Bull. Korean Math. Soc. 50 (2013), No. 1, pp. 233–240
http://dx.doi.org/10.4134/BKMS.2013.50.1.233
ON THE FOURTH POWER MEAN OF GENERALIZED
TWO-TERM EXPONENTIAL SUMS
Tingting Wang
Abstract. In this paper, we use the elementary method and the theory
of complex functions to study the computational problem of the fourth
power mean of the generalized two-term exponential sums, and give two
exact identities for them.
1. Introduction
Let q ≥ 3 be a positive integer. For any integers m and n, the generalized
two-term exponential sum C(m, n, k, χ; q) is defined by
q
X
mak + na
χ(a)e
C(m, n, k, χ; q) =
,
q
a=1
where χ denotes any Dirichlet character mod q, and e(y) = e2πiy .
The various properties of C(m, n, k, χ; q) were investigated by many authors
(see [2, 3, 4, 5] and [7]). For example, from the general result of Pigno and
Pinner [7] we can deduce that
1
|C(m, 0, 2, χ; q)| ≤ 2ω(q) q 2 ,
for (m, q) = 1, where ω(q) denotes the number of all distinct prime divisors
of q. In fact, |C(m, 0, 2, χ; q)| is a multiplicative function of q, and so if q =
αs
1 α2
pα
1 p2 · · · ps denotes the prime power decomposition of q, χ = χ1 χ2 · · · χs ,
αi
i
mi = q/pi , where χi is a multiplicative character mod pα
i , i = 1, 2, . . . , s.
Then for (m, q) = 1, we have
αi
Y
p
s X
s q 2 Y
i
√
mmi a i
= 2ω(q) · q.
χ
(a)e
|C(m, 0, 2, χ; q)| =
2
·
≤
pα
i
αi
i
p
i=1
i
i=1 a=1
Received July 22, 2011; Revised August 30, 2012.
2010 Mathematics Subject Classification. Primary 11L40, 11F20.
Key words and phrases. generalized two-term exponential sums, fourth power mean,
identity.
This work is supported by the N.S.F. (11071194) of P. R. China and G.I.C.F. (YZZ12065)
of NWU.
c
2013
The Korean Mathematical Society
233
234
TINGTING WANG
The case where q is a prime is due to Weil [8]. Zhang Wenpeng [9] proved
that for any odd prime p and integer n with (n, p) = 1, we have
X
4
|C(n, 0, 2, χ; p)|
χ mod p
=
(
h
√ i
p ,
(p − 1) 3p2 − 6p − 1 + 4 np
2
(p − 1)(3p − 6p − 1),
if p ≡ 1 mod 4;
if p ≡ 3 mod 4;
and
X
6
χ mod p
|G(n, 0, 2, χ; p)| = (p − 1)(10p3 − 25p2 − 4p − 1), if p ≡ 3 mod 4,
where np is the Legendre symbol.
The main purpose of this paper is to calculate exact formulae for the sums
p
X
X
4
|C(m, n, 3, χ; p)| ,
| C(m, n, 2, χ; p) |4 and
χ mod p
n=1
and give two computational formulae for them. For the sake of convenience, we
2
p
X
a
√
e
let C(1, p) denotes the quadratic Gauss sums, i.e., C(1, p) =
= p,
p
a=1
√
if p ≡ 1 mod 4; C(1, p) = i p, if p ≡ 3 mod 4, i2 = −1. Under these notations,
we shall prove the following:
Theorem 1. Let p be an odd prime. Then for any integers m and n with
(mn, p) = 1, we have the identity
X
4
|C(m, n, 2, χ; p)|
χ mod p
= 2p3 − 8p2 + 5p + 1
m
−4m · n2
−1
4m · n2
+ 2(p − 1)C(1, p)
e
+
e
,
p
p
p
p
where m
denotes the Legendre’s symbol, and m · m ≡ 1 mod p.
p
Theorem 2. Let p be an odd prime with p 6= 3a + 1. Then for any integer m
with (m, p) = 1, we have the identity
p
X
n=1
|C(m, n, 3, χ; p)|4

 p(2p2 − 3p − 3),
p2 (3p − 7) ,
=
 2
p (2p − 6) ,
if χ is the principal character mod p,
if χ is Legendre symbol mod p,
otherwise.
GENERALIZED TWO-TERM EXPONENTIAL SUMS
235
For general integer q ≥ 3, whether there exists a computational formula for
q
X
X
2r
2r
|C(m, n, k, χ; q)|
|C(m, n, k, χ; q)|
and
χ mod q
n=1
are two open problems, where k, r ≥ 3 are integers.
2. Several lemmas
In this section, we shall give several lemmas, which are necessary in the proof
of our theorems. First we have the following:
Lemma 1. Let p be an odd prime. Then for any integers m and n with
(mn, p) = 1, we have the identity
2
p
X
mb + nb
m
−4mn2
e
=
e
C(1, p),
p
p
p
b=1
where xp denotes the Legendre’s symbol, and mm ≡ 1 mod p.
Proof. Note that (mn, p) = 1, so from the properties of the complete residue
system mod p we have
2
X
p
p
X
mb + nb
m(mnb)2 + n(mnb)
e
e
(1)
=
p
p
b=1
b=1
p
X
mn2 (b2 + b)
e
=
p
b=1
p
X
4mn2 (4b2 + 4b)
e
=
p
b=1


