SIGN CHANGES OF KLOOSTERMAN SUMS AND EXCEPTIONAL
CHARACTERS
SARY DRAPPEAU AND JAMES MAYNARD

Abstract. We prove that theP existence of exceptional real zeroes of Dirichlet L-functions would leadto cancellations in the sum
Kl(1, p) of Kloosterman sums over primes, and also to sign changesp≤xof Kl(1, n), where n runs over integers with exactly two prime factors. Our arguments involve a variantof Bombieri’s sieve, bounds for twisted sums of Kloosterman sums, and work of Fouvry and Michel onsums of |Kl(1, n)|.

1. Introduction
Kloosterman sums. For n ∈ N>0 and a residue class a (mod n), define the normalized Kloosterman
sum as ν + aν̄ X1e,Kl(a, n) = √nnν (mod n)(ν,n)=1

where we write e(z) = e2πiz and ν̄ν ≡ 1 (mod n). These sums have a long history [Poi11, Klo27],
at the intersection of algebraic geometry and automorphic forms. The Weil bound [Wei48, Mat11]
yields |Kl(a, n)| ≤ 2ω(n) if 32 - n and |Kl(a, n)| ≤ 2ω(n)+1/2 in general, where ω(n) is the number ofdistinct prime factors of n. In particular, for a prime p|Kl(1, p)| ≤ 2.
Let θa,p ∈ [0, π] be such that Kl(a, p) = 2 cos(θa,p ). The “vertical” Sato-Tate law, due to Katz [Kat88],
asserts that the numbers{θa,p | 1 ≤ a < p}become equidistributed, as p → ∞, with respect to the Sato-Tate measure π2 sin(θ)2 dθ. The “horizontal”
Sato-Tate conjecture is the claim that the numbers{θ1,p | p ≤ x}
become equidistributed with respect to the same measure, as x → ∞. This would of course imply thatX
Kl(1, p) = o(π(x))(x → ∞).
p≤x

Unfortunately the horizontal Sato-Tato conjecture is still open, and very little is known about thissum. Fouvry and Michel [FM07] have obtained significant partial progress on replacing primes byalmost-primes: they show thatX
x(|Kl(1, n)| ± Kl(1, n)) 
.
log
x
n<x
p|n⇒p>x1/23.9

In particular, it follows that there are infinitely many sign changes in the set {Kl(1, n), ω(n) ≤ 23}.
After further work by many authors [FM03a, SF07, SF09, Mat11, Xi15a], the best know current result isdue to Xi [Xi15b] and shows that there are infinitely many sign changes in the set {Kl(1, n), ω(n) ≤ 7}.
We refer to the recent preprint [Xi18] for more references and related questions.
Date: March 14, 2018.
2010 Mathematics Subject Classification. 11L05, 11N36 (Primary); 11N75, 11L20, 11M20 (Secondary).
We thank É. Fouvry for helpful remarks concerning this work, and an anonymous referee for comments which helped
improve the manuscript. Part of this work was done during a visit of JM to Aix-Marseille university, supported by the
French-Austrian joint project MuDeRa (FWF I-1751-N26, ANR-14-CE34-0009).
1

