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J. Bajpai et al. Res. Number Theory (2022) 8:18
https://doi.org/10.1007/s40993-022-00310-3
RESEARCH
Exponential sums in prime fields for
modular forms
Jitendra Bajpai1, Subham Bhakta2* and Victor C. García3
*Correspondence:
subham.bhakta@mathematik.uni-
goettingen.de
12Mathematisches Institut,
Georg-August-Universität
Göttingen, Göttingen, Germany
Full list of author information is
available at the end of the article
Abstract
The main objective of this article is to study the exponential sums associated to Fourier
coefficients of modular forms supported at numbers having a fixed set of prime factors.
This is achieved by establishing an improvement on Shparlinski’s bound for exponential
sums attached to certain linear recurrence sequences over finite fields.
Keywords: Linear recurrence sequence, Exponential sums, Modular forms
Mathematics Subject Classification: Primary 11F30, 11L07; Secondary 11P05, 11B37,
11F80
Contents
1 Introduction ......................................... 2
1.1 Linear recurrence sequences and exponential sums ................. 3
1.2 Main results for exponential sum with modular forms ............... 5
1.3 Waring type problems over finite fields ....................... 7
2 Exponential sums with linear recurrence sequences ................... 8
2.1 Proof of Theorem 1 .................................. 10
3 Exponential sums for modular forms ............................ 14
3.1 Order of the roots of the characteristic polynomial ................. 14
3.2 Proof of Theorem 2 .................................. 15
Case 1. ∈P−1:....................................... 16
Case 2. ∈P1:........................................ 17
3.3 Consequences of Theorem 2 ............................. 19
4 Exponential sums for modular forms: beyond eigenforms ................ 20
4.1 GST: Beyond Sato-Tate ................................ 22
4.2 A consequence of GST ................................. 22
4.3 Proof of Theorem 3 .................................. 24
5 Exponential sums for modular forms: the inverse case .................. 24
5.1 Proof of Theorem 4 .................................. 24
5.2 Proof of Theorem 5 .................................. 26
6 Impact on Waring-type problems ............................. 27
6.1 Waring-type problems with linear recurrence sequences .............. 27
6.2 Waring-type problems for modular forms ...................... 30
6.3 Bound of non-linearity of a linear recurrence sequence .............. 30
References ............................................ 31
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18 Page 2 of 32 J. Bajpai et al. Res. Number Theory (2022) 8:18
1 Introduction
Let fbe a modular form of weight k∈2Zand level Nsuch that it has a Fourier expansion
f(z)=∞
n=1
a(n)e2πinz,(z)≥0,
with a(n)bethenth Fourier coefficient. In this article, we shall restrict to the family of
modular forms with rational coefficients, that is, f(z)witha(n)∈Qfor every n. We first
consider Hecke eigenforms or simply eigenforms in the space of cusp forms of weight k
for the congruence subgroup 1(N) with trivial nebentypus. When fis an eigenform with
integer Fourier coefficients, it follows from Deligne-Serre that for any prime ,there exists
a corresponding Galois representation
ρ()
f:GalQ/Q−→ GL2(Z)
such that tr(ρ()
f(Frobp)) =a(p),for any prime pN.For a quick reference about this
correspondence, we refer the interested reader to [8, Chapter 3].
In particular, a(p) (mod ) is determined by the trace of the corresponding Frobenius ele-
ment in GL2(Z/Z)=GL2(F).In certain cases, Chebotarev’s density theorem implies
that given any λ∈F,there exists a prime psuch that a(p)≡λ(mod ).However, the set
of such primes pcome with density strictly less than 1. So what about the other primes p?
In this context, we address the following Waring-type question.
Question Does there exist an absolute constant s such that for any given primes p and ,
any element of Fcan be written as a sum of at most s elements of the set {a(pn)}n≥1?
A related question was studied by Shparlinski in [24] for the Ramanujan’s τfunction,
where τ(n) is defined by the identity
(z)=q
n≥1
(1 −qn)24 =
n≥1
τ(n)qn,with q=exp(2πiz).
In [24], it is proved that the set {τ(n)}n≥1is an additive basis modulo any prime ,that is,
there exists an absolute constant ssuch that the Waring-type congruence
τ(n1)+···+τ(ns)≡λ(mod )
is solvable for any residue class λ(mod ).
Shparlinski’s work was later generalized by Garaev, García and Konyagin over the global
field Q.More precisely, in [10], the authors proved that for any λ∈Z, the equation
s
i=1
τ(ni)=λ
always has a solution for s=74,000.
Later García and Nicolae [12] extended this result for coefficients a(n) of normalized
Hecke eigenforms of weight kin Snew
k(0(N)). More precisely, they proved that for any
λ∈Z, the equation
s
i=1
a(ni)=λ
always has a solution for some s≤c(f)withc(f) satisfying
c(f)(2N3/8)k−1
2+εk3
16 k+O(1)+εlog(k+1).
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J. Bajpai et al. Res. Number Theory (2022) 8:18 Page 3 of 32 18
The proof of the above two results are connected to the identity a(p2)=a2(p)−pk−1and
the solubility of the equation
pk−1
1+···+pk−1
s=N, for primes p1,...,p
s.
We are studying the finite field version of this additivity problem by obtaining nontrivial
exponential sums associated with coefficients of modular forms, in the sense of [24].
We are working with the class of forms that García and Nicolae [12] considered, but
with Fourier coefficients evaluated only at prime powers. Our results are recorded in
Corollaries 15 and 16. Our main tool is Theorem 1which provides a nontrivial bound
for exponential sums with coefficients of modular forms. To study this problem, we shall
primarily focus on the exponential sums of type
max
ξ∈F∗
n≤τ
e(ξa(pn))
where p, are primes, and τis a suitable parameter which we shall specify later. This is
done in Theorems 2and 3.
