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Diophantine tuples and multiplicative structure of shifted multiplicative subgroups

Seoyoung Kim Mathematisches Institut, Georg-August Universität Göttingen, Bunsenstrasse 3-5, 37073 Göttingen, Germany [email protected] Chi Hoi Yip School of Mathematics
Georgia Institute of Technology
GA 30332
United States
[email protected]
 and  Semin Yoo Discrete Mathematics Group
Institute for Basic Science
55 Expo-ro Yuseong-gu, Daejeon 34126
South Korea
[email protected]
Abstract.

In this paper, we investigate the multiplicative structure of a shifted multiplicative subgroup and its connections with additive combinatorics and the theory of Diophantine equations. Among many new results, we highlight our main contributions as follows. First, we show that if a nontrivial shift of a multiplicative subgroup GG contains a product set ABAB, then |A||B||A||B| is essentially bounded by |G||G|, refining a well-known consequence of a classical result by Vinogradov. Second, we provide a sharper upper bound of Mk(n)M_{k}(n), the largest size of a set such that each pairwise product of its elements is nn less than a kk-th power, refining the recent result of Dixit, Kim, and Murty. One main ingredient in our proof is the first non-trivial upper bound on the maximum size of a generalized Diophantine tuple over a finite field. In addition, we determine the maximum size of an infinite family of generalized Diophantine tuples over finite fields with square order, which is of independent interest. We also make significant progress towards a conjecture of Sárközy on the multiplicative decompositions of shifted multiplicative subgroups. In particular, we prove that for almost all primes pp, the set {x21:x𝔽p}{0}\{x^{2}-1:x\in{\mathbb{F}}_{p}^{*}\}\setminus\{0\} cannot be decomposed as the product of two sets in 𝔽p{\mathbb{F}}_{p} non-trivially.

Key words and phrases:
Diophantine tuples, shifted multiplicative subgroup, multiplicative decomposition
2020 Mathematics Subject Classification:
Primary 11B30, 11D72; Secondary 11D45, 11N36, 11L40

1. Introduction

Let qq be a prime power, and let 𝔽q{\mathbb{F}}_{q} be the finite field with qq elements. Let GG be a multiplicative subgroup of 𝔽q{\mathbb{F}}_{q}. While GG itself has a “perfect” multiplicative structure, it is natural to ask if a (non-trivial additive) shift of GG still possesses some multiplicative structure. Indeed, as a fundamental question in additive combinatorics, this question has drawn the attention of many researchers and it is closely related to many questions in number theory. For example, a classical result of Vinogradov [36] states that for a prime pp and an integer nn such that pnp\nmid n, if A,B{1,2,,p1}A,B\subset\{1,2,\ldots,p-1\}, then

|aA,bB(ab+np)|p|A||B|.\bigg{|}\sum_{a\in A,\,b\in B}\bigg{(}\frac{ab+n}{p}\bigg{)}\bigg{|}\leq\sqrt{p|A||B|}. (1.1)

More generally, inequality 1.1 extends to all nontrivial multiplicative characters over all finite fields; see Proposition 3.1. Inequality 1.1 leads an estimate on the size of a product set contains in the set of shifted squares: if A,B𝔽pA,B\subset{\mathbb{F}}_{p}^{*}, λ𝔽p\lambda\in{\mathbb{F}}_{p}^{*}, and GG is the subgroup of 𝔽p{\mathbb{F}}_{p}^{*} of index 22 such that AB(G+λ)AB\subset(G+\lambda), then

|A||B|<(1+o(1))p.|A||B|<(1+o(1))p. (1.2)

For more recent works related to this question and its connection with other problems, we refer to [17, 27, 31, 37] and references therein. An analogue of this question over integers is closely related to the well-studied Diophantine tuples and their generalizations; see Subsection 1.1.

In this paper, we study the multiplicative structure of a shifted multiplicative subgroup following the spirit of the aforementioned works and discuss a few new applications in additive combinatorics and Diophantine equations. More precisely, one of our contributions is the following theorem.

Theorem 1.1.

Let d(q1)d\mid(q-1) with d2d\geq 2. Let Sd={xd:x𝔽q}S_{d}=\{x^{d}:x\in{\mathbb{F}}_{q}^{*}\}. Let A,B𝔽qA,B\subset{\mathbb{F}}_{q}^{*} and λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*} with |A|,|B|2|A|,|B|\geq 2. Assume further that (|A|1+q1d|A|)0(modp)\binom{|A|-1+\frac{q-1}{d}}{|A|}\not\equiv 0\pmod{p} if qpq\neq p. If AB+λSd{0}AB+\lambda\subset S_{d}\cup\{0\}, then

|A||B||Sd|+|B(λA1)|+|A|1.|A||B|\leq|S_{d}|+|B\cap(-\lambda A^{-1})|+|A|-1.

Moreover, when λSd\lambda\in S_{d}, we have a stronger upper bound:

|A||B||Sd|+|B(λA1)|1.|A||B|\leq|S_{d}|+|B\cap(-\lambda A^{-1})|-1.

Clearly, Theorem 1.1 improves inequality 1.2 implied by Vinogradov’s estimate 1.1 when d=2d=2, and the generalization of inequality 1.2 by Gyarmati [17, Theorem 8] for general dd, where the upper bound is given by (p+2)2(\sqrt{p}+2)^{2} when q=pq=p is a prime. We remark that in general the condition on the binomial coefficient in the statement of Theorem 1.1 cannot be dropped when qq is not a prime; see Theorem 1.6.

The proof of Theorem 1.1 is based on Stepanov’s method [35], and is motivated by a recent breakthrough of Hanson and Petridis [21]. In fact, Theorem 1.1 can be viewed as a multiplicative analog of their results. Going beyond the perspective of these multiplicative analogs, we provide new insights into the application of Stepanov’s method. For example, our technique applies to all finite fields while their technique only works over prime fields. We also prove a similar result for restricted product sets (see Theorem 4.2), whereas their technique appears to only lead to a weaker bound; see Remark 4.3.

Besides Theorem 1.1, we also provide three novel applications of Theorem 1.1 and its variants. These applications significantly improve on many previous results in the literature. Unsurprisingly, to achieve these applications, we need additional tools and insights from Diophantine approximation, sieve methods, additive combinatorics, and character sums. From here, we briefly mention what applications are about. In Subsection 1.1, we improve the upper bound of generalized Diophantine tuples over integers. Interestingly, Theorem 1.1 is closely related to a bipartite version of Diophantine tuples over finite fields. This new perspective yields a substantial improvement in the result of generalized Diophantine tuples over integers. In Subsection 1.2, we obtain the following nontrivial upper bounds on generalized Diophantine tuples and strong Diophantine tuples over finite fields. Moreover, some of our new bounds are sharp. Last but not least, in Subsection 1.3, we make significant progress towards a conjecture of Sárközy [31] on multiplicative decompositions of shifted multiplicative subgroups. We elaborate on the context of these applications in the next subsections.

1.1. Diophantine tuples over integers

A set {a1,a2,,am}\{a_{1},a_{2},\ldots,a_{m}\} of distinct positive integers is a Diophantine mm-tuple if the product of any two distinct elements in the set is one less than a square. The first known example of integral Diophantine 44-tuples is {1,3,8,120}\{1,3,8,120\} which was studied by Fermat. The Diophantine 44-tuple was extended by Euler to the rational 55-tuple {1,3,8,120,7774808288641}\{1,3,8,120,\frac{777480}{8288641}\}, and it had been conjectured that there is no Diophantine 55-tuple. The difficulty of extending Diophantine tuples can be explained by its connection to the problem of finding integral points on elliptic curves: if {a,b,c}\{a,b,c\} forms a Diophantine 33-tuple, in order to find a positive integer dd such that {a,b,c,d}\{a,b,c,d\} is a Diophantine 44-tuple, we need to solve the following simultaneous equation for dd:

ad+1=s2,bd+1=t2,cd+1=r2.ad+1=s^{2},\quad bd+1=t^{2},\quad cd+1=r^{2}.

This is related to the problem of finding an integral point (d,str)(d,str) on the following elliptic curve

y2=(ax+1)(bx+1)(cx+1).y^{2}=(ax+1)(bx+1)(cx+1).

From this, we can deduce that there are no infinite Diophantine mm-tuples by Siegel’s theorem on integral points. On the other hand, Siegel’s theorem is not sufficient to give an upper bound on the size of Diophantine tuples due to its ineffectivity. In the same vein, finding a Diophantine tuple of size greater than or equal to 55 is related to the problem of finding integral points on hyperelliptic curves of genus g2g\geq 2. Despite the aforementioned difficulties, the conjecture on the non-existence of Diophantine 55-tuples was recently proved to be true in the sequel of important papers by Dujella [9], and He, Togbé, and Ziegler [22].

The definition of Diophantine mm-tuples has been generalized and studied in various contexts. We refer to the recent book of Dujella [10] for a thorough list of known results on the topic and their reference. In this paper, we focus on the following generalization of Diophantine tuples: for each n1n\geq 1 and k2k\geq 2, we call a set {a1,a2,,am}\{a_{1},a_{2},\ldots,a_{m}\} of distinct positive integers a Diophantine mm-tuple with property Dk(n)D_{k}(n) if the product of any two distinct elements is nn less than a kk-th power. We write

Mk(n)=sup{|A|:A satisfies the property Dk(n)}.M_{k}(n)=\sup\{|A|\colon A\subset{\mathbb{N}}\text{ satisfies the property }D_{k}(n)\}.

Similar to the classical case, the problem of finding Mk(n)M_{k}(n) for k3k\geq 3 and n1n\geq 1 is related to the problem of counting the number of integral points of the superelliptic curve

yk=f(x)=(a1x+n)(a2x+n)(a3x+n)y^{k}=f(x)=(a_{1}x+n)(a_{2}x+n)(a_{3}x+n)

The theorem of Faltings [13] guarantees that the above curve has only finitely many integral points, and this, in turn, implies that a set with property Dk(n)D_{k}(n) must be finite. The known upper bounds for the number of integral points depend on the coefficients of f(x)f(x). The Caporaso-Harris-Mazur conjecture [6] implies that Mk(n)M_{k}(n) is uniformly bounded, independent of nn. For other conditional bounds, we refer the readers to Subsection 2.4.

Unconditionally, in [4], Bugeaud and Dujella showed that M3(1)7,M4(1)5,Mk(1)4M_{3}(1)\leq 7,M_{4}(1)\leq 5,M_{k}(1)\leq 4 for 5k1765\leq k\leq 176, and the uniform bound Mk(1)3M_{k}(1)\leq 3 for any k177k\geq 177. On the other hand, the best-known upper bound on M2(n)M_{2}(n) is (2+o(1))logn(2+o(1))\log n, due to Yip [39]. Very recently, Dixit, Kim, and Murty [7] studied the size of a generalized Diophantine mm-tuple with property Dk(n)D_{k}(n), improving the previously best-known upper bound Mk(n)2|n|5+3M_{k}(n)\leq 2|n|^{5}+3 for k5k\geq 5 given by Bérczes, Dujella, Hajdu and Luca [2] when nn\to\infty. They showed that if kk is fixed and nn\to\infty, then Mk(n)klognM_{k}(n)\ll_{k}\log n. Following their proof in [7], the bound can be more explicitly expressed as Mk(n)(3ϕ(k)+o(1))lognM_{k}(n)\leq(3\phi(k)+o(1))\log n when k3k\geq 3 is fixed, nn\to\infty, and ϕ\phi is the Euler phi function. Note that their upper bound on Mk(n)M_{k}(n) is perhaps not desirable. Indeed, it is natural to expect that Mk(n)M_{k}(n) would decrease if nn is fixed, and kk increases, since kk-th powers become sparser. Instead, our new upper bounds on Mk(n)M_{k}(n) support this heuristic.

In this paper, we provide a significant improvement on the upper bound of Mk(n)M_{k}(n) by using a novel combination of Stepanov’s method and Gallagher’s larger sieve inequality. In order to state our first result, we define the following constant

ηk=min||T2\eta_{k}=\min_{\mathcal{I}}\frac{|\mathcal{I}|}{T_{\mathcal{I}}^{2}} (1.3)

for each k2k\geq 2, where the minimum is taken over all nonempty subsets \mathcal{I} of the set

{1ik:gcd(i,k)=1,gcd(i1,k)>1},\{1\leq i\leq k:\gcd(i,k)=1,\gcd(i-1,k)>1\},

and T=igcd(i1,k)T_{\mathcal{I}}=\sum_{i\in\mathcal{I}}\sqrt{\gcd(i-1,k)}.

Theorem 1.2.

There is a constant c>0c^{\prime}>0, such that as nn\to\infty,

Mk(n)(2kk2+o(1))ηkϕ(k)logn,M_{k}(n)\leq\bigg{(}\frac{2k}{k-2}+o(1)\bigg{)}\ \eta_{k}\phi(k)\log n, (1.4)

holds uniformly for positive integers k,n3k,n\geq 3 such that logkcloglogn\log k\leq c^{\prime}\sqrt{\log\log n}.

The constant ηk\eta_{k} is essentially computed via the optimal collection of “admissible” residue classes when applying Gallagher’s larger sieve (see Section 5). Note that when ={1}\mathcal{I}=\{1\}, we have T=kT_{\mathcal{I}}=\sqrt{k}, and hence we have ηk1k\eta_{k}\leq\frac{1}{k}. In particular, if k3k\geq 3 is fixed and nn\to\infty, inequality 1.4 implies the upper bound

Mk(n)(2+o(1))ϕ(k)k2logn,M_{k}(n)\leq\frac{(2+o(1))\phi(k)}{k-2}\log n, (1.5)

which already improves the best-known upper bound Mk(n)(3ϕ(k)+o(1))lognM_{k}(n)\leq(3\phi(k)+o(1))\log n of [7] that we mentioned earlier substantially. In Appendix A, we illustrate the bound in inequality 1.4: for 2k10012\leq k\leq 1001, we compute the suggested upper bound

νk=2kk2ηkϕ(k)\nu_{k}=\frac{2k}{k-2}\eta_{k}\phi(k)

of γk=lim supnMk(n)logn\gamma_{k}=\limsup_{n\to\infty}\frac{M_{k}(n)}{\log n}. From Figure A.1, one can compare the bound of Mk(n)M_{k}(n) in Theorem 1.2 with the bound in [7]. From inequality 1.5, we see γk\gamma_{k} is uniformly bounded by 66. Table A.2 illustrates better upper bounds on γk\gamma_{k} for 2k2012\leq k\leq 201. In particular, we use a simple greedy algorithm to determine ηk\eta_{k} for a fixed kk. We also refer to Subsection 5.3 for a simple upper bound on ηk\eta_{k}, which well approximates ηk\eta_{k} empirically.

At first glance, Theorem 1.2 improves the bound in [7] of Mk(n)M_{k}(n) by only a constant multiplicative factor when kk is fixed. Nevertheless, note that Theorem 1.2 holds uniformly for kk and nn as long as logkcloglogn\log k\leq c^{\prime}\sqrt{\log\log n}. Thus, when kk is assumed to be a function of nn which increases as nn increases, we can break the “logn\log n barrier” in [7], that is, Mk(n)=Ok(logn)M_{k}(n)=O_{k}(\log n), and provide a dramatic improvement.