2

SARY DRAPPEAU AND JAMES MAYNARD

Landau-Siegel zeroes. In this work we study the implications of the existence of Landau-Siegel zeroeson this question. For D ≥ 3 an integer and χ (mod D) a real primitive character, define the valueηχ := L(1, χ) log D.
It would follow from GRH that ηχ  (log D)/ log log D and so ηχ → ∞ as D → ∞. Unconditionally,
however, Siegel’s non-effective lower-bound [Sie35] ηχ ε D−ε remains unsurpassed. It is as yet noteven known if ηχ is bounded away from 0; this is equivalent to the non-existence of real zeroes of L(s, χ)close to 1. On the other hand, if there were a sequence of characters χ1 (mod D1 ), χ2 (mod D2 ), . . .
such that ηχi → 0, then many desirable consequences would follow: the existence of twin primes [HB83],
equidistribution of primes in arithmetic progressions to large moduli [FI03], primes in very short intervals [FI04], or prime values of discriminants of elliptic curves [FI05, FI13], for example.
We are interested in obtaining cancellations in the sum
X
Kl(1, p)p≤xwhich go beyond the bound 2π(x) implied by the Weil bound should such exceptional characters exist.
Theorem 1.1. Let ε > 0. Then there are constants A, B > 0, depending only on ε, such that forD ≥ 3, x ≥ DA and any primitive real character χ (mod D), we haveX
Kl(1, p) ≤ π(x) ε + BL(1, χ) log x .
p<xThis statement is unconditional, but is only non-trivial if the value ηχ is suitably small. We note that
if there is a sequence of characters with ηχi → 0, then Theorem 1.1 shows that for a suitable sequenceof values of x,X
Kl(1, p) = o(π(x)),
p<xas predicted by the horizontal Sato-Tate conjecture.
Unfortunately we do not know unconditionally the expected lower boundX(1.1)|Kl(1, p)|  π(x).
p<xIn particular, even if there was a sequence of characters with ηχi → 0, we would not be able to conclude
from Theorem 1.1 that there are even infinitely many sign changes in the sequence Kl(1, p). If insteadof considering primes we consider products of exactly two primes, then the equivalent lower boundto (1.1) is known thanks to work of Fouvry-Kowalski-Michel [FKM14]. For technical reasons, whenworking with products of two primes we consider the variant
 pq 
X
log(pq)(log p)(log q) Kl(1, pq)
S(x) =
φxp,q prime
where φ : R+ → C is a smooth function compactly supported inside R∗+ . We note that the Weil boundimplies unconditionally that S(x)  x log2 x, whilst a variant of the horizontal Sato-Tate conjecture
would suggest that we should have S(x) = oφ (x log2 x).
Theorem 1.2. Let ε > 0. Then there exist A, B > 0, depending at most on ε, such that for any D ≥ 3,x ≥ DA and any primitive real character χ (mod D), we have
(1.2)
|S(x)| φ x(log x)2 ε + BL(1, χ) log x .
The implied constant depends only on the function φ.
As with Theorem 1.1, this is unconditional but non-trivial only if a sequence of exceptional characters
exist. Thus, in the presence of Siegel zeros, we are able to establish infinitely many sign changesof Kl(1, pq).
Corollary 1.3. For some absolute constants A, c > 0, if ηχ ≤ c, then every interval [x, 2x] ⊂[DA , D100A ] contains two numbers (n1 , n2 ) with ω(n1 ) = ω(n2 ) = 2, and Kl(1, n1 ) Kl(1, n2 ) < 0.
Proof. Choose φ to be real-valued with φ ≥ 1[1,2] , and consider the unsigned sum
XA(x) =(log pq)(log p)(log q) |Kl(1, pq)| .
p,qx<pq≤2x

SIGN CHANGES OF KLOOSTERMAN SUMS AND EXCEPTIONAL CHARACTERS

3

We have the following lower bound, due to Fouvry-Kowalski-Michel [FKM14, Proposition 5.1]:
A(x)  x(log x)2(1.3)(x ≥ 2).
Comparing (1.2) and (1.3) yields the claimed statement.

Notations. We denote by P − (n) (resp. P + (n)) the smallest (resp. largest) prime factor of n, withthe conventions P − (1) = ∞ and P + (1) = 1. The letter p denotes a prime.
2. Outline
If there is a character χ (mod D) such that L(1, χ) is very small, then ‘most’ primes p ∈ [DA , D100A ]have χ(p) = −1 = µ(p), for some suitable constant A. By multiplicativity, this means that the(poorly understood) Moebius function µ can be well-approximated by the character χ (which is periodic(mod D), and so better understood) on integers n < D100A with no small prime factors. In particular,
for n ≤ D100A
Λ(n) = (µ ∗ log)(n) ≈ (χ ∗ log)(n),(2.1)

and so we can approximate Λ by the convolution
Pof two simpler sequences. By applying the hyperbola
method, this would allow us to estimate a sum n<x Λ(n)an provided we could suitably estimate
n
X
X
X
X
n
χ(d)
an log , and
log d
an χ
(2.2)
d
d
n<x
d≤Y1

n≤x
d|n

d≤Y2

d|n

for some choice of Y1 , Y2 with Y1 Y2 = x = D100A . Often one can suitably estimate such sums forY1 = Y2 = x1/2−ε , which just falls short of this requirement. Much of the work on the distributionof primes under the assumption of a Siegel-Landau zero followed this strategy, and the key technicalchallenge is then to obtain a suitable estimate for one of the sums in (2.2) with Y1 or Y2 slightly beyond
x1/2 .
In our situation, an = Kl(1, n), and estimates for the two sums in (2.2) with Y1 = Y2 = x1/2−ε are
obtained in essentially the same way as Fouvry and Michel [FM07]. Unfortunately we do not know
how to extend
this work beyond x1/2 , and so this strategy fails. However, in the convolution identityPΛ(n) = d|n µ(d) log(n/d) it is only terms with d ∈ [x1/2−ε , x1/2+ε ] which we are unable to handle. We
note the alternative identity
√ 

