ON THE LEAST QUADRATIC NON-RESIDUE
YUK-KAM LAU AND JIE WU
Abstract. We prove that for almost all real primitive characters χd of modulus |d|,
the least positive integer nχd at which χd takes a value not equal to 0 and 1 satisfies
nχd log |d|, and give a quite precise estimate on the size of the exceptional set.
1. Introduction
Let q ≥ 2 be an integer and χ a non principal Dirichlet character modulo q. Here the
evaluation of the least integer nχ among all positive integers n for which χ(n) 6= 0, 1 is
referred as Linnik’s problem. In case χ coincides with the Legendre symbol, nχ is a least
quadratic non-residue. Concerning the size of nχ , Pólya-Vinogradov’s inequality
X
(1.1)
max χ(n) q 1/2 log q
x≥1
n≤x
implies trivially nχ q 1/2 log q. But for prime q, Vinogradov [21] proved the better bound
(1.2)
nχ q 1/(2
√
e)
(log q)2
by combining a simple argument with (1.1). He also conjectured that nχ ε q ε for all
integers q ≥ 2 and any ε > 0. Under the Generalized Riemann Hypothesis (GRH), Linnik
[16] settled this conjecture, and later Ankeny [1] gave a sharper estimate
(1.3)
nχ (log q)2
(still assuming GRH). Burgess ([3], [4], [5]) wrote a series of important papers on sharpening (1.1). His well known estimate on character sums is as follows: For any ε > 0, there
is δ(ε) > 0 such that
X
(1.4)
χ(n) ε xq −δ(ε)
n≤x
provided x ≥ q 1/3+ε . The last condition can be improved to x ≥ q 1/4+ε if q is cubefree.
When q is prime, Burgess deduced, via (1.4) and Vinogradov’s argument,
(1.5)
nχ ε q 1/(4
√
e)+ε
.
An entirely straightforward modification yields
(
√
q 1/(4 e)+ε if q is cubefree,
√
nχ ε
q 1/(3 e)+ε otherwise.
(However we would remark that we could not find this result in literature.)
2000 Mathematics Subject Classification. 11K31.
Key words and phrases. Special sequences.
1
2
YUK-KAM LAU AND JIE WU
Let us now focus on real primitive characters. Denote D (resp. D(Q)) to be the set of
fundamental discriminants d (resp. with |d| ≤ Q), that is, the set of non-zero integers d
which are products of coprime factors of the form −4, 8, −8, p0 where p0 := (−1)(p−1)/2 p
(p odd prime). Also, we write K (resp. K(Q)) for the set of real primitive characters (resp.
with modulus q ≤ Q). Then there is a bijection between D and K given by
d
d 7→ χd (·) =
· K
where d· K is the Kronecker symbol. Note that the modulus of χd equals |d| and
6
Q + O(Q1/2 ).
π2
In the opposite direction of (1.2), Fridlender [12], Salié [20] and Chowla & Turán (see
[10]) independently showed that there are infinitely many primes p for which
(1.6)
(1.7)
|D(Q)| = |K(Q)| =
nχp0 log p,
or in other words, nχp0 = Ω(log p). Under GRH, Montgomery [18] gave a stronger result
nχp0 = Ω(log p log2 p), where logk denotes the k-fold iterated logarithm. Without any
assumption Graham & Ringrose [13] obtained nχp0 = Ω(log p log3 p). In view of these
results, it is natural to wonder what is the size of the majority of nχp0 , or more generally
nχd . Indeed the density of p0 for which nχp0 satisfies (1.7) is low. This can be seen from
Erdős’ result [11],
1 X
nχp0 = constant,
(1.8)
lim
x→∞ π(x)
p≤x
where π(x) denotes the number of primes up to x. This result is extended and refined by
Elliott in [7] and [8]. Using (1.8) or its refinement in [7], it follows, for any fixed constant
δ > 0, that
X
x
.
(1.9)
1 δ
(log x)2
p≤x
nχp0 ≥δ log p
In [6], Duke & Kowalski indicated: Let α > 1 be given. Denote by N (Q, α) the number
of primitive characters χ (not necessarily real) of modulus q ≤ Q such that χ(n) = 1 for
all n ≤ (log Q)α and (n, q) = 1. Then one has
N (Q, α) ε Q2/α+ε
for all ε > 0. Therefore
|{|d| ≤ Q : nχd ≥ (log Q)α }| ε Q2/α+ε .