Theorem 1.3.

There is k=k(n)k=k(n) such that logkloglogn\log k\asymp\sqrt{\log\log n}, and

Mk(n)exp(c′′(loglogn)1/4logloglogn)logn,M_{k}(n)\ll\exp\bigg{(}{-}\frac{c^{\prime\prime}(\log\log n)^{1/4}}{\log\log\log n}\bigg{)}\log n,

where c′′>0c^{\prime\prime}>0 is an absolute constant.

The proofs of Theorem 1.2 and Theorem 1.3 require the study of (generalized) Diophantine tuples over finite fields, which we discuss below.

1.2. Diophantine tuples over finite fields

A Diophantine mm-tuple with property Dd(λ,𝔽q)D_{d}(\lambda,\mathbb{F}_{q}) is a set {a1,,am}\{a_{1},\ldots,a_{m}\} of mm distinct elements in 𝔽q\mathbb{F}_{q}^{*} such that aiaj+λa_{i}a_{j}+\lambda is a dd-th power in 𝔽q\mathbb{F}_{q}^{*} or 0 whenever iji\neq j. Moreover, we also define the strong Diophantine tuples in finite fields motivated by Dujella and Petričević [11]: a strong Diophantine mm-tuple with property SDd(λ,𝔽q)SD_{d}(\lambda,\mathbb{F}_{q}) is a set {a1,,am}\{a_{1},\ldots,a_{m}\} of mm distinct elements in 𝔽q\mathbb{F}_{q}^{*} such that aiaj+λa_{i}a_{j}+\lambda is a dd-th power in 𝔽q\mathbb{F}_{q}^{*} or 0 for any choice of ii and jj. Unlike the natural analog for the classical Diophantine tuples (of property D2(1)D_{2}(1)), it makes sense to talk about the strong Diophantine tuples in 𝔽q\mathbb{F}_{q}. The strong generalized Diophantine tuples with property Dk(n)D_{k}(n) in for general kk and nn are also meaningful to study: the problem of counting the explicit size of the strong generalized Diophantine tuples with property Dk(n)D_{k}(n) involves the problem of counting solutions of the equations appearing in the statement of Pillai’s conjecture. Theorem 1.2 can be improved for strong generalized Diophantine tuples with property Dk(n)D_{k}(n); see Theorem 5.2.

The generalizations of Diophantine tuples over finite fields are of independent interest. Perhaps the most interesting question to explore is the exact analog of estimating Mk(n)M_{k}(n) as discussed in Subsection 1.1. Indeed, estimating the size of the largest Diophantine tuple with property SDd(λ,𝔽q)SD_{d}(\lambda,\mathbb{F}_{q}) or with property Dd(λ,𝔽q)D_{d}(\lambda,\mathbb{F}_{q}) is of particular interest for the application of Diophantine tuples (over integers) as discussed in [1, 7, 8, 16, 24]. Similarly, we denote

MSDd(λ,𝔽q)\displaystyle MSD_{d}(\lambda,\mathbb{F}_{q}) =sup{|A|:A𝔽q satisfies property SDd(λ,𝔽q)},and\displaystyle=\sup\{|A|\colon A\subset{\mathbb{F}}_{q}^{*}\text{ satisfies property }SD_{d}(\lambda,\mathbb{F}_{q})\},\quad\text{and}
MDd(λ,𝔽q)\displaystyle MD_{d}(\lambda,\mathbb{F}_{q}) =sup{|A|:A𝔽q satisfies property Dd(λ,𝔽q)}.\displaystyle=\sup\{|A|\colon A\subset{\mathbb{F}}_{q}^{*}\text{ satisfies property }D_{d}(\lambda,\mathbb{F}_{q})\}.

Note that when λ=0\lambda=0, it is trivial that MSDd(λ,𝔽q)=MDd(λ,𝔽q)=q1dMSD_{d}(\lambda,\mathbb{F}_{q})=MD_{d}(\lambda,\mathbb{F}_{q})=\frac{q-1}{d}. Thus, we always assume λ0\lambda\neq 0 throughout the paper. In Section 3, we give an upper bound q+O(1)\sqrt{q}+O(1) of MSDd(λ,𝔽q)MSD_{d}(\lambda,\mathbb{F}_{q}) and MDd(λ,𝔽q)MD_{d}(\lambda,\mathbb{F}_{q}). More precisely, we prove the following proposition using a double character sum estimate. We refer to the bounds in the following proposition as the “trivial” upper bound.

Proposition 1.4 (Trivial upper bound).

Let d2d\geq 2 and let q1(modd)q\equiv 1\pmod{d} be a prime power. Let A𝔽qA\subset{\mathbb{F}}_{q}^{*} and λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}. Then MSDd(λ,𝔽q)4q3+12MSD_{d}(\lambda,\mathbb{F}_{q})\leq\frac{\sqrt{4q-3}+1}{2} and MDd(λ,𝔽q)q114+52MD_{d}(\lambda,\mathbb{F}_{q})\leq\sqrt{q-\frac{11}{4}}+\frac{5}{2}.

For the case q=pq=p, similar bounds of Proposition 1.4 are known previously in [1, 7, 17]. On the other hand, Proposition 1.4 gives an almost optimal bound of MSDd(λ,𝔽q)MSD_{d}(\lambda,\mathbb{F}_{q}) and MDd(λ,𝔽q)MD_{d}(\lambda,\mathbb{F}_{q}) when qq is a square (Theorem 1.6). Our next theorem improves the trivial upper bounds in Proposition 1.4 by a multiplicative constant factor 1/d\sqrt{1/d} or 2/d\sqrt{2/d} when q=pq=p is a prime.

Theorem 1.5.

Let d2d\geq 2. Let p1(modd)p\equiv 1\pmod{d} be a prime and let λ𝔽p\lambda\in{\mathbb{F}}_{p}^{*}. Then

  1. (1)

    MSDd(λ,𝔽p)(p1)/d+1MSD_{d}(\lambda,\mathbb{F}_{p})\leq\sqrt{(p-1)/d}+1. Moreover, if λ\lambda is a dd-th power in 𝔽p{\mathbb{F}}_{p}^{*}, then we have a stronger upper bound:

    MSDd(λ,𝔽p)p1d34+12.MSD_{d}(\lambda,\mathbb{F}_{p})\leq\sqrt{\frac{p-1}{d}-\frac{3}{4}}+\frac{1}{2}.
  2. (2)

    MDd(λ,𝔽p)2(p1)/d+4MD_{d}(\lambda,\mathbb{F}_{p})\leq\sqrt{2(p-1)/d}+4.

We remark that our new bound on MSDd(λ,𝔽p)MSD_{d}(\lambda,\mathbb{F}_{p}) is sometimes sharp. For example, we get a tight bound for a prime p{5,7,11,13,17,23,31,37,41,53,59,61,113}p\in\{5,7,11,13,17,23,31,37,41,53,59,61,113\} when d=2d=2 and λ=1\lambda=1. See also Theorem 4.7 and Remark 4.8 for a generalization of Theorem 1.5 over general finite fields with non-square order under some extra assumptions.

Nevertheless, in the case of finite fields of square order, we improve Proposition 1.4 by a little bit under some minor assumptions; see Theorem 4.5. Surprisingly, this tiny improvement turns out to be sharp for many infinite families of (q,d,λ)(q,d,\lambda). Equivalently, we determine MDd(λ,𝔽q)MD_{d}(\lambda,{\mathbb{F}}_{q}) and MSDd(λ,𝔽q)MSD_{d}(\lambda,{\mathbb{F}}_{q}) exactly in those families. In the following theorem, we give a sufficient condition so that MDd(λ,𝔽q)MD_{d}(\lambda,{\mathbb{F}}_{q}) and MSDd(λ,𝔽q)MSD_{d}(\lambda,{\mathbb{F}}_{q}) can be determined explicitly.

Theorem 1.6.

Let qq be a prime power and a square, d2d\geq 2, and d(q+1)d\mid(\sqrt{q}+1). Let Sd={xd:x𝔽q}S_{d}=\{x^{d}:x\in{\mathbb{F}}_{q}^{*}\}. Suppose that there is α𝔽q\alpha\in{\mathbb{F}}_{q} such that α2Sd\alpha^{2}\in S_{d} and λα2𝔽q\lambda\in\alpha^{2}{\mathbb{F}}_{\sqrt{q}}^{*} (for example, if α=1\alpha=1 and λ𝔽q)\lambda\in{\mathbb{F}}_{\sqrt{q}}^{*}). Suppose further that r(p1)qr\leq(p-1)\sqrt{q}, where rr is the remainder of q1d\frac{q-1}{d} divided by pqp\sqrt{q}. Then MSDd(λ,𝔽q)=q1MSD_{d}(\lambda,{\mathbb{F}}_{q})=\sqrt{q}-1. If q25q\geq 25 and d3d\geq 3, then we have the stronger conclusion that MDd(λ,𝔽q)=q1MD_{d}(\lambda,{\mathbb{F}}_{q})=\sqrt{q}-1.

Under the assumptions on Theorem 1.6, α𝔽q\alpha{\mathbb{F}}_{\sqrt{q}}^{*} satisfies the required property SDd(λ,𝔽q)SD_{d}(\lambda,\mathbb{F}_{q}) and Dd(λ,𝔽q)D_{d}(\lambda,\mathbb{F}_{q}). Compared to Theorem 1.5, it is tempting to conjecture that such an algebraic construction (which is unique to finite fields with proper prime power order) is the optimal one with the required property. Given Proposition 1.4, to show such construction is optimal, it suffices to rule out the possibility of a Diophantine tuple with property SDd(λ,𝔽q)SD_{d}(\lambda,\mathbb{F}_{q}) and Dd(λ,𝔽q)D_{d}(\lambda,\mathbb{F}_{q}) of size q\sqrt{q}. While this seems easy, it turned out that this requires non-trivial efforts.

Next, we give concrete examples where Theorem 1.6 applies.

Example 1.7.

Note that a Diophantine tuple with property SD2(1,𝔽q)SD_{2}(1,{\mathbb{F}}_{q}) corresponds to a strong Diophantine tuple over 𝔽q{\mathbb{F}}_{q}. If qq is an odd square, Theorem 1.6 implies that the largest size of a strong Diophantine tuple over 𝔽q{\mathbb{F}}_{q} is given by q1\sqrt{q}-1, which is achieved by 𝔽q{\mathbb{F}}_{\sqrt{q}}^{*}. Note that in this case we have r=pq12<(p1)qr=\frac{p\sqrt{q}-1}{2}<(p-1)\sqrt{q}.

We also consider the case that d=3d=3, d(q+1)d\mid(\sqrt{q}+1), and λ=1\lambda=1. In this case, Theorem 1.6 also applies. Note that 3(q+1)3\mid(\sqrt{q}+1) implies that p2(mod3)p\equiv 2\pmod{3}, in which case the base-pp representation of q13\frac{q-1}{3} only contains the digit p23\frac{p-2}{3} and 2p13\frac{2p-1}{3}, so that the condition r(p1)qr\leq(p-1)\sqrt{q} holds.

One key ingredient of the proof of Theorem 1.5 and Theorem 1.6 is Theorem 1.1. Indeed, Theorem 1.1 can also be viewed as on an upper bound of a bipartite version of Diophantine tuples over finite fields. For the applications to strong Diophantine tuples, Theorem 1.1 is sufficient. On the other hand, to obtain upper bounds on Diophantine tuples (which are not necessarily strong Diophantine tuples), we also need a version of Theorem 1.1 for restricted product sets, which can be found as Theorem 4.2. Indeed, Theorem 1.1 alone only implies a weaker bound of the form 2p/d2\sqrt{p/d} for MSDd(λ,𝔽p)MSD_{d}(\lambda,{\mathbb{F}}_{p}); see Remark 4.3.

1.3. Multiplicative decompositions of shifted multiplicative subgroups

A well-known conjecture of Sárközy [30] asserts that the set of nonzero squares S2={x2:x𝔽p}𝔽pS_{2}=\{x^{2}:x\in{\mathbb{F}}_{p}^{*}\}\subset{\mathbb{F}}_{p} cannot be written as S2=A+BS_{2}=A+B, where A,B𝔽pA,B\subset{\mathbb{F}}_{p} and |A|,|B|2|A|,|B|\geq 2, provided that pp is a sufficiently large prime. This conjecture essentially predicts that the set of quadratic residues in a prime field cannot have a rich additive structure. Similarly, one expects that any non-trivial shift of S2S_{2} cannot have a rich multiplicative structure. Indeed, this can be made precise via another interesting conjecture of Sárközy [31], which we make progress in the current paper.

Conjecture 1.8 (Sárközy).

If pp is a sufficiently large prime and λ𝔽p\lambda\in{\mathbb{F}}_{p}^{*}, then the shifted subgroup (S2λ){0}(S_{2}-\lambda)\setminus\{0\} cannot be written as the product ABAB, where A,B𝔽pA,B\subset{\mathbb{F}}_{p}^{*} and |A|,|B|2|A|,|B|\geq 2. In other words, (S2λ){0}(S_{2}-\lambda)\setminus\{0\} has no non-trivial multiplicative decomposition.

We note that it is necessary to take out the element 0 from the shifted subgroup, for otherwise one can always decompose S2λS_{2}-\lambda as {0,1}(S2λ)\{0,1\}\cdot(S_{2}-\lambda). It appears that this problem concerning multiplicative decompositions is more difficult than the one concerning additive decompositions stated previously, given that it might depend on the parameter λ\lambda. Inspired by 1.8, we formulate the following more general conjecture for any proper multiplicative subgroup. For simplicity, we denote Sd=Sd(𝔽q)={xd:x𝔽q}S_{d}=S_{d}({\mathbb{F}}_{q})=\{x^{d}:x\in{\mathbb{F}}_{q}^{*}\} to be the set of dd-th powers in 𝔽q{\mathbb{F}}_{q}^{*}, equivalently, the subgroup of 𝔽q{\mathbb{F}}_{q}^{*} with order q1d\frac{q-1}{d}.

Conjecture 1.9.

Let d2d\geq 2. If q1(modd)q\equiv 1\pmod{d} is a sufficiently large prime power, then for any λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}, (Sdλ){0}(S_{d}-\lambda)\setminus\{0\} does not admit a non-trivial multiplicative decomposition, that is, there do not exist two sets A,B𝔽qA,B\subset{\mathbb{F}}_{q}^{*} with |A|,|B|2|A|,|B|\geq 2, such that (Sdλ){0}=AB(S_{d}-\lambda)\setminus\{0\}=AB.

1.9 predicts that a shifted multiplicative subgroup of a finite field admits a non-trivial multiplicative decomposition only when it has a small size. We remark that the integer version of 1.9, namely, for each k2k\geq 2, a non-trivial shift of kk-th powers in integers has no non-trivial multiplicative decomposition, has been proved and strengthened in a series of papers by Hajdu and Sárközy [18, 19, 20]. On the other hand, to the best knowledge of the authors, 1.9 appears to be much harder and no partial progress has been made. For the analog of 1.9 on the additive decomposition of multiplicative subgroups, we refer to recent papers [21, 31, 33, 38] for an extensive discussion on partial progress.

