savastru-varbanets-0..

’iauliai Math. Semin.,
3 (11), 2008, 207219
ON THE REPRESENTATION OF INTEGERS
AS THE SUM OF kTH POWERS
IN ARITHMETIC PROGRESSIONS
Olga SAVASTRU, Pavel VARBANETS
Odessa National University, Dvoryanskaya st.2, Odessa, 65026, Ukraine,
e-mail: sav olga@bk.ru, varb@sana.od.ua
Abstract. We investigate the behaviour of integers represented as the
sum of k-th powers in arithmetic progressions. Using the Waring zeta-function
introduced by A. Vinogradov, we prove the asymptotic formula for the
summatory function of the number of representations of m as the sum of
k-th powers in arithmetic progressions.
Key words and phrases: asymptotic estimate, meromorphic continuation,
Waring zeta-function.
2000 Mathematics Subject Classication: 11N37.
1. Introduction
Dedicated to the 60th birthday
of Professor Antanas Laurin£ikas
Let α 0 , δ 0 ∈ (0, 1], α, δ ∈ (0, 1] n , and let n, k 2 be integers, k n
We denote
< n k.
Θ k (x; α 0 , δ 0 ) =
∞∑
e −πx(m+δ 0) k e 2πimα 0
.
m=0
For Re s > 1, consider the absolutely convergent series
ζ n,k (s; α, δ) =
∞∑
( n∑ ) −s
e 2πim α j +δ j )
j=1(m k , m α = m 1 α 1 +. . .+m n α n ,
m j =0
1jn

208 O. Savastru, P. Varbanets
which, for k = 2, n 2, coincides with the classical Epstein zeta-function,
and, for k = n 3, α = δ = (0, . . . , 0), becomes the Waring zeta-function
introduced for the rst time by A. Vinogradov [7]. Denote
Then we obtain
V n,k (λ; α, δ) =
ζ n,k (s; α, δ) = ∑ λ>0
∞∑
m j =0
n∑
(m j +δ j ) k =λ
j=1
e 2πi(m 1α 1 +...+m nα n) .
V n,k (λ; α, δ)
λ s , Re s > 1.
This function also will be called the Waring zeta-function.
For α = δ = (0, . . . , 0), the function V n,k (λ; α, δ) denes the number of
the representations of λ, λ ∈ N, as a sum of n kth powers of non-negative
integers. A. Vinogradov obtained [7] the asymptotic formula
∑
mN
V n,n (m; 0, 0) = P n
(N 1 n
)
+ O
(
N 3 4 log 2 N
where P n (u) is a polynomial of degree n with coecients depending on u.
C. Hooley (see, [2][4]) considered the problem on the representation of
positive integers as the sums of two or four h-powers. The Hooley method
is very specic, and it is dicult to apply it in general case. Using a result
of I. Vinogradov [8] on the famous Waring problem, S. Salerno studied [6]
∑
the distribution of square-free numbers n represented as r x k i in arithmetic
i=1
progression n ≡ 0( mod q) under condition r > (4+ε)k 2 log k, k > k 0 is large
enough.
In this paper, we study the function ζ n (s; α, δ), in particular, its behaviour
in the region 1 2 Re s 1, and applying the Waring zeta-function, we
construct the asymptotic formula for the summatory function of the number
of representations of natural numbers as a sum of n summands of kth powers
of non-negative integers.
In the proof of the main theorem, we shall use the following statement.
Let
ζ(s; 0, δ) =
∞∑
(n + δ) −s , Re s > 1,
n=0
denote the Hurwitz zeta-function.
)
,

