J OURNAL DE T HÉORIE DES N OMBRES DE B ORDEAUXJ EAN -L OUIS N ICOLAS
VARANASI S ITARAMAIAH
On a class of ψ-convolutions characterized by
the identical equation
Journal de Théorie des Nombres de Bordeaux, tome 14, no 2 (2002),
p. 561-583.
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561

On a class of 03C8-convolutions characterized
by the identical equation
par JEAN-LOUIS NICOLAS et VARANASI SITARAMAIAH*
dedicated to Michel Mend6s F’rance for his 60th birthdayRÉSUMÉ. Dans le cadre de la convolution de Dirichlet des fonctions arithmétiques, R. Vaidyanathaswamy a obtenu en 1931 une
formule de calcul de f(mn) valable pour toute fonction multiplicativef et tout couple d’entiers positifs m et n. Dans [7],cette formule a été généralisée aux 03C8-convolutions appelées convolutions de Lehmer-Narkiewicz, qui, entre autres, conservent lamultiplicativité. Dans cet article, nous démontrons la réciproque.
ABSTRACT. The identical equation for multiplicative functions established by R. Vaidyanathaswamy in the case of Dirichlet convolution in 1931 has been generalized to multiplicativity preserving
03C8-convolutions satisfying certain conditions (cf. [7]) which can becalled as Lehmer-Narkiewicz convolutions for some reasons. In
this paper we prove the converse.
1. Introduction

In 1930, R. Vaidyanathaswamy (see [9] and [10, Section VI]) establishedthe following remarkable identity valid for any multiplicative function andknown as the identical equation for multiplicative functions. If f is any
multiplicative function, then for any positive integers m and n, we havewhere f -1 is the inverse of f with respect to the familiar Dirichlet convolution so that

Manuscrit reçu le 11 mai 2000.
*
Research partially supported by CNRS, Institut Girard Desargues, UMR 5028 and IndoFrench cooperation in Mathematics.

562

for all positive integers m, whereandw(a) being the number of distinct prime factors of a and r(a) the productof distinct prime factors of a with w(1)0 and r(l)1.
In [7] the identical equation (1.1) has been generalized to a class of 0convolutions (which are certain binary operations in the set of arithmetic
functions introduced by D.H. Lehmer [2]): if (m, n) E T, where T is thedomain of 1/1, then for any multiplicative function f,==

f -1 being the inverse of f with respect to 0 (Section 2 contains undefinednotions used in this section); this class of 0-convolutions in which (1.4) has
been established is contained in the class of 0-convolutions preserving multiplicativity and satisfying ~(x, y) &#x3E; max{x, y} for x, y E T, as subclass:in addition to these, they satisfy certain other properties similar to thoseof regular A-convolutions [3]. It is interesting to note that these convolutions have been subsequently characterized (see [8], Corollary 4.1; also see§2 of the present paper) and have been named [8] as Lehmer-Narkiewicz
convolutions. Thus (1.4) holds when 1b is a Lehmer-Narkiewicz convolution.
The object of the present paper is to prove the converse. We show that(see Sections 3, 4 and 5) if o preserves
max{x, y}for all x, y E T and the identity (1.4) holds for all multiplicative functions,then 1b is a Lehmer-Narkiewicz convolution; in particular, if T = Z+ x Z+,
Z+ being the set of positive integers, it follows from Corollary 4.1 in [8](also, see Section 2, Lemma 2.16) that o is a Dirichlet convolution. Thus the
only 1b-convolution with domain T Z+ x 7G+, preserving multiplicativity,
for all
E T and with respect to which
satisfying
holds
all
for
multiplicative functions, is the Dirichlet convolution.
(1.4)
deals
with
§2
preliminaries. Sections 3-5 contain the main results.
For a brief discussion on different proofs and generalizations of the identical equation (1.1) and for certain special cases of (1.4), we refer to [7].
We are pleased to thank the referee for suggesting several improvementson the presentation of our paper.
=

563

2. Preliminaries: T 2013~ Z+ be a mapping satisfying theZ+denotesthe set of positive integers):
conditions(here
followingLet

For each n E Z+,(2.1)y) = n has a finite number of solutions.

