Research Article
R-Regular Integers Modulo
n
r
M. GaneshwarRaoaaChaitanya Bharathi Institute of Technology, Gandipet, Hyderabad, Telangana
Article History: Received: 11 January 2021; Accepted: 27 February 2021; Published online: 5 April 2021
_____________________________________________________________________________________________ Abstract: Introducing the notion of ar-regular integer modulo
r
n
we obtain some basic properties of such integers andarithmetic properties of certain functions related to them.
Keywords: r-regular integer modulo
,
r
n
unitary divisor, r-free integer, r-gcd of two integers___________________________________________________________________________
1. Introduction
Let r be a fixed positive integer. A positive integer a is said to be r-regular modulo
n
rif there is an integer xsuch that
(
)
1
mod
.
r r r
a
+x
a
n
The case
r =
1
gives the notion of aregular integer moduleon, introduced by (Morgado, J, 1972; Morgado J , 1974) who made an investigation of their properties.Clearly
a =
0
is r-regular modulor
n
for everyn
1.
Also if(
mod
)
r
a
b
n
then a and b are r-regular modulo
n
rsimultaneously. Further, if a and b are r-regular modulon
rthen so is ab.For positive integers a and b their greatest rth power common divisor is denoted by
(
a b
,
)
rand is called ther-gcd of a and b. Note that
(
a b
,
) (
1=
a b
,
)
,
the gcd ofa
and .
b
We recall the notions given in (McCarthy, 1985):A complete set of residues modulo
n
ris called a( )
n r
,
-residue system. ,
:1
r n r
C
=
a
a
n
is the
minimal
( )
n r
,
-residue system.The set of all a in an( )
n r
,
-residue system such that(
)
,
r1
r
a n
=
is called a
reduced
( )
n r
,
-residue system. ,
,(
)
:
,
r1
n r n r
r
R
=
a
C
a n
=
is the minimal reduced
( )
n r
,
-residue system.(V.L.Klee, 1948) defined a generalization
r of the Euler’s function by( )
#
:1
and
(
,
)
1
r
n
a
a
n
a n
r
=
=
and proved that
( )
( )
. ,
r r d nn
n
d
d
=
--- (1)Where
r is the r-analogue of the Mobius function
given by( ) ( )
(
1 2)
1 21 if
1
1 if
...
where
...
are primes
0 otherwise
r t r t tn
n
n
p p
p
p
p
p
=
= −
=
--- (2)Note that
1=
and that
r( )
n
=
#
R
n r,.
Let
Reg
( )
,: is r-regular modulo
r r
n
=
a
C
n ra
n
and( )
# Reg
( )
.
r rn
rn
=
Observe that any
a
R
n r, is inReg
r( )
n
.
In fact, ifa
R
n r, then(
)
,
r1
ra n
=
so that(
,
)
1
ra n
=
and therefore there is an integer
x
0 such that 01 mod
(
)
r
a x
n
which gives(
)
1 0mod
r r ra
+x
a
n
showing
a
Reg
r( )
n
.
Hence( ) ( )
r r r
r
n
rn
n
for every
n
1,
with( )
r r
r
n
n
=
if and only if n is squarefree.
Recently (Laszlo Toth, 2008; Yokesh, T.L., 2020) has studied several properties of the function
( )
n
:
1( )
n
.
=
In this paper we prove some basic properties of the integers in the set
Reg
r( )
n
and certain arithmetic properties of the function(
)
r r
n
2. Integers in Regr(n)
In all that follows
n
1
be of the canonical form:1 2
1 2
...
t t,
n
=
p
p
p
where
p
1
p
2
...
p
t are primes and
i are integers
1.
Theorem 1.For an integer
a
1
the following are equivalent:1.1
a
Reg
r( )
n
1.2 for every
i
1, 2, ...,
t
we have eitherp
i
|
a
orir r i
p
a
1.3(
)
,
r r,
ra n
n
(
d m
means thatd m
and,
m
1,
d
d
=
in which case d is called a unitary divisor ofm)
1.4
( )
r(
mod
)
r n r r ra
+
a
n
1.5 There is an integer
k
1
such that(
mod
)
.
k r r r
a
+
a
n
Proof: Suppose
a
Reg
r( )
n
so that(
)
1
0
mod
r r r
a
+x
a
n
for some integer
x
0.