On the representation of integers... 209
Lemma 1. Let s = σ + it, −2 σ 2, and let τ be a complex number with
arg τ = ( π
2 − |t|−1) sgn (Im s). Then there exists a constant t 0 > 1 such that
the following approximate equation
ζ(s; 0, δ) = 1 2
where
∑
n∈Z
|n+α|x log x
+π − 1−2s
2
+π − 1−2s
2
F (s; |n + δ|τ 1 2
√ π) + an F (1 − s; |n + δ|τ 1 2
√ π)
|n + δ| s
Γ ( 1
2
(1 − s)) ∑
2Γ( 1 2 s)
n∈Z
|n|y log y
Γ ( 1
2 (1 − s) + 1 ∑
2
2Γ( 1 2 s + 1 2 )
+O ( x −M) + O ( y −M) ,
a n =
)
n∈Z
|n|y log y
{
sgn(n) if n ≠ 0,
1 if n = 0,
F (1 − s; |n|τ − 1 2
√ π)e
2πinδ
|n| 1−s
b n F (−s; |n|τ − 1 2
√ π)
|n| 1−s
b n = −isgn(n)e 2πinδ ,
√
x = |t| 1 2 ( 2|τ|) −1 ,
√
y = |t| 1 2 |τ|( 2) −1 ,
and M > 0 is an arbitrary constant, holds. Moreover, uniformly in all parameters
t, τ, x, y, we have
(
F (ω, z) = l + O exp
(
− |z|2
|t|
)( ) |z| Re ω (
1 +
1
|t| 1 ∣
2
2 |t| 1 |z| 2
) −1 )
2 − |t| 1 ∣ ,
2
where l = 1 if |n + δ| x and |n| y, and l = 0 otherwise; for the integral
representation of F (ω, z), see [5].
The lemma follows easily from the Lavrik work [5].
Lemma 2. Let p be prime, n, k ∈ N, (p, k) = 1. Then the following estimates
hold.
∣
p−1
∑
e 2πi m 1 x 1 +...+mnxn
p
x 1 ,...,x n =0
n∑
x k i ≡a( mod p) i=1
∣
{(k − 1) n p n−1
2 if n = 3, 4,
(k − 1) n p n 2 if n > 4,

210 O. Savastru, P. Varbanets
Proof. The case n = 3 was considered by Hooley [3]. The case n = 4
can be investigated similarly, and the general case follows from the Deligne
work [1].
Corollary. Let (k, q) = 1. Then, for (m 1 , . . . , m n , q) = 1, we have
∑
e 2πi m 1 x 1 +...+mnxn
q
x 1 ,...,x n ( mod q)
n∑
x k i ≡a( mod q) i=1
≪ q κ τ(q),
where κ = n−1
2
if n = 3, 4, κ = n 2
if n 5, and the implied constant depends
only on k and n.
2. Analytic continuation of ζ n,k (s; α, δ)
For λ > 0, the following equality
λ −s =
πs
Γ(s)
∫ ∞
is true. Thus, we obtain that, for Re s > 1,
ζ n,k (s; α, δ) =
πs
Γ(s)
0
∫ ∞
0
x s−1 e −πxλ dx, Re s > 0,
x s−1 ( ∑
λ>0
V n,k (λ; α, δ)e −πxλ )
dx. (1)
Let σ l (α) denote an arbitrary collection of l elements of the set {α 1 , . . . ,
α n } (the elements α i dier not by values but by indices). Let ∑ (α) be the
l
set of all collections of σ l (α). It is clear that ∑ (α) contains ( n
) l
elements.
l
So, from the denitions of V n,k (λ; α, δ) and Θ k (x; α 0 , δ 0 ) we obtain that
∑
V n,k (λ; α, δ)e −πxλ =
λ>0
=
n∑ ∑
∏
l=1 σ l (α) α i ∈σ l (α)
n∏
j=1
Θ k (x; α i , δ i )
(Θ k (x; α j , δ j ) + 1) − 1, (2)
where ∑ means that the sum runs over all m of the form m = (m 1 +
m>0
δ 1 ) k + . . . + (m n + δ n ) k , and ∑
∑
denotes . Here we take into
σ l (α)
σ l (α)∈ ∑ (α)
l