The statements " (X 7 Y) E T, (~(~, y), z) E T"
(2.3) and "(y, z) E T, (x, V) (y, z)) E T" are equivalent; if one of these
conditions holds, we have ~(~(x, y), z) _ ~(~, ~(y, z)).

I

Let F denote the set of arithmetic functions (i.e. complex valued functions defined on 7L+). If f, 9 E F then the q#-product of f and g denotedby fog is defined byfor all n E 7G+.
The concept of 1/J-convolution given in (2.4) is due to D.H. Lehmer [2].
It is easily seen that (F, +, 1/J) is a commutative ring. Lety) xy for
Z+ x Z+ then o in (2.4) reduces to the Dirichletall (x, y) E T. If T
convolution. If T = {(x, y) E Z+ x Z+ : (x, y)
11, then 0 reduces to theunitary convolution [1]. More generally, if T =A(n)~,
where A is Narkiewicz’s regular convolution [3], then 0 reduces to the Aconvolution. Thus the binary operation in (2.4) is more general than thatof Narkiewicz’s A-convolution.
The following results (Lemmas 2.5 and 2.6) describe necessary and su%cient conditions concerning the existence of unity and inverses in (F, +, 1/J).
===

Lemma 2.5 (cf. [5, Theorem 2.2]). Let (F, +, 1/J) be a commutative ringand ~(x, y) &#x3E;
E T. Then (F, +, 0) possesses thefor all
unity if and only if for each k E Z+, ~(~, k) k has a solution. In such acase if g stands for the unity, then for each k E Z+,=

Lemma 2.6 (cf. [4], also see [10, Remark 1.1). Let o satisfy (2.1)&#x3E;
E T. For each k E Z+, let thefor all(2.3) andequationk) k have a solution so that the unity exists in (F, +, 1/J).

-

=

564

Let g denote the unity. Then f E F is invertible with respect to 0 if andonly if

for all k E 7G+.
computed by

In such a case, this inverse denoted by

can

be

..

and for lc &#x3E; 1,

The binary operation 1/; in (2.4) is called multiplicativity preserving
is multiplicative whenever f and g are ; as usual, f E F(cf. [6]) if
is called multiplicative if f (1)1 and f (mn)f (m) f (n) whenever=(~n,, n)==1.

The following results (Lemmas 2.7 and 2.9) give a characterization of&#x3E;
multiplicativity preserving 1/;-functions satisfying (i)allfork for all k E Z+:
x, y E T and (ii) ~(1, k)
Lemma 2.7 (cf. [6, Theorem 3.1~). Let o satisfy (2.1)(2.3), 1b(z, y) &#x3E;
all
k
k
7G+.
all
E
for
Suppose that thek)max{x, y} for x, y E T
inisbinary operation 0 (2.4) multiplicativity preserving . If x flr cti
and yfli where PI, P2, Pr are distinct primes, ai and ~3~ arenon-negative integers, we have
E T for i
1, 2, ... , r.
(a) (x, y) E T if and only if
For
and
each
prime pnon-negative integers 0:, f3 such that (PQ, PP) E(b)
T, there is a unique non-negative integers Bp(a, ~3) 2:: max{a"Q} such that
=

-

=

=

=

=

-

(c) If (x, y) E T, thenLemma 2.9 (cf. [6, Theorem 3.2]). Let T C Z+ x Z+ be such that(a) (1, x) e T for every x E 7L+.
(b) (x, y) e T if and only if (y, x) E T.
Lemma 2.7, then (x, y) E T if and only if(c) If x and y are as

Further, for each prime p and non-negcative integers 0:, (3 such that
E T, let 6~,(a"C3) be a non-negative integer satisfying

565(h) For non-negative integers a, ,8, ’Y and for any prime p, the statements
E T " and (pO:, pp) E T and
one
zuhen
are equivalent;of these conditions holds, we have«( p~e ~ p-y ) E T,

E T"