Therefore for eachi
(
1
i
t
)
,
p
a
(
ax
0−
1
)
.
r r i i Since(
a ax −
,
01
)
=
1
we have(
,
01
)
1,
ra
ax −
=
we have eithera
p
i
|
or r ip a
for each
i
,
and in the latter case it follows.
ir r
i
p
a
Thus
(i)
(ii).
Assume (ii). That is, a is an integer
1
such that eitherp
i
|
a
or.
ir r
i
p
a
We have to show
a
Reg
r( )
n
.
In casep
i|
a
then(
,
)
1
ir
i
a p
=
so that there is an integer
x
i with1 mod
(
)
ir i i
a x
p
and hence(
)
1mod
ir r r i ia
+x
a
p
.R-Regular Integers Modulo
n
In case
ir r
i
p
a
then for any integer
x
,(
)
r i r r
x
a
p
ia
+1
mod
holds. Thus(
)
1mod
ir r r ia
+x
a
p
is solvable for1 i
t
and hence(
)
r t r r r r
x
a
p
p
p
ta
mod
1.
2....
2 1 1
+is solvable, showing
a
Reg
r( )
n
.
Thus(ii)
(i).
Note that (ii) holds 0
.
,
r r
a
a d
=
where i i r r i p ad
=
p
and(
a
0,
n =
)
1
(
a
r,
n
r)
d
r,
=
which is a unitary divisor of
n
r(
,
r)
r r,
ra n
d
n
=
since(
) (
)
,
,
.
r r r ra
n
=
a n
Thus
(ii)
(iii).
(ii)
(iv).
Ifp
i ira
r then( )
(
)
mod
r r n r r ra
+
a
n
is obvious. If
p
i
|
a
, then by Euler-FermatTheorem,
( )
ir1 mod
(
)
i i p r ia
p
so that( )
( )
( )
( )
(
)
1 mod
,
r r i r i r r i i r i n p p n r ia
a
p
=
since( )
( )
( ) ( ) ( )
( )
( ) ( )
( )
r i r i r i j r j r r i r t r r r r r r i r r i i j i t ip
p
p
p
p
p
p
p
n
m
=
=
=
=...
:
2 1 2 1(
1)
1
p
i...
p
ir−.
= +
+ +
Mwhere( )
==
i j r j r jp
M
so that mis an integer.
Thus
( )
r(
mod
)
r n r r ir i
a
+
a
p
for
1
i
t
,
giving (iv)(iv)
(i).
If( )
(
mod
)
r r n r r r
a
+
a
n
then(
)
1 0.
mod
r r ra
+x
a
n
where( )
1 0 r r nx
=
a
− showinga
Reg
r( )
n
.
(iv)
(v)
is immediate withk
=
r( )
n
r.
Also if
(
mod
)
k r r r
a
+
a
n
for some
k
1
implies(
)
1 0.
mod
,
r r ra
+x
a
n
where 1 0,
kx
=
a
−showing
a
Reg
r( )
n
.
Thus(v)
(i).
3. The Function
ρ n .
r( )
rIn this section we study the function
( )
r
r
n
and its relation with
( )
.
r
r
n
Also we express the sum
S n
r( )
of the r-regular integers modulon
rin terms of( )
r r
n
Theorem 2: For every
n
1,
( )
( )
.
r r r r d nn
d
=
The function( )
r rn
is multiplicative and( )
(
1)
1,
r r r rp
p
p
=
−
−+
for any prime pand integer
1.
Proof: We give two proofs for the first part.
First Proof: Let
a
Reg
r( )
n
.
If
p
i
|
a
for1 i
t
then(
a n =
,
)
1
so that(
,
) (
,
)
1
r r r
r
a n
=
a
n
=
and the number of such as
( )
r rn
. Suppose ir r ip
a
for exactly one i so that
(
a p
,
j)
=
1
forj
i
and.
ir i
a b p
=
where1
i r r in
b
p
and,
1;
i r r in
b
p
=
the number of such a’s is i.
r r r i
n
p
Suppose ir r ip
a
and jr r jp
a
for
1
i
j
t
;
and fork
i j
,
(
p
k,
a =
)
1.
Then.
ir.
jr,
i ja
=
C p
p
where1
j i r r r i jn
C
p
p
and,
1;
j i r r r i jn
C
p
p
=
and the number of suchintegers is
;
.
j i r r r r i jn
p
p
and so on. Thus( ) ( )
+
+
+
+
=
t r r r i r r t j i r j r i r r t i r i r r r r r r i i j i tp
p
p
n
p
p
n
p
n
n
n
...