On the representation of integers... 211
account that the collections (α i1 , . . . , α il ) and (δ i1 , . . . , δ il ) are found in the
one-to-one correspondence.
Since the terms of the series for the Θ k decrease exponentially, we have
that the behaviour of the integrand in (1) on [1, ∞) guarantee an analytic
continuation of this part of integral to the whole complex plane. Therefore,
it remains to consider the integral
where
F (s) =
=
=
∫ 1
0
I(σ l (α)) =
x s−1 ( ∑
λ>0
n∑ ∑
∫ 1
l=1 σ l (α) 0
V n,k (λ; α, δ)e −πxλ )
dx
x s−1
∏
α i ∈σ l (α)
Θ k (x; α i , δ i )dx
n∑ ∑
I(σ l (α)), (3)
l=1 σ l (α)
∫ 1
0
x s−1
∏
α i ∈σ l (α)
Θ k (x; α i , δ i )dx. (4)
In order to investigate the behaviour of the function Θ n,k (x; α i , δ i ) on the
right of the point x = 0, we use the integral representation for this function.
Consider the Mellin pair
Γ(z) =
∫ ∞
0
e −y y z−1 dy, e −y = 1
2πi
∫
u+i∞
u−i∞
y −z Γ(z)dz, Re z = u > 0.
Taking y = πx(m + δ i ) k , u = 1 n
+ ε, ε > 0, we arrive at the relation
ε+ 1 n +i∞ ∫
Θ k (x; α 0 , δ 0 ) = 1 (πx) −z Γ(z)ζ(kz; α 0 , δ 0 )dz, (5)
2πi
ε+ 1 n −i∞
where ζ(w; α 0 , δ 0 ), 0 < δ 1, is the Lerch zeta-function dened, for Re w > 1,
by the Dirichlet series
ζ(w; α 0 , δ 0 ) =
∞∑
m=0
e 2πiα 0m
(m + δ 0 ) w .

212 O. Savastru, P. Varbanets
The function ζ(w; α 0 , δ 0 ) is entire, except for a simple pole at w = 1 with
residue 1 if α 0 ∈ Z.
In equality (5), move the contour of integration to the line Re z = −1+
2k.
1
The integrand has a simple pole at z = 1 k
(if α 0 ∈ Z), and a simple pole at
z = 0. Thus, we obtain that
Θ k (x; α 0 , δ 0 ) = ε(α 0 ) Γ( 1 k )
(
+ ζ(0; α
(πx) 1 0 , δ 0 ) + I − 1 + 1 )
k
2k ; x, α 0, δ 0 , (6)
where
I(b; x, α 0 , δ 0 ) = 1
2πi
∫
b+i∞
b−i∞
{ 0 if α
ε(α 0 ) =
0 ∉ Z,
1 if α 0 ∈ Z.
Since, for x → +0, ∫∞
I(b; x, α 0 , δ 0 ) ≪ x −b
≪
1
∫∞
x −b
≪ x −b ,
we obtain from (6)(7) that
Θ k (x; α 0 , δ 0 ) = ε(α 0 ) Γ( 1 k )
1
(πx) −z Γ(z)ζ(kz; α 0 , δ 0 )dz, (7)
|Γ(b + iv)||ζ(kb + ikv; α 0 , δ 0 )|dv
|Γ(b + iv)|v 1 2 −b |ζ(1 − kb − ikv; −δ 0 , α 0 )|dv
(πx) 1 k
Now from (3), (4) and (8) we have that
where
F (s) =
=
n∑ ∑
∫ 1
l=1 σ l (α) 0
× ∏
α i ∈σ l (α)
n∑
l=0
b k (l)
s − l k
b k (0) =
x s−1
(
ε(α i ) Γ( 1 k )
(πx) 1 k
(∣ ∣∣∣
+ O Re s + 1
2k ∣
n∑ ∑
l=1 σ l (α) α i ∈σ l (α)
+ ζ(0; α 0 , δ 0 ) + O(x 1− 1
2k ). (8)
+ ζ(0; α i , δ i ) + O ( x 1− 1
2k
) ) dx
∏
−1)
, (9)
ζ(0; α i , δ i ),