2
p
4mn2 (2b + 1) − 1
X

e
=
p
b=1
!
p
X
4mn2 b2 − 1
=
e
p
b=1
p
4mn2 b2
−4mn2 X
.
e
=e
p
p
b=1
Note that for any integer n with (n, p) = 1, from Theorem 7.5.4 of [6] we
2 p
X
2
nb
n
have C(n, p) =
= m
e
=
C(1, p), 4mn
p
p . Then from (1)
p
p
b=1
we may immediately deduce the identity
2
p
X
mb + nb
m
−4mn2
e
=
e
C(1, p).
p
p
p
b=1
236
TINGTING WANG
This proves Lemma 1.
Lemma 2. Let p be an odd prime with p 6= 3k + 1. Then for any character
χ mod p and integer m with (m, p) = 1, we have the identity
!2
p−1 p−1
p
X
X
X
mb3 (a3 − 1) + nb(a − 1)
e
χ(a)
p
n=1 a=2
b=1

if χ is the principal character mod p,
 p(p2 − p − 4),
p(2p2 − 5p − 1),
if
χ is the Legendre symbol mod p,
=

p p2 − 4p − 1 , otherwise.
Proof. Since p 6= 3k + 1, if b passes through a reduced residue system ( mod p),
then b3 also passes through a reduced residue system (mod p). Now for all
2 ≤ a ≤ p − 1, it is clear that (a − 1, p) = 1. Thus, from the well known
trigonometric sum
3 X
q
q
X
b n
bn
0, if (n, q) = 1,
=
=
e
e
q, if q | n,
q
q
b=1
b=1
we have
!2
p−1 X
mb3 (a3 − 1) + nb(a − 1)
e
χ(a)
(2)
p
n=1 a=2
b=1
!!2
p−1
p−1
p
3
X
X
X
mb3 a − 1 (a3 − 1) + nb
e
χ(a)
=
p
n=1 a=2
b=1


3
3
p−1
p−1
p−1
mb3 a − 1 (a3 − 1) − c − 1 (c3 − 1)
X
X
X

e
χ(c)
χ(a)
=p
p
c=2
a=2
b=1


3
3
p−1
p−1
p−1 X
mb a − 1 (a3 − 1) − c − 1 (c3 − 1)
X
X
.
e
χ(ac)
=p
p
a=2 c=2
p−1
p
X
X
b=1
3
3
Note that a − 1 (a3 − 1) ≡ c − 1 (c3 − 1) mod p if and only if a = c or a = c.
Also, p − 1 = p − 1. Thus if χ = χ0 is the principal character mod p, then
from (2) we have
!2
p−1
p
p−1
X
X
X mb3 (a3 − 1) + nb(a − 1) (3)
χ(a)
e
p
n=1 a=2
b=1


p−1
X 
X p−1

2

1
= p (p − 1)(2p − 5) −
 = p(p − p − 4).
a=2 c=2
a6=c, c
GENERALIZED TWO-TERM EXPONENTIAL SUMS
If χ =
(4)
∗
p
237
is the Legendre’s symbol, then we obtain from (2)
!2
p−1 p−1
p
X a X
X
mb3 (a3 − 1) + nb(a − 1)
e
p
p
n=1 a=2
b=1


p−1
X a
X p−1

c 

= p
(p
−
1)(2p
−
5)
−


p
p
a=2 c=2
a6=c, c

= p p(2p − 5) −
p−1 X
a=1
a
p
−1
!2 
= p(2p2 − 5p − 1).
If χ 6= χ0 and p∗ , then from (2) we have
(5)
p
X
p−1
X
p−1
X

mb3 (a3 − 1) + nb(a − 1)
χ(a)
e
p
n=1 a=2
b=1


p−1
p−1
p−1
p−1
XX

 XX
χ(a)χ(c)
χ(ac) −
= p p
= p p(p − 2) + p

= p p(p − 2) + p
= p p2 − 4p − 1 .
!2
a=2 c=2
a=2 c=2
a=c or c

p−2
X
a=2
p−1
X
a=1
χ(a2 ) −
p−1
X
a=2
χ2 (a) − 2p −
!2 
χ(a) 
p−1
X
a=1
χ(a) − 1
!2 

