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arXiv:2102.01949v1 [math.NT] 3 Feb 2021
ON SPARSITY OF REPRESENTATIONS OF
POLYNOMIALS AS LINEAR COMBINATIONS OF
EXPONENTIAL FUNCTIONS
DRAGOS GHIOCA, ALINA OSTAFE, SINA SALEH,
AND IGOR E. SHPARLINSKI
Abstract. Given an integer gand also some given integers m
(sufficiently large) and c1,...,cm, we show that the number of all
non-negative integers n6Mwith the property that there exist
non-negative integers k1,...,kmsuch that
n2=
m
X
i=1
cigki
is o(log M)m−1/2. We also obtain a similar bound when dealing
with more general inequalities
Q(n)−
m
X
i=1
ciλki6B,
where Q∈C[X] and also λ∈C(while Bis a real number).
1. Introduction
1.1. Set-up. Motivated by applications to the dynamical Mordell-Lang
conjecture (for more details on this open problem in arithmetic dynam-
ics, we refer the reader to [BGT16]), the authors [GOSS21] have re-
cently considered the question about representations of values of poly-
nomials Q∈Q[X] as fixed linear combinations of powers of a prime
p. In particular, it is shown in [GOSS21] that for fixed coefficients
c1,...,cm∈Qand integral exponents a1,...,amthe number of posi-
tive integers n6Nfor which Q(n) can be represented as
Q(n) =
m
X
i=1
cipaiki
with some k1, . . . , km∈Zis bounded by O((1 + log N)m) where the
implied constant depends only on the initial data. In fact it is easy
2010 Mathematics Subject Classification. 11A63, 11B57.
Key words and phrases. Squares,
1
2 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
to see that for Q(n) = n, this bound is tight. Furthermore, a similar
result is given in [GOSS21] for representations of the form
(1.1) Q(n) =
m
X
i=1
s
X
j=1
ci,jλaiki
j,
with algebraic integers λ1,...,λs, each one of them of absolute value
equal to qor √q(where qis a given power of a prime number).
Here we first consider representations of the form (1.1) with s= 1
but arbitrary complex (rather than algebraic) parameters. We also
generalise this to approximations of polynomials rather than precise
equalities, that is, we consider inequalities of the form
(1.2) Q(n)−
m
X
i=1
ciλki6B,
with Q∈C[X], c1,...,cm, λ ∈Cand some B∈R.
1.2. Notation. We now recall that the notations A=O(B), A≪B
and B>Aare all equivalent to the inequality |A|6cB with some
constant c. Throughout this work all implied constants may depend on
the polynomial Qand the parameters ci,i= 1,...,m and λin (1.2)
and also on gin (1.4) below.
For a finite set Swe use #Sto denote its cardinality.
1.3. New results. We remark that the argument of [GOSS21] is based
on a result of Laurent [Lau84, Th´eor`eme 6], which required all parame-
ters to be defined over a number field; furthermore, the result of [Lau84]
refers to equalities, not inequalities. Hence here we use a different ap-
proach to establish the following result.
Theorem 1.1. Let c1,...,cm, λ ∈C,B∈Rand let Q∈C[X]be a
non-constant polynomial. Then for N>2we have
#{n6N:(1.2) holds for some k1,...,km∈Z} ≪ (log N)m.
We observe that the implied constant in Theorem 1.1 is effectively
computable in terms of the sizes of the initial data, while in the result
of [GOSS21] it is not.
We also note (see Example 2.1) that if one considers inequalities of
the form
(1.3) Q(n)−
m
X
i=1
s
X
j=1
ci,jλaiki
j6B,
for some arbitrary complex numbers λj, then one cannot expect a
similar result as in Theorem 1.1. More precisely, there exists λ∈C
SPARSE REPRESENTATIONS OF SQUARES 3
such that for any ε > 0 and each sufficiently large integer n, there
exists some positive integer knwith the property that
n−i
π·2kn−λkn6ε,
see Example 2.1 for more details.
Furthermore, we consider the case of perfect squares and study rela-
tions of the form
(1.4) n2=
m
X
i=1
cigki,
with non-zero integer coefficients c1,...,cmand an integer basis g>
2. Using the square-sieve of Heath-Brown [H-B84] we improve the
exponent mof log N.
For m= 2, we write the equation (1.4) as n2=gk1c1+c2gk2−k1
(with k2>k1). Hence either c1+c2gk2−k1or c1g+c2gk2−k1+1 is a
perfect square j2for some j6n6N. Since the largest prime divisor
of j2+cfor any c6= 0 tends to infinity with j, see [Kea69], we see that
k2−k1can take only finitely many values. Hence for m= 2 we have
O(log N) solution to (1.4) with n6N. This bound is obviously the
best possible as the example of the numbers 2k1+ 2k2, with k2=k1+ 3
and even k1, shows. We also note that in [CGSZ, Theorem 5.1 (B)], it
is established even more generally the precise set of all positive integers
nfor which unis of the form c1gk1+c2gk2(for some given g,c1,c2),
where {un}n>1is an arbitrary linear recurrence sequence (the result
of [CGSZ, Theorem 5.1 (B)] is stated only when g=pis a prime
number, but as remarked in [CGSZ, Section 5], the method extends
verbatim to an arbitrary integer g).
So we are mostly interested in the case of m>3; furthermore, we
note that for m= 3 (and in some cases, depending on gand the ci,
even for m= 4), more precise results are available in the literature
(see [CoZa13]). However, when m>5, it is very difficult to find a
precise description of all n∈Nsuch that n2is of the form (1.4) (for
some given integers gand ci).
Theorem 1.2. Let m>3and let c1,...,cmand g>2be integers.
Then for N>2we have
#{n6N:(1.4) holds for some k1,...,km∈Z}6(log N)m−γm+o(1),
where
γ3=677
1969 and γm=677m
1323m+ 1354 for m>4.
4 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
We observe that γm→677/1323 = 0.5117 ... as m→ ∞. Thus for
large mTheorem 1.2 saves more than 1/2 compared to the general
bound of Theorem 1.1. More precisely, simple calculations show that
γm>1/2 for m>44.
We remark that the proof of Theorem 1.2 is based on some ideas
and results from [LuSh09], later enhanced in [BaSh17]. The numerical
constants come from the work of Baker and Harman [BaHa98] on large
prime divisors of shifted primes.
Furthermore, as in [BaSh17] we observe that under the Generalised
Riemann Hypothesis we can obtain a slightly larger value of γm.
On the other hand defining sas the largest integer with s(s+ 1)/26
mand considering numbers gh1+...+ghs2with
hi6log (N/s2)
2 log g, i = 1,...,s,
we see that for at least one choice c1,...,ct>0 with t6mand
c1+...+ct6s(s+ 1)/2 occurs at least C0(log N)stimes, for some
constant C0>0, which shows that the best possible exponent in any
result of the type of Theorem 1.2 must grow with m(at least as about
√2mfor large m).