However, in view of (1.6) this result is non-trivial only when α > 2 and it tells that
nχd ≤ (log |d|)2+ε for almost all fundamental discriminants d. Very recently Baier [2]
improved 2 + ε to 1 + ε by using the large sieve inequality of Heath-Brown [14] for real
primitive characters. However, the argument is unable to cover the case α = 1 or to
provide information on the sparsity of the primes p with nχp0 log p as in (1.9).
The purpose of this paper is two-fold. The first goal is to supplement the case α = 1,
using the large sieve inequality of Elliott-Montgomery-Vaughan (see [9] and [19]). We
obtain an almost all result, which is strong enough to yield a tighter estimate on the low
density of exceptional non-residues than in (1.9).
ON THE LEAST QUADRATIC NON-RESIDUE
3
Theorem 1.1. For 2 ≤ P ≤ Q, define
E(Q, P ) := {d ∈ D(Q) : χd (p) = 1 for P < p ≤ 2P and p - |d|}.
Then there are two absolute positive constants C and c such that
|E(Q, P )| Qe−c(log Q)/ log2 Q
(1.10)
holds uniformly for Q ≥ 10 and C log Q ≤ P ≤ (log Q)2 . In particular we have
nχd log |d|
(1.11)
for all but except O Qe−c(log Q)/ log2
Q
characters χd ∈ K(Q).
Sections 2 and 3 are devoted to the proof of Theorem 1.1.
Our second goal is to point out that the upper bound (1.10) for exceptional real primitive
characters set is quite optimal.
Theorem 1.2. For any fixed constant δ > 0, there are a sequence of positive real numbers
{Qn }∞
n=1 with Qn → ∞ and a positive constant c such that
X
1 δ Qn e−c(log Qn )/ log2 Qn .
(1.12)
1/2
Qn <p≤Qn
nχp ≥δ log p
Further if we assume that both L1 (s, Py ) and L4 (s, Py ) defined in (4.3) below have no
exceptional zeros in the region (4.4), then (1.12) holds for all Q ≥ 10.
Theorem 1.2 is essentially due to Graham & Ringrose [13]. Graham & Ringrose considered a problem of the quasi-random graphs (Paley graphs) which leads to study the lower
bound for the sum of the right-hand side of (5.5) below. Their work will be the core of our
proof. In Sections 4 and 5, we shall provide the salient points along the line of arguments
in [13] to prove Theorem 1.2.
2. A large sieve inequality of Montgomery-Vaughan
Our key tool for proving Theorem 1.1 is a large sieve inequality of Montgomery &
Vaughan in ([19], page 1050) following from ([19], Lemma 2). Here we state a slightly
refined version. Their original statement absorbs the factors (6/ log P )j and {6/(log P )2 }j
in the implied constant. We reproduce here their proof with a minuscule modification.
Lemma 2.1. We have
X (2.1)
X
d∈D(Q) P <p≤2P
j j
χd (p) 2j
6P
6j
Q
+
p P log P
(log P )2
uniformly for 2 ≤ P ≤ Q and j ≥ 1. The implied constant is absolute.
Proof. Since χd (n) is completely multiplicative on n, we can write
X
X
aj (m)
χd (p) j
=
χd (m),
p
m
j
j
P <p≤2P
P <m≤(2P )
where
aj (m) := |{(p1 , . . . , pj ) : p1 · · · pj = m, P < pi ≤ 2P }|.
By Lemma 2 of [19] with the choice of parameters
X = P j,
Y = (2P )j
and
am = aj (m)/m,
4
YUK-KAM LAU AND JIE WU
it follows that as aj (m1 )aj (m2 ) ≤ a2j (n2 ) for n2 = m1 m2 ,
X X χd (p) 2j
X
a2j (n2 )
(2.2)
+
Q
p n2
j
j
d∈D(Q) P <p≤2P
P <n≤(2P )
X
P <p≤2P
1
p1/2
2j
.
Writing n = pν11 · · · pνi i with ν1 + · · · + νi = j, we have
(2j)!
(2ν1 )! · · · (2νi )!
(2j)! ν1 !
νi !
