I nternat. J. Math. & Math. Sci.
(1978) 75-86
Vol
75
COMPLETE RESIDUE SYSTEMS_
IN THE QUADRATIC DOMAIN Z (e 211"i/3)
GERALD E. BERGUM
Department of Mathematics
South Dakota State University
Brookings, South Dakota 57006
(Received April 30, 1976)
ABSTRACT.
Several representations for a complete residue system in the
Euclidean domain Z() are presented in this paper.
i.
INTRODUCTION.
Throughout this paper, small case Latin letters with the exception of e and i
will represent rational integers.
The Latin letters e and i respectively repre-
sent the base for the natural logarithms and the imaginary unit.
m
e
2i/3 and Z(m)
{a
+ bmla,b
e Z}.
We let
The Greek letters e, 8, Y, 6, o, and
will always represent integers in Z().
We will illustrate the integers in Z() by the lattice points in a Cartesian
coordinate system formed by the intersections of the lines through the points
(x,0), x a real integer, and making angles of 60
or 120
with the x-axis.
This system is composed of equilateral-triangles.
In Uspensky and Heaslet (4), we find many representations for a complete
residue system modulo n.
Two of the most well-known complete residue systems
76
G.E. BERGUM
molulo n are the set of integers
n/2
{x
< x
<_ n/2}.
{0, i, 2, --.,n-l} and the
set of integers
The latter representation is the one with least absolute
values.
Jordan and Portratz (2) exhibit several representations for a complete residue system in the Gaussian integers and in Potratz (3) we find several represenrations for a complete residue system in the quadratic Euclidean domain
{a + b/Z
Z(//)
la,b
e Z}.
It is the purpose of this paper to exhibit several
representations for a complete residue system in the Euclidean domain Z(m).
We say that
-=
8(mod y)
el8 iff there
iffyl (e
e.
exists a 6 such that 8
Furthermore,
8). It is a trivial matter to show that congruence modulo
y is an equivalence relation on Z(m) and hence, as in the real case, it is
reasonable to define a complete residue system modulo y as a nonempty collection
S of elements in Z(m) such that (i) no two elements of S are congruent modulo
and (2) every element of Z(m) not in S is congruent to some element in S.
complete residue system modulo y is abbreviated as C.R.S. (mod y).
norm of y, denoted by N(y), as
a2
N()
2.
ab
I(I 2.
N()
If
a
+
b then
-
The first representation of a C.R.S. (mod y) appears to
be a natural generalization of {0, i, 2,-..,n-l} modulo n.
d- (a, b), and y
THEORE 2.i
2
d(a
is a C.R.S. (mod y).
el’ 2
I +
d.
bl
If d is even, T
1
i /i
1
{(x + )
+ (y
PROOF.
We define the
+ b 2.
REPRESENTATION I.
and T
+-)
I0
{x +
<_
x <
y/i
dll
2
I0
< x < d
<_
i, 0
y
Let y
ll
<_
(See Figure i).
d
a
el
e2 (md Y) then
there exists
a
2
+
b2
+ bm,
2
-i,
2}
0iYi
then T
It is a trivial matter to show that T is a subset of Z(m).
e T and
A
such that
},
TIUT 2
Suppose
-
If a
so that
COMPLETE RESIDUE SYSTEMS
x
I
I +
yl/i
(Yl Y2 )"
and a
x
2
ly I
But,
2
+
y21
y2/i
<
then
Yl
so that
x2 and
1
If
i
and a
If
l
and
d
(alb2
Y2
Yl
Since the smallest real number that y divides is d
we have x
77
l
2
+
and x
Y2
x
I
and
bla2 blb2)
y),
2(nod
IxI x21
< d
l
2
2"
1
2 are both in T 2 the same argument will again show that
I Xl + Yl/I and 2 (x2 + )
(Yl- (Y2 + -)) which is impossible since 21d.
2
are such that
then d/2 divides
1
+
al
2"
(Y2 + )
Hence, no two
distinct elements of T are congruent modulo y.