When fis a normalized eigenform, it is well known that a(n) is a multiplicative function
and for any prime pNsatisfies the relation
a(pn+2)=a(p)a(pn+1)−pk−1a(pn),n≥0.(1)
Moreover, we have a(pn)=a(p)nfor any prime p|N. These facts come from the
properties of Hecke operators, see [5, Proposition 5.8.5]. If a(p)∈Q,then one can
consider a(p) (mod )∈Fnaturally for any large enough prime .For instance, can be
taken to be any prime not dividing the denominators of the Fourier coefficients. On the
other hand, any cuspform can be uniquely written as a C-linear combination of pairwise
orthogonal eigenforms with Fourier coefficients coming from C.See[5, Chapter 5] for a
brief review of the Hecke theory of modular forms. However, here we are concerned with
all such cuspforms which can be uniquely written as a Q-linear combination of pairwise
orthogonal eigenforms with Fourier coefficients coming from Q. Note that, in this case,
the sequence {a(pn)}is a linear recurrence sequence of possibly higher degree. We now
turn to discuss the basic theory of linear recurrence sequences. We will also discuss the
bounds of their associated exponential sums.
1.1 Linear recurrence sequences and exponential sums
Let r≥1 be an integer and pbe an arbitrary prime number. A linear recurrence sequence
{sn}of order rin Fpconsists of a recursive relation
sn+r≡ar−1sn+r−1+···+a0sn(mod p),with n=0,1,2,... ,(2)
and initial values s0,...,s
r−1∈Fp.Here a0,...,a
r−1∈Fpare fixed. The characteristic
polynomial ω(x)associated to {sn}is
ω(x)=xr−ar−1xr−1−···−a1x−a0.
We see from Eq. (1) that {a(pn)}is a linear recurrence sequence of order 2 when fis an
eigenform. We shall prove the results in Sect. 2, by studying exponential sums associated
to a much more general class of linear recurrence sequences.
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18 Page 4 of 32 J. Bajpai et al. Res. Number Theory (2022) 8:18
Under certain assumptions, linear recurrence sequences become periodic modulo p,
see [15, Lemma 6.4] and [18, Theorem 6.11].
Let pbe a prime number and ω(x) be the characteristic polynomial of a linear recurrence
sequence {sn}defined by Eq. (2). If (a0,p)=1 and at least one of the s0,...,s
r−1are not
divisible by p, then the sequence {sn}is periodic modulo p, that is for some T≥1,
sn+T≡sn(mod p),n=0,1,2,... .
The least positive period is denoted by τ.Moreover, τ≤pr−1andτdivides Tfor any
period T≥1 of the sequence {sn}.
In 1953, Korobov [16] obtained bounds for rational exponential sums involving linear
recurrence sequences in residue classes. In particular, for the fields of order p, if {sn}is a
linear recurrence sequence of order rwith (a0,p)=1andperiodτ, it follows that
n≤τ
ep(sn)≤pr/2.(3)
Note that such a bound is nontrivial if pr/2<τand asymptotically effective only if
pr/2/τ →0asp→∞.Estimate (3) is optimal in general terms, indeed Korobov [15]
showed that there is a linear recurrence sequence {sn}with length rsatisfying
1
2pr/2<
n≤τ
ep(sn)≤pr/2.
In turn, for any given ε>0, it has been proved that there exists a class of linear recurrence
sequences with a better upper bound
n≤τ
ep(sn)≤τ1/2+ε.
However, the proof of the existence is ineffective in the sense that we do not know any
explicit characteristics of such family, see [7, Section 5.1].
The case when the associated polynomial ω(x) is irreducible in Fp[x],was widely studied.
In particular, from a more general result due to Katz [14, Theorem 4.1.1] it follows that if
ω(0) =1 then
n≤τ
ep(sn)≤p(r−1)/2.
Shparlinski [23] improved Korobov’s bound for all nonzero linear recurrence sequences
with irreducible characteristic polynomial ω(x)inFp[x].From [23, Theorem 3.1] we get
max
ξ∈F∗
p
n≤τ
ep(ξsn)≤τp−ε/(r−1) +r3/11τ8/11p(3r−1)/22,
for any given ε>0 and with period τsatisfying that
max
d<r
d|r
gcd(τ,p
d−1) <τp−ε.(4)
In particular, if ris fixed then the upper bound is non trivial for τ≥pr/2−1/6+ε.
We already pointed out that the inequality (3) is nontrivial for τ>pr/2+ε,so the most
important case occurs when τ≤pr/2+ε.Ifτ≤pr/2+ε,then condition (4) is needed
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J. Bajpai et al. Res. Number Theory (2022) 8:18 Page 5 of 32 18
to obtain a non trivial bound suggested by an example given in [23, Section 1]. In this
particular example, the exponential sums of type
(pm−1)/2
n=1
epTr(ag2n)=(pm−1)
2,
are considered for certain ain F∗
pmwith ga generator of F∗
pmand mbe any even integer.
It is worth noting that {Tr(ag2n)}is indeed a linear recurrence sequence of order min Fp.
Moreover, we consider the general case when the associated polynomial ω(x)isnot
necessarily irreducible, and deduce the following key result.