X
n
.
Λ(n) =µ(d) logdd|n√√The presence of the term log( n/d) means we expect that terms with d ≈ n to contribute less, and so
we might hope that these central values would be negligible. This is P
a variant of the idea that Bombieri
introduced in his asymptotic sieve [Bom76], where terms in Λ2 (n) = d|n µ(d) log2 (n/d) from d < n1−ε
could be handled by assumptions on equidistribution of congruence sums, and terms with d ∈ [n1−ε , n]could be bounded by virtue of the fact that log2 (n/d) was small in this range.
Unfortunately, as in Bombieri’s work, this strategy fails if we wish to count primes. To maintain thefeature that the support is essentially restricted to numbers with no small prime factors one multiplies
by a short sieve weight, which loses a factor ε2 from the two variables d and n/d.
√ This precisely cancels
out the gains of a factor ε2 coming from the range of d and from the size of log( n/d). Whilst this issue
might appear to be a technicality, at least in Bombieri’s work this is an expression of the fundamental
parity√ problem of sieve methods. If instead we counted with a weight involving a higher power oflog( n/d) (thereby counting products of a bounded number of primes), then this strategy can succeed.
In our case, we are interested in an = Kl(1, n). Although in general we expect the Weil bound|Kl(1, n)| ≤ 2ω(n) to be essentially sharp, for most integers n we expect |Kl(1, n)| is actually muchsmaller than this. Indeed, the ´horizontal Sato-Tate conjecture would predict that for any fixed a, theπaverage size of |Kl(a, p)| is π2 0 2| cos(t)| sin2 (t)dt = 8/3π < 1. By multiplicativity, we might then
expect |Kl(1, n)| ≈ (8/3π)ω(n) on average over n. Fouvry and Michel [FM03b, FM06] combined anargument of Hooley [Hoo64] based on the identity Kl(1, ab) = Kl(a2 , b) Kl(b2 , a) (for coprime a, b), withthe vertical Sato-Tate law, to show unconditionally that the factor 2ω(n) can be indeed improved to(8/3π)ω(n) when considering suitable averages. Since 8/3π < 1, numbers with a larger number of prime
factors contribute less to the problematic sums, and so there is less of a loss from being restricted to ashort sieve weight. This ultimately allows us to win an additional factor of (ε1−8/3π )2 for these sums

4

SARY DRAPPEAU AND JAMES MAYNARD

involving middle sized d, which
P is enough to conclude that such terms make a negligible contribution,and so we are able to bound p<x Kl(1, p).
3. Preparatory Lemmas
3.1. Level of distribution for twisted Kloosterman sums. Here we make precise the claim thatthe sums in (2.2) can be estimated with Y1 = Y2 = x1/2−ε by a variation of the work of Fouvry-Michel[FM07].
Proposition 3.1. Let ε > 0 be fixed. There exists η = η(ε) > 0, such that for any real x ≥ 2, positive
integer D ≤ xη , all characters χ (mod D), and any smooth function φ : R∗+ → C compactly supported
inside R∗+ , we havenXX
χ(n) Kl(1, n) ε,φ x1−η .
φ
(3.1)
x
1/2−ε
n≡0 (mod q)q≤xRemark. Note that the case D = 1 of the previous statement is a weaker form of Proposition 2.1
of [FM07].
Proof. We may plainly assume that the sums are restricted to (q, D) = (n, D) = 1. The bound we claim
is a variant of Proposition 2.1 of [FM07], which is based on the Kuznetsov formula [Kuz80, DI83], theWeil bound for Kloosterman sums, and a uniform bound θ ≤ 1/4 − ε towards Ramanujan-Petersson,which was first due to Luo-Rudnick-Sarnak [LRS95].
The difference in our case is the presence of the character. Recently, Blomer and Milićević [BM15]
have succeeded in analysing such sums in the context of modular forms with non-trivial nebentypus;
another argument was used in [Dra17], which is simpler for our purpose here. We will rely on work of
Topacogullari [Top15] to estimate the spectral sums.
In our case, we will use the notations and normalization described in section 4.1.2 of [Dra17]. Our aim
is to apply the Kuznetsov formula [Dra17, Lemma 4.5] for the group Γ0 (qD), nebentypus χ̄ (mod D),with cusps ∞ and 1/q, and parameters m ← D, n ← 1. For each q in the left-hand side of (3.1),
Lemma 4.3 of [Dra17], with the choice of scaling matrices (depending only on q and D) given there,
yields√χ(n) Kl(1, n) = χ(q)e(−q̄/D)S∞,1/q (D, 1; n D).
κLet κ ∈ {0,p 1} be such that χ(−1) = (−1) . The Kuznetsov formula with test function ψ(t) =φ(4π/(tx)) 4π/(tx) yields
n
X
√ 
χ(q)e(q̄/D)φ
χ(n) Kl(1, n) = x H + E + M ,xn≡0 (mod q)