If for (x, y) E T,y) is defined by (2.8), then (F, +, 1/J) is a commutative
is multiplicative
where
e is given by (1.2). Also,
withunity e,ring
are.
whenever f and g
Theorem 2.10 (cf. [7, Theorem]). Let T, 1/J and 8p be as in Lemma 2.9.
Further we assume that for each prime p, we have

If f is multiplicative, then the identity in (1.4) holds.
Definition 2.13. Let 0 be multiplicativity preserving withy) &#x3E;

k for all k E Z+. Let T andmax{x, y} for all (x, y) E T andconvolution or
2.9.
iscalleda
Lehmer-Narkiewicz
in
as
Lemma
Then wOp
andfor
all primes p.
convolution
satisfies
L-N
simply an(2.11)(2.12)
if 8p=

Definition 2.14 (see [3]). A binary operation B in the set of arithmetic
functions F is called a regular convolution if(a) (F, +, B) is a commutative ring with unity.
(b) B is multiplicativity preserving i.e. f Bg is multiplicative whenever f

and g are.

(c) The constant function 1 E F has an inverse which is 0 or -1 at prime
powers.

For each positive integer n, if A(n) is a non-empty subset of divisorsof ~, Narkiewicz [3] defines the binary operation A in F (called the A-

convolution) by

for all n E Z+. He then calls the convolution A given in (2.15) as regular
if A satisfies the definition (2.14). To make a distinction between a generalregular convolution as defined in (2.14) and that of the special operation

566

A defined in (2.15) which is regular, we call the later as regular Narkiewicz
convolution.
The following result gives a characterization of L-N convolutions :Lemma 2.16 (cf. [8, Corollary 4.1]). For each prime p, let ~rP denote aclass of subsets of non-negative integers such that(i) the union of all members of 7rp is the set of non-negative integers.
(ii) each members of 7rp contains zero.
(iii) no two members of 7rp contains a positive integer in common.
a2
al
ao
...,
If S E xp and S = {ao, al, az, ... ~ with 0
we define 8p(ai,aj) = ai+j, if ai, aj and ai+j E S (i and j need not tobe distinct). If 1/J and T are as given in Lemma 2.9 then 0 is an L-N
convodution and is also a regular convolutions. Also, every L-N convolution
can be obtained in this way.
=

It is clear from the above result that there are infinitely many L-N conx + y, for all x, y such that
volutions. If 8p(x,y)
E T, then ~~should consist of arithmetic progressions. Thus, Lemma 2.16 reduces tothe characterization Theorem on regular Narkiewicz’s convolutions (cf. [3],Theorem II).
It is interesting to note that a Lehmer-Narkiewicz convolution 0 with
Z+ x Z+,
domain T = Z+ x Z+ is the Dirichlet convolution. For, if T
the domain of 8p in Lemma 2.16 is (Z+ U 101) x (Z+ U {0}). Hence there
can be only one member S of 7rp viz: S{0,1, 2, ... } and 8p(i,j) = i + j
for all non-negative integers i and j. Hencey) xy for all positive
integers x and y so that w is the Dirichlet convolution.
=

=

=

=

Lemma 2.17.

Let 1b be as in Lemma 2.6. Then the following statements

equivalent:
multiplicative junction is invertible with respect to 1/J.
Every
(a)
k E Z+,Foreachk) k if and only if x 1.
(b)
is
holds
e
the
unity in (F, +, ~)).
(Thus, if (a)Proof. We assume (a). Since e (given in (1.2)) is multiplicative and isinvertible with respect to o by hypothesis,
0 for all k E Z+ by
are

=

=

Lemma 2.6. Since

= k.
= k
it is clear that (1, k) E T and
We now show that
= 1.
=
This
true
k
when
we
that
assumek&#x3E; 1.
1,
implies x
may
being
=
Assume that we can find xo &#x3E; 1 such thatk) k. Let 1 Xokxl ... zr be all the solutions ofk)(from (2.1), there are onlyLet
finitely many solutions).
=

567

where pi , ... , pt are distinct primes, al, a2, ... , at are positive integers and,since xo &#x3E; 1, t is positive. Let S’ denote the set of prime powers pi 1, ... ,We define the multiplicative function f by f ( 1 )1, f (prl) = -1 and=Fix j,

Clearly f (~o) _ -1.

q~&#x3E; 1 such thatq,8l1xj

1jr.