...
2 2 1 1 t t j i i j t i iy
y
y
y
y
y
y
y
y
y
...
...
2 1 1 1+
+
+
+
=
Where( )
ir i r iy
=
p
andy
=
y y
1 2... .
y
t Therefore( )
(
11
)(
21 ...
)
(
1
)
r rn
y
y
y
t
=
+
+
+
( )
(
1)
(
(
2)
)
(
( )
)
11
21 ...
t1
r r r rp
rp
rp
t
=
+
+
+
( )
.
r r r r r r r r r d n d nn
d
d
=
=
R-Regular Integers Modulo
n
Second Proof: Groupthe integers
a
C
n r, according to the value( )
,
r r.
ra n
=
d
Note that(
)
,
r r.
r ra n
=
d
=
a
j d
where1
r rn
j
d
and,
1.
r r rn
j
d
=
Hence the number of a’sin
C
n r, with(
)
,
r r ra n
=
d
is r r rn
d
. Thus( )
( )
.
r r r r r r r r r r r d n d nn
n
d
d
=
=
Now( )
( ) ( )
.
,
r r r r r D nn
D
D
=
--- (3)where
r( )
m
=
1
or 0 according as m is therth power of an integer or not. Therefore( )
(
)
( )
,
r r
r
n
r rI
n
=
where
I n
( )
1
for all n and is the unitary convolution of arithmetic functions discussed by (Eckford Cohen, 1960). Since unitary convolution preserves multiplicativity, we get( )
r rn
is multiplicative, because
r,
r and I are all multiplicative.Also
( ) ( )
(
1)
1
r1,
r r r rp
rp
p
p
=
+ =
−
−+
completing the proof of Theorem B.
Theorem 3.
( )
( )
, , 11
.
2
n r r r r r r a C a na
n
n
==
forn
1.
Proof: First observe that for positive integers a and b,
(
a b
,
)
r=
1
if and only if(
a b
,
)
is r-free (Recall that an integer not divisible by the rth power of any prime is said to be r-free). Letq m =
r( )
1
or 0 according as m is r-free or not. Then it is well-known (Apostol, 1998, problem 6, p.47; Ranjeeth 2020) that( )
( )
,
r r t mq
m
=
t
---(4) Where
is the Mobius functionNow, by (4) and (1), we get
( )
(
)
(
)
, 1 , 1.
,
r n r r r r r a C a n a na
a q
a n
==
=
=
r r r r n a n t a s tt
a
1)
(
( )
r r r r r t s n t nt s
t
=
( )
r r r r r n t n s tt t
s
=
( )
1
. .
1
2
r r r r r r r t nn
n
t t
t
t
=
+
( )
( )
2
r r2
r r r r r r t n t nn
n
n
t
t
t
=
+
( )
( )
.
2
r2
r r r r r t n t nn
n
n
t
t
t
=
+
( )
.
,
2
r r rn
n
=
since( )
0
r r t nt
=
forn
1.
Remark 1.The case
r =
1
of Theorem C is the well-known formula:(
)
( )
1 , 12
a n a nn
n
a
==
forn
1.
(For example see (Apostol, 1998, Problem 16, p.48)Theorem 4. If
( )
( ) Reg:
r n r aS
n
a
=
then( )
( )
1
2
1
+
=
r r rn
n
S
forn
1.
Proof: We have, by Theorem A, that
Reg
r( )
(
,
r)
r r.
ra
n
a n
=
d
n
R-Regular Integers Modulo
n
( )
( )
,( )
, , , r r r r r n r r r r n r r a C d n a n d a n n a CS
n
a
a
= =
=
,
1 , ,
= =
r r r r r r r d r n n d d n j C j rj
d
Since(
,
)
.
r r r ra n
=
d
=
a
j d
where1
r rn
j
d
and,
1.
r r rn
j
d
=
Now, in view of Theorem C and Theorem B, for
n
1
we have( )
1
. .
.
2
r r r r r r r r r r r r d n d nn
n
S
n
n
d
d
d
=
+
2
r r r r r r r r r d n d nn
n
n
d
=
+
( )
.
1
2
r r r rn
n
n
=
+
−
( )
1 ,
2
r r rn
n
=
+
proving the theorem.
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Zeitschr. 74, 66-80.
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