where
X

H=

X

ψ̇(k)Γ(k)ρf,∞ (D)ρf,1/q (1),k>κf ∈Bk (qD,χ̄)k≡κ (mod 2)E=

X 1 ˆ ∞ ψ̃(t)ρc,∞ (D, t)ρc,1/q (1, t)dt,
4π −∞ cosh(πt)
c sing.

M=

X
f ∈B(qD,χ̄)ψ̃(tf )
ρf,∞ (D)ρf,1/q (1).
cosh(πtf )Split M = M1 + M2 where M1 is the contribution of those f with tf ∈ R, and M2 is the remainder
contribution, which consists of f with tf ∈ [−i/4, i/4]. Consider first M2 . Using the bound |ψ̃(tf )| x2|tf | , we obtain by the Cauchy-Schwarz inequality
 X
1/2  X
1/2
M2 
x2|tf | |ρf,∞ (D)|2x2|tf | |ρf,1/q (1)|2.
f ∈B(qD,χ̄)√f ∈B(qD,χ̄)

We have q D ≤ x1/2−ε+η ≤ x1/2 , so that Lemma 2.9 of [Top15] may be applied to both sums, whichyields, as x → ∞,√ o
 √xD 4θ n
D
o(1)
(3.2)
M2  x1+
 x2η (x/q 2 )2θ .
qq

SIGN CHANGES OF KLOOSTERMAN SUMS AND EXCEPTIONAL CHARACTERS

5

The contribution of E, H and M1 is handled by similar standard arguments [DI83, page 267], using
instead Lemma 2.8 of [Top15]. We obtain
H + E + M1  xo(1) D1/2  xη

(3.3)

as x → ∞. Grouping our bounds (3.3) and (3.2), we findnXXχ(n) Kl(1, n) ε,φ x1+η−ε + x1+2η−(1/2−2θ)ε .
φ
x
1/2−ε
q≤x

n≡0 (mod q)

By work of Kim-Sarnak [Kim03], we are ensured that θ ≤ 7/64 < 1/4, and therefore our claimed boundfollows if Q ≤ x1/2−ε and 0 < η ≤ 3ε/16.
P
From Proposition 3.1 and the convolution µ2 (n) =d2 |n µ(d), we deduce the analogous boundwith n restricted to square-free numbers.
Corollary 3.2. In the setting and notations of Proposition 3.1, we have
n
X
X
(3.4)
µ2 (n)φχ(n) Kl(1, n) ε,φ x1−η
x
1/2−ε
q≤x

n≡0 (mod q)

for a possibly smaller value of η > 0, but depending on ε at most.
3.2. Sieve weights. To control the terms around the central point d ∈ [x1/2−ε , x1/2+ε ] we will introduce a short sieve weight to maintain the feature that our sums are essentially supported on integers
free of small prime factors. This is also crucial for carrying out the approximation (2.1) as effectively
as possible. We recall the construction of the β-sieve from [FI10], and a few of its relevant properties
for our application. LetXθ(n) =
ξdd|n

be an upper-bound β-sieve of level y and dimension κ ≥ 0, for the primes less than z. The parameters y,z and κ will be chosen later to be small powers of x with the condition that z 2 ≤ y. (We will ultimately
2choose z = xε/(log ε) and y = xε .) In particular, we have ξ1 = 1 and |ξd | ≤ 1; (ξd ) is supported on
integers up to y free of prime factor > z, and we have θ(n) ≥ 0 for all n ≥ 1. Moreover, for any
multiplicative function f with f (p) ∈ [0, κ], we haveX ξd f (d)Y
f (p) −1
(3.5)