S. Hence

We can find a prime power= 0.
This is true for

j = 1, 2, ... , r. Thus

f is not invertible by Lemma 2.2. This contradiction proves thatk) k implies x 1.
If we assume (b), then for any multiplicative function f,so that==0f is invertible, again by Lemma 2.2.
Lemma 2.18. Let T,1/1 and 8p be as in Lemma 2.9. We assume thateach multiplicative functions is invertible with respect to 1/1 and that (1.4) isvalid. We fix a prime p and write 0
Op. If (p~,~‘) E T, then for any
multiplicative function f,so that=wherE

Proof. Taking m=

p~ and n p’ in (1.4), we obtain=

From the definition of G given in (1.3), in the above sum either a = b
or a &#x3E; 0 and b &#x3E; 0. Hence

=

0

568

By Lemma 2.17,k) k if and only if x = 1 for each positive integer k.
It follows that O(x, y) x if and only if y = 0. Hence the conditions a &#x3E; 0
A andand b &#x3E; 0 in the sum on the right of (2.20) can be replaced
In this sum the term corresponding to (x, y) = (0, 0) is f -1
We separate this term from this sum and we obtain (2.19). This completes0
the proof of Lemma 2.18.
==

Remark 2.21. If each multiplicative function is invertible with respect to
1/1, as observed in the proof of Lemma 2.18, we have9(x, y) x if and only if y 0.
Let us recall also Property (e) of Lemma 2.9:
B(x, y) 0 if and only if x y 0.
(2.23)
Moreover, from Lemma 2.9 (g) and (h), the law A, p H 9(À,(2.22)

=

=

=

=

=

is commutative and associative so that we can define by induction for k &#x3E; 3,k} 2013~ {1,2,... , k} we haveand for any

By Lemma 2.6, for any prime power pt, t &#x3E; 0, we have

We frequently make use of (2.19) to (2.26) in the subsequent sections.
3. Theorem 3.1

In this section we prove

Theorem 3.l. Fix a prime p. Let T,1/1 and 0
Bp be as in Lemma 2.9.
Assume that the identical equation (1.4) holds for all multiplicative functions f . Then for any non-negative irttegers a, /3 and ~y,=Proof.

0(a, a)while,First we prove that if a and ,8 are non-negative integers with
0(a, ~3), then a (3. If a 0 then from (2.22) and (2.23)
=

=

=

0

=

569

~3 from (2.22).
a and (3 are positive integers with 0(a, a)that
0(a,,B) - Let
Supposewe deduce a contradiction. We define the multiplicative function fso that a = 0 ==bySince a # 0, it follows from (2.22) and (2.23) that 0(a, a) ~ {0, a} so thatf(Pe(a,a)) 0. Taking A = J-L a in (2.19) and with f as in (3.2), we
obtain==where

From the definition of f given in (3.2), in the sum defining :E(a,o:), thecorrespond to the values of x and y, where x E 10, a} andy E {0, a}, x ~ a, y # a and (~,~/) 7~ (0, 0). Since no such choice of (x, y)
is possible, it follows that :E( 0:,0:) is zero. Hence from (3.3),
non-zero terms

From (3.2), f W) = 0. Again by (2.19),

whereBut, from (3.2), in order that f (p-r)f (py) 54 0 we must have x E {0, a},= 1.
With the conditions (x, y) 0 (0, 0),y E {0, a}, and then
=
the
is
x
,(i,
only possibility
0, y = a. From (2.22), the only
x 54 a,solution in a of 0(a, 0) = a is a = a. So

However, from (2.24) and (2.25), 8(b, a) = /3 implies

570

so that, from (2.22), b = 0 and hence a = /3. Since~i is assumed itfollows that the sum on the right of (3.6) is an empty sum. Thus E (a, ~3)
0. Using this in (3.5), we obtain=

Since 0(a, a)=0(a, 0), (3.4) and (3.7) contradict each other. Hence a = ,Q.