Note that cycling over all gmchoices of
(c1,...,cm)∈ {0,...,g−1}m
we obtain from Theorem 1.2 a result about the sparsity of the values
of nfor which n2has at most mnon-zero digits to base g. Various
finiteness results on sparse digital representations of perfect powers
can be found in [BeBu14,BBM13,CoZa13,Mos21]. Note that as we
have just seen, in our setting of arbitrary mno finiteness result is
possible, and hence we can use Theorem 1.2 to provide a counting result
related to such representations. More precisely we have the following
straightforward consequence:
Corollary 1.3. Let m>3and let K>1and g>2be integers. Then
there are at most Km−γm+o(1) integer squares with g-ary expansion of
length Kand with at most mnon-zero digits.
2. Proof of Theorem 1.1
2.1. Counterexample to a possible extension to (1.3). Before
proceeding to the proof of Theorem 1.1, we provide the Example 2.1
(mentioned in Section 1.3), which shows that one cannot expect to gen-
eralise Theorem 1.1 to (1.3), that is, to the case when we approximate
Q(n) with a sum of powers of different λj.
SPARSE REPRESENTATIONS OF SQUARES 5
Example 2.1. We consider the sequence of positive integers {bj}j>2
given by
b2= 2 and bj+1 = 2bj+bj+ 1 for j>2.
We let λ= 2 ·e2π αi , where
α=
∞
X
j=2
j−1
2bj.
We let nbe a positive integer and show that
(2.1) n−i
π·22bn−λ2bn≪n
bn+1
.
Indeed, we first notice that
λ2bn= 22bn·e2πtni,
where
tn=
∞
X
j=n+1
j−1
2bj−bn.
Then
22bn−λ2bn= 22bn·((1 −cos(2πtn)) −isin (2πtn))
= 22bn·2 sin(πtn)·(sin(πtn)−icos(πtn)) ,
and so,
(2.2) i
π·22bn−λ2bn=21+2bnsin(πtn)
π·eπtni.
Now, by the definition of the rapidly increasing sequence {bj}j>2, we
have that
(2.3) 0 < tn−n
2bn+1−bn<1
22bn+1 ;
also, clearly, tn→0as n→ ∞. Furthermore, we know that when t
is close to 0, then
(2.4) |sin t−t|6t2.
So, using the inequalities (2.3) and (2.4), along with the fact that
bn+1 = 2bn+bn+ 1, we get that
(2.5) n−21+2bnsin (πtn)
π<2−bn≪1/bn+1
6 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
for all nsufficiently large. Also, for nlarge, using the inequality (2.3)
we have that
(2.6) 1−eπtni<2−bn≪1/bn+1 .
Recalling (2.2) and using (2.5) and (2.6), we derive (2.1).
Therefore, the conclusion of Theorem 1.1 cannot be generalised to
inequalities (1.3) where we approximate a polynomial Q(n) with sums
of powers of different λ1,...,λs.
Next we proceed to proving Theorem 1.1.
2.2. Preliminaries. We first note that if |λ|= 1, then the inequal-
ity (1.2) yields that |Q(n)|is uniformly bounded above and therefore,
we can only have finitely many n∈Nsatisfying such inequality since
Qis a non-constant polynomial. Furthermore, since the exponents ki
appearing in the inequality (1.2) are arbitrary integers, then without
loss of generality, we may assume from now on that |λ|>1.
Now, if some of the exponents ki,i= 1,...,m, from (1.2) were
non-positive, then the absolute value of the corresponding terms ciλki
is uniformly bounded above. So, at the expense of replacing Bby
a larger constant (but depending only on the absolute values of the
ci), we may assume from now on, that each exponent kifrom (1.2) is
positive.
Let Q(X) = adXd+···+a1X+a0for complex numbers a0,...,ad
with ad6= 0. There exists N0>0 (depending only on dand the
absolute values of the coefficients of Q) such that
(2.7) |Q(n)|62adnd6N0·ndfor each n>N0.
Furthermore, at the expense of replacing N0by a larger positive integer
(but still depending only on d, the absolute values of the coefficients
of Qand also depending on Bin this case), we may also assume that
(2.8) |Q(n1+n2)−Q(n1)|>2Bfor each n1, n2>N0.
2.3. Induction. We proceed to prove our desired result by induction
on m.
We prove first the base case m= 1 , which also constitutes the in-
spiration for our proof for the general case in Theorem 1.1. So, we
have that |Q(n)|=O(Nd) for each 0 6n6N(see also the inequal-
ity (2.7)). Therefore, for n6Nsatisfying the inequality
(2.9) Q(n)−c1λk16B
one has |λ|k1=ONdand thus, k1=O(log N) . On the other hand,
for a given k1, the inequality (2.9) is satisfied by O(1) non-negative
SPARSE REPRESENTATIONS OF SQUARES 7
integers n(see also (2.8)), thus proving the desired bound in the case
m= 1.
So, suppose that the result is true for m6sand we prove that
Theorem 1.1 holds when m=s+ 1 ; clearly we may assume each ci
for i= 1,...,s+ 1 is nonzero. Since there are mpowers of λin the
inequality (1.2), then in order to prove Theorem 1.1, it suffices to prove
that the set S0, consisting of all n∈Nfor which there exist integers
(2.10) 1 6k16k26···6ks+1
such that
(2.11) Q(n)−
s+1
X
j=1
cjλkj6B
satisfies
(2.12) {n∈S0:n6N} ≪ (log N)s+1 .
Let ∆ ∈Nbe sufficiently large (but depending only on |λ|, which
is larger than 1, and also depending on the absolute values of the
c1,...,cs+1 ) such that we have
(2.13) |cs+1|
2· |λ|ks+1 6
s+1
X
j=1
cjλkj,
for all integers ks+1 >··· >k1>0 satisfying the inequality (2.11)
along with the inequality ks+1 −ks>∆.
Now, we let Ube the subset of S0consisting of integers n∈N
for which one can find integers kjsatisfying (2.11) and in addition,
ks+1−ks<∆. Then the existence of such a solution tuple (k1,...,ks+1)
for each n∈Umeans that
Q(n)−c1λk1+···+cs−1λks−1+cs+cs+1λks+1 −ksλks6B.
Because ks+1 −ks∈ {0,...,∆−1}, applying the induction hypothesis
for each of the possible ∆ values of ks+1 −ks, we obtain the desired
conclusion regarding the asymptotic growth given by (2.12) (further-
more, we actually get that the exponent from the right-hand side of
the inequality (2.12) is s=m−1 not s+ 1 = m).
On the other hand, for each n∈S0\Usatisfying n>N0, we know
that there must exist some tuple of nonnegative integers (k1,...,ks+1)
satisfying (2.11) and in addition, ks+1 −ks>∆. Then using both (2.7)
and (2.13), we get
|cs+1|
2· |λ|ks+1 −B6c0+
s+1
X
j=1
cjλkj−B6|Q(n)|62|ad| · nd,
8 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
which implies that
(2.14) ks+1 6b0(1 + log n)
for some positive real number b0depending only on B,|λ|,d,|ad|
and |cs+1|.
