞ 1 0 ≤ 2 dt < 2 dt = 1 < ∞ 1 t 1 t Thus Z ∞ t − [t] 2 < ∞. 1 t Again note that

t−[t] Z ∞ Z ∞ 2 t − [t] 1 t − [t] 1 1 t = t−[t] < 1 ⇒ = O ⇒ dt = O dt = O 1 t2 t2 t2 t2 x t2 x x Hence equation (1.22) can be written as:

X 1 Z ∞ t − [t] Z ∞ t − [t] 1 = log x − dt + dt + 1 + O n t2 t2 x 1≤n≤x 1 x Z ∞ t − [t] 1 1 = log x + 1 − 2 dt +O + O 1 t x x | {z } B 1 = log x + B + O (1.23) x Taking x → ∞ in the above equation, we obtain:   X 1 lim  − log x = B x→∞ n 1≤n≤x

This implies that B = C. Hence equation (1.23) becomes:

X 1 1 = log x + C + O n x n≤x

(b) Putting y = 1 and f(t) = t−s, s > 0, s 6= 1 in the Euler’s summation formula, we obtain:

X 1 Z x 1 Z x x − [x] = dt + (t − [t])(−st−s−1)dt − ns ts xs 1

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x−s+1 1 Z x t − [t] x − [x] = − − s s+1 dt − s 1 − s 1 − s 1 t x x−s+1 1 Z x t − [t] 1 = − − s s+1 dt + O s 1 − s 1 − s 1 t x

Therefore,

X 1 x−s+1 1 Z x t − [t] 1 = + 1 − − s dt + O ns 1 − s 1 − s ts+1 xs 1≤n≤x 1 x1−s 1 Z ∞ t − [t] Z ∞ t − [t] 1 = − + 1 − s s+1 dt − s+1 dt + O s 1 − s 1 − s 1 t x t x x1−s 1 Z ∞ t − [t] 1 1 = − + 1 − s s+1 dt + O s + O s 1 − s 1 − s 1 t x x x1−s 1 Z ∞ t − [t] 1 = + 1 − − s s+1 dt +O s . (1.24) 1 − s 1 − s 1 t x | {z } B(s)

t − [t] s s =O ts+1 ts+1 Z ∞ t − [t] Z ∞ s ⇒ s s+1 dt = O s+1 dt x t x t 1 = O xs

Case 1: Suppose s > 1. Letting x → ∞ in the above equation, we obtain:

∞ X 1 = B(s) ns n=1

P 1 x1−s −s That is ζ(s) = B(s). Hence if s > 1, n≤x ns = 1−s + ζ(s) + O(x ). Thus equation (1.24) becomes:

X 1 x1−s = + ζ(s) + O(x−s). ns 1 − s n≤x

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Case 2: Suppose 0 < s < 1. Letting x → ∞ in equation (1.24), we obtain   X 1 x1−s lim  −  = B(s) x→∞ ns 1 − s n≤x

That is, ζ(s) = B(s) Hence equation (1.24) becomes:

X 1 x1−s = + ζ(s) + O(x−s). ns 1 − s n≤x (c)If s > 1,

∞ X 1 ζ(s) = ns n=1 X 1 X 1 = + ns ns n≤x n>x Therefore X X 1 = ζ(s) − ns n>x n≤x x1−s = ζ(s) − + ζ(s) + O(x−s) (by (b)) 1 − s x1−s = + O(x−s) s − 1 1−s 1−s −s 1−s = O(x ) + O(x )(∵ x ≤ x ) = O(x1−s)

(c)Taking f(t) = tα and y = 1 in the Euler formula obtain:

X 1 Z x Z x = tαdt + α {t}αtα−1dt − xα{x} ns 1

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xα+1 1 Z x = − + O α tα−1 + O(xα) + 1 α + 1 α + 1 1 xα+1 = + O(xα) α + 1

This completes the proof.

1.2.2 Partial sums of a Dirichlet product Theorem 1.2.4. If h = f ∗ g, let X X X H(x) = h(n),F (x) = f(n) and G(x) = g(n). n≤x n≤x n≤x

Then we have X x X x H(x) = f(n)G = g(n)F . n n n≤x n≤x

Proof. Let ( 0 if 0 < x < 1, U(x) = 1 if x ≥ 1. Claim 1:F = f ◦ U. Now X x (f ◦ U)(x) = f(n)U n n≤x X x = f(n) 0 < x ≤ n ⇒ > 1 ⇒ U(x/n) = 1. ∵ n n≤x = F (x).