We now return to the general case. Let a,# and -y be non-negative0,
integers with 9(a, ~3) 6(0:,,). We want to show that # -y. If
it easily follows that {3 = ~y. So, we may assume that a, ~3 and y are positive
integers. If a # or a 7, we obtain ,8 = 7 by the previous case. Hence
we can assume thatso that a, 0, ^t are pairwise/~ and a 54 -y.
distinct. We deduce a contradiction. Note that 0(a, 0) :0 0, a,,8, -y. We
define the multiplicative function f by
=

=

=

=

=

the values f (pl), f (~) and f(P1) remain to be fixed. From (2.19) with f
as in (3.8), we obtain

where

For positive integers m and n, letandWe have

571

it should be understood that x and y are fixed
where, for instance inand equal to 0 and a respectively so that

Remark 3.13. If 0(b, -y) = ,~ for some b, then from (2.24) and (2.25)

that, from (2.22), b 0 and y /3. Since -y 0 ~i is assumed, it follows
that ~(6,7) = # is not solvable. Therefore the sums in (3.12) corresponding
=

so

=

to y = 7 are empty sums, i.e.

E2, E. and E8 are empty sums.

We have, since, from (2.22), 0(a, 0)=aimplies a=a,Similarly,

a and 0(b, a) = /~, since 0(~c,!/) ~ maxfx, yl, it follows that
If 0(a,,8)aanda3 so that a ,a.Since
i is assumed it follows that
/3
=

=

572

Now,

If 8(a, 7)

=

a and

0(b, a) = /3, from (2.24) and (2.25) we have

0 and hence a = b = 0 so that q = a and athis is not the case, we obtainso that0(a, b)=={3. Since

From (3.9), (3.12), Remark 3.13, and (3.14)-(3.18), we obtain

Since 0(a, P) = 0(,8, a), interchanging the roles of a and ,Q in (3.19) we
obtainEquating (3.19) and (3.20) and cancelling the like terms we obtain

By Remark 3.13, the sum on the right hand side of (3.21) is an empty sum
Using this in (3.19), we getso that its value is zero.

In a similar way we can prove that

573

Weshallnowfind f-1(p«), f-1(p,e) and f-1(p7). Observing that, by (2.22),
a and 0athe conditions 9(x,y)
y10, al so that, from
(3-8), x E {{3, ’Y}, we have from (2.26)=

Also

From (2.22), the conditions imposed in the above sum imply that x rt. (0, # ) .
In order that the sum does not contain terms which become zero the only
possible choices are x = y or x = a. However, if x = ~y, by Remark 3.13
the sum corresponding to this choice will be empty. Hence

Similarly,
Making use of (3.24)-(3.26) in (3.22) and (3.23) and equating the two results
we obtain

We now take f(P7)the terms,

=

0 and

=

1 in (3.27). We get after transposing

To obtain a contradiction from (3.28) we require the following:

574

Lemma 3.29. Let À, it be two distinct positive integers and cp a multiplicative functions defined bycp(pa) and WV) being kept arbitrary. We set=p.

Then, if we define±pl for some positive integer r.

Proof. If F(A, p) is an empty sum its value is zero. We may assume that=
it is a non empty sum. In this case we show that+p~. We haveby (2.26)

a
In the inner most sum on the right hand side of (3.30), 0
al
p.
From
the
definitionofHence al 54 p.
the function cp, the only possible
choice of a¡ in order that this sum contains a non-zero term is a, = A, so
that

Clearly,

We note that Kl 0 0 implies that the sum defining K2 is an empty sum sothat K2 = 0. For, let K, = p so that p8(À, A ) . If K2 is not an empty
sum, we would have aO(A, bl), it 0(a, a), 0 b1 a, so that==and hence bl=

0 then K2 = 0 and hence
0. But b1 &#x3E; 0. Therefore if=
F(A, it) = -Kl - K2 -p. So let Kl 0 so that F(A, p) -K2. Again
=

=

=

575

by (2.26),

the inner most sum on the right

As before we can assume that
hand side above. So, we get

where

and

We write B28(À, A), A3 8(’B, ~2), A4 8(’B, As) and so on.
As before we note that K3 # 0 implies K4 is an empty sum so that
=