So, let Nbe an integer larger than N0; then for each integer N06
n6Ncontained in S0\U, we know there exists an (s+ 1) -tuple
of integers kisatisfying (2.10) and (2.11). Combining the fact that
16ki6ks+1 with the inequality (2.14), we get that there are at most
(b0(1 + log N))s+1 tuples (k1,...,ks+1) for which we could find some
n∈S0\Usatisfying the inequality N06n6N. However, since
n>N0, then the inequality (2.8) yields that for any such (s+ 1) -tuple
of integers ki, there are at most N0integers n∈(S0\U)∩[N0, N]
satisfying (2.11) with respect to the tuple (k1,...,ks+1). Hence, we
get the inequality
#{n6N:n∈S0\U}6N0·1 + (b0·(1 + log N))s+1.
for each positive integer N>N0. This concludes our proof of Theo-
rem 1.1.
3. Construction and properties of the sieving set of
primes
3.1. Multiplicative orders. Let τℓ(g) denote the multiplicative order
of an integer g>2 modulo a prime ℓ, that is, the smallest positive
integer τfor which gτ≡1 mod ℓ.
Let αbe a fixed real number such that
(3.1) # {ℓ6z:ℓis prime and P(ℓ−1) >ℓα} ≫ z
log z
for all sufficiently large z, where P(k) denotes the largest prime divisor
of an integer k>2 , and the implied constant depends only on α.
We recall the following well known result which follows from the
divisibility τℓ(g)|ℓ−1 (provided gcd(g, ℓ) = 1) and the bound
#ℓ6z:ℓis prime, P(ℓ−1) > ℓ1/2= (1 + o(1)) z
log z
as z→ ∞, which easily follows from a stronger result of Erd˝os and
Murty [ErMu96, Theorem 3]. Details can be found in the work of
Kurlberg and Pomerance [KuPo05, Lemma 20].
Lemma 3.1. For any fixed α>1/2satisfying (3.1) and any fixed
integer g>1we have
#{ℓ6z:ℓis prime, τℓ(g)>ℓα} ≫ z
log z
SPARSE REPRESENTATIONS OF SQUARES 9
as z→ ∞.
For an integer s>1 we denote by ν2(s) the 2-adic order of s, that
is, the largest power νsuch that 2ν|s.
Lemma 3.2. For any fixed α>1/2satisfying (3.1) and any fixed
integer g>1there are some absolute constants C1, C2>0, such that
for every sufficiently large real number z > 1, there exist some integer
u0and a set Lz⊆[z, C1z]of primes of cardinality
#Lz>C2z
log zlog log z
such that for every ℓ∈Lzwe have
P(ℓ−1) >zα, P (ℓ−1) |τℓ(g), ν2(τℓ(g)) = u0.
Proof. Lemma 3.1 obviously implies that for some absolute constants
C1, C3there are at least C3z/ log zprimes ℓ∈[z, C1z] satisfying only
the first two conditions, see also [LuSh09, Lemma 5.1]. Let Lzbe this
set. Trivially, there are at at most z/2v0primes ℓwith ν2(τℓ(g)) >v0.
Hence taking a sufficiently large C4, and v0=⌊C4log log z⌋, we see
that if we remove these primes from Lzwe obtain the set of f
Lz⊆Lz
of cardinality
#f
Lz>#Lz−z/2v0>0.5#Lz>0.5C3z/ log z.
Since obviously ν2(τℓ(g)) 6ν2(ℓ−1) 6v0, making a majority decision
we can find a set of Lzof cardinality
#Lz>#f
Lz
v0
>0.5C3z
v0log z
with ν2(τℓ(g)) = u0for some fixed u06v0for every ℓ∈Lz. Taking
C2= 0.5C3/C4we conclude the proof.
We note that the Brun-Titchmarsh theorem (see [IwKo04, Theo-
rem 6.6]) can be used to remove log log zin the bound on #Lzof
Lemma 3.2. However in our final result we do not try to optimise terms
of this order, so we ignore this and similar potential improvements.
3.2. Sieving set Lz.We see that using a result of Baker and Har-
man [BaHa98] one can take
(3.2) α= 0.677,
in Lemmas 3.1 and 3.2.
From now on, for any positive real number z, we fix a set Lzsatis-
fying the conclusion of Lemma 3.2 with αgiven by (3.2).
10 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
3.3. Bounds of some arithmetic sums. For an integer Kwe con-
sider the set
(3.3) K=Km(K)
where
(3.4) Km(K) = {0,...,K}m,
and for k= (k1,...,km)∈Kwe define
(3.5) F(k) =
m
X
i=1
cigki.
For a real z>2 let ωz(n) be the number of distinct prime factors
ℓ∈Lzof n.
Lemma 3.3. Let an integer Kand a real zbe sufficiently large. For
Kand F(k)as in (3.3) and (3.5), respectively, we have
X
k∈K
ωz(F(k)) ≪Kmz−α+Km−1#Lz.
Proof. We have
X
k∈K
ωz(F(k)) ≪X
k∈KX
ℓ∈Lz
ℓ|F(k)
1 = X
ℓ∈LzX
k∈K
ℓ|F(k)
1.
Clearly the last sum can be estimated as
X
k∈K
ℓ|F(k)
16(K+ 1)m−1K+ 1
τℓ(g)+ 1
≪Kmℓ−α+Km−16Kmz−α+Km−1,
and the result follows.
Remark 3.4. The proof of Lemma 3.3 appeals to essentially trivial
bound O(Km−1(K/τℓ(g) + 1)) on the number of solution to the con-
gruence F(k)≡0 (mod ℓ),k∈K. Using bounds of exponential
sums one can obtain a better bound, which however does not improve
our final result (see also our Appendix).
For a real κwe define the sums
Dκ=X
ℓ,r∈Lz
P(ℓ−1)6=P(r−1)
gcd (ℓ−1, r −1)κ.
SPARSE REPRESENTATIONS OF SQUARES 11
Lemma 3.5. Let a real zbe sufficiently large. Then for κ>1we
have
Dκ6zκ+α−ακ+1+o(1).
Proof. Clearly for each pair of primes (ℓ, r) in the sum Dκwe have
gcd (ℓ−1, r −1) 6(ℓ−1)/P (ℓ−1) 6Hfor some integer H≪z1−α.
Hence
Dκ6
H
X
h=1
hκX
ℓ,r∈Lz
P(ℓ−1)6=P(r−1)
gcd(ℓ−1,r−1)=h
1.
We estimate the inner sum trivially as O((z/h)2) and derive
Dκ6z2
H
X
h=1
hκ−26z2+o(1)Hκ−16z2+(κ−1)(1−α)+o(1) ,
and the desired result follows.
4. Bounds of character sums
4.1. Complete character sums with diagonal forms over finite
fields. Let qbe an odd prime power and let Fqbe the finite field of
qelements. We note that for the purpose of proving Theorem 1.2,
we only need to estimate the sums of this section over a prime finite
field. However, since our proofs work over arbitrary finite fields, we
present them in this more general setting with the hope they would be
of independent interest.