Our main contribution to 1.9 is the following two results. The first one is a corollary of Theorem 1.1.

Corollary 1.10.

Let d2d\geq 2 and pp be a prime such that d(p1)d\mid(p-1). Let λSd\lambda\in S_{d}. If (Sdλ){0}(S_{d}-\lambda)\setminus\{0\} can be multiplicative decomposed as the product of two sets A,B𝔽pA,B\subset{\mathbb{F}}_{p}^{*} with |A|,|B|2|A|,|B|\geq 2, then we must have |A||B|=|Sd|1|A||B|=|S_{d}|-1, that is, all products abab are distinct. In other words, A,BA,B are multiplicatively co-Sidon.

In particular, Corollary 1.10 confirms 1.9 under the assumption that qq is a prime, λSd\lambda\in S_{d}, and |Sd|1|S_{d}|-1 is a prime. The second result provides a partial answer to 1.9 asymptotically.

Theorem 1.11.

Let d2d\geq 2 be fixed and nn be a positive integer. As xx\to\infty, the number of primes pxp\leq x such that p1(modd)p\equiv 1\pmod{d} and (Sd(𝔽p)n){0}(S_{d}({\mathbb{F}}_{p})-n)\setminus\{0\} has no non-trivial multiplicative decomposition is at least

(1[(e2πi/d,n1/d):]o(1))π(x).\bigg{(}\frac{1}{[\mathbb{Q}(e^{2\pi i/d},n^{1/d}):\mathbb{Q}]}-o(1)\bigg{)}\pi(x).

In particular, by setting n=1n=1 and d=2d=2, our result has the following significant implication to Sárközy’s conjecture [31]: for almost all odd primes pp, the shifted multiplicative subgroup (S2(𝔽p)1){0}(S_{2}({\mathbb{F}}_{p})-1)\setminus\{0\} has no non-trivial multiplicative decomposition. In other words, if λ=1\lambda=1, then Sárközy’s conjecture holds for almost all primes pp; see Theorem 6.1 for a precise statement.

Some partial progress has been made for 1.9 when the multiplicative decomposition is assumed to have special forms [31, 33]. We also make progress in this direction in Subsection 6.2. In particular, in Theorem 6.6, we confirm the ternary version of 1.9 in a strong sense, which generalizes [31, Theorem 2].

Notations. We follow standard notations from analytic number theory. We use π\pi and θ\theta to denote the standard prime-counting functions. We adopt standard asymptotic notations O,o,O,o,\asymp. We also follow the Vinogradov notation \ll: we write XYX\ll Y if there is an absolute constant C>0C>0 so that |X|CY|X|\leq CY.

Throughout the paper, let pp be a prime and qq a power of pp. Let 𝔽q{\mathbb{F}}_{q} be the finite field with qq elements and let 𝔽q=𝔽q{0}{\mathbb{F}}_{q}^{*}={\mathbb{F}}_{q}\setminus\{0\}. We always assume that d(q1)d\mid(q-1) with d2d\geq 2, and denote Sd(𝔽q)={xd:x𝔽q}S_{d}({\mathbb{F}}_{q})=\{x^{d}:x\in{\mathbb{F}}_{q}^{*}\} to be the subgroup of 𝔽q{\mathbb{F}}_{q}^{*} with order q1d\frac{q-1}{d}. If qq is assumed to be fixed, for brevity, we simply write SdS_{d} instead of Sd(𝔽q)S_{d}({\mathbb{F}}_{q}).

We also need some notations for arithmetic operations among sets. Given two sets AA and BB, we write the product set AB={ab:aA,bB}AB=\{ab:a\in A,b\in B\}, and the sumset A+B={a+b:aA,bB}A+B=\{a+b:a\in A,b\in B\}. Given the definition of Diophantine tuples, it is also useful to define the restricted product set of AA, that is, A×^A={ab:a,bA,ab}A\hat{\times}A=\{ab:a,b\in A,a\neq b\}.

Structure of the paper. In Section 2, we introduce more background. In Section 3, using Gauss sums and Weil’s bound, we give an upper bound on the size of the set which satisfies various multiplicative properties based on character sum estimates. In particular, we prove Proposition 1.4. In Section 4, we first prove Theorem 1.1 using Stepanov’s method. At the end of the section, we deduce applications of Theorem 1.1 to Diophantine tuples and prove Theorem 1.5 and Theorem 1.6. Via Gallagher’s larger sieve inequality and other tools from analytic number theory, in Section 5, we prove Theorem 1.2 and Theorem 1.3. In Section 6, we study multiplicative decompositions and prove Theorem 1.11.

2. Background

2.1. Stepanov’s method

We first describe Stepanov’s method [35]. If we can construct a low degree non-zero auxiliary polynomial that vanishes on each element of a set AA with high multiplicity, then we can give an upper bound on |A||A| based on the degree of the polynomial. It turns out that the most challenging part of our proofs is to show that the auxiliary polynomial constructed is not identically zero.

To check that each root has a high multiplicity, standard derivatives might not work since we are working in a field with characteristic pp. To resolve this issue, we need the following notation of derivatives, known as the Hasse derivatives or hyper-derivatives; see [26, Section 6.4].

Definition 2.1.

Let c0,c1,cd𝔽qc_{0},c_{1},\ldots c_{d}\in{\mathbb{F}}_{q}. If nn is a non-negative integer, then the nn-th hyper-derivative of f(x)=j=0dcjxjf(x)=\sum_{j=0}^{d}c_{j}x^{j} is

E(n)(f)=j=0d(jn)cjxjn,E^{(n)}(f)=\sum_{j=0}^{d}\binom{j}{n}c_{j}x^{j-n},

where we follow the standard convention that (jn)=0\binom{j}{n}=0 for j<nj<n, so that the nn-th hyper-derivative is a polynomial.

Following the definition, we have E(0)f=fE^{(0)}f=f. We also need the next three lemmas.

Lemma 2.2 ([26, Lemma 6.47]).

If f,g𝔽q[x]f,g\in{\mathbb{F}}_{q}[x], then E(n)(fg)=k=0nE(k)(f)E(nk)(g).E^{(n)}(fg)=\sum_{k=0}^{n}E^{(k)}(f)E^{(n-k)}(g).

Lemma 2.3 ([26, Corollary 6.48]).

Let n,dn,d be positive integers. If a𝔽qa\in{\mathbb{F}}^{*}_{q} and c𝔽qc\in{\mathbb{F}}_{q}, then we have

E(n)((ax+c)d)=an(dn)(ax+c)dn.E^{(n)}\big{(}(ax+c)^{d}\big{)}=a^{n}\binom{d}{n}(ax+c)^{d-n}. (2.1)
Lemma 2.4 ([26, Lemma 6.51]).

Let ff be a non-zero polynomial in 𝔽q[x]{\mathbb{F}}_{q}[x]. If cc is a root of E(k)(f)E^{(k)}(f) for k=0,1,,m1k=0,1,\ldots,m-1, then cc is a root of multiplicity at least mm.

2.2. Gallagher’s larger sieve inequality

In this subsection, we introduce Gallagher’s larger sieve inequality and provide necessary estimations from it. Gallagher’s larger sieve inequality will be one of the main ingredients for the proof of Theorem 1.2. In 1971, Gallagher [15] discovered the following sieve inequality.

Theorem 2.5.

(Gallagher’s larger sieve inequality) Let NN be a natural number and A{1,2,,N}A\subset\{1,2,\ldots,N\}. Let 𝒫{\mathcal{P}} be a set of primes. For each prime p𝒫p\in{\mathcal{P}}, let Ap=A(modp)A_{p}=A\pmod{p}. For any 1<QN1<Q\leq N, we have

|A|pQ,p𝒫logplogNpQ,p𝒫logp|Ap|logN,|A|\leq\frac{\underset{p\leq Q,p\in\mathcal{P}}{\sum}\log p-\log N}{\underset{p\leq Q,p\in\mathcal{P}}{\sum}\frac{\log p}{|A_{p}|}-\log N}, (2.2)

provided that the denominator is positive.

As a preparation to apply Gallagher’s larger sieve in our proof, we need to establish a few estimates related to primes in arithmetic progressions. For (a,k)=1(a,k)=1, we follow the standard notation

θ(x;k,a)=pxpa mod klogp.\theta(x;k,a)=\sum_{\begin{subarray}{c}p\leq x\\ p\equiv a\text{ mod }k\end{subarray}}\log p.

For our purposes, logk\log k could be as large as loglogx\sqrt{\log\log x} (for example, see Theorem 1.2), and we need the Siegel-Walfisz theorem to estimate θ(x;k,a)\theta(x;k,a).

Lemma 2.6 ([28, Corollary 11.21]).

Let A>0A>0 be a constant. There is a constant c1>0c_{1}>0 such that

θ(Q;k,a)=Qϕ(k)+OA(Qexp(c1logQ))\theta(Q;k,a)=\frac{Q}{\phi(k)}+O_{A}\bigg{(}Q\exp(-c_{1}\sqrt{\log Q})\bigg{)}

holds uniformly for k(logQ)Ak\leq(\log Q)^{A} and (a,k)=1(a,k)=1.

A standard application of partial summation with Lemma 2.6 leads to the following corollary.

Corollary 2.7.

There is a constant c>0c>0, such that

pQpa mod klogpp=2Qϕ(k)+O(Qexp(clogQ))\sum_{\begin{subarray}{c}p\leq Q\\ p\equiv a\textup{ mod }k\end{subarray}}\frac{\log p}{\sqrt{p}}=\frac{2\sqrt{Q}}{\phi(k)}+O\left(\sqrt{Q}\exp(-c\sqrt{\log Q})\right)

holds uniformly for klogQk\leq\log Q and (a,k)=1(a,k)=1.

We also need the following lemma.

Lemma 2.8 ([1, page 72]).

Let nn be a positive integer. Then

pnlogpp(logn)1/2.\sum_{p\mid n}\frac{\log p}{\sqrt{p}}\ll(\log n)^{1/2}.

2.3. An effective estimate for Mk(n,kk2)M_{k}(n,\frac{k}{k-2})

Following [7], for each real number L>0L>0, we write

Mk(n;L):=sup{|S[nL,)|:S satisfies property Dk(n)}.M_{k}(n;L):=\sup\{|S\cap[n^{L},\infty)|:S\text{ satisfies property }D_{k}(n)\}.

It is shown in [7] that Mk(n,3)k1M_{k}(n,3)\ll_{k}1 as nn\to\infty. For our application, we show a stronger result that Mk(n,kk2)k1M_{k}(n,\frac{k}{k-2})\ll_{k}1 and we will make this estimate explicit and effective. We follow the proof in [7] closely and prove the following proposition, which will be used later in the proof of Theorem 1.2.

Proposition 2.9.

If k3k\geq 3 and n>(2ke)k2n>(2ke)^{k^{2}}, then Mk(n,kk2)logkloglogkM_{k}(n,\frac{k}{k-2})\ll\log k\log\log k, where the implicit constant is absolute.

Let k3k\geq 3. Let m=Mk(n,kk2)m=M_{k}(n,\frac{k}{k-2}) and A={a1,a2,,am}A=\{a_{1},a_{2},\ldots,a_{m}\} be a generalized mm-tuple with property Dk(n)D_{k}(n) and nkk2<a1<a2<<amn^{\frac{k}{k-2}}<a_{1}<a_{2}<\cdots<a_{m}. Consider the system of equations

{a1x+n=uka2x+n=vk.\left\{\begin{array}[]{ll}a_{1}x+n=u^{k}&\\ a_{2}x+n=v^{k}.&\end{array}\right. (2.3)

Clearly, for each i3i\geq 3, x=aix=a_{i} is a solution to this system, and we denote ui,viu_{i},v_{i} so that a1ai+n=uika_{1}a_{i}+n=u_{i}^{k} and a2ai+n=vika_{2}a_{i}+n=v_{i}^{k}. Let α:=(a1/a2)1/k\alpha:=(a_{1}/a_{2})^{1/k}.

The following lemma is a generalization of [7, Lemma 3.1], showing that ui/viu_{i}/v_{i} provides a “good” rational approximation to α\alpha if nn is large. Note that in [7, Lemma 3.1], it was further assumed that kk is odd and L=3L=3. Nevertheless, an almost identical proof works, and we skip the proof.

Lemma 2.10.

Let

c(k):=j=1(k1)/2(sin2πjk)2.c(k):=\prod_{j=1}^{\lfloor(k-1)/2\rfloor}\left(\sin\frac{2\pi j}{k}\right)^{2}.

Assume that n>(2/c(k))(k2)/2n>(2/c(k))^{(k-2)/2}. Then for each 3im3\leq i\leq m, we have

|uiviα|a22vik.\bigg{|}\frac{u_{i}}{v_{i}}-\alpha\bigg{|}\leq\frac{a_{2}}{2v_{i}^{k}}. (2.4)
Corollary 2.11.

Assume that n>(2/c(k))(k2)/2n>(2/c(k))^{(k-2)/2}. Then via24v_{i}\geq a_{2}^{4} for each 14im14\leq i\leq m and

|uiviα|<1vik1/2.\bigg{|}\frac{u_{i}}{v_{i}}-\alpha\bigg{|}<\frac{1}{v_{i}^{k-1/2}}. (2.5)
Proof.

Let 2im32\leq i\leq m-3. Applying the gap principle from [7, Lemma 2.4] to ai,ai+1,ai+2,ai+3a_{i},a_{i+1},a_{i+2},a_{i+3}, we have

ai+1ai+3kknk(aiai+2)k1kknk(aiai+1)k1.a_{i+1}a_{i+3}\geq k^{k}n^{-k}(a_{i}a_{i+2})^{k-1}\geq k^{k}n^{-k}(a_{i}a_{i+1})^{k-1}.

It follows that

ai+3aik1ai+1k2nkaik1.a_{i+3}\geq a_{i}^{k-1}a_{i+1}^{k-2}n^{-k}\geq a_{i}^{k-1}.