Claim 2: G = g ◦ U. Now X x (g ◦ U)(x) = g(n)U n n≤x X x = g(n) 0 < x ≤ n ⇒ > 1 ⇒ U(x/n) = 1. ∵ n n≤x = G(x).

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Claim 3: h ◦ U = H.

X x (h ◦ U)(x) = h(n)U n n≤x X = h(n) = H(x). n≤x Claim 4: H = f ◦ G. Now f ◦ G = f ◦ (g ◦ U) = (f ∗ g) ◦ U = h ◦ U = H. Claim 5: H = g ◦ F . Now g ◦ F = g ◦ (f ◦ U) = (g ∗ f) ◦ U = h ◦ U = H. This completes the proof. P Theorem 1.2.5. thm33 If F (x) = n≤x f(n) we have X X X hxi X x f(d) = f(n) = F . n n n≤x d|n n≤x n≤x Proof. From the above theorem, we have

H = f ◦ G, and H = g ◦ F.

If g = 1, then we have: X G(x) = 1 = [x]. n≤x Now

H(x) = (f ◦ G)(x) X x = f(n)G n n≤x X hxi = f(n) (1.25) n n≤x Again

H(x) = (g ◦ F )(x)

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X x = g(n)F n n≤x X x = F (1.26) n n≤x Also X H(x) = h(n) n≤x X = (f ∗ g)(n) n≤x X X x = f(n)g . (1.27) n n≤x d|n From equations (1.25), (1.26) and (1.27), we have X X X hxi X x f(d) = f(n) = F . n n n≤x d|n n≤x n≤x This completes the proof.

1.2.3 Applications to µ(n) and Λ(n) Theorem 1.2.6. For n ≥ 1 we have X hxi µ(n) = 1 n n≤x and X hxi Λ(n) = log[x]!. n n≤x Proof. Putting f(n) = µ(n) theorem we obtain:

X hxi X X X 1 µ(n) = µ(d) = = 1. n n n≤x n≤x d|n n≤x Again putting f(n) = Λ(n) theorem we obtain:

X hxi X X X Λ(n) = Λ(d) = log n = log[x]!. n n≤x n≤x d|n n≤x This completes the proof.

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Theorem 1.2.7. For all x ≥ 1 we have

X µ(n) ≤ 1 n n≤x with equality holding only if x < 2. Proof. Here we consider two cases: Case1: Suppose x < 2. In this case X µ(n) µ(1) = = 1. n 1 n≤x Note that in this case equality occurs. Case 2: Suppose x ≥ 2. For each real number y, let

{y} := y − [y].

From theorem 1.2.3 we have: X hxi 1 = µ(n) n n≤x X x nxo = µ(n) − n n n≤x X µ(n) X nxo = x − µ(n) n n n≤x n≤x This implies that X µ(n) X nxo x = 1 + µ(n) n n n≤x n≤x Therefore

X µ(n) X nxo x = 1 + µ(n) n n n≤x n≤x X nxo ≤ 1 + n n≤x X nxo = 1 + {x} + n 2≤n≤x

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X < 1 + {x} + 1 (∵ 0 ≤ {y} < 1) 2≤n≤x   X = 1 + {x} + [x] − 1 ∵ 1 = [x] n≤x = 1 + x − [x] + [x] − 1 = x.

Dividing by x we obtain:

X µ(n) < 1 n n≤x This completes the proof.