We let m>1 and d>2 be integers with dcoprime with q.
Let Xdenote the set of multiplicative characters of F∗
qand let
X∗=X\ {χ0}be the set of non-principal characters, we refer
to [IwKo04, Chapter 3] for a background on characters. We also denote
by η∈X∗the quadratic characters (that is η2=χ0).
We recall that the implied constant may depend on m(but not on
d,qand other parameters).
We start with ‘pure’ bounds of sums of quadratic characters.
We note that in our next result we have an additional condition of
dbeing an even integer.
Lemma 4.1. Assume that the integer d>2satisfies gcd(d, q) = 1
and is even. Let a1,...,am∈F∗
q. Then for
S=X
x1,...,xm∈Fq
ηa1xd
1+···+amxd
m
12 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
we have
|S|6dm−1(q−1)q(m−1)/2.
Proof. The proof follows by induction on m. For m= 1, since dis
even, the sum becomes X
x1∈Fq
ηa1xd
1=q−1.
We assume the bound true for m−1 and we prove it for m. We have
S=X
xm∈F∗
qX
x1,...,xm−1∈Fq
ηa1xd
1+···+amxd
m
+X
x1,...,xm−1∈Fq
ηa1xd
1+···+am−1xd
m−1.
By the induction hypothesis, the second sum in the above is bounded
by dm−2(q−1)q(m−2)/2. Hence, we have
(4.1) |S|6|S∗|+dm−2(q−1)q(m−2)/2,
where
S∗=X
xm∈F∗
qX
x1,...,xm−1∈Fq
ηa1xd
1+···+amxd
m,
to which we apply [Kat02, Theorem 2.1]. Indeed, since xm6= 0 in S∗,
we make the transformation xi→xixm,i= 1,...,m−1, which does
not change the sum. Moreover, since again dis even and η(xd
m) = 1,
we obtain
S∗=X
xm∈F∗
qX
x1,...,xm−1∈Fq
ηa1xd
1+···+am−1xd
m−1+am
= (q−1) X
x1,...,xm−1∈Fq
ηa1xd
1+···+am−1xd
m−1+am.
(4.2)
Let now
F(X1,...,Xm−1) = a1Xd
1+···+am−1Xd
m−1+am∈Fq[X1,...,Xm−1].
We note that the equation F(X1,...,Xm−1) = 0 defines a smooth
hypersurface in the affine space Am−1(Fq). Indeed, considering the
partial derivatives of Fwith respect to each variable Xi, we ob-
tain that the only possible singular point would be (0,...,0). How-
ever, since am6= 0, this point does not belong to the hypersurface
F(X1,...,Xm−1) = 0.
SPARSE REPRESENTATIONS OF SQUARES 13
Similarly, the equation given by the leading homogenous part of F,
a1Xd
1+···+am−1Xd
m−1= 0, defines a smooth hypersurface in the
projective space Pm−2(Fq).
Applying now [Kat02, Theorem 2.1], we conclude from (4.2) that
(4.3) |S∗|6(d−1)m−1(q−1)q(m−1)/2.
Substituting (4.3) in (4.1), we obtain
|S|6(d−1)m−1(q−1)q(m−1)/2+dm−2(q−1)q(m−2)/2
6dm−2(q−1)q(m−2)/2(d−1)q1
2+ 1.
Since (d−1)q1
2+ 1 < dq1/2, we conclude the proof.
Next we need the following bound on multidimensional sum of qua-
dratic characters, twisted by arbitrary characters. In the next result
we do not use that dis even.
Lemma 4.2. Assume that the integer d>2satisfies gcd(d, q) = 1.
Let a1,...,am∈F∗
q. Then for any χ1,...,χm∈Xwe have
X
x1,...,xm∈Fq
ηa1xd
1+···+amxd
mχ1(x1). . . χm(xm)≪dmq(m+1)/2.
Proof. First we note that if each χiis equal to the principal character,
then the result follows from Lemma 4.1. So, from now on, we assume
that not all of the characters χiare equal to the principal character.
We have
(4.4) X
x1,...,xm∈Fq
ηa1xd
1+···+amxd
mχ1(x1). . . χm(xm) = S1−S0,
where
S0=X
x1,...,xm∈Fq
χ1(x1). . . χm(xm)
and
S1=X
y∈F∗
qX
x1,...,xm∈Fq
a1xd
1+···+amxd
m=y2
χ1(x1). . . χm(xm).
Indeed, we observe that each vector (x1,...,xm)∈Fm
qcontributes
2χ1(x1). . . χm(xm) to the sum S1. It is also easy to see that S0vanishes
unless χ1=... =χm=χ0; therefore, due to our assumption from
above, we get that S0= 0. We now fix a nontrivial additive character
14 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
ψof Fq. By the orthogonality relation,
1
qX
λ∈Fq
ψ(λu) = (1,if u= 0,
0,if u∈F∗
q,
see [IwKo04, Section 3.1]. Hence we write
S1=X
x1,...,xm∈FqX
y∈F∗
q
1
qX
λ∈Fq
ψλa1xd
1+···+amxd
m−y2χ1(x1). . . χm(xm)
=1
qX
λ∈FqX
y∈F∗
q
ψ−λy2
X
x1,...,xm∈Fq
ψλa1xd
1+···+amxd
mχ1(x1). . . χm(xm)
=1
qX
λ∈FqX
y∈F∗
q
ψ−λy2m
Y
i=1 X
xi∈Fq
ψλaixd
iχi(xi).
The contribution from the terms corresponding to λ= 0 is obviously
equal to q−1
q·S0= 0 since S0= 0 (because not all of the characters
χiare equal to the principal character). Hence
(4.5) S1=W,
where
W=1
qX
λ∈F∗
qX
y∈F∗
q
ψ−λy2m
Y
i=1 X
xi∈Fq
ψλaixd
iχi(xi).
Now the sum over ydiffers from the classical Gauss sums by only one
term corresponding to y= 0, and so we have
(4.6) X
y∈F∗
q
ψ−λy2≪q1/2,
see [IwKo04, Theorem 3.4]. For the remaining sums, using that λai∈
F∗
qwe apply the Weil bound [Wei74, Appendix 5, Example 12] of mixed
sums of additive and multiplicative characters which implies
(4.7) X
xi∈Fq
ψλaixd
iχi(xi)≪dq1/2,
SPARSE REPRESENTATIONS OF SQUARES 15
see also [Li96, Chapter 6, Theorem 3]. Therefore, the bounds (4.6)
and (4.7), combined together yield
W≪dmq(m+1)/2.
and together with (4.4) and (4.5) we conclude the proof.
We remark that some, or all, of the characters χ1,...,χm∈X
can be principal and that the implied constant from the conclusion of
Lemma 4.2 depends only on m.