Theorem 1 Let p be a large prime number and ε>ε
>0. Suppose that {sn}is a
nonzero linear recurrence sequence with positive order and period τin Fpsuch that its
characteristic polynomial ω(x)has distinct roots in its splitting field, and (ω(0),p)=1.Set
ω(x)=ν
iωi(x)as a product of distinct irreducible polynomials in Fp[x], and for each i, αi
denotes a root of ωi(x). If all polynomials ωi(x)have the same degree, i.e. deg ωi(x)=r>1,
and the system τi=ord αi, satisfies
(a)max
d<r
d|r
gcd(τi,p
d−1) <τ
ip−ε,for any 1 ≤i≤ν,
(b)gcd(τi,τj)<pε,for some pair i= jalongwithFp(αi)∼
=Fp(αj),(5)
then there exists a δ=δ(ε,ε)>0such that
max
ξ∈F∗
p
n≤τ
ep(ξsn)≤τp−δ.(6)
It turns out that, this extends [2, Corollary] due to Bourgain, where all of the irreducible
factors have degree r=1,while Theorem 1deals with the case r≥2.
This will be of immense use in what follows, roughly because the characteristic polyno-
mial associated to {a(pn)}have degree two.
Theorem 1will be essential to establish Theorem 2and Corollaries 12 and 17.Our
approach, which relies on the sum-product phenomenon, provides an improvement over
Theorem 3.1 of [23] for the same class of linear recurrence sequences, obtaining non trivial
exponential sums in a larger range. To be more precise, if p(r) denotes the least prime
divisor of rthen any τ>pr/p(r)+εsatisfies
τp−ε>pr/p(r)≥max
d<r
d|r
gcd(τ,p
d−1).
In particular, our result works for any τ>pr/p(r)+ε,while bound in [23] is nontrivial if
τ>pr/2−1/6+ε.This is an improvement if p(r)>2,more precisely when ris odd.
1.2 Main results for exponential sum with modular forms
We now quickly discuss the main results obtained in this article. In the list, our first result
is the following:
Theorem 2 Let f (z)be an eigenform with rational coefficients a(n).LetPbe the set of
primes p such that a(pu)= 0for any u ∈N. Then the following is true.
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18 Page 6 of 32 J. Bajpai et al. Res. Number Theory (2022) 8:18
(i) The set of primes Psatisfies that given p ∈P,forany0<ε<1/2there exists a
δ=δ(ε)>0such that the following estimate
max
ξ∈F∗
n≤τ
e(ξa(pn))≤τ
−δ,(7)
holds for π(y)+O(y2ε)many primes ≤y, where the least period τof the linear
recurrence sequence {a(pn)}(mod )depends on both p and ,andπ(y)denotes the
number of primes up to y which is asymptotically equivalent to y
log y.
(ii) For the exceptional set of primes p /∈P, let u be the least natural number such that
a(pu)=0.Thenforany0<ε<1/2, there exists a δ=δ(ε)>0such that the
following estimate
max
ξ∈F∗
n≤τ
e(ξa(pn))=τ
u+1+O(τ
−δ+u).(8)
holds for π(y)+O(y2ε)many primes ≤y.
Roughly speaking, a newform of level Nis a normalized eigenform which is not a
cuspform of level Nfor any proper divisor Nof N. For details and basics on modular
forms, we refer the reader to [5]. A newform is said to have complex multiplication (CM)
by a quadratic Dirichlet character φif f=f⊗φ, where we define the twist as
f⊗φ=∞
n=1
a(n)φ(n)qn.
In part (i)ofTheorem2, the condition a(pu)= 0 holds for almost all prime pprovided
that fis a newform without CM. This is a consequence of Sato-Tate conjecture and we
shall discuss this again in the proof of Lemma 11. In particular, we have a non trivial
estimate for the following exponential sum
max
ξ∈F∗
n≤τ
e(ξa(pn)).(9)
Let us recall that any general cusp form fcan be uniquely written as C-linear combination
of eigenforms. These eigenforms will be called as components of f. We then have the
following result.
Theorem 3 Let f (z)be a cusp form which is not necessarily an eigenform, and can be
written as a Q-linear combination of newforms with rational coefficients. Suppose that
there are r2many components with CM, then under the assumption of GST hypothesis1
there exists a set of primes p with density at least 2−r2such that for any 0<ε<1/2there
exists a δ=δ(ε)>0for which the following estimate
max
ξ∈F∗
n≤τ
e(ξa(pn))≤τ
−δ,(10)
holds for cfπ(y)+O(y2ε)many primes ≤y, where cf>0is a constant.
1See Sect. 4.1 for the discussion about GST hypothesis.
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J. Bajpai et al. Res. Number Theory (2022) 8:18 Page 7 of 32 18
In both of the above theorems, we took a fixed prime pand looked for primes for
which a non trivial estimate to (9) holds. However, these results are valid for almost all
primes ,and we do not know explicitly which of the primes are being excluded in this
process. Thus, one may naturally ask, what if we now fix a prime and find out for how
many primes pthe sum at (9) is non trivial. In this regard, we have the following results.
Theorem 4 Let f (z)be a newform of weight k, without CM, and with integer Fourier
coefficients. Consider the set P=prime |(k−1,−1) =1. Then, for any fixed ε>0
and any large enough ∈P, the set of primes p satisfying
max
ξ∈F∗
n≤τ
e(ξa(pn))≤τ
−δ(11)
have density at least 1+Oε1
1−3ε,whereδ=δ(ε)is same as in Theorem 1.
Intuitively, this theorem can be regarded as the inverse of Theorem 2, and in this
analogy, the following result as the inverse of Theorem 3. Just for the sake of simplicity
we are assuming (k−1,−1) =1,which can be easily avoided and will be evident from
the proof of the following theorem.