In particular a14a2(k1)4a24ka_{14}\geq a_{2}^{(k-1)^{4}}\geq a_{2}^{4k}. Thus, if i14i\geq 14, then viai1/ka141/ka24v_{i}\geq a_{i}^{1/k}\geq a_{14}^{1/k}\geq a_{2}^{4} and inequality (2.5) follows from Lemma 2.10. ∎

Now we are ready to prove Proposition 2.9. Recall we have the inequality sinx2xπ\sin x\geq\frac{2x}{\pi} for x[0,π2]x\in[0,\frac{\pi}{2}], and the inequality s!(s/e)ss!\geq(s/e)^{s} for all positive integers ss. It follows that

c(k)=j=1(k1)/2sin2πjk=j=1(k1)/2sinπjkj=1(k1)/22jk=((k1)/2)!(k/2)(k1)/2(k1ke)(k1)/2\sqrt{c(k)}=\prod_{j=1}^{(k-1)/2}\sin\frac{2\pi j}{k}=\prod_{j=1}^{(k-1)/2}\sin\frac{\pi j}{k}\geq\prod_{j=1}^{(k-1)/2}\frac{2j}{k}=\frac{((k-1)/2)!}{(k/2)^{(k-1)/2}}\geq\bigg{(}\frac{k-1}{ke}\bigg{)}^{(k-1)/2}

when kk is odd, and

(c(k))1/4=j=1(k2)/2sin2πjk=j=1k/4sinπjk/2j=1k/44jk=(k/4)!(k/4)k/4(4k/4ke)k/4(c(k))^{1/4}=\sqrt{\prod_{j=1}^{(k-2)/2}\sin\frac{2\pi j}{k}}=\prod_{j=1}^{\lfloor k/4\rfloor}\sin\frac{\pi j}{k/2}\geq\prod_{j=1}^{\lfloor k/4\rfloor}\frac{4j}{k}=\frac{(\lfloor k/4\rfloor)!}{(k/4)^{\lfloor k/4\rfloor}}\geq\bigg{(}\frac{4\lfloor k/4\rfloor}{ke}\bigg{)}^{\lfloor k/4\rfloor}

when kk is even. Thus, when k3k\geq 3, we always have

2c(k)2(kek2)k2(ke)k.\frac{2}{c(k)}\leq 2\bigg{(}\frac{ke}{k-2}\bigg{)}^{k}\leq 2(ke)^{k}.

Therefore, when n>(2ke)k2n>(2ke)^{k^{2}}, we can apply Lemma 2.10. Note that the absolute height of α\alpha is H(α)a21/kH(\alpha)\leq a_{2}^{1/k}. Since k3k\geq 3, for 14im14\leq i\leq m, Lemma 2.10 implies that

|uiviα|1vik1/21vi2.5;\left|\frac{u_{i}}{v_{i}}-\alpha\right|\leq\frac{1}{v_{i}^{k-1/2}}\leq\frac{1}{v_{i}^{2.5}};

moreover, max(ui,vi)=vi>a21/kmax(H(α),2)\max(u_{i},v_{i})=v_{i}>a_{2}^{1/k}\geq\max(H(\alpha),2). Therefore, we can apply the quantitative Roth’s theorem due to Evertse [12] (see also [7, Theorem 2.2]) to conclude that

m13+228log(2k)log(2log2k)logkloglogk,m\leq 13+2^{28}\log(2k)\log(2\log 2k)\ll\log k\log\log k,

where the implicit constant is absolute.

2.4. Implications of the Paley graph conjecture

The Paley graph conjecture on double character sums implies many results of the present paper related to the estimation of character sums. We record the statement of the conjecture (see for example [7, 16]).

Conjecture 2.12 (Paley graph conjecture).

Let ϵ>0\epsilon>0 be a real number. Then there is p0=p0(ϵ)p_{0}=p_{0}(\epsilon) and δ=δ(ϵ)>0\delta=\delta(\epsilon)>0 such that for any prime p>p0p>p_{0}, any A,B𝔽pA,B\subseteq\mathbb{F}_{p} with |A|,|B|>pϵ|A|,|B|>p^{\epsilon}, and any non-trivial multiplicative character χ\chi of 𝔽p{\mathbb{F}}_{p}, the following inequality holds:

|aA,bBχ(a+b)|pδ|A||B|.\bigg{|}\sum_{a\in A,\,b\in B}\chi(a+b)\bigg{|}\leq p^{-\delta}|A||B|.

The connection between the Paley graph conjecture and the problem of bounding the size of Diophantine tuples was first observed by Güloğlu and Murty in [16]. Let d2d\geq 2 be fixed, λ𝔽p\lambda\in{\mathbb{F}}_{p}^{*}, where p1(modd)p\equiv 1\pmod{d} is a prime. The Paley graph conjecture trivially implies MDd(λ,𝔽p)=po(1)MD_{d}(\lambda,{\mathbb{F}}_{p})=p^{o(1)} and MSDd(λ,𝔽p)=po(1)MSD_{d}(\lambda,{\mathbb{F}}_{p})=p^{o(1)} as pp\to\infty. Also, the bound on Mk(n)M_{k}(n) in Theorem 1.2 can be improved to (logn)o(1)(\log n)^{o(1)} (see [7], [16]) when kk is fixed and nn\to\infty. Furthermore, the Paley graph conjecture also immediately implies Sárközy’s conjecture (1.8) in view of Proposition 3.4. However, the Paley graph conjecture itself remains widely open, and our results are unconditional. We refer to [32] and the references therein for recent progress on the Paley graph conjecture assuming A,BA,B have small doubling.

3. Preliminary estimations for product sets in shifted multiplicative subgroups

3.1. Character sum estimate and the square root upper bound

The purpose of this subsection is to prove Proposition 1.4 by establishing an upper bound on the double character sum in Proposition 3.1 using basic properties of characters and Gauss sums. For any prime pp, and any x𝔽px\in{\mathbb{F}}_{p}, we follow the standard notation that ep(x)=exp(2πix/p)e_{p}(x)=\exp(2\pi ix/p), where we embed 𝔽p{\mathbb{F}}_{p} into {\mathbb{Z}}. We refer the reader to [26, Chapter 5] for more results related to Gauss sums and character sums.

We refer to [1, Section 2] for a historical discussion of Vinogradov’s inequality 1.1. Gyarmati [17, Theorem 7] and Becker and Murty [1, Proposition 2.7] independently showed that the Legendre symbol in inequality 1.1 can be replaced with any non-trivial Dirichlet character modulo pp. For our purposes, we extend Vinogradov’s inequality 1.1 to all finite fields 𝔽q{\mathbb{F}}_{q} and all nontrivial multiplicative characters χ\chi of 𝔽q{\mathbb{F}}_{q}, with a slightly improved upper bound.

Proposition 3.1.

Let χ\chi be a non-trivial multiplicative character of 𝔽q{\mathbb{F}}_{q} and λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}. For any A,B𝔽qA,B\subset{\mathbb{F}}_{q}^{*}, we have

|aA,bBχ(ab+λ)|q|A||B|(1max{|A|,|B|}q)1/2.\bigg{|}\sum_{a\in A,\,b\in B}\chi(ab+\lambda)\bigg{|}\leq\sqrt{q|A||B|}\bigg{(}1-\frac{\max\{|A|,|B|\}}{q}\bigg{)}^{1/2}.

Before proving Proposition 3.1, we need some preliminary estimates. Let χ\chi be a multiplicative character of 𝔽q{\mathbb{F}}_{q}; then the Gauss sum associated to χ\chi is defined to be

G(χ)=c𝔽qχ(c)ep(Tr𝔽q(c)),G(\chi)=\sum_{c\in{\mathbb{F}}_{q}}\chi(c)e_{p}\big{(}\operatorname{Tr}_{{\mathbb{F}}_{q}}(c)\big{)},

where Tr𝔽q:𝔽q𝔽p\operatorname{Tr}_{{\mathbb{F}}_{q}}:{\mathbb{F}}_{q}\to{\mathbb{F}}_{p} is the absolute trace map.

Lemma 3.2 ([26, Theorem 5.12]).

Let χ\chi be a multiplicative character of 𝔽q{\mathbb{F}}_{q}. Then for any a𝔽qa\in{\mathbb{F}}_{q},

χ(a)¯=1G(χ)c𝔽qχ(c)ep(Tr𝔽q(ac)).\overline{\chi(a)}=\frac{1}{G(\chi)}\sum_{c\in{\mathbb{F}}_{q}}\chi(c)e_{p}\big{(}\operatorname{Tr}_{{\mathbb{F}}_{q}}(ac)\big{)}.

Now we are ready to prove Proposition 3.1.

Proof of Proposition 3.1.

By Lemma 3.2, we can write

aA,bBχ(ab+λ)¯\displaystyle\sum_{a\in A,\,b\in B}\overline{\chi(ab+\lambda)} =aA,bBχ(b)¯χ(a+λb1)¯\displaystyle=\sum_{a\in A,\,b\in B}\overline{\chi(b)}\overline{\chi(a+\lambda b^{-1})}
=1G(χ)c𝔽qχ(c)aA,bBχ(b)¯ep(Tr((a+λb1)c)).\displaystyle=\frac{1}{G(\chi)}\sum_{c\in{\mathbb{F}}_{q}}\chi(c)\sum_{a\in A,\,b\in B}\overline{\chi(b)}e_{p}\big{(}\operatorname{Tr}((a+\lambda b^{-1})c)\big{)}.

It is well-known that |G(χ)|=q|G(\chi)|=\sqrt{q} (see for example [26, Theorem 5.11]). Since |χ(c)|=1|\chi(c)|=1 for each c𝔽qc\in{\mathbb{F}}_{q}^{*}, we can apply the triangle inequality and Cauchy-Schwarz inequality to obtain

|aA,bBχ(ab+λ)|\displaystyle\bigg{|}\sum_{a\in A,\,b\in B}\chi(ab+\lambda)\bigg{|} 1qc𝔽q|aA,bBχ(b)¯ep(Tr((a+λb1)c))|\displaystyle\leq\frac{1}{\sqrt{q}}\sum_{c\in{\mathbb{F}}_{q}^{*}}\bigg{|}\sum_{a\in A,\,b\in B}\overline{\chi(b)}e_{p}\big{(}\operatorname{Tr}((a+\lambda b^{-1})c)\big{)}\bigg{|}
1q(c𝔽q|aAep(Tr(ac))|2)1/2(c𝔽q|bBχ(b)¯ep(Tr(b1c))|2)1/2.\displaystyle\leq\frac{1}{\sqrt{q}}\bigg{(}\sum_{c\in{\mathbb{F}}_{q}^{*}}\bigg{|}\sum_{a\in A}e_{p}\big{(}\operatorname{Tr}(ac)\big{)}\bigg{|}^{2}\bigg{)}^{1/2}\bigg{(}\sum_{c\in{\mathbb{F}}_{q}^{*}}\bigg{|}\sum_{b\in B}\overline{\chi(b)}e_{p}\big{(}\operatorname{Tr}(b^{-1}c)\big{)}\bigg{|}^{2}\bigg{)}^{1/2}.

By orthogonality relations, we have

c𝔽q|aAep(Tr(ac))|2=q|A||A|2,c𝔽q|bBχ(b)¯ep(Tr(b1c))|2q|B|.\sum_{c\in{\mathbb{F}}_{q}^{*}}\bigg{|}\sum_{a\in A}e_{p}\big{(}\operatorname{Tr}(ac)\big{)}\bigg{|}^{2}=q|A|-|A|^{2},\quad\sum_{c\in{\mathbb{F}}_{q}^{*}}\bigg{|}\sum_{b\in B}\overline{\chi(b)}e_{p}\big{(}\operatorname{Tr}(b^{-1}c)\big{)}\bigg{|}^{2}\leq q|B|.

Thus, we have

|aA,bBχ(ab+λ)|q|A||B|(1|A|q)1/2.\bigg{|}\sum_{a\in A,\,b\in B}\chi(ab+\lambda)\bigg{|}\leq\sqrt{q|A||B|}\bigg{(}1-\frac{|A|}{q}\bigg{)}^{1/2}.

By switching the roles of AA and BB, we obtain the required character sum estimate. ∎

Let d(q1)d\mid(q-1) such that d2d\geq 2 and denote Sd={xd:x𝔽q}S_{d}=\{x^{d}\colon x\in\mathbb{F}_{q}^{*}\} with order q1d\frac{q-1}{d}. Then we prove Proposition 1.4.

Proof of Proposition 1.4.

Let χ\chi be a multiplicative character of order dd.

Let A𝔽qA\subset{\mathbb{F}}_{q}^{*} with property SDd(λ,𝔽q)SD_{d}(\lambda,{\mathbb{F}}_{q}), that is, AA+λSd{0}AA+\lambda\subset S_{d}\cup\{0\}. Note that χ(ab+λ)=1\chi(ab+\lambda)=1 for each a,bAa,b\in A, unless ab+λ=0ab+\lambda=0. Note that given aAa\in A, there is at most one bAb\in A such that ab+λ=0ab+\lambda=0. Therefore, by Proposition 3.1, we have

|A|2|A||a,bAχ(ab+λ)|q|A|(1|A|q)1/2.|A|^{2}-|A|\leq\bigg{|}\sum_{a,b\in A}\chi(ab+\lambda)\bigg{|}\leq\sqrt{q}|A|\bigg{(}1-\frac{|A|}{q}\bigg{)}^{1/2}.

It follows that

(|A|1)2q|A||A|4q3+12.(|A|-1)^{2}\leq q-|A|\implies|A|\leq\frac{\sqrt{4q-3}+1}{2}.

Next we work under the weaker assumption A×^A+λSd{0}A\hat{\times}A+\lambda\subset S_{d}\cup\{0\}. In this case, note that χ(ab+λ)=1\chi(ab+\lambda)=1 for each a,bAa,b\in A such that aba\neq b, unless ab+λ=0ab+\lambda=0. Proposition 3.1 then implies that

|A|23|A||a,bAχ(ab+λ)|q|A|(1|A|q)1/2|A|^{2}-3|A|\leq\bigg{|}\sum_{a,b\in A}\chi(ab+\lambda)\bigg{|}\leq\sqrt{q}|A|\bigg{(}1-\frac{|A|}{q}\bigg{)}^{1/2}

and it follows that |A|q114+52|A|\leq\sqrt{q-\frac{11}{4}}+\frac{5}{2}. ∎

3.2. Estimates on |A||A| and |B||B| if AB=(Sdλ){0}AB=(S_{d}-\lambda)\setminus\{0\}

Let A,B𝔽qA,B\subset{\mathbb{F}}_{q}^{*} and λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}. In this subsection, we provide several useful estimates on |A||A| and |B||B| when AB=(Sdλ){0}AB=(S_{d}-\lambda)\setminus\{0\}, which will be used in Section 6.

We need to use the following lemma, due to Karatsuba [23].

Lemma 3.3.

Let A,B𝔽qA,B\subset{\mathbb{F}}_{q}^{*} and λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}. Then for any non-trivial multiplicative character χ\chi of 𝔽q{\mathbb{F}}_{q} and any positive integer ν\nu, we have

aAbBχ(ab+λ)ν|A|(2ν1)/2ν(|B|1/2q1/2ν+|B|q1/4ν).\sum_{\begin{subarray}{c}a\in A\\ b\in B\end{subarray}}\chi(ab+\lambda)\ll_{\nu}|A|^{(2\nu-1)/2\nu}(|B|^{1/2}q^{1/2\nu}+|B|q^{1/4\nu}).

The following proposition improves and generalizes [31, Theorem 1]. It also improves [33, Lemma 17] (see Remark 3.7).

Proposition 3.4.

Let ϵ>0\epsilon>0. Let d(q1)d\mid(q-1) such that 2dq1/2ϵ2\leq d\leq q^{1/2-\epsilon} and λ𝔽q\lambda\in{\mathbb{F}}^{*}_{q}. If AB=(Sdλ){0}AB=(S_{d}-\lambda)\setminus\{0\} for some A,B𝔽qA,B\subset{\mathbb{F}}_{q}^{*} with |A|,|B|2|A|,|B|\geq 2, then

qdmin{|A|,|B|}max{|A|,|B|}q1/2.\frac{\sqrt{q}}{d}\ll\min\{|A|,|B|\}\leq\max\{|A|,|B|\}\ll q^{1/2}.
Proof.