=

=

If K3
K4. For a given a with 0(a, A) = J.L, this procedure can0, F(a,not be continued indefinitely since we obtain a strictly decreasing sequenceof positive integers with ~&#x3E;a&#x3E;&#x26;i&#x3E;&#x26;2&#x3E;-"&#x3E;0. Hence for someandpositive integer r(a) we get that
=

=

576

Since O(A, A) 8(À, p) implies A = p (which has been proved earlier), Ak =ae implies k E. Hence there can be at most one term in the above sumDwhich completes the proof of Lemma 3.29.
so that F(A, p) ===We now return to the equation (3.28). We apply Lemma 3.29 with A == ,8,= 0 inpa, cpf (on noting that we have chosen(3.8)) so that= 0 or
for
some
r.
If
positive integer:i:(f(¡l))r
F(,Q, a) = 0, (3.28)
F(#, a)reduces to==Since the coefficient of f (p~) is positive in the above equation, we canwhich does not satisfy the equation (3.31). If
easily assign a value toweobtainfrom (3.28)
F(~i, a) :!:(f(pp)r,=a) ~ 0 implies N(,Q, a) &#x3E; 0, we can regard equation (3.32) as a
If f (p,8) is chosen
polynomial equation of degree r+2 in the variablein such a way that
equals none of the r+2 roots of the equation (3.32),
Since

we obtain a contradiction.

This completes the proof of Theorem 3.1.

0

4. Theorem 4.1

In this section we prove

Theorem 4.1. (With the hypothesis of Theorem 3.1). If a,,8,, and 6 are
non-negative integers such thatthen either a = 0(1, c) for some c &#x3E; 0 or ,~=d) for some d &#x3E; 0.

Proof. Most of the arguments we shall use in this proof were already usedin the proof of Theorem 3.1, so that we shall omit some details.
We first prove that if a and (3 are non-negative integers, then

(4.1)

0 (a, a)=~(/3, ~i) implies a=/3.

We may assume that a and ,Q are positive integers. Let a =1=,8. We deduce
a contradiction. Define the multiplicative function f by

577

First, we observe from (2.22) and (2.23) that 0(a, a) 0(,8, P) 0 10,,Bl
a in (2.19) and f as in (4.2), we
so that f(p8(a,a») = 0. Taking A = p==obtainwhere

We have, as in (3.12)

If 0(c, #)=a for some c, we have from

(2.23) and (2.24)(~ is assumed,
6(c, c) 0 and hence c = 0 so that a {3. Since
it follows that 0(c, #) = a has no solutions. Hence each sum on the righthand side of (4.4) is empty so that ~(a, a) = 0. Using this in (4.3), we
obtainso that==

We shall now evaluate f -1 (pe(«, «)~ directly from (2.26). We obtainIn the above sum, from (4.2), the non-zero terms correspond to x = 0 or
x = ~3; but, from (2.22), x = 0 and 9(x,y) = 0(a, a) implies y = 0(a, a)which is forbidden, so/3. By Theorem 3.1, x (3, 9(x, y) _Hence9(a, a) 0 (,8, (3) implies y /3.
==

Again by (2.26),

=

578

since, in the above sum, the non-zero terms correspond to x E 10, 01 which
are forbidden.

Hence from (4.7),(4.5) and (4.8) contradict each other. This, proves (4.1).
Suppose now that 0:, (3, T and J are non-negative integers such that
0 (a, (3) 9(7, 6). We prove that=

(4.9) easily follows if
positive integers.
Let a~i. If q

0. So we may assume that a,,Q, q and 6 are

J, then we have 0 (a, a) = 9(~, q) so that by (4.1),
Hence a = 0(y, 0). Let 77~. We now have=a = q.