=
···
aj (n).
j! (2ν1 )!
(2νi )!
a2j (n2 ) =
¿From this, it is easy to see a2j (n2 ) ≤ j j aj (n), and thus
X
X
a2j (n2 )
j
≤
j
n2
j
j
j
P <n≤(2P )j
P <n≤(2P )
1
= j
p2
P <p≤2P
j
6j
.
≤
P log P
X
aj (n)
n2
j
Inserting this into (2.2) and using the estimate
X
1
6P 1/2
,
≤
log P
p1/2
P <p≤2P
we obtain the required result (2.1).
3. Proof of Theorem 1.1
Define
E ∗ (Q, P ) := d ∈ D(Q) : Q1/2 ≤ |d| ≤ Q and χd (p) = 1 (P < p ≤ 2P, p - |d|) .
Let C log Q ≤ P ≤ (log Q)2 . For d ∈ E ∗ (Q, P ), we invoke the prime number theorem to
deduce
X χd (p)
X 1
X
1
=
−
p
p
p
P <p≤2P
P <p≤2P
P <p≤2P, p||d|
log 2 + o(1) {1 + o(1)} log Q
−
log P
P log2 Q
log 2 − 2/C + o(1)
≥
log P
1
>
,
2 log P
provided C is sufficiently large. It is apparent from (2.1) that
X X χd (p) 2j
|E ∗ (Q, P )|
≤
(2 log P )2j
p ≥
d∈D(Q) P <p≤2P
Q
6j
P log P
j
+
6P
(log P )2
j
.
ON THE LEAST QUADRATIC NON-RESIDUE
5
Hence we obtain
12j log P j
|E (Q, P )| Q
+ (12P )j
P
∗
uniformly for C log Q ≤ P ≤ (log Q)2 and j ≥ 1. Taking
log Q
j=
+ 1,
48 log P
a simple calculation shows that
|E ∗ (Q, P )| Qe−c(log Q)/ log2 Q
with c = (log 2)/48. This implies (1.10).
Finally let
E ∗ (Q) := d ∈ D(Q) : d ≤ Q1/2 ∪ E ∗ (Q, C log Q).
Then by (1.10), we have
|E ∗ (Q)| Qe−c(log Q)/ log2 Q ;
and for any d ∈ D(Q) \ E ∗ (Q) there is a prime number p log Q log |d| such that
χd (p) 6= 1, which implies (1.11). The proof is complete.
4. Graham-Ringrose’s method
In this section, we shall state and extend the main results of ([13], Theorems 2, 3 and
4) for our purposes. For characters of certain moduli, Graham & Ringrose [13] obtained
a wide zero-free region and good zero density estimates for the corresponding Dirichlet
L-functions. The main ingredient of their method is an q-analogue of van der Corput’s
result, which can be stated as follows: Suppose that q = 2ν r, where 0 ≤ ν ≤ 3 and r is
an odd squarefree integer, and that χ is a non-prinicipal character mod q. Let p be the
largest prime factor of q. Suppose that k is a non-negative integer, and K = 2k . Finally,
assume that N ≤ M . Then
X
k+3
k2 +3k+4
32k2 +11k+8
k+3
1
(4.1)
χ(n) M 1− 8K−2 p 32K−8 q 8K−2 d(q) 16K−4 (log q) 8K−2 σ−1 (q),
M <n≤M +N
where σa (q) :=
P
d|q
da and d(q) := σ0 (q). The implied constant is absolute.
Recall that for any odd prime p,
2
q
q
χ8 (p) =
,
χq0 (p) =
=
p
p K
p
(q odd prime, q 0 := (−1)(q−1)/2 q)
Q
by definition. For squarefree m ≥ 2, the character χm := p|m χp0 for odd m or χm :=
χ8 χm0 for m = 2m0 is a real primitive of modulus m or 4m, respectively. By convention, we
set χ1 ≡ 1. Moreover, if χ4 is the real primitive character mod 4, i.e. χ4 (n) = (−1)(n−1)/2
for odd n, then χ4m := χ4 χm is also a real primitive character of modulus 4m.