Combining (3), (4) and (5) we may immediately deduce Lemma 2.
3. Proof of the theorems
In this section, we shall use the lemmas from Section 2 to complete the proof
of our theorems. First we prove Theorem 1. For any character χ mod p, note
that
p−1
X
X p−1
m(a2 − b2 ) + n(a − b)
2
χ (ab) e
|C(m, n, 2, χ; p)| =
p
a=1 b=1
p−1
p−1
X
X
mb2 (a2 − 1) + nb(a − 1)
e
χ(a)
(6)
=
.
p
a=1
b=1
238
TINGTING WANG
Then from (6) and the orthogonality relation for character sums mod p we
have
X
(7)
|C(m, n, 2, χ; p)|4
χ mod p
=
=
=
=
2
p−1
X p−1
X mb2 (a2 − 1) + nb(a − 1) X
e
χ(a)
p
χ mod p a=1
b=1
p−1 2
p−1
X X
mb2 (a2 − 1) + nb(a − 1) e
(p − 1)
p
a=1 b=1
2
p−2
X p−1
X mb2 (a2 − 1) + nb(a − 1) 3
(p − 1) + (p − 1) + (p − 1)
e
p
a=2 b=1
2
p−3
X p−1
X mb2 (a2 + 2a) + nba 3
(p − 1) + (p − 1) + (p − 1)
e
.
p
a=1 b=1
Then applying Lemma 1 and (7) we have
X
4
(8)
|C(m, n, 2, χ; p)|
χ mod p
= (p − 1)3 + (p − 1)
p−3 X
m(a2 + 2a)
e
+ (p − 1)
p
a=1
= (p − 1)3 + (p − 1)
p−3 X
m(1 + 2a)
+ (p − 1)
e
p
a=1
= (p − 1)3 + (p − 1)
p−3 X
m1 + 2a
+ (p − 1)
e
p
a=1
−4m(a2 + 2a)(na)2
p
−4m(1 + 2a)n2
p
!
!
2
C(1, p) − 1
2
C(1, p) − 1
2
!
−4m(1 + 2a)n2
C(1, p) − 1
p
2
p−2
X ma
−4m · an2
3
= (p − 1) + (p − 1) + (p − 1)
C(1, p) − 1
p e
p
a=2
3
= (p − 1) + (p − 1) + (p − 1)(p + 1)(p − 3) − (p − 1) R + R ,
where
p−2 X
ma
−4m · an2
e
C(1, p).
R=
p
p
a=2
GENERALIZED TWO-TERM EXPONENTIAL SUMS
239
It is clear that
(9)
R = C(1, p)
"
p−1
X
a=1
ma
p
#
4mn2
−4man2
m
−4mn2
−m
e
−
e
−
e
p
p
p
p
p
4m · n2
−1
m
−4m · n2
−m
C(1, p) −
e
−
e
p
p
p
p
p
2
2
−4m · n
−1
4m · n
m
e
+
e
,
= p − C(1, p)
p
p
p
p
= C(1, p)
and
m
−4m · n2
−1
4m · n2
−1
(10) R = p − C(1, p)
e
+
e
.
p
p
p
p
p
Note that C 2 (1, p) = −1
p and C(1, p) · C(1, p) = p (see Theorem 9.16 of
p
[1]), so C(1, p) = −1
C(1, p). Then from (8), (9) and (10) we have
p
X
χ mod p
4
|C(m, n, 2, χ; p)|
= (p − 1)3 + (p − 1) + (p − 1)(p + 1)(p − 3) − 2(p − 1)R
= (p − 1)3 + (p − 1) + (p − 1)(p + 1)(p − 3) − 2(p − 1)p
−4m · n2
−1
m
4m · n2
e
+
e
+ 2(p − 1)C(1, p)
p
p
p
p
= 2p3 − 8p2 + 5p + 1
4m · n2
m
−4m · n2
−1
+ 2(p − 1)C(1, p)
e
+
e
.
p
p
p
p
This proves Theorem 1.
Now we prove Theorem 2. Note that for any character χ mod p, we have
(11)
|C(m, n, 3, χ; p)|2
p−1 p−1
X
X
mb3 (a3 − 1) + nb(a − 1)
e
χ(a)
=
p
a=1
b=1
p−1
p−1
X
X
mb3 (a3 − 1) + nb(a − 1)
e
χ(a)
= (p − 1) +
p
a=2
b=1
!
p−1
p−1
3
X
X
mb3 a − 1 (a3 − 1) + nb
e
χ(a)
= p−1+
.
p
a=2
b=1
240
TINGTING WANG
So from Lemma 2 and (11) we have
p
X
n=1
=
p
X
n=1
4
|C(m, n, 3, χ; p)|
(p − 1) +
2
= p(p − 1) +
p
X
n=1
p−1
X
χ(a)
a=2
p−1
X
a=2

 p(2p2 − 3p − 3),
p2 (3p − 7) ,
=
 2
p (2p − 6) ,
p−1
X
3
mb3 a − 1 (a3 − 1) + nb
p
e
b=1
χ(a)
p−1
X
3
e
b=1
!!2
mb3 a − 1 (a3 − 1) + nb
p
!!2
if χ is the principal character mod p,
if χ is the Legendre symbol mod p,
otherwise.
This completes the proof of Theorem 2.
Acknowledgment. The author would like to thank the referee for his very
helpful and detailed comments which have significantly improved the presentation of this paper.
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Department of Mathematics
Northwest University
Xi’an, Shaanxi, P. R. China
E-mail address: [email protected]