==In this case we must prove (4.9) with (3 = a. First, we observe that, if 6
can be written 6 = 0(a, d’), then from (2.24) and (2.25),

which implies by Theorem 3.1, a8(~y, d’) so that (4.9) is satisfied. The
if
we
assumesame conclusion holds,
that, = 0(a, c’). Thus, if (4.9) does
not hold, none of the equations=

is solvable. We now define the multiplicative function f by

the values f(Po.), f(P1) and f (pb) are left to be chosen. Taking A = /-tin (2.19) with f as in (4.11), we obtain

=

a

where

By our assumption on the solvability of the equations mentioned above, inthe sum on the right hand side of E (a, a), x or y can not be either -y or
J. The allowable choice is (x, y) = (0, ) which is again not possible since(~!/) 7~ (0, ). Hence ¿(o:,a) 0. Using this in (4.12) we obtain=

579

By (2.26), (4.11) and by Theorem 3.1,

In the above sum, if x is equal to a, q, 6 respectively, then from Theorem 3.1,
y is equal to a, 6, 1. Further, these three values of y satisfy 0y
9 (a, a) .
For instance, y :5 ~(7?~) = 0(a, a) and ’Y = 0(a, a) is forbidden since1
8(a, c) is not solvable. Therefore=

By (2.26) and by our assumption,

since "f=0(6, y) &#x3E; y implies yq, and with (2.22), y 0 -y. Similarly,

Making use of (4.15)-(4.17) in (4.14), we obtain after simplification,

Comparing (4.13) and (4.18), we obtain
We choose f (pa)0. Not both the sums F(,,8) and F(8, 7) can be nonIf
is1.
empty. F(-y, 8) non-empty, so that0, we choose
=

=

=

580

0 or tl. In this case, (4.19) reduces to
As in Lemma 3.29, F(y, 6)
0 2 f (~b) + F(y, J), so that by choosing~), we obtain acontradiction. If F(b, 7) is a non-empty sum (so that
6) 0) a similar
contradiction can be obtained. If both8) and F(b, ’1) are empty sums,
Thus thetoa contradiction.
chooseobtainwe0
simplyaresults
inThus
contradiction
inallcases.
(4.9) is true
equation (4.19)=

f (pa) ~ -2F(y,==We now complete the remaining cases. We can assume now that 0(a, ,8)8(’1, 6), where a =1= (3. If y = b, we obtain 8(",) 0 (a, (3). Henceby the previous considerations, we have ’1 0 (a, c) or y 0 (p, c). So wehave a = 8(’Y,¿) or (36(y, d’) 0(6, d’), by Theorem 3.1. So we may
assume that a ~ (3 and ’1 =I 8. If aq, then a0(,y, 0). If a J
then (by Theorem 3.1) (3 y. In this case we must show that a 9(,8, c)or ,Q0(a, d). If these two possibilities do not occur, we may take themultiplicative function f defined by
=

=

=

=

=

=

=

=

=

=

=

=

By (2.19) with this f , we can show thatof (2.26) we obtain that=

1 and making use

2 leading to an absurdity.
a, #, q, 6 are pairwise distinct positive
Finally,with
0 (a, #) 9(" 8). Suppose that the conclusion of Theorem 4.1
integers
does not hold; this together with Theorem 3.1 implies that none of the four=

we can now assume that=

equations
is solvable.
Let f be the multiplicative function defined byandremain to be fixed. Since
the values offrom (2.19), with f as in (4.21) we obtain

where

=

f(¡l)

=

0,

581

In the sum on the right hand side of E (a, 0), by the assumption made on
6 are notthe solvability of equations in (4.20), the choices x = 7 or yHencepermissible.
=sothat, as in (3.12)

Remark 4.24. We may note here that ~(c~/3) B(-y, b) and Theorem 3.1
imply that 0(b, q) =,8 if and only if 0(b, a) J. In particular N(7"Q) _N(a, b) and F(y, /3) F(a, J). Similarly ~(a, /3) = 7 if and only if 0(a, J) =a. Thus N(,Q, 7)
N (6, a) and F(~i, ’Y) F(6, a).
=

=

=

=

=

Now by Remark 4.24,

and

0 and hence ab = 0, so that 6 = a and q = /3. This is
not possible since a, ~3, -y and 6 are pairwise distinct. It follows thatso that0(a, b)==Making use of (4.25)-(4.27) in (4.23) and replacing the resulting value of¿(0:,{3) in (4.22) we obtain
From (2.26), we can show that