Theorem 5 If f (z)is a cuspform, and can be written as Qlinear combination of r many
newforms without CM and with integer coefficients, such that all of these components
satisfies GST hypothesis. Then, for any fixed ε>0and large enough ,thesetofprimesp
satisfying
max
ξ∈F∗
n≤τ
e(ξa(pn))≤τ
−δ(12)
have density at least 2−r+Oε1
1−2ε,whereδ=δ(ε)is same as in Theorem 1.
1.3 Waring type problems over finite fields
Given a sequence {xn}, one of the classical questions are to decide whether {xn}is an
additive basis. More precisely, is there an absolute constant k≥1 such that any residue
class λmodulo pcan be represented as
xn1+···+xnk≡λ(mod p),
for infinitely many primes p?
In this article, we are concerned about the case when {xn}is a linear recurrence sequence
in Fp. For a simple sequence 2n(mod p),it follows combining a result of Erd˝os and Murty [6]
and a result of Glibichuk [13] that for almost all primes p, every residue class modulo p
can be represented in the following form
2n1+···+2n8(mod p),
for certain positive integers n1,···,n
8.
Note that 2n(mod p) is a linear recurrence sequence of order 1.Forhigherordercases,
one can ask about the classical case of Fibonacci sequences. The third author proved
in [11, Theorem 2.2], that given a parameter N→∞,for π(N)(1 +o(1)) primes p≤N,
every residue class modulo pcan be written as
Fn1+···+Fn16 ≡λ(mod p),
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18 Page 8 of 32 J. Bajpai et al. Res. Number Theory (2022) 8:18
provided that n1,...,n
16 ≤N1/2+o(1).The method is based on the distribution properties
of sparse sequences for almost all primes and particular identities of Lucas sequence. It
does not seem easy to extend such ideas for general linear recurrence sequences. In this
article, we prove that if {sn}is a linear recurrence sequence in Z,whose characteristic
polynomial ω(x)∈Z[x] is monic, irreducible, and having prime degree, then there exists
an absolute constant ksuch that every residue class λ(mod p) can be represented as
sn1+···+snk≡λ(mod p),
for a set of primes pwith positive density. We record this in Theorem 13 in Sect. 6.
2 Exponential sums with linear recurrence sequences
In this section, our main goal is to prove Theorem 1, which is one of our key tool in
establishing several important results of this article. Recalling the example of Shparlinski
in [23, Section 1], we already noticed in Sect. 1.1 that, condition (a) of Theorem 1is needed
if ω(x) is irreducible in Fp[x].We shall discuss more about this condition later in Remark 1.
Now, we illustrate with an example that all of the gcd(τi,τj)scannot be too large. In
other words, we need condition (b) (or some other condition) to obtain a non trivial
bound in Theorem 1. For example, let r=2andgbe a generator of F∗
2.Then, consider
the sequence
sn=Tr gn(2+1)/2−gn,
with characteristic polynomial (x−g)(x−g)(x−g(2+1)/2)(x−g(2+1)/2). Note that
τ2=ord g=2−1andτ1=ord g(2+1)/2=2−1
gcd(2−1,(2+1)/2) .
It is easy to see that gcd(2−1,(2+1)/2) =1,so gcd(τ1,τ2)=2−1.On another hand
we note that gcd(τ1,−1) =−1.Then, one can show that
2−1
n=1
e(sn)=
2−1
n=1
eTr gn(2+1)/2−gn
=
(2−1)/2
n=1
eTr g2n(2+1)/2−g2n
+
(2−1)/2
n=1
eTr g(2n−1)(2+1)/2−g2n−1
=2−1
2+
(2−1)/2
n=1
eTr −2g2n−1=2−1
2+
h∈H
e(Tr (−2gh)) ,
where H=g2.
Let pbe any prime and qbe any power of p. Then, the classical theorem about additive
sums for one-variable polynomial, due to A. Weil (see [17, Theorem 3.2]), states that, for
a given polynomial f(x)∈Fq[x]withdegreed, d <q, gcd(d, q)=1 and a nontrivial
additive character ψin Fq,we have
x∈Fq
ψ(f(x))≤(d−1)√q. (13)
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J. Bajpai et al. Res. Number Theory (2022) 8:18 Page 9 of 32 18
Consider
1+2
h∈H
e(Tr (−2gh)) =
x∈F2
ψ(x2),
where ψ(ω)=e(Tr (−2gω)) is a nonzero additive character of F2.Applying (13)with
f(x)=x2, it follows that
h∈H
e(Tr (−2gh))≤
x∈F2
ψ(x2)≤.
Therefore, the linear recurrence sequence {sn}satisfies
2−1
n=1
e(sn)=2−1
2+O().
We now need to discuss some necessary background. Let Kbe a finite field of characteristic
pand Fbe an extension of Kwith [F:K]=r. The trace function TrF/K:F→Kis
defined by
TrF/K(z)=z+zp+···+zpr−1,z∈F.
The following properties of TrF/K(z) are well known.
TrF/K(az +w)=aTrF/K(z)+TrF/K(w),for all a∈K, z, w ∈F. (14)
TrF/K(a)=ra, for any a∈K. (15)
TrF/K(zp)=TrF/K(z),for any z∈F. (16)
Throughout this section, F=Fq,K=Fpwith q=prand we will simply write Tr (z)
instead TrF/K(z).
Let {sn}be a linear recurrence sequence of order r≥1inFpwith characteristic polyno-
mial ω(x)inFp[x].It is well known that nth-term can be written in terms of the roots of the
characteristic polynomial, see Theorem 6.21 in [18]. Therefore, if the roots α0,...,αr−1
of ω(x) are all distinct in its splitting field, then
sn=
r−1
i=0
βiαn
i,for n=0,1,2,...,(17)
where β0,...,βr−1are uniquely determined by initial values s0,...,s
r−1,and belong to
the splitting field of ω(x) over Fp.If the characteristic polynomial ω(x) is irreducible and
αis a root, then its rdistinct conjugates are
α,αp,...,αpr−2,αpr−1.