1+
.
d
p
p≤z

d

For some technical simplifications, we will work with the smoothed version
n X nX
θ0 (n) :=ξd log
=
θ
Λ(d).
d
d
d|n

d|n

Note that θ0 (n) = log n if P − (n) > z.
The key property we use of θ0 is the following estimate on averages of multiplicative functions weighted
by the sieve weight.
Lemma 3.3. Let the multiplicative function f be supported on squarefree numbers, with 0 ≤ f (p) ≤ κ.
Then
X
Y 
f (p) (3.6)
θ0 (n)f (n)  x
1+
.
p
n≤x

z<p≤x

Proof. Denote S the sum on the left-hand side of (3.6). By definition of θ0 , we have
n
XX
X
X
S=
f (n)θ
log p ≤
f (n)θ(n)f (p) log ppn≤x p|nn≤xby positivity of the summand. Therefore,X θ(n)f (n)S κ x
≤x
n
n≤x

p≤x/n

X
P + (n)≤x

θ(n)f (n).
n6SARY DRAPPEAU AND JAMES MAYNARD

Opening the convolution in θ(n), we arrive at
X ξd f (d)S κ xd+
P (d)≤x

(3.7)
=x

Y

1+

p≤x

By (3.5), the sum over d above is κf (p) pX
P + (n)≤x(n,d)=1

X
P + (d)≤xQp≤z (1 +f (n)nξd f (d) Y 
f (p) −1
1+
.
d
p
p|d

f (p) −1, and our claimed bound follows.
p )

4. Argument for the upper boundIn what follows, we let ε > 0 be fixed, and η > 0 be a small parameter, to be chosen in terms of ε.
We assume that D ≤ xη .
4.1. Initial setup. For a positive integer r, define the arithmetic functions Λ̃r (n) by
√ r

X
n
.
Λ̃r (n) =µ(d) logdd|n

We note that Λ̃1 (n) = Λ(n) and Λ̃2 (n) = Λ2 (n) − Λ(n) log n. On squarefree n, Λ̃1 is supported onprimes, and Λ̃2 is supported on products of exactly two primes.
Let φ be smooth bounded compactly supported function approximating the characteristic function 1[0,1] of the interval [0, 1] from below, such that φ and 1[0,1] differ only on [0, δ] and [1 − δ, 1], forsome small constant δ > 0. If we can show that for any ε > 0n
Xµ2 (n)φ(4.1)(log n)Λ̃(n) Kl(1, n) φ εx log x + Oε (xL(1, χ) log2 x),xnthen, using the Weil bound to control the contribution from n < δx or n > (1 − δ)x, we see from partial
summation thatX
Kl(1, p)  δπ(x) + Oδ (επ(x)) + Oδ,ε (L(1, χ) log x).
p≤x

Thus, by choosing δ small and then ε sufficiently small in terms of δ, we see it suffices to show (4.1) for
any fixed ε > 0.
Since µ2 (n)Λ̃(n) is only supported on primes, we have
n
n
X
X
(log n)Λ̃(n) Kl(1, n) =µ2 (n)φθ0 (n)Λ̃(n) Kl(1, n) + Oε (xε ).
µ2 (n)φxxn(n,D)=1

For products of two primes, we note that for all x ≥ 2,X
(4.2)
Λ̃2 (n)  x,n≤xn6=pq for all p6=q,p,q primewith the main contribution arising from pairs n = pq 2 (p, q primes), whileX(4.3)
Λ̃2 (n)  x log log xn≤x(n,D)>1

for 2 ≤ D ≤ x. Thus pq 
X
φ(log pq)(log p)(log q) Kl(1, pq)xp,qp,q prime

X

=

φ

p,q
p,q prime

(4.4)

=

1
2

X
(n,D)=1

 pq 
θ0 (pq)(log p)(log q) Kl(1, pq) + O(x(log x)(log z))xµ2 (n)φn
θ0 (n)Λ̃2 (n) Kl(1, n) + O(x(log x)(log z)).
x

We have used both (4.2) and (4.3) in the last line.

SIGN CHANGES OF KLOOSTERMAN SUMS AND EXCEPTIONAL CHARACTERS

7

Thus, we see that to establish Theorem 1.1 and Theorem 1.2 it is sufficient to show for every ε > 0,
every smooth compactly supported φ and each r ∈ {1, 2} thatnX
θ0 (n)Λ̃r (n) Kl(1, n) φ εx logr x + Oε (xL(1, χ) logr+1 x),(4.5) S(x) = S(x, r) =µ2 (n)φx(n,D)=1

where D and x satisfy the hypotheses of Theorem 1.1 and χ is any real primitive character (mod D).
LetX
ν(n) = (µ ∗ µχ)(n) =µ(u)µ(v)χ(v).
n=uv