Let a
d
Since
0 < n
n
2
I
dql +
ql and r I such that y
Find
r
< d.
I where 0 <_ r I
(a, b), there exists u and v such that au + bv
q2 and n 2 such that x
find
x
+ y.
x
< d
l
i
2
If r
au
av
2n
I
+
n
au
I
av
+
bu
d
lvl
2
dql.
q2 +
n2
If r
where
<
I + i find q2 and n 2 such that 0 n 2
bu
dlvl 2 q2 + n2"
When r
2nl,
I
2nI
I
< d
l
2
where
we flnd that
=x+y
dll12q2 +
n
2
+
n
li
(v + u(1 + (o))y + n2 +
(nod y)
so that a is congruent to an element of T
nl’’i
On the other hand, if r
I
I.
2n
I + I,
we find that
a=x+y
d]12q2 +
-=
i
(v + u(l + ))y + (n 2 + )
+ (nI
i
i /i
+ (n I + )
(mod Y
(n 2 +)
so that a is congruent to an element of T
+_i
In either case,
2.
is congruent to
an element of T and T is a C.R.S. (nod y).
THEOREM 2.2.
and
r2
If d is odd, T
I
{x +
i
{(x + )
+ (y + i
is a C.R.S. (nod y).
y/il0
)/i[0 < x < dll
(See Figure 2).
2
< x <
d[l
I, 0 <y
2
<---3}
d
i, 0 < y <
then T
dl}
TIUT 2
78
G.E. B ERGUM
The proof of Theorem 2.2 is very similar to that of Theorem 2.1 and hence
Furthermore, examining the results of Theorems 2.1 and 2.2,
has been omitted.
we see that the following is true.
171
The cardinallty of a C.R.S. (mod y) is
COROLLARY 2.1.
+
2
Let T
of points
I be the collection
)7/2,
inside the rhombus ABCD whose vertices are respectively (i + )7/2, (i
3.
Let y
REPRESENTATION II.
a
(-i -)7/2, and (-i + )7/2.
b.
Let T 2 be the collection of points on the half-
open line segments (+(-i + )7/2, (-I
THEOREM 3.1.
Let T
PROOF.
--If e
I
a
el
--+
i
+
TIUT 2
)7/2).
then T is a C.R.S. (mod 7).
(See Figures 3, 4,
5).
7
I
s
rl,
.
function then C
and
1 <
S
+
Sl) 7
i
<
1
e
1
+
+
N(7)
2
[D I +
I
then
bl
(ala- alb + blb)/N
Let CI
C
I +
(Y), D I
21--],
S
DI el/y
Hence, e I
1
DI
# 0, and CASE 3.
b
a
+
b
Sl)7
is equal
to-1/2.
1
I
[C I + ],
(rood y).
a
# 0, 2a
+
# 0, 2a
b
1
(-, )
and (R
1
+
1
# 0, CASE 2.
b
+
SI)7
Si)7
is in T
is in T
2
I
if R
I
s
I
+
b
if eigher
S
I and I
P1
or S
1
Hence, every element of Z() is congruent to some element of T.
Using the equations for the sides of ABCD, it can be shown that if e
then r
a
it can be shown by using the
0:
b
equations for the sides of the rhombus that (R
are in the open interval
rI
s
For each of the following; CASE I.
2a
alb)/N(7),
(ab I
i
+
N(7)
is the greatest integer
I where
i
i <
R <
(r I + Sl) + (RI + SI) where
I
(r I + Sl)7 + ( + Sl)7 so that
(R1 +
Z() and e I
+
0.
I
e T
0,
COMPLETE RESIDUE SYSTEMS
Let
such that
where-
i
a
and
b
3
R2
0 or
3
i
a
However,
2
al
i, 2 so
for j
and S
3
Ul
S
I
b
2
(rood 7) then there exists
a
3
+
b3
( Sl)7
2 (R2 + S2)7
that (
R2) + (S I S2) a 3 + b3.