On the representation of integers... 213
b k (n) =
r =
( ( ) ) 1 r
Γ (π) − 1 k ,
k
n∑
i=1
α i =1
1, (10)
and b k (l), 0 < l < n, are polynomials in Γ( 1 k
) with coecients which depend
on α 1 , . . . , α n and δ 1 , . . . , δ n .
Equality (9) gives meromorphic continuation of the Waring zeta-function
to the half-plane Re s −1 +
2k. 1 In view of (9), we see that ζ n,k (s; α, δ) has
poles at the points s = k, l l = 0, 1, . . . , n.
Now we will estimate the function ζ n,k (s; 0, δ) for 1 2
< Re s 1 + ε. In
(1), the integrand is holomorphic for Re s > 0. Therefore, if |t| 1, we can
turn by the angle ϕ = ( π
2 − |t|−1) sgn(t) the ray of integration x ∈ [0, +∞).
Hence, we have
πs
ζ n,k (s; α, δ) =
Γ(s)
∫ ∞
x=0
z s−1 ( ∑
λ>0
V n,k (λ; α, δ)e −πzλ )
dz, (11)
z = xe iϕ . This choice of the ray of integration shows that the Waring zetafunction
has the power growth for |t| → ∞. We put
ζ n,k (s; α, δ) =
def
π s
Γ(s)
= ζ (1)
n,k
⎛
⎝
∫ 1
Now, using the well-known estimate
we nd that
ζ (2)
n,k (s; α, δ) = ∑ λ>0
0
I λ (s) = 1
Γ(s)
By a trivial estimate ∑
deduce from (14) that
+
∫ ∞
1
⎞ (
⎠
z s−1 ∑ λ>0
V n,k (λ; α, δ)e −πzλ )
dz
(s; α, δ) + ζ(2)
n,k
(s; α, δ). (12)
∫ ∞
x=λ
z=xe iϕ
V n,k (λ; α, δ)
m s I m (s) ≪ ∑ λ>0
λx
∫∞
ζ (2)
n,k (s; α, δ) ≪ ∑
V n,k (λ; α, δ) ≪ x n k
λ 0
λx
z s−1 e −z dz ≪ e − λ
|t|
, (13)
V n,k (λ; α, δ)
λ σ e − λ
|t|
. (14)
and partial summation, we
(
V n,k (λ; α, δ)
λ σ e − x
|t|
1 + x ) dx
|t| x σ+1

214 O. Savastru, P. Varbanets
≪
min
(
λ n k −σ
)
0
n
k − σ , |t| n k −σ log |t| + |t| n k −σ , (15)
λ 0
λ 0 = δ k 1 + . . . + δk n. Hence, in view of (3)(5), we obtain that, for Re s > 1 2,
where
n∑
ζ (1)
n,k
(s; α, δ) =
∑
l=1 σ l (α)
π s
Γ(s)
∫ 1
x=0
x s−1 e iϕs G(x, σ l (α))dz,
and
G(x, σ l (α)) =
∏
α j ∈σ l (α) 1
1
2k +i∞ ∫
ζ (1)
n,k (s; α, δ) = πs e iϕs
Γ(s)
2k −i∞ e iw j
(πx) −w j
Γ(w j )ζ(kw j ; α j , δ j )dw j ,
×
l∏
j=1
n∑ ∑
∫
∫
· · ·
l=1 σ l (α)
Re w j = 1
2k
1jl
e iϕw j
Γ(w j )
ζ(kw j ; α j , δ j ) dw 1 . . . dw l
s − w 1 − w l
. (16)
The integrals on w j , j = 1, . . . , l, can be calculated by the method of stationary
phase using the Stirling formula for Γ(w) and the approximate functional
equation for ζ(s; α, δ), see Lemma 1. We have
n∑
ζ (1)
n,k
(s; α, δ) ≪
∑
∑
l=1 σ l (α) α∈σ l (α)
∑
V l,k (λ; α, δ)
(17)
(σ − 1 λλ 0
2 ) + ||t| − λ|.
The main contribution in (16) is given by the summand with l = n , and
then σ l (α) = (α 1 , . . . , α l ), λ 0 = δ k 1 + . . . + δk n. Hence, and from (12), (15)
and (17) we obtain that, for Re s > 1 2,
( )
ζ n,k (s; α, δ) ≪ |t| n 1
k −σ min
n
k − σ , log(|t| + 3)
+ ∑ V n,k (m; α, δ) λ
(σ − 1 m 2
) + ||t| − m|e−
|t|
. (18)