We write(4.6)

Λ̃r (n) =


1 X
ab r
ν(b)χ(a) log
,
r
2
c
n=abc

so that for r ∈ {1, 2},(4.7)

S(x) =

1
r2r

Xµ2 (abc)φ(a,b,c)∈N3(abc,D)=1

 abc xab r
θ0 (abc)ν(b)χ(a) log
Kl(1, abc).
c

We split the sum over (a, b, c) into b > 1 or b = 1 and min{a, c} ≤ x1/2 /y 3 , or b = 1 and min{a, c} >x1/2 /y 3 . This gives
r2r S(x) = SL (x) + SN (x) + SC (x),

(4.8)
with
(4.9)

X

SL (x) =µ2 (abc)φ(a,b,c)∈N3
b>1(abc,D)=1(4.10)SN (x) =X abc ab rKl(1, abc),
θ0 (abc)ν(b)χ(a) logxcµ2 (ac)φ(a,c)∈N2√
min{a,c}≤ x/y 3(ac,D)=1(4.11)SC (x) =X(a,c)∈N2√
min{a,c}> x/y 3(ac,D)=1µ2 (ac)φa r
Kl(1, ac),θ0 (ac)χ(a) logxc ac a r
θ0 (ac)χ(a) log
Kl(1, ac).
xc

 ac 

We can control SL (x) by L(1, χ), since µ2 ν is supported on integers with all prime factors satisfying
χ(p) = 1 and θ0 is concentrated on integers free of small prime factors. Thus if L(1, χ) is small, we
expect the support to be a lacunary sequence and so the sum SL (x) to be small.
We can control SN (x) by the level of distribution estimates of Section 3.1, since the sum over one ofa or c is a long sum.
Finally, SC is the contribution near the central point, and we will show this is small since | log(a/c)| ≤6 log y + Oφ (1), which is small compared with log x, a is resticted to a short range on the logarithmic
scale, and Kl(1, n) is typically small on numbers with many prime factors.
4.2. Bounding the lacunary sum SL . We begin by bounding SL (x) in terms of L(1, χ).
22

Lemma 4.1. For all small enough ε > 0, and D8 ≤ xε ≤ z ≤ y 1/(log ε) , we haveSL (x) φ x(log x)r ε−κ−2 L(1, χ) log x + ε ,Proof. We note that for squarefree n, we have |ν(n)| = (1 ∗ χ)(n), and we recall that θ0 (n) = (Λ ∗ θ)(n).
Thus, bounding the summand of SL (x) and letting d = ac, and then expanding the definition of θ, we8SARY DRAPPEAU AND JAMES MAYNARD

find
SL (x)  (log x)rX(1 ∗ χ)(b)θ0 (bd)τ (bd)2 µ(bd)2

dbx
b>1

= (log x)rX(1 ∗ χ)(b)θ(be)Λ(f )τ (bef )2 µ(bef )2 .

bef xb>1

When f > z is prime we see θ(be) = θ(bef ). Thus the contribution from f > z can be bounded byX(4.12)(log x)r+1(1 ∗ χ)(b)θ(bg)µ2 (bg)τ (bg)4 .
bgx
b>1

The contribution from f ≤ z can be bounded by
X
X(4.13)(log x)rΛ(f )(1 ∗ χ)(b)θ(be)µ2 (be)τ (be)2 .
f zbex/fb>1

These are precisely sums of the type considered in Section 24.7 of [FI10]. Indeed, there it is established[FI10, equation (24.58)] that if D8 ≤ z ≤ y ≤ Y 1/8 and ρ is a completely multiplicative function with0 ≤ ρ(p) ≤ κ/6 then log Y κ/2+1X
Y L(1, χ) log Y + e−t(1 ∗ χ)(b)θ(ab)ρ(ab) 
,
log Y
log z
ab<Y
b>1

where t = log y/ log z. Since µ2 (n)τ (n)k ≤ 2kΩ(n) (k = 2, 4), we may apply this to the bounds (4.12)
and (4.13) above, and find that provided D8 ≤ z ≤ y ≤ x1/9 and κ > 6 × 24 , we have log x κ/2+1
X,(log x)r+1(1 ∗ χ)(b)θ(bg)τ (bg)4  x(log x)r L(1, χ) log x + e−t
log z
bgx
b>1

(log x)r

X

X

Λ(a)

f z

bex/fb>1 log x κ/2+1 X Λ(f )(1 ∗ χ)(b)θ(be)τ (be)2  (log x)r−1 L(1, χ) log x + e−t
log z
f
f z


 log x κ/2
 x(log x)r L(1, χ) log x + e−t.
log z
Putting this together, we find log x κ/2+1
SL (x)  x(log x)r L(1, χ) log x + e−t
.
log z
2

2

If xε ≤ z ≤ y 1/(log ε) , then this givesSL (x)  x(log x)r ε−κ−2 L(1, χ) log x + ε ,(4.14)

as required.
4.3. Bounding the non-central sum SN .
Lemma 4.2. Let y ≥ xε . Then there is a constant η > 0 depending only on ε such thatSN (x) ε x1−η .