4.2. Incomplete character sums with exponential functions.
We now extend the definition of τℓ(g) to orders modulo any composite
moduli qwith gcd(g, q) = 1 . We also use (u/q) to denote the Jacobi
symbol modulo an odd q.
Here we need to obtain multidimensional analogues of the result on
character sums from [BaSh17, Section 3]. Although this does not re-
quire new ideas and can be achieved at the cost of merely typographical
changes we present some short proofs of these results.
As usual, we write e(t) = exp(2πit) for all t∈R.
We use the following variant of the result of [BaSh17, Lemma 3.1],
which in turn is based on some ideas of Korobov [Kor70, Theorem 3].
Lemma 4.3. Let a1, b1...,am, bm∈Zand let ϑ∈Zwith ϑ>2. Let
ℓand rbe distinct primes with
tℓ=τℓ(ϑ), tr=τr(ϑ), t =τℓr(ϑ)
and such that
gcd (ℓr, a1. . . amϑ) = gcd (tℓ, tr) = 1.
We define integers bi,ℓ and bi,r by the conditions
bi,ℓtr+bi,r tℓ≡bi(mod t),06bi,ℓ < tℓ,06bi,r < tr,
for i= 1,...,m. Then, for
S=
t
X
k1,...,km=1 a1ϑk1+...+amϑkm
ℓr eb1k1+...+bmkm
t
we have
S=SℓSr
16 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
where
Sℓ=
tℓ
X
x1,...,xm=1 a1ϑx1+...+amϑxm
ℓeb1,ℓx1+...+bm,ℓxm
tℓ,
Sr=
tr
X
y1,...,ym=1 a1ϑy1+...+amϑym
reb1,ry1+...+bm,r ym
tr.
Proof. As in the proof of [BaSh17, Lemma 3.1], using that gcd(tℓ, tr) =
1, we see that the integers
xtr+ytℓ,06x < tℓ,06y < tr,
run through the complete residue system modulo
t=tℓtr.
Moreover,
(4.8) ϑxtr+ytℓ≡ϑxtr(mod ℓ), ϑxtr+ytℓ≡ϑytℓ(mod r),
and
(4.9) e(b(xtr+ytℓ)/t) = e(bx/tℓ)e(by/tr).
Hence,
S=
tℓ
X
x1,...,xm=1
tr
X
y1,...,ym=1 a1ϑx1tr+y1tℓ+...+amϑxmtr+ymtℓ
ℓr
eb1(x1tr+y1tℓ) + ...+bm(xmtr+ymtℓ)
t.
(4.10)
Using the multiplicativity of the Jacobi symbol, and recalling the
congruences (4.8), we derive
a1ϑx1tr+y1tℓ+...+amϑxmtr+ymtℓ
ℓr
=a1ϑx1tr+y1tℓ+...+amϑxmtr+ymtℓ
ℓ
a1ϑx1tr+y1tℓ+...+amϑxmtr+ymtℓ
r
=a1ϑx1tr+...+amϑxmtr
ℓ
a1ϑy1tℓ+...+amϑymtℓ
r.
(4.11)
SPARSE REPRESENTATIONS OF SQUARES 17
Furthermore, by (4.9) we have
eb1(x1tr+y1tℓ) + ...+bm(xmtr+ymtℓ)
t
=eb1x1+...+bmxm
tℓeb1y1+...+bmtm
tr.
(4.12)
Using (4.11) and (4.12) in (4.10), we see that the sum Scan be de-
composed into a product of two sums as follows
S=
tℓ
X
x1,...,xm=1 a1ϑx1tr+...+amϑxmtr
ℓeb1x1+...+bmxm
tℓ
tr
X
y1,...,ym=1 a1ϑy1tℓ+...+amϑymtℓ
reb1y1+...+bmtm
tr.
We now replace xiwith xit−1
r(mod tℓ) and yiwith yit−1
ℓ(mod tr),
and take into account that
bit−1
r≡bi,ℓ (mod tℓ) and bit−1
ℓ≡bi,r (mod tr),
for i= 1,...,m. This concludes the proof.
Next we estimate the sums Sℓand Srwhich appear in Lemma 4.3.
Namely we now establish an analogue of [BaSh17, Lemma 3.2].
Lemma 4.4. Let a1, b1...,am, bm∈Zand let ϑ∈Zwith ϑ>2. Let
ℓbe a prime with
tℓ=τℓ(ϑ)
and such that
gcd (ℓ, a1...amϑ) = 1 and gcd(tℓ,2) = 1.
Then for
Sℓ=
tℓ
X
x1,...,xm=1 a1ϑx1+...+amϑxm
ℓeb1x1+...+bmxm
tℓ
we have
Sℓ≪(ℓ(m+1)/2,for arbitrary b1,...,bm,
tℓℓ(m−1)/2,for b1=...=bm= 0.
18 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
Proof. Denoting d= (ℓ−1)/tℓ, we can write ϑ=ρdwith some prim-
itive root ρmodulo ℓ. Then,
Sℓ=1
dm
ℓ−1
X
x1,...,xm=1 a1ρdx1+...+amρdxm
ℓ
ed(b1x1+...+bmxm)
ℓ−1
=1
dm
ℓ−1
X
w1,...,wm=1 a1wd
1+...+amwd
m
ℓχ1(w1). . . χm(wm),
(4.13)
where for w∈Fℓwe define χiby
χi(w) = e(bidx/(ℓ−1)) , i = 1,...,m,
where xis any integer for which w≡ρx(mod ℓ).
As in the proof of [BaSh17, Lemma 3.2] we observe χiis a multi-
plicative character of Fℓfor each i= 1,...,m. Recalling Lemma 4.2,
we derive from (4.13)
Sℓ≪1
dmdmℓ(m+1)/2=ℓ(m+1)/2,
which establishes the desired bound for arbitrary b1,...,bm∈Z.
For b1=... =bm= 0 we observe that since tℓis odd, dis even
and hence we can use Lemma 4.1 instead of Lemma 4.2. Thus in this
case (4.13) implies
Sℓ≪1
dmdm−1ℓ(m+1)/2=1
dℓ(m+1)/2≪tℓℓ(m−1)/2,
which concludes the proof.
Lemmas 4.3 and 4.4 combined together imply the following bound.
Corollary 4.5. Let a1, b1...,am, bm∈Zand let ϑ∈Zwith ϑ>2.
Let ℓand rbe distinct primes with
tℓ=τℓ(ϑ), tr=τr(ϑ), t =τℓr(ϑ)
and such that
gcd (ℓr, a1. . . amϑ) = gcd (tℓ, tr) = gcd(tℓtr,2) = 1.
Then, for
S=
t
X
k1,...,km=1 a1ϑk1+...+amϑkm
ℓr eb1k1+...+bmkm
t
SPARSE REPRESENTATIONS OF SQUARES 19
we have
S≪((ℓr)(m+1)/2,for arbitrary b1,...,bm,
t(ℓr)(m−1)/2,for b1=...=bm= 0.