On the representation of integers... 215
3. Sum of n kth powers in arithmetic progression
Recall that V n,k (m) denotes the number of representations of a natural
m as a sum of n kth powers of non-negative integers, i.e.,
∑
V n,k (m) =
1. (1)
x 1 ,...,x n0
x k 1 +...+xk n=m
Since we have not an acceptable estimate for V n,k (m), we cannot use the
lemma on partial sums of Dirichlet series. Therefore, we apply the Perron
formula for calculation of the weighted mean for V n,k (m). We have that
∑
m≡a( mod q)
mN
V n,k (m)(1 − m N ) = 1
2πi
where δ = ( l 1
q
, . . . , ln q ), ∑
n∑
li k
i=1
≡ a( mod q).
(a,q)
c+i∞ ∫
c−i∞
1 ∑
q ks
(a,q)
N s
ζ n,k (s; 0, δ) ds, (2)
s(s + 1)
denotes a summing over l 1 , . . . , l n ( mod q), and
We consider only the case k 2 < n k because for n k 2
we cannot dene
the main term in the asymptotic formula.
By (14) and (16) it is clear that we can move the contour of integration
in (2) to the line Re s = σ 0 > 2. 1 This gives
∑
m≡a( mod q)
mN
V n,k (m)
= N n,k (a, q) ∑
+ 1
2πi
∫
σ 0 +i∞
(
1 − m N
)
( ) l
N k
a n,k (l)
q k k
2

216 O. Savastru, P. Varbanets
Let, for Re s > 1,
Obviously,
Z n,k (s; a, q) =
∞∑
m=1
m≡a( mod q)
Z n,k (s; a, q) = 1
q ks ∑
(a,q)
V n,k (m)
m s .
ζ n,k (s; 0, δ).
Then, similarly to the proof of relation (12), we can obtain that, for Re s 1 2,
|Im s| > 1,
where
and
Z n,k (s; a, q) = Z (1)
n,k
(s; a, q) + Z(2)
n,k
(s; a, q), (4)
∫ 1
Z (1)
πs ∑
n,k
(s; a, q) = x s−1 e iϕs q −ks
Γ(s)
(a,q) 0
n∑ ( n
) ( ( ) 1
× Γ (πxe iϕ ) − 1 k
l k
l=1
( )) 1
l
+I
2k ; x, 0, δ i dx, (5)
Z (2)
∞ n,k
(s; a, q) =
∑
m=1
m≡a( mod q)
V n,k (m)
m s I m (s). (6)
The denitions of I(b; x, α 0 , δ 0 ) and I m (s) are given in (7) and (13), respectively.
After some simple transformations of the integrand in (5), we obtain that
Z (1)
πs
n,k
(s; a, q) =
Γ(s) eiϕs q −ks
( π
s
+O
Γ(s)
( ∑
×
(a,q)
n∑ a n,k (l)
s − l N n,k (a, q)
l=1 k
n∑
∫ 1
∣ x s−1 e iϕs q −ks
l=1
0
l∏
( 1
I
2k ; x, 0, l ))
)
i
dx
q ∣
i=1
. (7)

On the representation of integers... 217
The main contribution in the error term in (7) is given by the summand with
l = n. In order to estimate the integral
Φ(s) =
∫ 1
0
π s e iϕs
Γ(s) xs−1 q −ks ⎛
we use (7). This leads to the estimate
where
∫
Φ(s) ≪
∣
∫
· · ·
Re w j = 1
2k
1jn
q −ks πs e iϕs
Γ(s)
∑ ∑
(ζ) = ζ
(a,q)
⎝ ∑ (a,q)
n∏
i=1
Γ(w 1 ) . . . Γ(w n )
π (w 1+...+w n )
( 1
I
2k ; x, 0, l ) ⎞
i
⎠ dx
q
(
kw 1 ; 0, l ) (
1
. . . ζ kw n ; 0, l )
n
.
q
q
∑
(ζ)
dw 1 . . . dw n
s − w 1 − w n
∣ ∣∣∣∣∣∣∣∣∣
, (8)
Applying the approximate functional equation for the Hurwitz zeta-function
ζ(kw; 0, δ) with |τ| = |t| − 1 2 log −1 (|t| + 3), see Lemma 1, we easily nd that
Φ(s)
≪
∞∑
m j =1
1jn
∫
× ∣
∣ eiϕs
∣∣ ∑ Γ(s)
(a,q)
∫
· · ·
Re w j = 1
2k
1jn
n∏
j=1
e 2πi m 1 l 1 +...+mnln
q
Γ(w j )
π w jm w jk
j
∣
dw 1 . . . dw n
s − w 1 − w n
∣ ∣∣.
Computing the integrals by the method of stationary phase, we obtain
Hence,
Φ(s)
≪
≪
∞∑
m j =1
1jn
∣ ∑ (a,q)
e 2πi m 1 l 1 +...+mnln
q
e − (mk 1 +...+mk n )
|t|
×
(σ − 1 2 ) + ||s| − mk 1 − . . . − mk n| q−kσ
∞∑
m=1
∣
V n,k (m)e − m |t|
(σ − 1 2 ) + ||t| − m| qκ−kσ . (9)
Z (1)
n,k (s; a, q) = N n,k(a, q) πs
Γ(s) eiϕs q −ks
n∑ a n,k (l)
l=1
s − l k
N l q