Theorem 1.2.8 (Legendre’s identity). For every x ≥ 1 we have Y [x]! = pα(p), p≤x where the product is extended over all primes ≤ x, and

∞ X x α(p) = pm m=1 Proof. We have X hxi Λ(n) = log[x]! (see theorem 1.2.3) n n≤x Putting n = pm in the above equation, we obtain:

X x log[x]! = Λ(pm) pm pm≤x X h x i X h x i = Λ(2m) + Λ(3m) + ··· 2m 3m 2,22,...2m≤x 3,32,...3m≤x   ∞ X X x =  Λ(pm)   pm  p≤x m=1 pm≤x   ∞ X X x =  log p   pm  p≤x m=1 pm≤x

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  ∞ X X x = log p    pm  p≤x m=1 pm≤x ∞ X X x = α(p) log p, α(p) = pm p≤x m=1 pm≤x X = log pα(p) p≤x   Y α(p) = log  p  p≤x

Q α(p) Hence [x]! = p≤x p . This completes the proof. Theorem 1.2.9. If x ≥ 2 we have

log[x]! = x log x − x + O(log x), and hence X hxi Λ(n) = x log x − x + O(logx). n n≤x Proof. Putting f(t) = log t in Euler’s summation formula with y = 1 we obtain: X Z x Z x 1 log n = log tdt + (t − [t]) dt + log x([x] − x) − 0 t 1

log x([x] − x) = |[x] − x| < 1 ⇒ log x([x] − x) = O(log x) log x

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Adding 1 on both sides of equation (1.28) we obtain: X log n = x log x − x + 2 + O(log x) + O(log x) n≤x = x log x − x + O(log x)

That is

log[x]! = x log x − x + O(log x).

From the light of theorem 1.2.3 it follows that X hxi Λ(n) = x log x − x + O(log x).. n n≤x

This completes the proof.

Theorem 1.2.10. For x ≥ 2 we have

X x log p = x log x + O(x), p p≤x where the sum is extended over all primes ≤ x.

Proof. We have X hxi log[x]! = Λ(n) . n n≤x Therefore,

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  ∞ X X x = log p    pm  p≤x m=1 pm≤x ∞ X x X X x = Λ(p) + Λ(pm) p pm p≤x p≤x m=2 ∞ X x X X x = Λ(p) + log p . p pm p≤x p≤x m=2

Therefore

∞ X x X X x Λ(p) = log[x]! − log p p pm p≤x p≤x m=2 ∞ X X x = x log x − x + O(log x) − log p (by theorem 1.2.9) pm p≤x m=2 (1.29)

P P∞ h x i Claim : p≤x m=2 log p pm = O(x).

∞ ∞ X X x X X x log p ≤ log p pm pm p≤x m=2 p≤x m=2 ∞ X X x = log p pm p≤x m=2 ∞ X X 1 = x log p pm p≤x m=2 X 1 1 = x log p + + ··· p2 p3 p≤x X 1 1 = x log p 1 + + ··· p2 p p≤x X 1 1 = x log p p2 1 − 1/p p≤x X 1 p = x log p p2 p − 1 p≤x

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X log n ≤ x n(n − 1) n≤x | <{z∞ } This implies that ∞ X X x log p = O(x). pm p≤x m=2 Hence equation (1.29) can be written as: X x Λ(p) = x log x − x + O(log x) − O(x) p p≤x = x log x + O(x). This completes the proof.

1.2.4 Another identity for the partial sums of a Dirichlet product Theorem 1.2.11. If a and b are positive real numbers such that ab = x, then X X x X x f(d)g(q) = f(n)G + g(n)F − F (a)G(b). n n q,d n≤a n≤b qd≤x Proof. Let X F (x) = f(n), n≤x X G(x) = g(n) and n≤x X H(x) = h(x), h = f ∗ g. (1.30) n≤x Then X X n H(x) = f(d)g d n≤x d|n X = f(d)g(q). (1.31) q,d qd≤x Since a and b are positive real numbers such that ab = x it follows that (a, b) is a point on the rectangular hyperbola qd = x.

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The sum H(x) in equation (1.31) is extended over all lattice points in the ﬁrst quadrant of (q, d)- plane, below the rectangular hyperbola qd = x between the two lines q = 1 and d = 1. Let D = {(d, q): q ≥ 1, d ≥ 1, qd ≤ x} A = {(d, q): d ≥ 1, q > b, qd ≤ x} B = {(d, q): q ∈ [1, a], d ∈ [1, b]} C = {(d, q): d > a, q ≥ 1, qd ≤ x}. Then D = A ∪ B ∪ C. Now X X X = f(d)g(q) (d,q)∈A∪B d≤a q≤x/d X X = f(d) g(q) d≤a q≤x/d

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X x = f(d)G d d≤a X x = f(n)G n n≤a

X X X f(d)g(q) = f(d)g(q) (d,q)∈B∪C q≤b d≤x/q X X = g(q) f(d) q≤b d≤x/q X x = g(q)F q q≤b X x = g(n)F . n n≤b

X X X = f(d)g(q) (d,q)∈B d≤a q≤b X X = f(d) g(q) d≤a q≤b = F (a)G(b).