Proof. Denote by χ0 (n) the principal character (mod D). Opening the summation in θ0 (ac), we obtainn a2 r XX Xnµ2 (n)φχ0 (n) log
log
Kl(1, n)SN (x) ≤xnd√3d≤y a< x/yn≡0 (mod [a,d]) c2 r n
χ(n) log
log
Kl(1, n)xnd√d≤y c< x/y 3 n≡0 (mod [c,d])nXX(4.15) τ (q)(log x)r+1µ2 (n)φχ1 (n)(log(n/x))j Kl(1, n) .
sup
sup
x
√
χ1 (mod D) 0≤j≤r+1
+

X

X

X

q≤ x/yµ2 (n)φnn≡0 (mod q)

SIGN CHANGES OF KLOOSTERMAN SUMS AND EXCEPTIONAL CHARACTERS

9

Recalling that we assume y ≥ xε and applying Corollary 3.2, we obtain
SN (x) ε x1−η(4.16)
for some η depending at most on ε.

4.4. Bounding the central point sum SC .
2

2

Lemma 4.3. Let xε ≤ z ≤ y 1/(log ε) and y ≤ x2ε . Then we have
 log y 2
SC (x)  ε10 x(log x)r + ε3/10 x(log x)r.
log z
Proof. Recall that the sum SC (x) is defined in (4.11). By construction, for all (a, c) in the summation
range, we have | log(a/c)|  log y.
Therefore, by the triangle inequality and the support condition on φ, our task is to boundX(log y)rµ2 (ac)θ0 (ac) |Kl(1, ac)| .
x/y 3 ≤axy 3acx

Following an idea of Hooley [Hoo64], we let a parameter u ∈ [y, x1/4 ] be given, and decompose accordinglya = a1 a2 ,P + (a1 ) ≤ u,
P − (a2 ) > u,and similarly let c = c1 c2 . Notice that
θ0 (ac) = θ(a1 c1 ) log(a2 c2 ) + θ0 (a1 c1 )  θ(a1 c1 ) log x + θ0 (a1 c1 )since P − (a2 c2 ) > u > z. We writeX(4.17)µ2 (a1 a2 c1 c2 )θ0 (a1 a2 c1 c2 ) |Kl(1, a1 a2 c1 c2 )| = S1 + S2 ,a1 a2 c1 c2 xx1/2 /y 3 <a1 a2 x1/2 y 3
P + (a1 c1 )≤u<P − (a2 c2 )where S1 is the contribution of a1 c1 ≤ x1/10 , and S2 is the contribution of a1 c1 > x1/10 .
Concerning S1 , we haveX
S1 
µ2 (a1 c1 ){θ(a1 c1 ) log x + θ0 (a1 c1 )}T (a1 , c1 ),a1 c1 ≤x1/10
P + (a1 c1 )≤uwithX

T (a1 , c1 ) =µ2 (a2 c2 ) |Kl(1, a1 a2 c1 c2 )| .

a2 c2 x/a1 c1
P − (a2 c2 )>ux1/2 /y 3 <a1 a2 x1/2 y 3

By a reasoning identical to page 277 of [FM03b], for a1 c1 ≤ x1/10 , we haveXX
T (a1 , c1 ) ≤Kl(b2 , a1 c1 )µ2 (a2 c2 )2ω(a2 c2 )a2 c2 x/a1 c1
x1/2 /a1 y 3 <a2 x1/2 y 3 /a1
P − (a2 c2 )>uba2 c2 ≡1 (mod a1 c1 )b (mod a1 c1 )(b,a1 c1 )=1Xb (mod a1 c1 )(b,a1 c1 )=1