218 O. Savastru, P. Varbanets
+
Moreover, using the estimates
and
∑
m≡a( mod q)
mX
V n,k (m) = ∑ (a,q)
∑
m≡a( mod q)
mX
∞∑
m=1
V n,k (m)e − m |t|
(σ − 1 2 ) + ||t| − m| qκ−kσ . (10)
∑
m i
∑ (mi +δ i ) k
V n,k (m) e− m |t|
we obtain by partial summation that
Z (2)
n,k
(
) 1
k
X
q k
(
n
X
)
k
1 ≪ N n,k (a, q)
q n + 1
m σ ≪ q n k −σ if |t| q, (11)
(s; a, q)
⎧
⎨q n k −σ if |t| q,
≪ ( n
⎩ |t|
−σ
)
k
q
+ |t| −σ min ( 1
n n
k −σ , log(|t| + 3)N n,k(a, q) ) if |t| < q.
Thus, taking σ = 1 2 + 1
log N
, we obtain by (4), (10) and (12) that
∑
m≡a( mod q)
mN
(
V n,k (m) 1 − m )
N
= N n,k (a, q)
(
+O
n∑
l=1
a n,k (l)
)
q l N l q + O
(N 1 n
2 q k − 1 2 log 2 N
)
N n,k (a, q)N 1 2 q
− 3 2 log 2 N
)
+O
(N 1 2 q
κ− k 2 τ(q) log N
(12)
, (13)
where a n,k (n) =
(
1
π Γ( 1 k ) ) n
, the constants a n,k (l), 1 l n, are computable,
and the O-constants do not depend on N and q. From this, by standard way,
we obtain the following statement.
Theorem. Let V n,k (m) denote the number of the representations of m as the
sum of n kth powers, and let a, q, N be natural numbers. Then the following
asymptotic formula

On the representation of integers... 219
∑
m≡a( mod q)
mN
V n,k (m)
= N n,k (a, q)
+O
n∑ a n,k (l)
N l q + O
(N 3 2n−k
4 q 2k
q l
l=1 )
(N 3 4 q
κ− k 2 τ(q) log N
)
− 1
2 log 2 N
(
)
+ O N n,k (a, q)N 3 4 q
− 3 2 log 2 N
holds. Here N n,k (a, q) = #sol{x k 1 + . . . + xk n ≡ a( mod q), 0 x i q − 1},
{ n−1
κ = 2
if n = 3, 4,
n
2
if n 5,
and τ(q) is the number of divisors of q.
References
[1] P. Deligne, La conjecture de Weil. I, II, IHES Publ. Math. 43 (1974), 273307,
52 (1980), 137252.
[2] C. Hooley, On the representation of a number as the sum of four cubes. I,
Proc. London Math. Soc.(3) 36 (1978), 117140.
[3] C. Hooley, On the numbers that are representable as the sum of two cubes. I,
Reine Angew. Math. 314 (1980), 146173.
[4] C. Hooley, On another sieve method and the numbers that are a sum of two
hth powers, Proc. London Math. Soc. (3) 43 (1981), 73109.
[5] A. Lavrik, Approximate functional equation for the Dirichlet L-functions, Tr.
Mosk. Mat. O.-va 18 (1968), 91104 (in Russian).
[6] S. Salerno, On the distribution of x k 1 + . . . + x k s in the arithmetic progression,
Studio Scient. Math. Hung. 20 (1985), 357374.
[7] A. I. Vinogradov, Waring zeta-function, in: Collection of Papers Dedicated
to the 60th Birthday of Prof. V. Sprindzuk, 3340, 1997 (in Russian).
[8] I. M. Vinogradov, The Method of Trigonometric Sums in the Theory of Numbers,
Interscience, London, 1954.
Received
31 January 2008