+
and
Considering the possible values for
3.
i, 2, and using the fact that a 3 and b are integers, we have
3
for j
Sj
<
Rj, Sj
Therefore, RI
Rj
+
Ul/7 u2/7
<_
.
e T be such that
al, a2
79
Ul
u2
so that T is a C.R.S. (rood 7).
It is interesting to observe when T 2 is empty.
To do this, we first note
that the following is true.
THEOREM 3.2.
If 7
+
a
b then
21N(7)
iff
217.
We are now able to show
THEOREM 3.3.
PROOF.
--Let
on CD then N(7)
CASE 2.
a
whereas
Ul
2a
if
0.
b
al
+
b
a
+b
and T
TIUT 2.
The set T
2
is empty iff
al + bl" CASE I. a + b # 0 and 2a b # 0.
2(alb ala blb) while N(7) 2(alb ab I) if"u I
2
b # 0. If
0 and 2a
i is on CD then N(7) 3a
i
If
27.
al
is
is on BC.
2a(b
I
2a
I)
2(alb abl). CASE 3. a + b # 0 and
i is on CD then N(7) 2(alb ala blb) and N(7) 2a(2aI b I)
implies that 217.
Therefore, T 2 #
it is easy to see that if 217 then the vertex C of the rhombus
on BC implies that N(7)
If
is on BC.
Conversely,
is a point of T
4.
Let 7
2 since the value of C I in Theorem 3.1 is
REPRESENTATION III.
-.
i
In Hardman and Jordan (i) and Potratz (3) we find a
discussion of a "better" or "absolute minimal representation
of a C.R.S. (mod 7).
For consistency, we have
DFINITION 4.1.
A representation T of a C.R.S. (rood 7) is said to be an
absolute ml representation iff for any representation R of a C.R.S. (mod 7),
we have
aT
R
80
G.E. BERGUM
It is the purpose of this section to exhibit a representation of a C.R.S.
As before,
(rood ) which is a "better" or "absolute minimal representation".
we let
a
+
b.
Let TI be the set of points interior to the hexagon ABCDEF whose vertices
ki/3 where i < k
are given respectively by
)e
_< 6. Let T 2 be the set of
3(i
points on the llne segments [-
3(i
)e 5i/3] and
THEOREM 4.1.
),
(i
)e 5i/3
[3(i
(i
)ei/3],
(i
m)).
(i
)e
TIOT 2.
Let T
(See Figures
The set T described above is a C.R.S. (mod y).
6,7,8,9).
Let a I
PROOF.
a
I +
In Theorem 3.1, it was shown that there exist
bl.
integers r I and s I together with rationals R1 and S 1 such that
(r I +
i/Y
where-i
<_
CASE i.
-i <
CASE 3.
0 <
2R
I
< i and-i
<_
2S
< i.
I
Sl)
+ (R1 +
Sl)
Consider the follwoing cases:
2 < 0 and -I < 2SI < 0, CASE 2. -i
2 < i and -i <_ 2S I _< 0, and CASE 4.
< 2R
I
0 <
< 0 and 0 < 2S
2
< i,
I
< i and 0 < 2S
I
<
I.
It can be shown in each case that there exist integers r and s together with
rationals R and S such that
(r + s) + (R + S)
Cl/’
where (i) -i
<_ R +
S < i, (2) -i < R
2S
<_
i, and (3) -i < S
shall say that a number in this form is in standard form.
is an element of Z() and that a
I
# 0, and a # b, CASE 2.
CASE 4.
a
# 0, b # 0, and
<_
We
i.
Note that (R
+ S)y
(R + S)y (mod ).
Let us now consider the following possibilities for y:
b
2R
a
0 and b # 0, CASE 3.
a
b.
a
CASE i.
# 0 and b
a
# 0,
0, and
As in the proof of Theorem 3.1, if we use the
equations for the sides of the hexagon together with the restrictions placed on
a number a
case that (R + S)y e T and in
I in standard form we find in each
COMPLETE RESIDUE SYSTEMS
particular that it is on DE if R
R- 2S
Let
r
i
+
s
+
-i, on CD if S
Sle
T in standard form.