Hence, the coefficients snare given by
sn=
r−1
i=0
βiαpin,n=0,1,2,3,... .
One of our main tools is the bound for Gauss sum in finite fields given by Bourgain and
Chang [3, Theorem 2]. This will be required to prove Theorem 1. Assume that for a given
α∈Fqand ε>0,
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18 Page 10 of 32 J. Bajpai et al. Res. Number Theory (2022) 8:18
such that ord α=tsatisfies
t>pεand max
1≤d<r
d|r
gcd(t, pd−1) <tp−ε.(18)
Then, there exists a δ=δ(ε)>0 such that for any nontrivial additive character ψof Fq,
we have
n≤t
ψ(αn)≤tp−δ.
Note that the second assumption in (18) implies the first one whenever r≥2.
2.1 Proof of Theorem 1
We proceed by induction over ν.Before that, following properties (14)and(15)oftrace
function we write
sn=Tr r−1sn=r−1Tr ν
i=1
(βi,0αn
i+···+βi,r −1αpr−1n
i)
=r−1
ν
i=1
r−1
j=0
Tr βi,jαpjn
i.
By the assumption, [Fp(αi):Fp]=rfor any 1 ≤i≤ν.In other words, any such αiis in
Fpr.We then have, r=[Fp(α1,...,αν):Fp]andzpr=zfor any z∈Fp(α1,...,αν).In
addition, from (16) it follows that, Tr (zp)=Tr (z)for any z∈Fp(α1,...,αν).Then, for
each pair (i, j),raising each argument βi,j αpjn
ito the power pr−j
Tr βi,jαpjn
i=Tr βpr−j
i,j αpjn·pr−j
i=Tr βpr−j
i,j αprn
i=Tr βpr−j
i,j αn
i.
This implies that
sn=r−1
ν
i=1
r−1
j=0
Tr βpr−j
i,j αn
i=r−1
ν
i=1
Tr ⎛
⎝⎛
⎝
r−1
j=0
βpr−i
i,j ⎞
⎠αn
i⎞
⎠
=Tr γ1αn
1+···+ Tr γναn
ν,(19)
where γi=r−1r−1
j=0βpr−i
i,j ,for each 1 ≤i≤ν.
The case ν=1 follows from Bourgain and Chang [3, Theorem 2]. We shall now proceed
inductively, and ν=2 will be the base case. We start by denoting h=gcd(τ1,τ2).It is
clear that lcm(τ1,τ2)=τ1τ2/his a period of sn,then
n≤τ
ep(ξsn)=τ
τ1τ2/h
n≤τ1τ2
h
ep(ξsn)
.
Hence, it is enough to prove that
n≤τ1τ2
h
ep(ξsn)≤τ1τ2
hp−δ,with (ξ,p)=1,
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J. Bajpai et al. Res. Number Theory (2022) 8:18 Page 11 of 32 18
for some δ=δ(ε)>0.Dividing the range of the sum n≤τ1τ2/hinto the form n=mh+u0
with m≤τ1τ2/h2and 0 ≤u0≤h−1,we have
n≤τ1τ2
h
ep(ξsn)=
h−1
u0=0
n≤τ1τ2
h2
epξsnh+u0≤
h−1
u0=0
n≤τ1τ2
h2
epξsnh+u0
≤h×max
0≤u0≤h−1
n≤τ1τ2/h2
epξsnh+u0.(20)
Let (n1,n
2)beatuplewithni≤τi
h.Since gcd( τ1
h,τ2
h)=1, by Chinese remainder theorem,
there exist integers m1,m
2with gcd(m1,τ1
h)=gcd(m2,τ2
h)=1,such that
nmod τ1τ2
h2:1≤n≤τ1τ2
h2=n1m1τ2
h+n2m2τ1
hmod τ1τ2
h2:1≤ni≤τi
h.
(21)
Moreover, the pair (m1,m
2) has the following property: given (n1,n
2),with 1 ≤ni≤τi/h,
then n=n1m1τ2
h+n2m2τ1
hsatisfies
n≡n1mod τ1
hand n≡n2mod τ2
h,
and nis unique modulo τ1τ2
h2.Since τ1
h=ord αh
1and τ2
h=ord αh
2, then
αhn
i=αhn1m1τ2
h+n2m2τ1
h
i=αhni
i,1≤i≤2.(22)
Combining (21)and(22), we have
n≤τ1τ2
h2
epξsnh+u0=
n1≤τ1
h
epTr ξγ1αn1h+u0
1
×
n2≤τ2
h
epTr ξγ2αn2h+u0
2
=
n1≤τ1
h
epTr γ
1αn1h
1×
n2≤τ2
h
epTr γ
2αn2h
2
,(23)
with γ
1=ξγ1αu0
1,γ
2=ξγ2αu0
2in Fp(α1,α2).Since {sn}is a nonzero sequence, therefore
γ
i= 0,at least for some 1 ≤i≤2.First, let us assume that γ
1,γ
2= 0.
Each epTr ξγ
izcorresponds to a nontrivial additive character, say ψi(z),in Fp(αi)=
Fpr.In order to satisfy condition (18), we first recall assumptions h<pε,ε>ε
>0
and maxd<r
d|r
gcd(τi,p
d−1) <τ
ip−εfor some i∈{1,2}.Without loss of generality, let us
assume that i=1.Then, for any d|rwith 1 ≤d<r,wehave
gcd τ1
h,p
d−1≤gcd(τ1,p
d−1) <τ
1p−ε<τ1
hp−(ε−ε).