Let A,B𝔽qA,B\subset{\mathbb{F}}_{q}^{*} and λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*} such that AB=(Sdλ){0}AB=(S_{d}-\lambda)\setminus\{0\} with |A|,|B|2|A|,|B|\geq 2. Without loss of generality, we assume that |A||B||A|\geq|B|. We first establish a weaker lower bound that |B|qϵ/2|B|\gg q^{\epsilon/2}.

When d=2d=2, Sárközy [31] has shown that |B|q3logq|B|\gg\frac{\sqrt{q}}{3\log q}. While he only proved this estimate when q=pq=p is a prime [31, Theorem 1], it is clear that the same proof extends to all finite fields 𝔽q{\mathbb{F}}_{q}.

Next, assume that d3d\geq 3. Let B={b1,b2,,bk}B=\{b_{1},b_{2},\ldots,b_{k}\} and |B|=k|B|=k. Since ABSdλAB\subset S_{d}-\lambda, we have AB+λSdAB+\lambda\subset S_{d}. Let χ\chi be a multiplicative character of 𝔽q{\mathbb{F}}_{q} with order dd. Then it follows that for each aAa\in A, we have χ(a+λbi1)=1/χ(bi)\chi(a+\lambda b_{i}^{-1})=1/\chi(b_{i}) for each 1ik1\leq i\leq k. Therefore, by a well-known consequence of Weil’s bound (see for example [26, Exercise 5.66]),

|A|qdk+(k1kd+1dk)q+kd<qdk+kq.|A|\leq\frac{q}{d^{k}}+\bigg{(}k-1-\frac{k}{d}+\frac{1}{d^{k}}\bigg{)}\sqrt{q}+\frac{k}{d}<\frac{q}{d^{k}}+k\sqrt{q}.

On the other hand, since AB=(Sdλ){0}AB=(S_{d}-\lambda)\setminus\{0\}, we have

|A||B||AB||Sd|1=q1d1.|A||B|\geq|AB|\geq|S_{d}|-1=\frac{q-1}{d}-1.

Combining the above two inequalities, we obtain that

2qd2+k2qkqdk+k2q>|A||B|q1d1.\frac{2q}{d^{2}}+k^{2}\sqrt{q}\geq\frac{kq}{d^{k}}+k^{2}\sqrt{q}>|A||B|\geq\frac{q-1}{d}-1.

Since d3d\geq 3, it follows that k2qqdk^{2}\sqrt{q}\gg\frac{q}{d} and thus

|B|=kq1/4dqϵ/2.|B|=k\gg\frac{q^{1/4}}{\sqrt{d}}\gg q^{\epsilon/2}.

Let ν=2/ϵ\nu=\lceil 2/\epsilon\rceil. By Lemma 3.3, and as |B|q1/ν|B|\gg q^{1/\nu}, we have

|A||B|=aAbBχ(ab+λ)|A|(2ν1)/2ν(|B|1/2q1/2ν+|B|q1/4ν)|A|(2ν1)/2ν|B|q1/4ν.|A||B|=\sum_{\begin{subarray}{c}a\in A\\ b\in B\end{subarray}}\chi(ab+\lambda)\ll|A|^{(2\nu-1)/2\nu}(|B|^{1/2}q^{1/2\nu}+|B|q^{1/4\nu})\ll|A|^{(2\nu-1)/2\nu}|B|q^{1/4\nu}.

It follows that |A|q1/2|A|\ll q^{1/2}. Thus, |B||Sd|/|A|q1/2/d|B|\gg|S_{d}|/|A|\gg q^{1/2}/d. ∎

Remark 3.5.

We remark that the same method could be used to refine a similar result for the additive decomposition of multiplicative subgroups, which improves a result of Shparlinski [34, Theorem 6.1] (moreover, our proof appears to be much simpler than his proof). More precisely, we can prove the following:

Let ϵ>0\epsilon>0. Let d(q1)d\mid(q-1) such that 2dq1/2ϵ2\leq d\leq q^{1/2-\epsilon}. If A+B=SdA+B=S_{d} for some A,B𝔽qA,B\subset{\mathbb{F}}_{q} with |A|,|B|2|A|,|B|\geq 2, then

qdmin{|A|,|B|}max{|A|,|B|}q1/2.\frac{\sqrt{q}}{d}\ll\min\{|A|,|B|\}\leq\max\{|A|,|B|\}\ll q^{1/2}.

Note that Proposition 3.4 only applies to multiplicative subgroups G=SdG=S_{d} with |G|>q|G|>\sqrt{q}. When q=pq=p is a prime, and GG is a non-trivial multiplicative subgroup of 𝔽p{\mathbb{F}}_{p} (in particular, |G|<p|G|<\sqrt{p} is allowed), we have the following estimate, due to Shkredov [33].

Lemma 3.6.

If GG is a proper multiplicative subgroup of 𝔽p{\mathbb{F}}_{p} such that AB=(Gλ){0}AB=(G-\lambda)\setminus\{0\} for some A,B𝔽pA,B\subset{\mathbb{F}}_{p}^{*} with |A|,|B|2|A|,|B|\geq 2 and some λ𝔽p\lambda\in{\mathbb{F}}_{p}^{*}, then

|G|1/2+o(1)=min{|A|,|B|}max{|A|,|B|}=|G|1/2+o(1)|G|^{1/2+o(1)}=\min\{|A|,|B|\}\leq\max\{|A|,|B|\}=|G|^{1/2+o(1)}

as |G||G|\to\infty.

Proof.

If 0Gλ0\in G-\lambda, let A=A{0}A^{\prime}=A\cup\{0\}; otherwise, let A=AA^{\prime}=A. Then we have AB=GλA^{\prime}B=G-\lambda, or equivalently, A/(B1)=GλA^{\prime}/(B^{-1})=G-\lambda. Note that |A{0}|=|A|2|A^{\prime}\setminus\{0\}|=|A|\geq 2 and |B1|=|B|2|B^{-1}|=|B|\geq 2, thus, the lemma then follows immediately from [33, Lemma 17]. ∎

Remark 3.7.

Let AB=(Sdλ){0}AB=(S_{d}-\lambda)\setminus\{0\}, where A,B𝔽pA,B\subset{\mathbb{F}}_{p}^{*} with |A|,|B|2|A|,|B|\geq 2 and λ𝔽p\lambda\in{\mathbb{F}}_{p}^{*}. When dd is a constant, and pp\to\infty, Proposition 3.4 is better than Lemma 3.6. Indeed, in [33, Lemma 17], Shkredov showed that max{|A|,|B|}plogp\max\{|A|,|B|\}\ll\sqrt{p}\log p, while Proposition 3.4 showed the stronger result that max{|A|,|B|}p\max\{|A|,|B|\}\ll\sqrt{p}, removing the logp\log p factor. This stronger bound would be crucial in proving results related to multiplicative decompositions (for example, Theorem 6.1). Instead, Lemma 3.6 will be useful for applications in ternary decompositions (Theorem 6.6).

Remark 3.8.

Lemma 3.6 fails to extend to 𝔽q{\mathbb{F}}_{q}, where qq is a proper prime power. Let q=r2q=r^{2} be a square, G=𝔽rG={\mathbb{F}}_{r}^{*}, and λ=1\lambda=-1. Let AA be a subset of 𝔽r{\mathbb{F}}_{r}^{*} with size (r1)/2\lfloor(r-1)/2\rfloor. Let B=𝔽rA1B={\mathbb{F}}_{r}^{*}\setminus A^{-1}. Then we have AB=𝔽r{1}=(G+1){0}AB={\mathbb{F}}_{r}^{*}\setminus\{1\}=(G+1)\setminus\{0\} while |A|,|B||G||A|,|B|\gg|G|.

Remark 3.9.

Let d2d\geq 2 be fixed. Let q1(modd)q\equiv 1\pmod{d} be a prime power and let λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}, we define N(q,λ)N(q,\lambda) be the total number of pairs (A,B)(A,B) of sets A,B𝔽qA,B\subseteq\mathbb{F}_{q} with |A|,|B|2|A|,|B|\geq 2 such that AB=(Sdλ){0}AB=(S_{d}-\lambda)\setminus\{0\}. Note that 1.9 implies N(q,λ)=0N(q,\lambda)=0 when qq is sufficiently large, but it seems out of reach in general. Instead, one can find a non-trivial upper bound of N(q,λ)N(q,\lambda). Using Proposition 3.4 and following the same strategy of the proof in [3, Theorem 1], we have a non-trivial upper bound of N(q,λ)N(q,\lambda) as follows:

N(q,λ)exp(O(q1/2)).N(q,\lambda)\leq\exp\left(O(q^{1/2})\right).

We also note that by using Corollary 1.10, we confirmed N(q,λ)=0N(q,\lambda)=0 if qq is a prime, λSd\lambda\in S_{d}, and |Sd|1|S_{d}|-1 is a prime. In addition, Theorem 1.11 gives the same result for N(p,n)N(p,n) asymptotically for a subset of primes pp with lower density at least 1[(e2πi/d,n1/d):]\frac{1}{[\mathbb{Q}(e^{2\pi i/d},n^{1/d}):\mathbb{Q}]}.

4. Products and restricted products in shifted multiplicative subgroups

In this section, we use Stepanov’s method to study product sets and restricted product sets that are contained in shifted multiplicative subgroups. Our proofs are inspired by Stepanov’s original paper [35], and the recent breakthrough of Hanson and Pertidis [21].

Throughout the section, we assume d2d\geq 2 and q1(modd)q\equiv 1\pmod{d} is a prime power. Recall that Sd=Sd(𝔽q)={xd:x𝔽q}S_{d}=S_{d}({\mathbb{F}}_{q})=\{x^{d}:x\in{\mathbb{F}}_{q}^{*}\}.

4.1. Product set in a shifted multiplicative subgroup

In this subsection, we prove Theorem 1.1, which can be viewed as the bipartite version of Diophantine tuples over finite fields. As a corollary of Theorem 1.1, we prove Corollary 1.10. Besides it, Theorem 1.1 will be also repeatedly used to prove several of our main results in the present paper.

Proof of Theorem 1.1.

Let r=|B(λA1)|r=|B\cap(-\lambda A^{-1})|. Let A={a1,a2,,an}A=\{a_{1},a_{2},\ldots,a_{n}\} and B={b1,b2,,bm}B=\{b_{1},b_{2},\ldots,b_{m}\} such that br+1,,bm(λA1)b_{r+1},\ldots,b_{m}\notin(-\lambda A^{-1}). Since AB+λSd{0}AB+\lambda\subset S_{d}\cup\{0\}, we have

(aibj+λ)q1d+1=aibj+λ(a_{i}b_{j}+\lambda)^{\frac{q-1}{d}+1}=a_{i}b_{j}+\lambda

for each 1in1\leq i\leq n and 1jm1\leq j\leq m. This simple observation will be used repeatedly in the following computation.

Let c1,c2,,cn𝔽qc_{1},c_{2},...,c_{n}\in{\mathbb{F}}_{q} be the unique solution of the following system of equations:

{\TABbinary\tabbedCenterstack[l]i=1nci=1i=1nciaij=0,1jn1.\left\{\TABbinary\tabbedCenterstack[l]{\sum_{i=1}^{n}c_{i}=1\\ \\ \sum_{i=1}^{n}c_{i}a_{i}^{j}=0,\quad 1\leq j\leq n-1}.\right. (4.1)

This is justified by the invertibility of the coefficient matrix of the system (a Vandermonde matrix). We claim that i=1nciain0\sum_{i=1}^{n}c_{i}a_{i}^{n}\neq 0. Suppose otherwise that i=1nciain=0\sum_{i=1}^{n}c_{i}a_{i}^{n}=0, then ci=0c_{i}=0 for all ii, violating the assumption i=1nci=1\sum_{i=1}^{n}c_{i}=1 in equation (4.1). Indeed, the generalized Vandermonde matrix (aij)1in,1jn(a_{i}^{j})_{1\leq i\leq n,1\leq j\leq n} is non-singular since it has determinant

a1a2ani<j(ajai)0.a_{1}a_{2}\ldots a_{n}\prod_{i<j}(a_{j}-a_{i})\neq 0.

Consider the following auxiliary polynomial

f(x)=λn1+i=1nci(aix+λ)n1+q1d𝔽q[x].f(x)=-\lambda^{n-1}+\sum_{i=1}^{n}c_{i}(a_{i}x+\lambda)^{n-1+\frac{q-1}{d}}\in{\mathbb{F}}_{q}[x]. (4.2)

Note that n=|A||Sd{0}|q1d+1n=|A|\leq|S_{d}\cup\{0\}|\leq\frac{q-1}{d}+1. Thus, if q=pq=p is a prime, then n1+p1d2(p1)dp1n-1+\frac{p-1}{d}\leq\frac{2(p-1)}{d}\leq p-1 and thus the condition (n1+q1dn)0(modp)\binom{n-1+\frac{q-1}{d}}{n}\not\equiv 0\pmod{p} is automatically satisfied. Then ff is a non-zero polynomial since the coefficient of xnx^{n} in ff is

(n1+q1dn)λq1d1i=1nciain0\binom{n-1+\frac{q-1}{d}}{n}\cdot\lambda^{\frac{q-1}{d}-1}\cdot\sum_{i=1}^{n}c_{i}a_{i}^{n}\neq 0

by the assumption on the binomial coefficient. Also, it is clear that the degree of ff is at most n1+q1dn-1+\frac{q-1}{d}.

Next, we compute the derivatives of ff on BB. For each 1jm1\leq j\leq m, system 4.1 implies that

E(0)f(bj)=λn1+i=1nci(aibj+λ)n1=λn1+=0n1(n1)λn1(i=1nciai)bj=0.\displaystyle E^{(0)}f(b_{j})=-\lambda^{n-1}+\sum_{i=1}^{n}c_{i}(a_{i}b_{j}+\lambda)^{n-1}=-\lambda^{n-1}+\sum_{\ell=0}^{n-1}\binom{n-1}{\ell}\lambda^{n-1-\ell}\bigg{(}\sum_{i=1}^{n}c_{i}a_{i}^{\ell}\bigg{)}b_{j}^{\ell}=0.

For each 1jm1\leq j\leq m and 1kn21\leq k\leq n-2, we have that

E(k)f(bj)\displaystyle E^{(k)}f(b_{j}) =(n1+q1dk)i=1nciaik(aibj+λ)n1+q1dk\displaystyle=\binom{n-1+\frac{q-1}{d}}{k}\sum_{i=1}^{n}c_{i}a_{i}^{k}(a_{i}b_{j}+\lambda)^{n-1+\frac{q-1}{d}-k}
=(n1+q1dk)i=1nciaik(aibj+λ)n1k\displaystyle=\binom{n-1+\frac{q-1}{d}}{k}\sum_{i=1}^{n}c_{i}a_{i}^{k}(a_{i}b_{j}+\lambda)^{n-1-k}
=(n1+q1dk)=0n1k(n1k)λn1k(i=1nciaik+)bj=0,\displaystyle=\binom{n-1+\frac{q-1}{d}}{k}\sum_{\ell=0}^{n-1-k}\binom{n-1-k}{\ell}\lambda^{n-1-k-\ell}\bigg{(}\sum_{i=1}^{n}c_{i}a_{i}^{k+\ell}\bigg{)}b_{j}^{\ell}=0,

where we use Lemma 2.3 and the assumptions in system 4.1.