Hence X X f(d)g(q) = f(d)g(q) d,q (d,q)∈D qd≤x X X X = f(d)g(q) + − (d,q)∈A∪B (d,q)∈B∪C (d,q)∈B X x X x = f(n)G + g(n)F − F (a)G(b). n n n≤a n≤b

1.2.5 Exercises 1. Use Euler’s summation formula to deduce the following for x ≥ 2:

P log n 1 2 log x (a) n≤x n = 2 log x + A + O x , where A is constant.

66 School of Distance Education, University of Calicut Anil Kumar V. Number Theory

P 1 1 (b) 2≤n≤x n log n = log(log x) + B + O x log x , where B is a constant. (a) Prove that [2x] − [x] is either 0 or 1.

67 School of Distance Education, University of Calicut Anil Kumar V. Number Theory

68 School of Distance Education, University of Calicut Bibliography

[1] Apostol T.M., Introduction to Analytic Number Theory, Narosa Pub- lishing House, New Delhi, 1990.

[2] G. H. Hardy and E.M. Wright: Introduction to the theory of numbers; Oxford International Edn; 1985

[3] A. Hurwitz & N. Kritiko: Lectures on Number Theory; Springer Verlag ,Universitext; 1986

[4] T. Koshy: Elementary Number Theory with Applications; Harcourt / Academic Press; 2002

[5] D. Redmond: Number Theory; Monographs & Texts in Mathematics No: 220; Marcel Dekker Inc.; 1994

[6] P. Ribenboim: The little book of Big Primes; Springer-Verlag, New York; 1991

[7] K.H. Rosen: Elementary Number Theory and its applications(3rd Edn.); Addison Wesley Pub Co.; 1993

[8] W. Stallings: Cryptography and Network Security-Principles and Practices; PHI; 2004

[9] D.R. Stinson: Cryptography- Theory and Practice(2nd Edn.); Chap- man & Hall / CRC (214. Simon Sing : The Code Book The Fourth Estate London); 1999

[10] J. Stopple: A Primer of Analytic Number Theory-From Pythagorus to Riemann; Cambridge Univ Press; 2003

[11] S.Y. Yan: Number Theroy for Computing(2nd Edn.); Springer-Verlag; 2002

69 Anil Kumar V. Number Theory

1.A Syllabus

Semester 1

MTH1CO5: NUMBER THEORY No. of Credits: 4 No. of hours of Lectures/week: 5

TEXT 1 : APOSTOL T.M., INTRODUCTION TO ANALYTIC NUM- BER THEORY, Narosa Publishing House, New Delhi, 1990.

TEXT 2: KOBLITZ NEAL A., COURSE IN NUMBER THEROY AND CRYPTOGRAPHY, SpringerVerlag, NewYork, 1987.

Module 1 Arithmetical functions and Dirichlet multiplication; Averages of arithmetical functions [Chapter 2: sections 2.1 to 2.14, 2.18, 2.19; Chapter 3: sections 3.1 to 3.4, 3.9 to 3.12 of Text 1]

Module 2 Some elementary theorems on the distribution of prime numbers [Chap- ter 4: Sections 4.1 to 4.10 of Text 1]

Module 3 Quadratic residues and quadratic reciprocity law [Chapter 9: sections 9.1 to 9.8 of Text 1] Cryptography, Public key [Chapters 3 ; Chapter 4 sections 1 and 2 of Text 2.]

70 School of Distance Education, University of Calicut Anil Kumar V. Number Theory

1.A Notations

N = the set of natural numbers C = the set of complex numbers [x] = the greatest integer less than or equal to x {x} = x − [x] d|n = d divides n X = summation over all divisors of n d|n X = summation over all prime divisors of n p|n #(A) = the cardinality of the set A (c, d) = the greatest common divisor of c and d

71 School of Distance Education, University of Calicut