Clearly in Lemma 4.4 and Corollary 4.5 the parity condition on mul-
tiplicative orders is important only in the case where b1=...=bm= 0 ,
as only these parts appeal to Lemma 4.1 (which required dto be even).
Combining Corollary 4.5 with the completing method, see [IwKo04,
Section 12.2], we derive an analogue of [BaSh17, Lemma 3.4], which is
our main technical tool.
Lemma 4.6. Let a1,...,am∈Zand let ϑ∈Zwith ϑ>2. Let ℓand
rbe distinct primes with
gcd (ℓr, a1. . . amϑ) = gcd (τℓ(ϑ), τr(ϑ)) = gcd(τℓ(ϑ)τr(ϑ),2) = 1.
Then, for any integers L1,...Lm>1, we have
L1
X
k1=1
...
Lm
X
km=1 a1ϑk1+...+amϑkm
ℓr
≪L1. . . Lmt−m+1(ℓr)(m−1)/2
+Lm−1t−m+1 + 1(ℓr)(m+1)/2(log(ℓr))m,
where
L= max{L1,...Lm}and t=τℓr(ϑ),
and the implied constant is absolute.
Proof. Clearly we can split the above sum into ⌊L1/t⌋ · ·· · · ⌊Lm/t⌋
complete sums, where each variable runs over the complete residue
system modulo tand into at most O((L/t)m−1+ 1) incomplete sums
over a complete residue system modulo ℓr .
By Corollary 4.5 each of these complete sums can be estimated as
Ot(ℓr)(m−1)/2, so they contribute OL1. . . Lmt−m+1(ℓr)(m−1)/2in
total.
By the standard completing techniques, see, for example, [IwKo04,
Section 12.2], we derive from Corollary 4.5 that each incomplete sum
can be estimated as O(ℓr)(m+1)/2(log(ℓr))m. Therefore, in total they
contribute O(Lm−1t−m+1 + 1) (ℓr)(m+1)/2(log(ℓr))m.
Combining both contributions together, we conclude the proof.
5. Proof of Theorem 1.2
5.1. Preliminary transformations. We can always assume that
km>...>k1.
20 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
We note that there is an integer constant h0depending only on the
initial data such that if km>km−1+h0then
n2=
m
X
i=1
cigki>0.5gkm
and hence for some
(5.1) K≪log N
we have
(5.2) km= max{k1,...,km} ≪ K.
On the other hand, for km< km−1+h0, writing km=km−1+h,
h= 0,...,h0−1 and
n2=
m
X
i=1
cigki=
m−2
X
i=1
cigki+ (cm−1+cmgh)gkm−1
by Theorem 1.1 we obtain at most O((log N)m−1) solutions n6N.
Hence we now estimate the number of solutions to (1.4) with (5.2)
(for Kas in (5.1)).
We recall the notation (3.3) and (3.5), and we write (1.4) as
(5.3) n2=F(k).
We also recall the definitions of the set Lzin Section 3.2 and of ωz(n)
from Section 3.3.
To simplify the exposition everywhere below we replace logarithmic,
and double logarithmic factors of zwith zo(1) (implicitly assuming that
z→ ∞). In particular we simply write
(5.4) #Lz=z1+o(1).
Since zo(1) also absorbs all implied constants, we use 6instead of ≪
in the corresponding bounds.
5.2. Sieving. Note that if F(k) is a perfect square, then we always
have X
ℓ∈LzF(k)
ℓ= #Lz−ωz(F(k)) .
Hence
#Lz6X
ℓ∈LzF(k)
ℓ+ωz(F(k)) .
SPARSE REPRESENTATIONS OF SQUARES 21
Denote by Mthe set of values of k∈Ksatisfying (5.3) and let
M= #Mbe its cardinality. Invoking Lemma 3.3, we obtain
M#Lz6X
k∈MX
ℓ∈LzF(k)
ℓ+X
k∈M
ωz(F(k))
6X
k∈MX
ℓ∈LzF(k)
ℓ+X
k∈K
ωz(F(k))
≪X
k∈MX
ℓ∈LzF(k)
ℓ+Kmz−α+Km−1#Lz.
Therefore either
(5.5) M#Lz≪X
k∈MX
ℓ∈LzF(k)
ℓ
or
(5.6) M≪Kmz−α+Km−1.
Assuming that (5.5) holds, by the Cauchy inequality
(M#Lz)26MX
k∈MX
ℓ∈LzF(k)
ℓ
2
and extending summation back to all k∈Kand using (5.4), we
obtain
(5.7) M6z−2+o(1)W,
where
W=X
k∈KX
ℓ∈LzF(k)
ℓ
2
=X
k∈KX
ℓ,r∈LzF(k)
ℓr .
Combining (5.6) and (5.7), we see that in any case we have
(5.8) M6Kmz−α+Km−1+z−2Wzo(1).
We further split the sum Winto two sums as W=U+V, where
U=X
ℓ,r∈Lz
P(ℓ−1)=P(r−1) X
k∈KF(k)
ℓr ,
V=X
ℓ,r∈Lz
P(ℓ−1)6=P(r−1) X
k∈KF(k)
ℓr .
(5.9)
22 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
To estimate U(which also includes the diagonal case ℓ=r), we use
the trivial bound (K+ 1)mon each inner sum, deriving that
U6(K+ 1)mX
ℓ,r∈Lz
P(ℓ−1)=P(r−1)
1
6(K+ 1)mX
d>zαX
ℓ,r∈Lz
ℓ≡r≡1 mod d
1≪KmX
d>zα
z2
d2≪Kmz2−α.
Hence we can write (5.8) as
(5.10) M6Kmz−α+Km−1+z−2Vzo(1).
5.3. Bounds of character sums. To estimate V, we first observe
that for every ℓ∈Lzthe inequality τℓ(g)>P(ℓ−1) >ℓαimplies
(since α > 1
2) that P(ℓ−1) |τℓ(g).
Fix a pair (ℓ, r)∈Lz×Lzwith P(ℓ−1) 6=P(r−1) and define
h= gcd (τℓ(g), τr(g)) and ϑ=gh.
We observe that
τℓ(ϑ) = τℓ(g)/h and τr(ϑ) = τr(g)/h.
Furthermore, due to our choice of the set Lzin Section 3.2 we have
ν2(τℓ(g)) = ν2(τr(g)) = ν2(h)
and hence both τℓ(ϑ) and τr(ϑ) are odd.
We now write
(5.11) X
k∈KF(k)
ℓr =
h
X
j1,...,jm=1
Tℓ,r (j1,...,jm),
where
Tℓ,r (j1,...,jm) = X
16k16(K−j1)/h
... X
16km6(K−jm)/h c1gk1h+j1+...+cmgkmh+jm
ℓr
=X
16k16(K−j1)/h
... X
16km6(K−jm)/h c1gj1ϑk1+...+cmgjmϑkm
ℓr .