Let
Y
(4.2)
Py :=
p = e{1+o(1)}y
(y → ∞),
p≤y
and define for ` = 1 or 4,
(4.3)
L` (s, Py ) :=
Y
m|Py
L(s, χ`m ),
6
YUK-KAM LAU AND JIE WU
where L(s, χ`m ) is the Dirichlet L-function associated to χ`m . Denote by N` (α) the number
of zeros of L` (s, Py ) in the rectangle
α≤σ≤1
and
|τ | ≤ log Py .
Here and in the sequel we implicitly define the real numbers σ and τ by the relation
s = σ + iτ .
The next lemmas 4.1, 4.2 and 4.3 are trivial extensions of Theorems 2, 3 and 4 of [13],
respectively.
Lemma 4.1.
Q Let y ≥ 100. Then there is an absolute positive constant C1 such that the
L-function `=1,4 L` (s, Py ) has at most one zero in the region
C1 (log2 Py )1/2
log Py
The exceptional zero, if exists, is real.
(4.4)
σ ≥1−
and
|τ | ≤ log Py .
Proof. As the crucial estimate (4.1) holds for all non-principal primitive characters of
modulus q = 2ν r ≥ 2 with 0 ≤ ν ≤ 3 and r being odd squarefree. Consider the case
ν = 0 or 3, and ν = 2 or 3, respectively. We see that (4.1) applies to χm and χ4m for
any m|Py . It follows that ([13], Lemma 6.1) is valid for these characters.
Q Proceeding
with the same argument, we have ([13], Lemma 6.2) for our L-function `=1,4 L` (s, Py )
in place of L(s, Py ) there. Then the same proof of ([13], Theorem 2) will give the desired
result. (Note that the value of φ suffers a negligible change when Py is replaced by 4Py
or 8Py .) The exceptional zero must be real, for otherwise, its conjugate is another zero in
the specified region.
Lemma 4.2. Let C1 be as in Lemma 4.1. There is a sequence of positive real numbers
{yn }∞
n=1 with yn → ∞ such that both L1 (s, Pyn ) and L4 (s, Pyn ) have no zeros in the region
(4.5)
σ ≥ 1 − η(yn )
where
η(y) :=
and
|τ | ≤ log Pyn ,
C1 (log2 Py )1/2
.
2 log Py
Proof. Similar to ([13], Theorem 3), our proof is also based on an interesting argument
attributed to Maier [17]. Suppose that for some y, the product L1 (s, Py )L4 (s, Py ) has an
exceptional zero in the region (4.4). That is, it has a real zero β > 1 − 2η(y). In view of
(4.2), we can take yn ≥ y such that
η(yn ) < 1 − β < 2η(yn ).
Q
By Lemma 4.1, β is the only exceptional zero of `=1,4 L` (s, Pyn ) in the region
(4.6)
σ > 1 − 2η(yn )
|τ | ≤ log Pyn .
Q
Together with the first inequality in (4.6), this forces `=1,4 L` (s, Pyn ) to have no zero in
the region (4.5). It follows that we can find a sequence of positive real numbers {yn }∞
n=1
with yn → ∞ such that both L1 (s, Pyn ) and L4 (s, Pyn ) have no zero in this region.
and
Lemma 4.3. Let ` = 1 or 4 and y ≥ 100. Then there is an absolute constant C2 such
that

log3 Py
C2 (1 − α) log Py


p
exp
+
if α ≥ 1 − η1 (y),


2
log2 Py
(4.7)
N` (α) 

C2 (1 − α) log Py

exp
if α < 1 − η1 (y),
log(1/(1 − α))
ON THE LEAST QUADRATIC NON-RESIDUE
7
where
k0 (y) := (log2 Py )1/2
and
η1 (y) :=
k0 (y)
k
2(2 0 (y) −
2)
.
Proof. The case of ` = 1 has been done in ([13], Sections 7 and 8) and N4 (α) can be
treated in the same way by applying (4.1) to our χ4m .
5. Proof of Theorem 1.2
In this section, we denote by p and q prime numbers. Define
Py := {p : p ≡ 1 (mod 4) and χp (q) = 1 for all q ≤ y}.
Clearly we have nχp > y for any p ∈ Py . We shall first show that the set Py is not too
small for suitable y.
Proposition 5.1. Let δ > 0 be a fixed small constant and y(x) be an strictly increasing
function defined on [120, ∞) satisfying
(log x)e−δ(log2 x)
(5.1)
1/2
≤ y(x) ≤ δ(log x) log3 x.