Therefore, by Bourgain and Chang [3, Theorem 2] it follows that
n1≤τ1/h
epTr γ
1αn1h
1=
n1≤τ1/h
ψ1(αn1h
1)≤τ1
hp−δ.
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18 Page 12 of 32 J. Bajpai et al. Res. Number Theory (2022) 8:18
On the other hand, bounding trivially we have
n2≤τ2/h
epTr γ
2αn2h
2=
n2≤τ2/h
ψ2(αn2h
2)≤τ2
h.
Thus, combining above equations with (20)and(23)weget
max
ξ∈F∗
p
n≤τ1τ2
h
ep(ξsn)≤h×τ1τ2
h2p−δ=τ1τ2
hp−δ.
Now, let us assume that one of the λ
i=0,say for i=2. Arguing exactly as few lines
above, it follows from assumption (a) that
n1≤τ1/h
epTr γ
1αn1h
1≤τ1
hp−δ,and
n2≤τ2/h
epTr γ
2αn2h
2=τ2
h.
Hence, the desired bound follows. This conclude the case ν=2.
Now, we proceed by induction over ν,and assume Theorem 1to be true up to ν−1.
We follow the idea due to Garaev [9, Section 4.4]. Considering (19) and periodicity, for
any t≥1weget
τ
n≤τ
ep(ξsn)
2t
=
m≤τ
n≤τ
ep(ξsm+n)
2t
=
m≤τ
n≤τ
epξ(Trγ1αm+n
1+···+ Tr γναm+n
ν)
2t
≤
n1≤τ···
n2t≤τ
m≤τ
epξ
ν
i=1
Tr γiαm
iαn1
i+···−αn2t
i.
Raising to the power 2t, and applying Cauchy–Schwarz, we have
τ2t
n≤τ
ep(ξsn)
4t2
≤τ2t(2t−1)
×
n1≤τ···
n2t≤τ
m≤τ
epξ
ν
i=1
Tr γiαm
iαn1
i+···−αn2t
i
2t
.
Given (λ1,···,λν)∈Fν
q,let Jt(λ1,···,λν) denote the number of solutions of the system
⎧
⎪
⎪
⎨
⎪
⎪
⎩
αn1
1+···+αnt
1=αnt+1
1+···+αn2t
1+λ1
.
.
..
.
..
.
..
.
..
.
.
αn1
ν+···+αnt
ν=αnt+1
ν+···+αn2t
ν+λν
with 1 ≤n1,···,n
2t≤τ.Therefore,
n≤τ
ep(ξsn)
4t2
≤τ4t2−4t
λ1∈Fq···
λν∈Fq
Jt(λ1,···,λν)
×
m≤τ
epξ
ν
i=1
Tr γiλiαm
i
2t
.(24)
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J. Bajpai et al. Res. Number Theory (2022) 8:18 Page 13 of 32 18
Note that writing Jν(λ1···,λν) in terms of character sums, it follows that
Jt(λ1···,λν)=1
qν
x1∈Fq···
xν∈Fq
n≤τ
epTr x1αn
1···epTr xναn
ν
2t
×ep(Tr (x1λ1)) ···epTr xναn
ν
≤1
qν
x1∈Fq···
xν∈Fq
n≤τ
epTr x1αn
1···epTr xναn
ν
2t
≤Jt(0,...,0) =:Jt,ν.
In particular, we note that Jt,ν≤Jt,ν−1.From (24), it follows that
n≤τ
ep(ξsn)
4t2
≤τ4t2−4tJt,ν
m1≤τ···
m2t≤τ
λ1∈Fq···
λν∈Fq
epν
i=1
Tr ξγiλi(αm1
i+···−αm2t
i).
Note that aγλ,with aγ= 0,runs over λ∈Fq, then ep(Tr (aθλz)) runs through all
additive characters ψin
Fq,evaluated at z. Then, the above expression can be written as
n≤τ
ep(ξsn)
4t2
≤τ4t2−4tJt,ν
m1≤τ···
m2t≤τ
ν
i=1⎛
⎝
x∈Fq
epx(αm1
i+···−αm2t
i)⎞
⎠
≤τ4t2−4tqνJ2
t,ν≤τ4t2−4tqνJ2
t,ν−1.(25)
We now require an estimate for Jt,ν−1,and write
Jt,ν−1=1
qν−1
λ1∈Fq···
λν−1∈Fq
m≤τ
epTr λ1αm
1+···+λν−1αm
ν−1
2t
=τ2t
qν−1+O⎛
⎜
⎜
⎝⎛
⎜
⎜
⎝max
(λ1,...,λν−1)∈Fν−1
q
(λ1,...,λν−1)=0
m≤τ
epTr λ1αm
1+···+λν−1αm
ν−1⎞
⎟
⎟
⎠
2t⎞
⎟
⎟
⎠.
(26)
Finally, we note that s
m=Tr λ1αm
1+···+λν−1αm
ν−1defines a linear recurrence
sequence with period τdividing τ,which in particular satisfies induction hypothesis.
Therefore
m≤τ
epTr λ1αm
1+···+λν−1αm
ν−1≤τp−δ,
for some δ=δ(ε)>0.Now, taking t>d(ν−1)/2δ(where d=[Fq:Fp]) and
combining with (26), we get
Jt,ν−1τ2t
qν−1.
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18 Page 14 of 32 J. Bajpai et al. Res. Number Theory (2022) 8:18
We conclude the proof combining the above estimate with (25)toget
2
max
ξ∈F∗
p
n≤τ
ep(ξsn)≤τp−δ,with δ=d(ν−2)
4t2.