=

Kl(b2 , a1 c1 )1  log x 2
xa1 c1 ϕ(a1 c1 ) log x log u
1/2x
log y  log x 4
a1 c1 ϕ(a1 c1 ) log2 x log uxXX/a1 y 3 <a2 x1/2 y 3 /a1
P − (a2 )>uKl(b2 , a1 c1 )b (mod a1 c1 )(b,a1 c1 )=1x log y  log x 4 2ω(a1 c1 ) κ(a1 c1 )a1 c1
log2 x log u

for a multiplicative function κ. Fouvry and Michel [FM03b, equation (3.4)] show that4
κ(p) =+ O(p−1/4 ),
3π

2ω(a2 )a2

10

SARY DRAPPEAU AND JAMES MAYNARD

which they deduce from Katz’s Sato-Tate law for Kloosterman sums [Kat88]. We insert this back into
S1 , and we relax the summation conditions a1 c1 ≤ x1/10 . Letting
f (n) := µ2 (n)4ω(n) κ(n),we obtain
S1 (4.18)

x log y  log x 4 X (θ(n) log x + θ0 (n))f (n).
nlog2 x log u P + (n)≤uNote that16+ O(p−1/4 )(p ≤ u).
3π
Taking Euler products, computations similar to (3.7) yield log u 16/(3π)
X θ(n)f (n)Y 
f (p) 1+
(4.19)

.
n
p
log z
+
f (p) =z<p≤u

P (n)≤uOn the other hand,X θ0 (n)f (n)P + (n)≤un=

X f (q) log qqq≤uq primeX
P + (n)≤u(n,q)=1

 log u 16/(3π)θ(n)f (n) (log u)nlog z

by dropping the condition (n, q) = 1 and using (4.19). Inserting in (4.18), we find
 log y  log x 4  log u 16/(3π).
(4.20)
S1  xlog x log u
log zConsider now the contribution S2 to (4.17). Using the Weil bound |Kl(1, n)| ≤ 2ω(n) , we haveXS2 ≤
θ0 (n)g0 (n),
nx

where

(
µ2 (n)4ω(n)g0 (n) :=0Q
if p|n,p≤u p > x1/4 ,otherwise.

By Rankin’s inequality, we have g0 (n) ≤ x−1/(4 log u) g1 (n), whereY
g1 (n) := µ2 (n)4ω(n)p1/ log u .
p|n,p≤uNote that the function g1 is multiplicative with(
4p1/ log ug(p) =
4

if p ≤ u
if p > u.

In any case we have g(p) ≤ 4e. Assuming κ ≥ 4e, we obtain log x 4  log u 4enXlog x oexp −.
S2 ≤ x−1/(4 log u)
θ0 (n)g1 (n)  x
log u
log z
4 log unx22

We now choose u = x1/(log ε) . Recalling that we have assumed xε ≤ z, we see that this gives
S2 (x)  ε10 x.

(4.21)
2

Similarly, we substitute u = x1/(log ε) into (4.20). Using the assumption z ≤ y ≤ x2ε and noting that2 − 16/3π > 3/10, we find
1 4  log y 2−16/3π  log y 16/3π(log y)S1  x log x log
ε
log x
log z
 log y 2
(4.22).
 ε3/10 x log x
log z
Recalling that SC (x)  (S1 + S2 )(log y)r , that r ∈ {1, 2}, and inserting the bounds (4.22) and (4.21),
we obtain
 log y 2
SC (x)  ε10 x(log x)r + ε3/10 x(log x)r,log z
as required.
SIGN CHANGES OF KLOOSTERMAN SUMS AND EXCEPTIONAL CHARACTERS

11

4.5. Conclusion. We choose
y = xε ,2z = xε/(log ε) .

With this choice, we see that y and z satisfy the assumptions of lemmas 4.1, 4.2 and 4.3. Thus, provided
2
D8 ≤ xε we may apply these Lemmas, giving
S(x)  SL (x) + SN (x) + SC (x)1 4 ε−κ−2 xL(1, χ)(log x)r+1 + εx(log x)r + Oε (x1−η ) + ε3/10 x(log x)r log
εfor some quantity η > 0 depending only on ε. Reinterpreting ε, we obtain the claimed statement (4.5),
which then gives Theorems 1.1 and Theorem 1.2.

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[BM15]

12

SARY DRAPPEAU AND JAMES MAYNARD

Aix Marseille Université, CNRS, Centrale Marseille, I2M UMR 7373, 13453 Marseille, France
E-mail address: sary-aurelien.drappeau@univ-amu.fr
Mathematical Institute, Radcliffe Observatory Quarter, Woodstock Road, Oxford OX2 6GG, United
KingdomE-mail address: james.alexander.maynard@gmail.com