For each r+1jmr+1\leq j\leq m, by the assumption, bj(λA1)b_{j}\notin(-\lambda A^{-1}), that is, aibj+λ0a_{i}b_{j}+\lambda\neq 0 for each 1in1\leq i\leq n. Thus, for each r+1jmr+1\leq j\leq m, we additionally have

E(n1)f(bj)\displaystyle E^{(n-1)}f(b_{j}) =(n1+q1dn1)i=1nciain1(aibj+λ)q1d=(n1+q1dn1)i=1nciain1=0.\displaystyle=\binom{n-1+\frac{q-1}{d}}{n-1}\sum_{i=1}^{n}c_{i}a_{i}^{n-1}(a_{i}b_{j}+\lambda)^{\frac{q-1}{d}}=\binom{n-1+\frac{q-1}{d}}{n-1}\sum_{i=1}^{n}c_{i}a_{i}^{n-1}=0.

Therefore, Lemma 2.4 allows us to conclude that each of b1,b2,brb_{1},b_{2},\ldots b_{r} is a root of ff with multiplicity at least n1n-1, and each of br+1,br+2,bmb_{r+1},b_{r+2},\ldots b_{m} is a root of ff with multiplicity at least nn. It follows that

r(n1)+(mr)n=mnrdegfq1d+n1.r(n-1)+(m-r)n=mn-r\leq\operatorname{deg}f\leq\frac{q-1}{d}+n-1.

Finally, assuming that λSd\lambda\in S_{d}. In this case,

f(0)=λn1+λn1+q1di=1nci=λn1+λn1=0.f(0)=-\lambda^{n-1}+\lambda^{n-1+\frac{q-1}{d}}\sum_{i=1}^{n}c_{i}=-\lambda^{n-1}+\lambda^{n-1}=0.

And the coefficient of xjx^{j} of ff is 0 for each 1jn11\leq j\leq n-1 by the assumptions on cic_{i}’s. It follows that 0 is also a root of ff with multiplicity nn. Since 0B0\notin B, we have the stronger estimate that mnr+nq1d+n1.mn-r+n\leq\frac{q-1}{d}+n-1.

Remark 4.1.

More generally, one can study the same question if AB+λAB+\lambda is instead contained in a coset of SdS_{d}. However, note that this more general case can be always reduced to the special case studied in Theorem 1.1. Indeed, if AB+λξSd{0}AB+\lambda\subset\xi S_{d}\cup\{0\} with ξ𝔽q\xi\in{\mathbb{F}}_{q}^{*}, then AB+λ/ξSd{0}A^{\prime}B+\lambda/\xi\subset S_{d}\cup\{0\}, where A=A/ξA^{\prime}=A/\xi.

Next, we prove Corollary 1.10, an important corollary of Theorem 1.1. It would be crucial for proving results in Section 6.

Proof of Corollary 1.10.

Since 0AB+λ0\notin AB+\lambda, we have B(λA1)=B\cap(-\lambda A^{-1})=\emptyset, thus Theorem 1.1 implies that |A||B||Sd|1|A||B|\leq|S_{d}|-1. On the other hand, since (Sdλ){0}=AB(S_{d}-\lambda)\setminus\{0\}=AB, it follows that |A||B||AB|=|(Sdλ){0}|=|Sd|1|A||B|\geq|AB|=|(S_{d}-\lambda)\setminus\{0\}|=|S_{d}|-1. Therefore, we have |A||B|=|AB|=|Sd|1|A||B|=|AB|=|S_{d}|-1. ∎

4.2. Restricted product set in a shifted multiplicative subgroup

Recall that A𝔽qA^{\prime}\subset{\mathbb{F}}_{q}^{*} has property Dd(λ,𝔽q)D_{d}(\lambda,{\mathbb{F}}_{q}) if and only if ab+λSd{0}ab+\lambda\in S_{d}\cup\{0\} for each a,bAa,b\in A^{\prime} such that aba\neq b. In other words, A×^A+λSd{0}A^{\prime}\hat{\times}A^{\prime}+\lambda\subset S_{d}\cup\{0\}. Thus, in this subsection, we are led to study restricted product sets and we establish the following restricted product analog of Theorem 1.1. In the next subsection, Theorem 1.1 and Theorem 4.2 will be applied together to prove Theorem 1.5 in the case that qq is a prime and Theorem 1.6 in the case that qq is a square.

Theorem 4.2.

Let d2d\geq 2 and let q1(modd)q\equiv 1\pmod{d} be a prime power. Let A𝔽qA^{\prime}\subset{\mathbb{F}}_{q}^{*} and λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}. If A×^A+λSd{0}A^{\prime}\hat{\times}A^{\prime}+\lambda\subset S_{d}\cup\{0\} while AA+λSd{0}A^{\prime}A^{\prime}+\lambda\not\subset S_{d}\cup\{0\}, then |A|2(q1)/d+4|A^{\prime}|\leq\sqrt{2(q-1)/d}+4.

The proof of Theorem 4.2 is similar to Theorem 1.1, but it is more delicate. In particular, the choice of the auxiliary polynomial 4.4 needs to be modified from that of the proof 4.2 of Theorem 1.1. In view of Theorem 1.1, we can further assume that AA+λSd{0}A^{\prime}A^{\prime}+\lambda\not\subset S_{d}\cup\{0\}, for otherwise we already have a good bound on |A||A^{\prime}|; we refer to Subsection 4.3 for details. It turns out that this additional assumption (which we get for free) is crucial in our proof since it guarantees that the auxiliary polynomial we constructed is not identically zero.

Proof of Theorem 4.2.

Since AA+λSd{0}A^{\prime}A^{\prime}+\lambda\not\subset S_{d}\cup\{0\}, there is bAb\in A^{\prime} such that b2+λSd{0}b^{2}+\lambda\not\subset S_{d}\cup\{0\}. Let A′′=A{λ/b}A^{\prime\prime}=A^{\prime}\setminus\{-\lambda/b\}. If |A′′|=1|A^{\prime\prime}|=1, then we are done. Otherwise, if |A′′||A^{\prime\prime}| is even, let A=A′′A=A^{\prime\prime}; if |A′′||A^{\prime\prime}| is odd, let A=A′′{b}A=A^{\prime\prime}\setminus\{b^{\prime}\}, where bA′′b^{\prime}\in A^{\prime\prime} is an arbitrary element such that bbb^{\prime}\neq b. Then we have bAb\in A and |A||A| is even. Note that |A||A|2|A|\geq|A^{\prime}|-2, thus it suffices to show |A|2(q1)/d+2|A|\leq\sqrt{2(q-1)/d}+2.

Let |A|=n|A|=n, where nn is even. Write A={a1,a2,,an}A=\{a_{1},a_{2},\ldots,a_{n}\}. Without loss of generality, we may assume that a1=ba_{1}=b. Let m=n/21m=n/2-1. Let c1,c2,,cn𝔽qc_{1},c_{2},...,c_{n}\in{\mathbb{F}}_{q} be the unique solution of the following system of equations:

{\TABbinary\tabbedCenterstack[l]i=1nciaij=0,mjmi=1nciaim+1=1.\left\{\TABbinary\tabbedCenterstack[l]{\sum_{i=1}^{n}c_{i}a_{i}^{j}=0,\quad-m\leq j\leq m\\ \\ \sum_{i=1}^{n}c_{i}a_{i}^{m+1}=1.}\right. (4.3)

Indeed, that coefficient matrix of the system is the generalized Vandermonde matrix (aij)1in,mjm+1,(a_{i}^{j})_{1\leq i\leq n,-m\leq j\leq m+1}, which is non-singular since it has nonzero determinant (a1a2an)mi<j(ajai)0.(a_{1}a_{2}\ldots a_{n})^{-m}\prod_{i<j}(a_{j}-a_{i})\neq 0. Note that c10c_{1}\neq 0; for otherwise c1=0c_{1}=0 and we must have c1=c2==cn=0c_{1}=c_{2}=\ldots=c_{n}=0 in view of the first n1n-1 equations in system 4.3, which contradicts the last equation in system 4.3.

Consider the following auxiliary polynomial

f(x)=i=1nci(aix+λ)m+q1d(ai1x1)m𝔽q[x].f(x)=\sum_{i=1}^{n}c_{i}(a_{i}x+\lambda)^{m+\frac{q-1}{d}}(a_{i}^{-1}x-1)^{m}\in{\mathbb{F}}_{q}[x]. (4.4)

It is clear that the degree of ff is at most 2m+q1d2m+\frac{q-1}{d}. Since A×^A+λSd{0}A\hat{\times}A+\lambda\subset S_{d}\cup\{0\}, we have

(aiaj+λ)q1d+1(ai1aj1)=(aiaj+λ)(ai1aj1)(a_{i}a_{j}+\lambda)^{\frac{q-1}{d}+1}(a_{i}^{-1}a_{j}-1)=(a_{i}a_{j}+\lambda)(a_{i}^{-1}a_{j}-1)

for each 1i,jn1\leq i,j\leq n. This simple observation will be used repeatedly in the following computation.

First, we claim that for each 0k1<m0\leq k_{1}<m, 0k2<m0\leq k_{2}<m, and 1jn1\leq j\leq n, we have

i=1nciE(k1)[(aix+λ)m+q1d](aj)E(k2)[(ai1x1)m](aj)=0.\sum_{i=1}^{n}c_{i}E^{(k_{1})}[(a_{i}x+\lambda)^{m+\frac{q-1}{d}}](a_{j})\cdot E^{(k_{2})}[(a_{i}^{-1}x-1)^{m}](a_{j})=0. (4.5)

Indeed, by Lemma 2.3, we have

i=1nciE(k1)[(aix+λ)m+q1d](aj)E(k2)[(ai1x1)m](aj)\displaystyle\sum_{i=1}^{n}c_{i}E^{(k_{1})}[(a_{i}x+\lambda)^{m+\frac{q-1}{d}}](a_{j})\cdot E^{(k_{2})}[(a_{i}^{-1}x-1)^{m}](a_{j})
=(m+q1dk1)(mk2)(i=1nciaik1k2(ajai+λ)mk1(ai1aj1)mk2)\displaystyle=\binom{m+\frac{q-1}{d}}{k_{1}}\binom{m}{k_{2}}\bigg{(}\sum_{i=1}^{n}c_{i}a_{i}^{k_{1}-k_{2}}(a_{j}a_{i}+\lambda)^{m-k_{1}}(a_{i}^{-1}a_{j}-1)^{m-k_{2}}\bigg{)}
=(m+q1dk1)(mk2)1=0mk12=0mk2(mk11)(mk22)(i=1nciaik1k2(ajai)1λmk11(ai1aj)2(1)mk22)\displaystyle=\binom{m+\frac{q-1}{d}}{k_{1}}\binom{m}{k_{2}}\sum_{\ell_{1}=0}^{m-k_{1}}\sum_{\ell_{2}=0}^{m-k_{2}}\binom{m-k_{1}}{\ell_{1}}\binom{m-k_{2}}{\ell_{2}}\bigg{(}\sum_{i=1}^{n}c_{i}a_{i}^{k_{1}-k_{2}}(a_{j}a_{i})^{\ell_{1}}\lambda^{m-k_{1}-\ell_{1}}(a_{i}^{-1}a_{j})^{\ell_{2}}(-1)^{m-k_{2}-\ell_{2}}\bigg{)}
=(m+q1dk1)(mk2)1=0mk12=0mk2(mk11)(mk22)aj1+2λmk11(1)mk22(i=1nciai(k1+1)(k2+2)).\displaystyle=\binom{m+\frac{q-1}{d}}{k_{1}}\binom{m}{k_{2}}\sum_{\ell_{1}=0}^{m-k_{1}}\sum_{\ell_{2}=0}^{m-k_{2}}\binom{m-k_{1}}{\ell_{1}}\binom{m-k_{2}}{\ell_{2}}a_{j}^{\ell_{1}+\ell_{2}}\lambda^{m-k_{1}-\ell_{1}}(-1)^{m-k_{2}-\ell_{2}}\bigg{(}\sum_{i=1}^{n}c_{i}a_{i}^{(k_{1}+\ell_{1})-(k_{2}+\ell_{2})}\bigg{)}.

Note that in the exponent of the last summand, we always have 0k1+1m0\leq k_{1}+\ell_{1}\leq m and 0k2+2m0\leq k_{2}+\ell_{2}\leq m so that m(k1+1)(k2+2)m-m\leq(k_{1}+\ell_{1})-(k_{2}+\ell_{2})\leq m, and thus

i=1nciai(k1+1)(k2+2)=0\sum_{i=1}^{n}c_{i}a_{i}^{(k_{1}+\ell_{1})-(k_{2}+\ell_{2})}=0

by the assumptions in system 4.3. This proves the claim.

Then, for each 1jn1\leq j\leq n and 0rm10\leq r\leq m-1, we apply Lemma 2.2 and equation 4.5 in the above claim to obtain that

E(r)f(aj)=i=1nci(k=0rE(k)[(aix+λ)m+q1d](aj)E(rk)[(ai1x1)m](aj))=0.\displaystyle E^{(r)}f(a_{j})=\sum_{i=1}^{n}c_{i}\bigg{(}\sum_{k=0}^{r}E^{(k)}[(a_{i}x+\lambda)^{m+\frac{q-1}{d}}](a_{j})\cdot E^{(r-k)}[(a_{i}^{-1}x-1)^{m}](a_{j})\bigg{)}=0.

Similarly, using Lemma 2.2, Lemma 2.3, system 4.3, and equation 4.5, we can compute

E(m)f(a1)\displaystyle E^{(m)}f(a_{1}) =i=1nci(k=0mE(k)[(aix+λ)m+q1d](a1)E(mk)[(ai1x1)m](a1))\displaystyle=\sum_{i=1}^{n}c_{i}\bigg{(}\sum_{k=0}^{m}E^{(k)}[(a_{i}x+\lambda)^{m+\frac{q-1}{d}}](a_{1})\cdot E^{(m-k)}[(a_{i}^{-1}x-1)^{m}](a_{1})\bigg{)}
=i=1nci(E(0)[(aix+λ)m+q1d](a1)E(m)[(ai1x1)m](a1))\displaystyle=\sum_{i=1}^{n}c_{i}\bigg{(}E^{(0)}[(a_{i}x+\lambda)^{m+\frac{q-1}{d}}](a_{1})\cdot E^{(m)}[(a_{i}^{-1}x-1)^{m}](a_{1})\bigg{)}
+i=1nci(E(m)[(aix+λ)m+q1d](a1)E(0)[(ai1x1)m](a1))\displaystyle\quad+\sum_{i=1}^{n}c_{i}\bigg{(}E^{(m)}[(a_{i}x+\lambda)^{m+\frac{q-1}{d}}](a_{1})\cdot E^{(0)}[(a_{i}^{-1}x-1)^{m}](a_{1})\bigg{)}
=i=1nci(a1ai+λ)m+q1daim+(m+q1dm)(i=1nciaim(a1ai+λ)q1d(ai1a11)m).\displaystyle=\sum_{i=1}^{n}c_{i}(a_{1}a_{i}+\lambda)^{m+\frac{q-1}{d}}a_{i}^{-m}+\binom{m+\frac{q-1}{d}}{m}\bigg{(}\sum_{i=1}^{n}c_{i}a_{i}^{m}(a_{1}a_{i}+\lambda)^{\frac{q-1}{d}}(a_{i}^{-1}a_{1}-1)^{m}\bigg{)}.