582

and

Making use of (4.29) and (4.30) in (4.28), after simplification we obtain

Now, by (2.26), by Theorem 3.1 and since

=

f V)

=

0, we obtain

Making use of (4.29) and (4.30) in (4.32) we obtain after simplification
Equating (4.31) and (4.33) and transposing the terms, we get

By Theorem 3.1, it is clear that the N-functions take the value 0 or 1.
Also, N(a, J) and N(6, a) can not be simultaneously unity. It follows thatis positive. If the sum F(y, 6) is non-empty, so
the coefficients ofwechoosethat F(6, q)0,f(p7) 1. Applying Lemma 3.29 with A y,f (on noting that f (pcl) f(pP) 0) yields F(7, ð) 0, +1.
p _ ~, cp
=

=

=

=

=

=

=

In this case (4.34) reduces tof (pb)

not satisfying (4.35). A similar contradiction
It is easy to chooseif
can be obtainedF(b, y) is non-empty. If F(~y, 8) and F(8, y) are bothThis obviously gives aempty sums, (4.34) reduces to 0 =contradiction if
0. This completes the proof of Theorem 4.1.
D

5. Final results

From Lemma 2.17, Theorems 3.1 and 4.1 and the definition given in(2.13), we obtain

Theorem 5.1. If 1/1 preserves multiplicativity,y) &#x3E; max(z, yl for allisinvertible
with respect to 1/1 andeachE
multiplicative function(x, y) T,
the identical equation (1.4) holds for all multiplicative functions, then V) is
a Lehmer-Narkiewicz convolution.

583

It may be noted that the multiplicativity preserving property of ’0 is
not a necessary condition for the validity of (1.4). For example, if T == n for all n E
andE
Z+, then

1 (1, n), (n, 1) : n(1.4) holds trivially for all arithmetic functions f with f (1)=

1. However,~ does not preserve multiplicativity.
References[1] E. COHEN, Arithmetical functions associated with the unitary divisors of an integer. Math.
Z. 74 (1960), 66-80.
[2] D. H. LEHMER, Arithmetic of double series. Trans. Amer. Math. Soc. 33 (1931), 945-957.
[3] W. NARKIEWICZ, On a class of arithmetical convolutions. Colloq. Math. 10 (1963), 81-94.
[4] V. SITARAMAIAH, On the 03C8-product of D. H. Lehmer. Indian J. Pure and Appl. Math. 16(1985), 994-1008.
[5] V. SITARAMAIAH, On the existence of unity in Lehmer’s 03C8-product ring. Indian J. Pure andAppl. Math. 20 (1989), 1184-1190.
[6] V. SITARAMAIAH, M. V. SUBBARAO, On a class of 03C8-products preserving multiplicativity.
Indian J. Pure and Appl. Math. 22 (1991), 819-832.
[7] V. SITARAMAIAH, M. V. SUBBARAO, The identical equation in 03C8-products. Proc. Amer.
Math. Soc. 124 (1996), 361-369.
[8] V. SITARAMAIAH, M. V. SUBBARAO, On regular 03C8-convolutions. J. Indian Math. Soc. 64(1997), 131-150.
[9] R. VAIDYANATHASWAMY, The identical equation of the multiplicative functions. Bull. Amer.
Math. Soc. 36 (1930), 762-772.
[10] R. VAIDYANATHASWAMY, The theory of multiplicative arithmetic functions. Trans. Amer.
Math. Soc. 33 (1931), 579-662. (=[11], 326-414.)[11] R. VAIDYANATHASWAMY, The collected papers of Prof. R. Vaidyanathaswamy. Madras University, 1957.
Jean-Louis NICOLAS
Institut Girard Desargues
Mathématiques, Bit. 101
U niversité Claude Bernard (Lyon 1)
69622 Villeurbanne cédex, France.

E-mail: jlnicolaGin2p3.fr
Varanasi SITARAMAIAH
Department of Mathematics

Pondicherry Engineering College
Pondicherry 605014, India
E-mail : vsrmCsatyam.net.in