+
Since e
0, or r + s
s
i.
is between BC and EF while s
I is between AB and DE, we have
2r equals I, 0, or -i since e is between CD
I
Examining the twenty-seven possibilities, it is easy to see that all
Hence, r
0 and e
s
Suppose
2
el
2s equals -I, 0, or i since
Similarly, r
cases lead to a contradiction except where r
If
+
el
e2
+i +
is not solvable in T for these
el
0, and r
(R + S)y and
el
(mod y) then there exists a 6 such that
e2 + Y6
s
0.
2s
(R + S)y.
I
e T in standard form where
el, e2
0, 2r
s
standard form restrictions, it can be shown that 6
However,
i, and on EF if
2R
Therefore, every element of z() is congruent to some element of T.
i.
-i, r
and FA.
S
81
’s
Y6"
,
+,
(U + V)y.
e2
Using the
-+i, or 0.
0; therefore,
unless 6
any two distinct elements are incongruent modulo y and T is a C.R.S. modulo y.
In a paper to follow, the author will investigate necessary and sufficient
conditions for the boundary of the complete residue system to be empty.
If -I < a < 1 or -i < a < i and r is any integer then 0 < r 2
LEMMA 4.1.
+
The proof of Lemma 4.1 is straight forward and hence the details have been
omitted.
LEMMA 4.2.
PROOF.
therefore
Let
e T then
Let /y
8/Y
IBIYI
R
+
el
<
181
for all
Now,
S be in standard form.
s
+ y for
(R + S) + (c + d) where c and d are integers.
[(R +
c) 2
Rc
cS
d)2]
(R + c)(S + d) + (S +
+
--ID-I-l/y12]
c2
1/2
cd
+
2Sd
Rd
+
d2
+
1/2
Hence,
some
ar.
G.E. BEEGUM
82
d then D
Suppose c
I
then-c
Rd
D
d2
+ d(2S
R- c)
D
d2
+ d(2S
R) > 0.
+
c2
+
Sd
d2
d2
c < -d or -d 2 >
R- c < 1
< 2S
+
+ c(2R- S)
> 0.
+ d(R + S)
(2S
> 0.
R- c)d.
If c > d and c
If c > d and c > 0 then -c < -d
or (2R- S
d)c + c 2 > 0.
Similarly, D
>_
Hence, D
c2
0 if c < d and either c
+ c(2R-
S
If c > d and c < 0
Therefore,
0 then we have
i < 2R- S
d) + d 2 + d(2S
0, c > 0 or c < 0.
d < 1
R) > 0.
Hence,
That T is an absolute minimal representation of a C.R.S. (rood y) is an
immediate consequence of Lemma 4.2.
REFERENCES
I.
Hardman, N. R. and J. H. Jordan. A Minimum Problem Connected with Complete
Residue Systems in the Gaussian Integers, Amer. Math. Monthly 74 (1967)
559-561, No. 5.
2.
Jordan, J. H. and C. J. Potratz. Complete Residue Systems in the Gaussian
Integers, Math. Mag. 38 (1965) 1-12.
3.
Potratz, C. J.
4.
Uspensky, J. V. and M. A. Heaslet.
New York, 1939.
Character Sums in Z(/)/(p).
tion, Washington State University, 1966.
Unpublished Ph.D. disserta-
Elementary Number Theory, McGraw-Hill,
KEF ORDS AND PHRASES. Complete Ridue Systems, Repre of Complete
Ridue
AMS(MOS) SUBJECT CLASSIFICATIONS (1970).
12F05.
d
COMPLETE RESIDUE SYSTEMS
FIG. 1 C.E.S. (d 4
+ 8)
FIG. 2 C.R.S. (mod 5
FIG. 3 C.R.S. (rood 4
+ 60)
+ i0)
83
G.E. BERGUM
84
FIG. 6 C.R.S. (mod 5
+ 10o)
COMPLETE ISIDUE SYSTtS
.
--
.
X
,To
FIG. 8 C.R.S. (mod 7)
X
FIG. 9 C.R.S. (rood 5
+ 5)
85