SPARSE REPRESENTATIONS OF SQUARES 23
We can certainly assume that zis large enough so that
gcd(c1···cm, ℓr) = 1
for ℓ, r ∈Lz. Therefore Lemma 4.6 applies to Tℓ,r (j1,...,jm) and
implies
Tℓ,r (j1,...,jm)
≪(K/h)m(τℓ(ϑ)τr(ϑ))−m+1 (ℓr)(m−1)/2
+(K/h)m−1(τℓ(ϑ)τr(ϑ))−m+1 + 1(ℓr)(m+1)/2(log(ℓr))m.
Using
τℓ(ϑ)τr(ϑ)≫z2αand ℓr ≪z2
we see that
Tℓ,r (j1,...,jm)
6(K/h)mz(m−1)(1−2α)+(K/h)m−1z(m−1)(1−2α)+2 +zm+1zo(1).
Therefore, after the substitution in (5.11) we obtain
X
k∈KF(k)
ℓr
6Kmz(m−1)(1−2α)+Km−1hz(m−1)(1−2α)+2 +hmzm+1zo(1).
Since obviously h6gcd(ℓ−1, r −1), from the definition of Vin (5.9)
and using (5.4), we now derive
V6Kmz(m−1)(1−2α)+D1Km−1z(m−1)(1−2α)+Dmzm−1z2+o(1),
where D1and Dmare as in Lemma 3.5 and thus we get
D16z2+o(1) and Dm6zm+α−αm+1+o(1).
Therefore
V6Kmz(m−1)(1−2α)+Km−1z(m−1)(1−2α)+2 +z2m+α−αmz2+o(1) .
which after the substitution in (5.10) yields
M6(Kmz−α+Km−1+Kmz(m−1)(1−2α)
+Km−1z(m−1)(1−2α)+2 +z2m+α−αm)zo(1).
24 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
5.4. Optimisation. Clearly we can assume that
(5.12) z6K1/(2−α)
as otherwise the last term z2m+α−αm exceeds the trivial bound Km.
Furthermore, for m>3 and αas in (3.2), we have
(5.13) (m−1)(2α−1) > α
and hence (using that z(m−1)(1−2α)< z−αdue to the inequality (5.13)),
we can simplify the above bound as follows:
M6Kmz−α+Km−1+Km−1z2−(m−1)(2α−1) +z2m+α−αmzo(1).
Moreover, since we have (5.12) and α < 1, we see that
Kmz−α> Km−1
which means that
(5.14) M6Kmz−α+Km−1z2−(m−1)(2α−1) +z2m+α−αmzo(1).
First we note that for m>5, with our choice of α= 0.677 in (3.2)
along with our assumption (5.12), we have that
Kmz−α> Km−1z2−(m−1)(2α−1)
and thus the second term in (5.14) never dominates and we choose
z=Km/(2m+2α−αm)
to balance the first and the third terms. Hence for if m>5, we obtain:
(5.15) M6Km−mα/(2m+2α−αm)+o(1).
For m= 4 and with (3.2), direct calculations show that as for m>5,
it is better to balance the first and the third terms in (5.14) (rather
than the first and the second terms), hence (5.15) also holds for m= 4.
Finally, for m= 3 one checks that balancing the first and the second
terms in (5.14) with z=K1/(4−3α)leads to an optimal result
M6K3−α/(4−3α)+K(6−2α)/(4−3α)Ko(1) 6K3−α/(4−3α)+o(1),
which concludes the proof (see also (5.1)).
6. Comments
The proof of Theorem 1.2, depends on the bound
X
k∈K
ωz(F(k)) 6X
ℓ∈Lz
Tm(K, ℓ),
where Tm(K, ℓ) is the number of solutions to the congruence
F(k)≡0 (mod ℓ),k∈Km(K),
SPARSE REPRESENTATIONS OF SQUARES 25
where Km(K) is given by (3.4), see the proof of Lemma 3.3. In fact,
in the proof of Lemma 3.3 we use the trivial bound
(6.1) Tm(K, ℓ)6(K+ 1)m−1K+ 1
τℓ(g)+ 1≪Kmz−α+Km−1,
which holds for any ℓ∈Lzand is the best possible for m= 2 . For
m>3 we get a better bound using exponential sum. This does not
improve our final result, however since it can be of independent interest
and since it maybe becomes important if better bounds of Win (5.7)
become available (or maybe with some other modifications of the argu-
ment) we present such a better bound in Appendix A, see Lemma A.2.
The method of the proof of Theorem 1.2 also works for relations of
the form
n2=
m
X
i=1
cigki
i,
with integer coefficients c1,...,cmof the same sign and arbitrary in-
teger bases g1, . . . , gm>2. Indeed, in this case we still have a bound
O(log N) on the exponents k1,...,km, which is important for our
method. It is an interesting open question to establish such a bound
for arbitrary c1,...,cm. Similarly, our method can also be used to
estimate the number of n6Nwhich can be represented as
n2=u1+...+um
for some S-units u1,...,um, that is, as a sum of mintegers which
have all their prime factors from a prescribed finite set of primes S.
Again, if negative values of u1,...,umare allowed then some additional
arguments are needed to bound the powers of primes in each S-unit.
Furthermore, as in [BaSh17] we observe that under the Generalised
Riemann Hypothesis we can obtain a slightly large value of γm. We
now recall that pis called a Sophie Germain prime if pand 2p+ 1
are both prime. Under the assumption of the existence of the expected
number of Sophie Germain primes in intervals, or in fact of just z1+o(1)
such primes up to z, we can choose a set Lzin the argument of the
proof of Theorem 1.2 with any α < 1 and we see that under this
assumption we can take γm=m/(m+ 2) for m>3.
Finally, we note that other perfect powers nνfor a fixed ν= 3,4,...,
can be investigated by our method. However, one needs a version of a
result of Baker and Harman [BaHa98] for primes ℓin the arithmetic
progression ℓ≡1 (mod ν), so that there are multiplicative characters
modulo ℓof order ν.
26 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
Appendix A. Congruences with exponential functions
First we recall the following special case of a classical result of Ko-
robov [Kor72, Lemma 2].
Lemma A.1. Let a∈Zand let ϑ∈Zwith ϑ>2. Let ℓbe a prime
with
t=τℓ(ϑ),
and such that
gcd (ℓ, aϑ) = 1.
Then, we have
t
X
k=1
eaϑk/t6ℓ1/2.
We now have a bound on Tm(K, ℓ) which improves (6.1) in some
ranges.
Lemma A.2. Let m>3. Then for K>zand ℓ∈Lz, where Lzis
as in Section 3.2, we have
Tm(K, ℓ)≪Kmz−1+Kmzm/2−α(m−1)−1.
Proof. Let t=τℓ(g) ; then since ℓ∈Lz, we know that t>zα. We
also let
Tm(ℓ) = Tm(t−1, ℓ).
First we observe that K>z≫tand thus
(A.1) Tm(K, ℓ)6K+ 1
t+ 1m
Tm(ℓ)≪Kmt−mTm(ℓ).