Then there is a positive constant c = c(δ) and a sequence of positive real numbers {xn }∞
n=1
with xn → ∞ such that
X
(5.2)
1 xn e−cy(xn )/ log y(xn ) .
1/2
xn <p≤xn log xn
p∈Py(xn )
Further if we assume that both L1 (s, Py ) and L4 (s, Py ) have no zeros in the region (4.4)
for all y ≥ 100, then there is a positive constant c such that for all x ≥ 100 we have
X
(5.3)
1 x e−cy(x)/ log y(x) .
x1/2 <p≤x log x
p∈Py
Proof. First let 10 ≤ y ≤ x1/2 . As usual, π(y) denotes the number of prime numbers ≤ y.
Clearly we have
(
Y
1 if p ∈ Py ,
(5.4)
2−π(y)−1 1 + χ4 (p)
1 + χp (q) =
0 if p ∈
/ Py .
q≤y
When p and q are odd primes with p ≡ 1 (mod 4), i.e. χ4 (p) = 1, we infer by quadratic
reciprocity law that
p
q
χp (q) =
=
= χq0 (p) (q 0 := (−1)(q−1)/2 q).
q
p
Note also for odd prime p,
p
2
χp (2) =
=
= χ8 (p).
2 K
p
Thus we can replace χp (q) by qp in (5.4) to write
X
x1/2 <p≤x log x
p∈Py
1=
1
2π(y)+1
X
x1/2 <p≤x log x
Y
q
1 + χ4 (p)
1+
.
p
q≤y
8
YUK-KAM LAU AND JIE WU
It is convenient to introduce the weight factor (log p) e−p/(2x) − e−p/x to the summands,
X
X
1
1 ≥ π(y)+2
(log p) e−p/(2x) − e−p/x ×
2
log x 1/2
1/2
x
x
<p≤x log x
p∈Py
<p≤x log x
Y
q
.
1+
× 1 + χ4 (p)
p
q≤y
We want to relax the range of the sum over p. To this end, we observe that by the prime
number theorem and integration by parts,
X
Y
1
q
−p/(2x)
−p/x
1+
(log p) e
−e
1 + χ4 (p)
π(y)
p
2
log x
q≤y
x log x<p≤x2
X
−p/(2x)
−p/x
e
−e
x log x<p≤x2
x1/2 / log x.
Combining this with the preceeding inequality, we obtain
1/2 X
X
1
x
(5.5)
1≥
,
S
(m)
+
S
(4m)
+
O
x
x
π(y)+2
log
x
(log
x)2
1/2
x
m|Py
<p≤x log x
p∈Py
where ` = 1 or 4, and
Sx (`m) :=
X
(log p) e−p/(2x) − e−p/x χ`m (p).
x1/2 <p≤x2
By the Perron formula, we can write
Z 2+i∞
1
L0
(5.6)
Sx (`m) =
− (s, χ`m )(2s − 1)Γ(s)xs ds + O x1/2 log x .
2πi 2−i∞
L
We shift the line of integration to σ = −34. The function (2s − 1)Γ(s)xs has no pole in the
strip −34 ≤ σ ≤ 2 since the pole of Γ(s) at s = 0 is canceled by the zero of (2s − 1). Thus
the only poles of the integrand in (5.6) occur at s = 1 if `m = 1 (note that L(s, χ1 ) is the
Riemann ζ-function), or at the zeros ρ(`m) = β(`m) + iγ(`m) of L(s, χ`m ). It follows that
X
Sx (`m) = δ`m,1 x −
(2ρ(`m) − 1)Γ(ρ(`m))xρ(`m) + O x1/2 log x ,
ρ(`m)
where δj,1 = 1 if j = 1 and 0 otherwise, and the sum is over all zeros with 0 ≤ β(`m) < 1.