Since a1ai+λ0a_{1}a_{i}+\lambda\neq 0 for each 1in1\leq i\leq n, we have (a1ai+λ)q1d=1(a_{1}a_{i}+\lambda)^{\frac{q-1}{d}}=1 for i>1i>1, and thus

(a1ai+λ)q1d(ai1a11)=ai1a11(a_{1}a_{i}+\lambda)^{\frac{q-1}{d}}(a_{i}^{-1}a_{1}-1)=a_{i}^{-1}a_{1}-1

for all ii. Since a12+λSd{0}a_{1}^{2}+\lambda\notin S_{d}\cup\{0\}, we have

(a12+λ)m((a12+λ)q1d1)0.(a_{1}^{2}+\lambda)^{m}\bigg{(}(a_{1}^{2}+\lambda)^{\frac{q-1}{d}}-1\bigg{)}\neq 0.

Putting these altogether into the computation of E(m)f(a1)E^{(m)}f(a_{1}), we have

E(m)f(a1)\displaystyle E^{(m)}f(a_{1}) =i=1nci(a1ai+λ)m+q1daim+(m+q1dm)i=1nciaim(ai1a11)m\displaystyle=\sum_{i=1}^{n}c_{i}(a_{1}a_{i}+\lambda)^{m+\frac{q-1}{d}}a_{i}^{-m}+\binom{m+\frac{q-1}{d}}{m}\sum_{i=1}^{n}c_{i}a_{i}^{m}(a_{i}^{-1}a_{1}-1)^{m}
=c1a1m((a12+λ)m+q1d(a12+λ)m)+i=1nci(a1ai+λ)maim\displaystyle=c_{1}a_{1}^{-m}\bigg{(}(a_{1}^{2}+\lambda)^{m+\frac{q-1}{d}}-(a_{1}^{2}+\lambda)^{m}\bigg{)}+\sum_{i=1}^{n}c_{i}(a_{1}a_{i}+\lambda)^{m}a_{i}^{-m}
+(m+q1dm)k=0m(mk)a1mk(1)k(i=1nciaik)\displaystyle+\binom{m+\frac{q-1}{d}}{m}\sum_{k=0}^{m}\binom{m}{k}a_{1}^{m-k}(-1)^{k}\bigg{(}\sum_{i=1}^{n}c_{i}a_{i}^{k}\bigg{)}
=c1a1m(a12+λ)m((a12+λ)q1d1)+k=0m(mk)a1kλmk(i=1nciaikm)\displaystyle=c_{1}a_{1}^{-m}(a_{1}^{2}+\lambda)^{m}\bigg{(}(a_{1}^{2}+\lambda)^{\frac{q-1}{d}}-1\bigg{)}+\sum_{k=0}^{m}\binom{m}{k}a_{1}^{k}\lambda^{m-k}\bigg{(}\sum_{i=1}^{n}c_{i}a_{i}^{k-m}\bigg{)}
=c1a1m(a12+λ)m((a12+λ)q1d1)0,\displaystyle=c_{1}a_{1}^{-m}(a_{1}^{2}+\lambda)^{m}\bigg{(}(a_{1}^{2}+\lambda)^{\frac{q-1}{d}}-1\bigg{)}\neq 0,

where we used the fact c10c_{1}\neq 0. In particular, ff is not identically zero.

In conclusion, ff is a non-zero polynomial with degree at most q1d+2m\frac{q-1}{d}+2m, and Lemma 2.4 implies that each of a1,a2,ana_{1},a_{2},\ldots a_{n} is a root of ff with multiplicity at least mm. Recall that m=n/21m=n/2-1. It follows that

n(n2)2=mndegfq1d+2m=q1d+n2,\frac{n(n-2)}{2}=mn\leq\operatorname{deg}f\leq\frac{q-1}{d}+2m=\frac{q-1}{d}+n-2,

that is, we have (n2)22(q1)d.(n-2)^{2}\leq\frac{2(q-1)}{d}. Therefore, n2(q1)/d+2n\leq\sqrt{2(q-1)/d}+2. This finishes the proof. ∎

4.3. Applications to generalized Diophantine tuples over finite fields

In this subsection, we illustrate how to apply Theorem 1.1 and Theorem 4.2 for obtaining improved upper bounds on the size of a generalized Diophantine tuple or a strong generalized Diophantine tuple over 𝔽q\mathbb{F}_{q}, when q=pq=p is a prime and qq is a square.

Proof of Theorem 1.5.

(1) Let A𝔽pA\subset{\mathbb{F}}_{p}^{*} with property SDd(λ,𝔽p)SD_{d}(\lambda,{\mathbb{F}}_{p}), that is, AA+λSd{0}AA+\lambda\subset S_{d}\cup\{0\}. Theorem 1.1 implies that

|A|2|Sd|+|A(λA1)|+|A|1|Sd|+2|A|1.|A|^{2}\leq|S_{d}|+|A\cap(-\lambda A^{-1})|+|A|-1\leq|S_{d}|+2|A|-1.

It follows that (|A|1)2|Sd|(|A|-1)^{2}\leq|S_{d}|. If λSd\lambda\in S_{d}, we have a stronger upper bound:

|A|2|Sd|+|A(λA1)|1|Sd|+|A|1.|A|^{2}\leq|S_{d}|+|A\cap(-\lambda A^{-1})|-1\leq|S_{d}|+|A|-1.

It follows that (|A|12)2|Sd|34(|A|-\frac{1}{2})^{2}\leq|S_{d}|-\frac{3}{4}.

(2) Let A𝔽pA\subset{\mathbb{F}}_{p}^{*} with property Dd(λ,𝔽p)D_{d}(\lambda,{\mathbb{F}}_{p}), that is, A×^A+λSd{0}A\hat{\times}A+\lambda\subset S_{d}\cup\{0\}. If AA+λSd{0}AA+\lambda\subset S_{d}\cup\{0\}, then (1) implies that |A|p/d+1|A|\leq\sqrt{p/d}+1 and we are done. If AA+λSd{0}AA+\lambda\not\subset S_{d}\cup\{0\}, then Theorem 4.2 implies the required upper bound. ∎

Remark 4.3.

Theorem 1.1 can be used to deduce a weaker upper bound of the form 2p/d+O(1)2\sqrt{p/d}+O(1) for Theorem 1.5 (2). Let A𝔽pA\subset{\mathbb{F}}_{p}^{*} such that A×^A+λSd{0}A\hat{\times}A+\lambda\subset S_{d}\cup\{0\}. We can write A=BCA=B\sqcup C such that |B||B| and |C||C| differ by at most 11. Note that since BB and CC are disjoint, we have BC+λA×^A+λSd{0}BC+\lambda\subset A\hat{\times}A+\lambda\subset S_{d}\cup\{0\} and thus Theorem 1.1 implies that |B||C|p/d+|B|+|C||B||C|\leq p/d+|B|+|C|, which further implies that |A|2p/d+O(1)|A|\leq 2\sqrt{p/d}+O(1). Note that such a weaker upper bound is worse than the trivial upper bound from character sums (Proposition 1.4) when d=2,3d=2,3, and this is one of our main motivations for establishing the bound 2p/d+O(1)\sqrt{2p/d}+O(1) in Theorem 1.5 (2).

Next, we consider the case qq is a square. First we establish a non-trivial upper bound on MSDd(λ,𝔽q)MSD_{d}(\lambda,{\mathbb{F}}_{q}) and MDd(λ,𝔽q)MD_{d}(\lambda,{\mathbb{F}}_{q}) under some minor assumption. While these new bounds only improve the trivial upper bound from character sums (Proposition 1.4) slightly, we will see these new bounds are sometimes sharp in the proof of Theorem 1.6. To achieve our goal, we need the following special case of Kummer’s theorem [25].

Lemma 4.4.

Let pp be a prime and m,nm,n be positive integers. If there is no carry between the addition of mm and nn in base-pp, then (m+nn)\binom{m+n}{n} is not divisible by pp.

Theorem 4.5.

Let qq be a prime power and a square, and let λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}.

  1. (1)

    Let d2d\geq 2 be a divisor of (q1)(q-1). Let rr be the remainder of q1d\frac{q-1}{d} divided by pqp\sqrt{q}. If r(p1)qr\leq(p-1)\sqrt{q}, then MSDd(λ,𝔽q)q1MSD_{d}(\lambda,{\mathbb{F}}_{q})\leq\sqrt{q}-1.

  2. (2)

    Let q25q\geq 25 and let d3d\geq 3 be a divisor of (q1)(q-1). Let rr be the remainder of q1d\frac{q-1}{d} divided by pqp\sqrt{q}. If r(p1)qr\leq(p-1)\sqrt{q}, then MDd(λ,𝔽q)q1MD_{d}(\lambda,{\mathbb{F}}_{q})\leq\sqrt{q}-1.

Proof.

(1) Since r(p1)qr\leq(p-1)\sqrt{q}, there is no carry between the addition of r1r-1 and q\sqrt{q} in base-pp. Thus, there is no carry between the addition of q1d1\frac{q-1}{d}-1 and q\sqrt{q} in base-pp. It follows from Lemma 4.4 that

(q1+q1dq)0(modp).\binom{\sqrt{q}-1+\frac{q-1}{d}}{\sqrt{q}}\not\equiv 0\pmod{p}.

Let A𝔽qA\subset{\mathbb{F}}_{q}^{*} with property SDd(λ,𝔽q)SD_{d}(\lambda,{\mathbb{F}}_{q}) such that |A|=MSDd(λ,𝔽q)|A|=MSD_{d}(\lambda,{\mathbb{F}}_{q}). Note that Proposition 1.4 implies that |A|q|A|\leq\sqrt{q}. For the sake of contradiction, assume that |A|=q|A|=\sqrt{q}. Note that AA+λSd{0}AA+\lambda\subset S_{d}\cup\{0\} and

(|A|1+q1d|A|)=(q1+q1dq)0(modp),\binom{|A|-1+\frac{q-1}{d}}{|A|}=\binom{\sqrt{q}-1+\frac{q-1}{d}}{\sqrt{q}}\not\equiv 0\pmod{p},

it follows from Theorem 1.1 that

|A|2|Sd|+|A(λA1)|+|A|1|Sd|+2|A|1,|A|^{2}\leq|S_{d}|+|A\cap(-\lambda A^{-1})|+|A|-1\leq|S_{d}|+2|A|-1,

that is, |A||Sd|+1<q|A|\leq\sqrt{|S_{d}|}+1<\sqrt{q}, a contradiction. This completes the proof.

(2) Let A𝔽qA\subset{\mathbb{F}}_{q}^{*} with property Dd(λ,𝔽q)D_{d}(\lambda,{\mathbb{F}}_{q}) such that |A|=MDd(λ,𝔽q)|A|=MD_{d}(\lambda,{\mathbb{F}}_{q}). Then A×^A+λSd{0}A\hat{\times}A+\lambda\subset S_{d}\cup\{0\}. If AA+λSd{0}AA+\lambda\subset S_{d}\cup\{0\}, we just apply (1). Next assume that AA+λSd{0}AA+\lambda\not\subset S_{d}\cup\{0\}, then Theorem 4.2 implies that

|A|2(q1)d+42(q1)3+4q1,|A|\leq\sqrt{\frac{2(q-1)}{d}}+4\leq\sqrt{\frac{2(q-1)}{3}}+4\leq\sqrt{q}-1,

provided that q738q\geq 738. When 25q73725\leq q\leq 737, we have used SageMath to verify the theorem. ∎

Now we are ready to prove Theorem 1.6, which determines the maximum size of an infinitely family of generalized Diophantine tuples and strong generalized Diophantine tuples over finite fields.

Proof of Theorem 1.6.

In both cases, the upper bound q1\sqrt{q}-1 follows from Theorem 4.5. To show that q1\sqrt{q}-1 is a lower bound, we observe that A=α𝔽qA=\alpha{\mathbb{F}}_{\sqrt{q}}^{*} has property SDd(λ,𝔽q)SD_{d}(\lambda,\mathbb{F}_{q}) (and therefore Dd(λ,𝔽q)D_{d}(\lambda,\mathbb{F}_{q})). Indeed, AA+λ=α2𝔽q+λα2𝔽qSd{0}AA+\lambda=\alpha^{2}{\mathbb{F}}_{\sqrt{q}}^{*}+\lambda\subset\alpha^{2}{\mathbb{F}}_{\sqrt{q}}\subset S_{d}\cup\{0\} since α2Sd\alpha^{2}\in S_{d} and 𝔽qSd{\mathbb{F}}_{\sqrt{q}}^{*}\subset S_{d} (from the assumption d(q+1)d\mid(\sqrt{q}+1)). ∎

Remark 4.6.

Our SageMath code indicates that the last statement of Theorem 1.6 does not hold when d=2d=2 and q=9,25,49q=9,25,49, when d=3d=3 and q=4,16q=4,16, and when d=4d=4 and q=9q=9. We conjecture the same statement holds for d=2d=2, provided that qq is sufficiently large.

So far we have only considered special cases of applying Theorem 1.1. In general, to apply Theorem 1.1, the assumption on the binomial coefficient in the statement of Theorem 1.1 might be tricky to analyze. However, if the base-pp representation of q1d\frac{q-1}{d} behaves “nicely” (for example, if the order of pp modulo dd is small, then the base-pp representation is periodic with a small period), then it is still convenient to apply Theorem 1.1. As a further illustration, we prove the following theorem. Note that the new bound is of the same shape as that in Theorem 1.5 (2), so it can be viewed as a generalization of Theorem 1.5 (2) as changing a prime pp to an arbitrary power of pp, provided that d(p1)d\mid(p-1).

Theorem 4.7.

Let d2d\geq 2, and let qq be a power of pp such that d(p1)d\mid(p-1). Then MDd(λ,𝔽q)2(q1)/d+4MD_{d}(\lambda,{\mathbb{F}}_{q})\leq\sqrt{2(q-1)/d}+4 for any λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}.

Proof.