Now, using the orthogonality of exponential functions, we write
Tm(ℓ) = 1
ℓX
k∈Km(t−1)
ℓ−1
X
a=0
e(aF (k)/ℓ),
where Km(t−1) consists of all m-tuples (k1,...,km) of non-negative
integers ki< t. Now changing the order of summation, we obtain
Tm(ℓ) = 1
ℓ
ℓ−1
X
a=0 X
k∈Km(t−1)
e(aF (k)/ℓ) = 1
ℓ
ℓ−1
X
a=0
m
Y
i=1
t−1
X
ki=0
eacigki/ℓ.
The term corresponding to a= 0 is equal to tm/ℓ. We can as-
sume that zis large enough (as otherwise the bound is trivial) so
SPARSE REPRESENTATIONS OF SQUARES 27
that gcd(c1···cm, ℓ) = 1 for ℓ∈Lz. We apply now the bound of
Lemma A.1 to m−2 sums over k3,...,kmand derive
(A.2) Tm(ℓ)6tm/ℓ +ℓ(m−2)/21
ℓR,
where
R=
ℓ−1
X
a=0
t−1
X
k1=0
eac1gk1/ℓ·
t−1
X
k2=0
eac2gk2/ℓ
(note that after an application of Lemma A.1 we have added the term
corresponding to a= 0 back to the sum). By the Cauchy inequality
R26
ℓ−1
X
a=0
t−1
X
k1=0
eac1gk1/ℓ
2
·
ℓ−1
X
a=0
t−1
X
k2=0
eac2gk2/ℓ
2
.
Using the orthogonality of exponential functions again, we derive
ℓ−1
X
a=0
t−1
X
k1=0
eac1gk1/ℓ
2
=ℓt
and similarly for the sum over k2. Hence R6ℓt, and after substitution
in (A.2) we derive
Tm(ℓ)6tm/ℓ +ℓ(m−2)/2t,
which together with the inequality (A.1) and the fact that t>zα
concludes the proof.
Acknowledgements
D. G. and S. S. were partially supported by a Discovery Grant from
NSERC, A. O. by ARC Grants DP180100201 and DP200100355, and
I. S. by an ARC Grant DP200100355.
References
[BaHa98] R. C. Baker and G. Harman, ‘Shifted primes without large prime factors’,
Acta Arith.,83 (1998), 331–361. 4,9,25
[BaSh17] W. D. Banks and I. E. Shparlinski, ‘On coincidences among quadratic
fields generated by the Shanks sequence’, Quart. J Math.,68 (2017),
465–484. 4,15,16,17,18,19,25
[BGT16] J. P. Bell, D. Ghioca and T. J. Tucker, ‘The dynamical Mordell-
Lang conjecture’, Mathematical Surveys and Monographs 210, American
Mathematical Society, Providence, RI, 2016. xiii+280 pp. 1
[BeBu14] M. A. Bennett and Y. Bugeaud, ‘Perfect powers with three digits’, Math-
ematika,60 (2014), 66–84. 4
28 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
[BBM13] M. A. Bennett, Y. Bugeaud and M. Mignotte, ‘Perfect powers with few
binary digits and related Diophantine problems’, Ann. Sc. Norm. Super.
Pisa Cl. Sci.,12 (2013), 941–953. 4
[CGSZ] P. Corvaja, D. Ghioca, T. Scanlon and U. Zannier, ‘The Dynamical
Mordell-Lang Conjecture for endomorphisms of semiabelian varieties de-
fined over fields of positive characteristic’, J. Inst. Math. Jussieu, 30 pp.
(2019), to appear. 3
[CoZa13] P. Corvaja and U. Zannier, ‘Finiteness of odd perfect powers with four
nonzero binary digits’, Ann. Inst. Fourier (Grenoble),63 (2013), 715–
731. 3,4
[ErMu96] P. Erd˝os and M. R. Murty, ‘On the order of a(mod p)’, Number theory
(Ottawa, ON, 1996), CRM Proc. Lecture Notes, 19, Amer. Math. Soc.,
Providence, RI, 1999, 87–97, 8
[GOSS21] D. Ghioca, A. Ostafe, S. Saleh and I. E. Shparlinski, ‘A sparsity result
for the Dynamical Mordell-Lang Conjecture in positive characteristic’,
Bull. Aust. Math. Soc., 2021, 8 pp., to appear. 1,2
[H-B84] D. R. Heath-Brown, ‘The square sieve and consecutive square-free num-
bers’, Math. Ann.,266 (1984), 251–259. 3
[IwKo04] H. Iwaniec and E. Kowalski, Analytic number theory, Amer. Math. Soc.,
Providence, RI, 2004. 9,11,14,19
[Kat02] N. Katz, ‘Estimates for nonsingular multiplicative character sums’, Int.
Math. Res. Not.,7(2002), 333–349. 12,13
[Kea69] M. Keates, ‘On the greatest prime factor of a polynomial, Proc. Edin-
burgh Math. Soc.,16 (1969), 301–303. 3
[Kor70] N. M. Korobov, ‘Trigonometric sums with exponential functions, and the
distribution of the digits in periodic fractions’, Mat. Zametki ,8(1970)
641–652 (in Russian); Translation in English: Mathem. Notes ,8(1970),
831–837. 15
[Kor72] N. Korobov, ‘On the distribution of digits in periodic fractions’, Matem.
Sbornik,89 (1972), 654–670; translation in English: Mathem. USSR
Sbornik,18 (1972), 659–676. 26
[KuPo05] P. Kurlberg and C. Pomerance, ‘On the period of the linear congruential
and power generators’, Acta Arith.,119 (2005), 149–169. 8
[Lau84] M. Laurent, ‘´
Equations diophantiennes exponentielles’, Invent. Math.,
78 (1984), 299–327. 2
[Li96] W.-C. W. Li, Number theory with applications, World Scientific, Singa-
pore, 1996. 15
[LuSh09] F. Luca and I. E. Shparlinski, ‘On quadratic fields generated by the
Shanks sequence’, Proc. Edinb. Math. Soc.,52 (2009), 719–729. 4,9
[Mos21] A. Moscariello, ‘On sparse perfect powers’, Preprint, 2021 (available
from http://arxiv.org/abs/2101.10415). 4
[Wei74] A. Weil, Basic number theory, Springer-Verlag, New York, 1974. 14
Dragos Ghioca, Department of Mathematics, University of British
Columbia, Vancouver, BC V6T 1Z2, Canada
Email address:dghioca@math.ubc.ca
SPARSE REPRESENTATIONS OF SQUARES 29
Alina Ostafe, School of Mathematics and Statistics, University of
New South Wales, Sydney NSW 2052, Australia
Email address:alina.ostafe@unsw.edu.au
Sina Saleh, Department of Mathematics, University of British Co-
lumbia, Vancouver, BC V6T 1Z2, Canada
Email address:sinas@math.ubc.ca
Igor E. Shparlinski, School of Mathematics and Statistics, Univer-
sity of New South Wales, Sydney NSW 2052, Australia
Email address:igor.shparlinski@unsw.edu.au