We write N (T, χ`m ) for the number of zeros of L(s, χ`m ) in the rectangle 0 < β(`m) < 1
and |γ| ≤ T . Then we have the classical bound
(5.7)
N (T, χ`m ) T log(T m),
which implies, for any α ∈ (0, 1),
(5.8)
N` (α) ≤
X
N (log Py , χ`m ) 2π(y) y 2 .
m|Py
On the other hand, by means of (2s − 1)Γ(s)xs xσ |τ |e−(π/2)|τ | , the contribution of the
zeros with |γ(`m)| ≥ log Py to Sx (`m) is 1. Let ε be an arbitrarily small positive
number. The zeros with β(`m) ≤ 1 − ε and |γ(`m)| ≤ log Py contribute
x1−ε N (log Py , χ`m ) x1−ε (log Py )2 x1−ε y 2 .
ON THE LEAST QUADRATIC NON-RESIDUE
9
Combining these with (5.5), we conclude
X
x
T1 (x, y) + T4 (x, y)
1−ε π(y) 2
1≥
(5.9)
+O x 2
y +
(log x)2π(y)+2
(log x)2π(y)
1/2
x
<p≤x log x
p∈Py
uniformly for x ≥ 10 and 1 ≤ y ≤ x1/2 , where
X
X
T` (x, y) :=
m|Py
Z
xβ(`m)
ρ(`m)
β(`m)≥1−ε, |γ(`m)|≤log Py
1
xα dN` (α).
=−
1−ε
It remains to estimate T` (x, y). ¿From now on we take y = y(x). By integration by
parts and by using (5.8), we can deduce
T` (x, y) x1−ε 2π(y) y 2 + x(log x)I` ,
(5.10)
where
Z
I` :=
ε
x−β N` (1 − β)dβ.
0
Let η = η(y) and η1 = η1 (y) be defined as in Lemmas 3 and 4, respectively. Set η2 :=
2y(x)/(log x) log y. It is easy to verify that 0 < η < η1 < η2 < ε. (The inequality η1 < η2
governs the lower bound of y(x) in (5.1).) Thus we can divide the interval [0, ε] into four
subintervals [0, η], [η, η1 ], [η1 , η2 ] and [η2 , ε], and denote by I`,0 , I`,1 , I`,2 and I`,3 the
corresponding contribution to I` . Plainly we have
1
η
log3 Py ≤ log x,
2
4
C2 log Py
1
p
≤ log x,
4
log2 Py
C2 log Py
1
≤ log x,
log(1/η2 )
2
y
η2
=
log x.
log y
2
(The third inequality governs the upper bound of y(x) in (5.1).) ¿From Lemma 4.3 and
(5.8), we deduce that
Z η1
C2 β log Py
1
x−η/2
I`,1 exp − β log x + p
+ log3 Py dβ ,
2
log x
log2 Py
η
Z η2
C2 β log Py
x−η1 /2
dβ ,
I`,2 exp − β log x +
log(1/β)
log x
η1
Z ε
y
x−η2 /2
I`,3 exp − β log x +
dβ .
log y
log x
η2
Hence, all of them satisfy
I`,i = o (log x)−1
(i = 1, 2, 3).
If we assume that both L1 (s, Py ) and L4 (s, Py ) have no zeros in the region (4.4) for all
y ≥ 100, then I`,0 = 0. Otherwise we use Lemma 4.2 to ensure the existence of {yn }∞
n=1
such that I`,0 = 0.
With (5.10), our conclusion is
xn
(n → ∞),
T` (xn , yn ) = o
(log xn )2π(yn )
or
x
T` (x, y) = o
(x → ∞)
(log x)2π(y)
10
YUK-KAM LAU AND JIE WU
under the assumption that both L1 (s, Py ) and L4 (s, Py ) have no exceptional zeros. Clearly
this and (5.9) imply the required result. This completes the proof of Proposition 5.1. Now we are ready to prove Theorem 1.2.
Taking Qn = xn log xn and y(x) = 100δ log x in Proposition 5.1 and noticing that
p ∈ Py ⇒ nχp ≥ y, we have
X
1 Qn e−c1 (log Qn )/ log2 Qn .
(Qn / log Qn )1/2 <p≤Qn
nχp ≥100δ log Qn
It implies the first assertion of Theorem 1.2, and the second one can be treated similarly.
This concludes Theorem 1.2.
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Department of Mathematics, The University of Hong Kong, Pokfulam Road, Hong Kong
E-mail address: [email protected]
Institut Elie Cartan, Université Henri Poincaré (Nancy 1), 54506 Vandœuvre-lès-Nancy,
France
E-mail address: [email protected]