The following is an immediate corollary of this theorem which will be quite handy in
establishing several results in Sects. 3and 6.
Corollary 6 Suppose that {sn}is a nonzero linear recurrence sequence of order r ≥2such
that its characteristic polynomial ω(x)is irreducible in Fp[x]. If its period τsatisfies
max
d<r
d|r
gcd(τ,p
d−1) <τp−ε,
then there exists a δ=δ(ε)>0such that
max
ξ∈F∗
p
n≤τ
ep(ξsn)≤τp−δ.
Remark 1 It is possible to relax the condition (a) by assuming that
max
d<r
d|r
gcd(τi,p
d−1) <τ
ip−ε
holds for some 1 ≤i≤νfor which λ
i= 0,where λ
iis defined in the proof of Theorem 1.
Also, note that λ
i=0ifandonlyifλi=0.
Since {sn}is a nonzero linear recurrence sequence, there exists some 1 ≤i≤νfor
which λi= 0.We discussed in Sect. 1.1 that why (a) (or some other condition) is needed
to prove the irreducible case of Theorem 2. Now, for the reducible case, some of the λi
could be 0.For the worst case scenario, let us assume that only one of them is nonzero,
say for i=1.Then, it follows from (19) that, we are back to considering the irreducible
case and then we need the condition (a) for i=1.In particular, we need (a) (or some
other condition) for each irreducible component of the underlying ω(x).
3 Exponential sums for modular forms
In this section, we study the effect of linear recurrence sequence and Theorem 1in the
behaviour of the exponential sums associated with certain Fourier coefficients of modular
forms. As a consequence, we obtain interesting results which have been summarized
earlier in the form of Theorems 2and 3.
3.1 Order of the roots of the characteristic polynomial
In the case of normalized eigenforms, the sequence {a(pn)}defines a linear recurrence
sequence of order two when pN, otherwise it is of order one. This is one of the tools
for Theorem 2. However, we do not need to assume that the form is normalized because
the normalizing factor is in Q,and we can realize that to be an element of F∗
for any large
enough prime .Before going into the proof of this theorem, we develop a tool which will
be quite useful throughout. We state it in the form of following lemma.
2To get a non trivial estimate, we must have a non zero δ.This is true when ν>2.Hence our induction step starts
from ν=2.
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J. Bajpai et al. Res. Number Theory (2022) 8:18 Page 15 of 32 18
Lemma 7 Let ω(x)=x2+ax +b∈Z[x]be a quadratic polynomial with b = 0and let
α,βbe its roots such that none of the α,βor αβ−1is a root of unity. For any prime , let
α,βbe its roots in the splitting field of ω(x)over F.
Then, given 0<ε<1/2,forπ(y)+O(y2ε)many primes ≤y, we have
ord α>
ε,ord β>
εand ord (αβ−1
)>
ε.
Proof Given a large positive parameter T, we begin by considering the polynomial
GT(x)=
t≤T
(xt−1)(x2t−bt)∈Z[x].
It is clear that ω(x) (mod ) has distinct roots for all but finitely many primes ,since
a2−4b= 0.For any such prime ,letαand βbe the distinct roots in its splitting field.
We now consider the resultant Res(ω(x),G
T(x)), and note that
Res(ω(x),G
T(x)) (mod )=
1≤i≤3T
(α−μi)(β−μi),
where each μiis a root of GT(x) in its splitting field over F.
In particular, Res(ω(x),G
T(x)) ≡0 (mod )ifandonlyifω(x) (mod )andGT(x)(mod )
have common roots in some finite extension of F.Additionally, since αβ=b, it follows
that ord (αβ−1
)≤Tif and only if α2t
−bt=0(orβ2t
−bt=0), for some t≤T.
Therefore, α(or β) is a common root of ω(x) (mod )andGT(x) (mod )if ordαor
ord (αβ−1
) (or ord βor ord (αβ−1
)) is less than T. Now, the Sylvester matrix of ω(x)
and GT(x) is a square matrix of order 2 +deg(GT(x)) T2,and entries bounded by an
absolute constant M(which depends on a, b and not on or the parameter T). Then, by
Hadamard’s inequality, the determinant
Res(ω(x),G
T(x)) ≤TT2×MT2M2T2log T.
Note that Res(ω(x),G
T(x)) is zero if and only if αt=1,βt=1or(αβ−1)t=1 for some
t≤T, which, following our assumption, can not happen. In particular, the resultant has
at most OT2many distinct prime divisors. This shows that
|{prime |ord α≤Tor ord β≤Tor ord αβ−1≤T}| = O(T2).
Choosing T=yε, the number of primes ≤ysuch that
ord α≤εor ord β≤εor ord (αβ−1
)≤ε
is Oy2ε.
Let us now proceed to prove the main result of this section.
3.2 Proof of Theorem 2
If p|N, then a(pn)=a(p)nfor any n. We only need to consider
max
ξ∈F∗
n≤τ
e(ξa(p)n).(27)
If p/∈P,then there exists usuch that a(pu)=0.Since p|N, we have a(p)=0.In this
case, the sum is O(1) because we have τ=1.
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18 Page 16 of 32 J. Bajpai et al. Res. Number Theory (2022) 8:18
On the other hand, if p∈P,then for any prime large enough τis simply the order of
a(p) (mod )inF∗
.Due to Lemma 7, we may assume that τ>pεholds for π(y)+O(y2ε)
many primes <y. Hence, this case is settled down by [4, Theorem 6].
Let us now consider the case pN. The characteristic polynomial of (1)is
ω(x)=x2−a(p)x+pk−1,(28)
and has discriminant a2(p)−4pk−1.We note that in our case the discriminant does
not vanish, otherwise |a(p)|=2p(k−1)/2is absurd, with a(p) being integer and p(k−1)/2
irrational. Let Pbe the set of all primes. We divide the proof for primes p∈Pand
p∈P\P.Since a2(p)−4pk−1= 0,for any p∈P,we write a2(p)−4pk−1=u2Dp,with
Dp<0 square-free and u= 0.Let us split the cases according to Dp(mod ) is quadratic
residue, zero or non quadratic residue modulo .Set
P=P0∪P1∪P−1,where Pν=∈P:Dp
=ν.
For ν=0,1,−1, we also define
Pν(x)=Pν∩[1,x],πν(x)=Pν(x)and κν=lim
x→∞
πν(x)
π(x).
It is clear that πν(x)=π(x)(κν+o(1)),and κ0+κ1+κ−1=1.
Note that for a given prime p, the associated polynomial ω(x)(mod ) has a single root in
Fif and only if u2Dp≡0(mod ).Since such equation has finitely many solutions for ,
we get κ0=0.On the other hand, Chebotarev’s density theorem implies that the uniform
distribution of primes such that ω(x) (mod ) is irreducible or has distinct roots in F.
Equivalently, the primes satisfying Dp
=±1 are distributed in the same proportion,
therefore κ−1=κ1=1/2.We now turn to establish nontrivial exponential sums for
{a(pn)}(mod )with∈Pν,for ν=±1.
Case 1. ∈P−1:
we want to show that the inequality (7) is satisfied by π(y)
2+O(y2ε) many primes ≤y
in P−1.In this case the associated polynomial (28) is irreducible modulo ,then the idea
is to employ Corollary 6.Letαand β=αbe the conjugate roots of (28) in its splitting
field F(α).For a given ε>0,from Lemma 7it follows that for π(y)+O(y2ε) many primes
≤y, the following inequalities
ord α=ord α>
εand ord αβ−1=ord α1−>
ε(29)
hold. Combining the identity
ord α−1=ord α
gcd( ord α,−1)
with the second inequality of (29), we get
gcd( ord α,−1) =ord α
ord α−1=ord α
ord α1−<(ordα)−ε.
Applying Corollary 6we complete the proof of this case.
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J. Bajpai et al. Res. Number Theory (2022) 8:18 Page 17 of 32 18
Case 2. ∈P1:
let α,βbe the roots of ω(x)(mod )insideF∗
.From (17) it follows that for n≥0,a(pn)≡
cαn+dβn(mod ),for some constants c, d in F,with (α,β)= (0,0).It is clear that −1
is a period of the sequence a(pn)(mod ),and hence τdivides −1.We have
n≤τ
e(ξa(pn))=τ
−1
n≤−1
e(ξa(pn))=τ
−1
n≤−1
e(ξ(cαn+bβn)).
From Lemma 7, there is a subset of P1with π(y)
2+O(y2ε) many primes ≤ysuch that
ord α,ord βand ord (αβ−1) are bigger than ε.It follows from [2, Corollary, page 479]
that there exists a δ=δ(ε)>0 such that
max
(c,d)∈F×F
(c,d)=(0,0)
n≤−1
e(cαn+dβn)≤1−δ.
Hence, (i) of Theorem 2holds. Now, assume that pbelongs to the exceptional set P\P,
that is a(pu)=0 for some u≥1.We consider u=u(p) to be the least such integer. Since
the discriminant is nonzero (the roots αand βof (28) are distinct), we get3
a(pu)=αu+1−βu+1
α−β=0.
Set b(u+1) =a(pu), then it follows that for all n≥1wehave
b(n(u+1)) =a(pn(u+1)−1)=αn(u+1) −βn(u+1)
α−β=0.
Therefore,
n≤τ
e(ξa(pn))=
τ−1
n=0
e(ξb(n+1))=⎛
⎝τ/(u+1)
n=0
u
e=0
e(ξb(n(u+1) +e))⎞
⎠
+O(u)
=⎛
⎝τ/(u+1)
n=0
e(ξb(n(u+1)))+
u
e=1
τ/(u+1)
n=0
e(ξb(n(u+1) +e))⎞
⎠
+O(u)
= τ
u+1+⎛
⎝
u
e=1
τ/(u+1)
n=0
e(ξb(n(u+1) +e))⎞
⎠+O(u).(30)
First of all observe that uis odd. As otherwise, if uis even then we would get
αu+1+βu+1=2αu+1=±2p(u+1)(k−1)
2,
which is absurd as αu+1+βu+1is a rational, but p(u+1)(k−1)
2is not. Now, for any 0 <e<u+1
we have
b((u+1)n+e)=α(u+1)n(αe−βe)
α−β=±p(u+1)(k−1)
2n
a(pe−1),
where the sign on the right hand side above depends on the sign of αu+1.Without loss
of generality, we are assuming that this sign is negative. It is easy to see that our next
3The explicit expression of a(pu)can be obtained by using induction on ualong with the fact that α+β=a(p),αβ =
pk−1and the recurrence relation at (1).
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18 Page 18 of 32 J. Bajpai et al. Res. Number Theory (2022) 8:18
argument applies to the positive sign case as well. Since uis fixed, so are all the e’s up to
u−1.In particular, we may consider large primes for which all of the a(pe)≡ 0(mod )
for any 1 ≤e≤u−1.Then,wehave
τ/(u+1)
n=0
e(ξb(n(u+1) +e))=
τ/(u+1)
n=0
eξ