Let B𝔽qB\subset{\mathbb{F}}_{q}^{*} with property Dd(λ,𝔽q)D_{d}(\lambda,{\mathbb{F}}_{q}), that is, B×^B+λSd{0}B\hat{\times}B+\lambda\subset S_{d}\cup\{0\}. If BB+λSd{0}BB+\lambda\nsubseteq S_{d}\cup\{0\}, we are done by Theorem 4.2. Thus, we may assume that BB+λSd{0}BB+\lambda\subset S_{d}\cup\{0\}. It suffices to show |B|2(q1)/d+4|B|\leq\sqrt{2(q-1)/d}+4. To achieve that, we try to find an arbitrary subset AA of BB such that (|A|1+q1d|A|)0(modp)\binom{|A|-1+\frac{q-1}{d}}{|A|}\not\equiv 0\pmod{p} and |A||A| is as large as possible. With such a subset AA, we have AB+λSd{0}AB+\lambda\subset S_{d}\cup\{0\} so that we can apply Theorem 1.1. In the rest of the proof, we aim to find such an AA with |A||B|/2|A|\geq|B|/2 so that, from Theorem 1.1, we can deduce

|B|22q1d+2|B|1|B|2(q1)d+2+2<2(q1)d+4.\frac{|B|^{2}}{2}\leq\frac{q-1}{d}+2|B|-1\implies|B|\leq\sqrt{\frac{2(q-1)}{d}+2}+2<\sqrt{\frac{2(q-1)}{d}}+4.

Write |B|1=(ck,ck1,,c1,c0)p|B|-1=(c_{k},c_{k-1},\ldots,c_{1},c_{0})_{p} in base-pp, that is, |B|1=i=0kcipi|B|-1=\sum_{i=0}^{k}c_{i}p^{i} with 0cip10\leq c_{i}\leq p-1 for each 0ik0\leq i\leq k and ck1c_{k}\geq 1. Next, we construct AA according to the size of ckc_{k}.

Case 1. ckp1p1dc_{k}\leq p-1-\frac{p-1}{d}. In this case, let AA be an arbitrary subset of BB with |A|1=(ck,0,,0)p|A|-1=(c_{k},0,\ldots,0)_{p}, that is, |A|=ckpk+1|A|=c_{k}p^{k}+1. It is easy to verify that (|A|1+q1d|A|)0(modp)\binom{|A|-1+\frac{q-1}{d}}{|A|}\not\equiv 0\pmod{p} using Lemma 4.4. Since |B|(ck+1)pk|B|\leq(c_{k}+1)p^{k}, it also follows readily that |A||B|/2|A|\geq|B|/2.

Case 2. ck>p1p1dc_{k}>p-1-\frac{p-1}{d}. In this case, let AA be an arbitrary subset of BB with

|A|1=((d1)(p1)d,(d1)(p1)d,,(d1)(p1)d)p,|A|-1=\bigg{(}\frac{(d-1)(p-1)}{d},\frac{(d-1)(p-1)}{d},\ldots,\frac{(d-1)(p-1)}{d}\bigg{)}_{p},

that is, |A|=(d1)(p1)di=0kpi+1|A|=\frac{(d-1)(p-1)}{d}\cdot\sum_{i=0}^{k}p^{i}+1. Again, it is easy to verify that (|A|1+q1d|A|)0(modp)\binom{|A|-1+\frac{q-1}{d}}{|A|}\not\equiv 0\pmod{p} using Lemma 4.4. Since d2d\geq 2, it follows that 2|A|(p1)i=0kpi+2=pk+1+1>|B|2|A|\geq(p-1)\sum_{i=0}^{k}p^{i}+2=p^{k+1}+1>|B|. ∎

Remark 4.8.

Under the same assumption, the proof of Theorem 4.7 can be refined to obtain improved upper bounds on MSDd(λ,𝔽q)MSD_{d}(\lambda,{\mathbb{F}}_{q}). In particular, if d,r2d,r\geq 2 are fixed, and p1(modd)p\equiv 1\pmod{d} is a prime, then as pp\to\infty, we can show that MSDd(λ,𝔽p2r1)(1+o(1))p2r1/dMSD_{d}(\lambda,{\mathbb{F}}_{p^{2r-1}})\leq(1+o(1))\sqrt{p^{2r-1}/d} uniformly among λ𝔽p2r1\lambda\in{\mathbb{F}}_{p^{2r-1}}^{*}. Indeed, if B𝔽qB\subset{\mathbb{F}}_{q}^{*} with property Dd(λ,𝔽q)D_{d}(\lambda,{\mathbb{F}}_{q}) with q=p2r1q=p^{2r-1} and λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}, then we can assume without loss of generality that q/d<|B|\sqrt{q/d}<|B|. Otherwise, we are done. Note that |B|<q+O(1)|B|<\sqrt{q}+O(1) by Proposition 1.4. Following the notations used in the proof of Theorem 4.7, we have p/d1ckp\sqrt{p/d}-1\leq c_{k}\leq\sqrt{p} and thus we are always in Case 1, and the same construction of AA gives |A|=(1o(1))|B||A|=(1-o(1))|B| as pp\to\infty. Thus, Theorem 1.1 gives |B|(1+o(1))q/d|B|\leq(1+o(1))\sqrt{q/d}.

5. Improved upper bounds on the largest size of generalized Diophantine tuples over integers

5.1. Proof of Theorem 1.2

In this subsection, we improve the upper bounds on the largest size of generalized Diophantine tuples with property Dk(n)D_{k}(n). We first recall that for each n1n\geq 1 and k2k\geq 2,

Mk(n)=sup{|A|:A satisfies property Dk(n)}.M_{k}(n)=\sup\{|A|\colon A\text{ satisfies property }D_{k}(n)\}.

For k2k\geq 2, we defined the constant in the introduction

ηk=min||T2,\eta_{k}=\min_{\mathcal{I}}\frac{|\mathcal{I}|}{T_{\mathcal{I}}^{2}}, (5.1)

where the minimum is taken over all nonempty subset \mathcal{I} of

{1ik:gcd(i,k)=1,gcd(i1,k)>1},\{1\leq i\leq k:\gcd(i,k)=1,\gcd(i-1,k)>1\},

and

T=igcd(i1,k).T_{\mathcal{I}}=\sum_{i\in\mathcal{I}}\sqrt{\gcd(i-1,k)}. (5.2)

Here is the proof of our main theorem, Theorem 1.2.

Proof of Theorem 1.2.

Let A={a1,a2,,am}A=\{a_{1},a_{2},\ldots,a_{m}\} be a generalized Diophantine mm-tuple with property Dk(n)D_{k}(n) and k3k\geq 3. Given the assumption that logk=O(loglogn)\log k=O(\sqrt{\log\log n}), Proposition 2.9 implies that the contribution of aia_{i} with ai>nkk2a_{i}>n^{\frac{k}{k-2}} is |A(nkk2,)|=O(logkloglogk)|A\cap(n^{\frac{k}{k-2}},\infty)|=O(\log k\log\log k) is negligible. Thus, we can assume that A[1,nkk2]A\subset[1,n^{\frac{k}{k-2}}]. Let \mathcal{I} be a nonempty subset of {1ik:gcd(i,k)=1,gcd(i1,k)>1}\{1\leq i\leq k:\gcd(i,k)=1,\gcd(i-1,k)>1\}, such that the ratio ||/T2|\mathcal{I}|/T_{\mathcal{I}}^{2} in equation 5.1 is minimized by \mathcal{I}. In other words, we have ηk=||/T2\eta_{k}=|\mathcal{I}|/T_{\mathcal{I}}^{2}, where

T=T=igcd(i1,k).T=T_{\mathcal{I}}=\sum_{i\in\mathcal{I}}\sqrt{\gcd(i-1,k)}.

To apply the Gallagher sieve inequality (Theorem 2.5), we set N=nkk2N=n^{\frac{k}{k-2}} and define the set of primes

𝒫={p:pi(modk) for some i}{p:pn}.\mathcal{P}=\{p:p\equiv i\pmod{k}\text{ for some }i\in\mathcal{I}\}\setminus\{p:p\mid n\}.

For each prime p𝒫p\in\mathcal{P}, denote by ApA_{p} the image of A(modp)A\pmod{p} and let Ap=Ap{0}A_{p}^{*}=A_{p}\setminus\{0\}.

Let p𝒫p\in\mathcal{P}. We can naturally view ApA_{p}^{*} as a subset of 𝔽p{\mathbb{F}}_{p}^{*}. Since AA has property Dk(n)D_{k}(n), it follows that Ap×^Ap+n{xk:x𝔽p}{0}A_{p}^{*}\hat{\times}A_{p}^{*}+n\subset\{x^{k}:x\in{\mathbb{F}}_{p}^{*}\}\cup\{0\}. Note that {xk:x𝔽p}\{x^{k}:x\in{\mathbb{F}}_{p}^{*}\} is the multiplicative subgroup of 𝔽p{\mathbb{F}}_{p}^{*} with order p1gcd(p1,k)\frac{p-1}{\gcd(p-1,k)}. Since gcd(p1,k)>1\gcd(p-1,k)>1 and pnp\nmid n, Theorem 1.5 (2) implies that

|Ap||Ap|+12(p1)gcd(p1,k)+5.|A_{p}|\leq|A_{p}^{*}|+1\leq\sqrt{\frac{2(p-1)}{\gcd(p-1,k)}}+5.

Set Q=2(ϕ(k)logNT)2Q=2(\frac{\phi(k)\log N}{T})^{2}. Applying Gallagher’s larger sieve, we obtain that

|A|p𝒫,pQlogplogNp𝒫,pQlogp|Ap|logN.|A|\leq\frac{\sum_{p\in\mathcal{P},p\leq Q}\log p-\log N}{\sum_{p\in\mathcal{P},p\leq Q}\frac{\log p}{|A_{p}|}-\log N}. (5.3)

Let cc be the constant from Corollary 2.7. For the numerator on the right-hand side of inequality 5.3, we have

p𝒫,pQlogplogN\displaystyle\sum_{p\in\mathcal{P},p\leq Q}\log p-\log N i(pi mod k,pQlogp)logN\displaystyle\leq\sum_{i\in\mathcal{I}}\bigg{(}\sum_{\begin{subarray}{c}p\equiv i\text{ mod }k,\\ p\leq Q\end{subarray}}\log p\bigg{)}-\log N
=||Qϕ(k)+O(||Qexp(clogQ))logN.\displaystyle=\frac{|\mathcal{I}|Q}{\phi(k)}+O\bigg{(}|\mathcal{I}|Q\exp(-c\sqrt{\log Q})\bigg{)}-\log N.

Next, we estimate the denominator on the right-hand side of inequality 5.3. Note that ||T=igcd(i1,k)|\mathcal{I}|\leq T=\sum_{i\in\mathcal{I}}\sqrt{\gcd(i-1,k)}. Then we have T||kϕ(k)kT\leq|\mathcal{I}|\sqrt{k}\leq\phi(k)\sqrt{k}, and so ϕ(k)/T1/k\phi(k)/T\geq 1/\sqrt{k}. Since k=(logN)o(1)k=(\log N)^{o(1)}, we deduce Q>2(logN)2o(1)Q>2(\log N)^{2-o(1)}. Thus we have k=Qo(1)k=Q^{o(1)}. This, together with Corollary 2.7 and Lemma 2.8, deduces that for each ii\in\mathcal{I},

p𝒫,pQpi mod klogp|Ap|\displaystyle\sum_{\begin{subarray}{c}p\in\mathcal{P},p\leq Q\\ p\equiv i\text{ mod }k\end{subarray}}\frac{\log p}{|A_{p}|} p𝒫,pQpi mod klogp2(p1)gcd(i1,k)+5\displaystyle\geq\sum_{\begin{subarray}{c}p\in\mathcal{P},p\leq Q\\ p\equiv i\text{ mod }k\end{subarray}}\frac{\log p}{\sqrt{\frac{2(p-1)}{\gcd(i-1,k)}}+5}
=pQpi mod klogp2pgcd(i1,k)+O(pQklogpp)+O(pnklogpp)\displaystyle=\sum_{\begin{subarray}{c}p\leq Q\\ p\equiv i\text{ mod }k\end{subarray}}\frac{\log p}{\sqrt{\frac{2p}{\gcd(i-1,k)}}}+O\bigg{(}\sum_{p\leq Q}\frac{k\log p}{p}\bigg{)}+O\bigg{(}\sum_{p\mid n}\frac{\sqrt{k}\log p}{\sqrt{p}}\bigg{)}
=gcd(i1,k)2pQpi mod klogpp+O(klogQ)+O(k(logn)1/2)\displaystyle=\sqrt{\frac{\gcd(i-1,k)}{2}}\sum_{\begin{subarray}{c}p\leq Q\\ p\equiv i\text{ mod }k\end{subarray}}\frac{\log p}{\sqrt{p}}+O(k\log Q)+O(k(\log n)^{1/2})
=2Qgcd(i1,k)ϕ(k)+O(Qgcd(i1,k)exp(clogQ)).\displaystyle=\frac{\sqrt{2Q\gcd(i-1,k)}}{\phi(k)}+O\left(\sqrt{Q}\sqrt{\gcd(i-1,k)}\exp(-c\sqrt{\log Q})\right).

Thus we have

|A|\displaystyle|A| p𝒫,pQlogplogNp𝒫,pQlogp|Ap|logN\displaystyle\leq\frac{\sum_{p\in\mathcal{P},p\leq Q}\log p-\log N}{\sum_{p\in\mathcal{P},p\leq Q}\frac{\log p}{|A_{p}|}-\log N}
||Qϕ(k)+O(||Qexp(clogQ))logNi(p𝒫,pQpi mod klogp|Ap|)logN\displaystyle\leq\frac{\frac{|\mathcal{I}|Q}{\phi(k)}+O(|\mathcal{I}|Q\exp(-c\sqrt{\log Q}))-\log N}{\sum_{i\in\mathcal{I}}\bigg{(}\sum_{\begin{subarray}{c}p\in\mathcal{P},p\leq Q\\ p\equiv i\text{ mod }k\end{subarray}}\frac{\log p}{|A_{p}|}\bigg{)}-\log N}
||Qϕ(k)+O(||Qexp(clogQ))logNT2Qϕ(k)+O(TQexp(clogQ))logN.\displaystyle\leq\frac{\frac{|\mathcal{I}|Q}{\phi(k)}+O(|\mathcal{I}|Q\exp(-c\sqrt{\log Q}))-\log N}{\frac{T\sqrt{2Q}}{\phi(k)}+O\left(T\sqrt{Q}\exp(-c\sqrt{\log Q})\right)-\log N}.

Finally, recall that Q=2(ϕ(k)logNT)2Q=2(\frac{\phi(k)\log N}{T})^{2}. It follows that

|A|\displaystyle|A| 2||ϕ(k)(logNT)2+O(||(ϕ(k)logNT)2exp(clog(ϕ(k)logNT)))2logN+O(ϕ(k)logNexp(clog(ϕ(k)logNT)))logN\displaystyle\leq\frac{2|\mathcal{I}|\phi(k)(\frac{\log N}{T})^{2}+O\left(|\mathcal{I}|(\frac{\phi(k)\log N}{T})^{2}\exp\Big{(}-c\sqrt{\log(\frac{\phi(k)\log N}{T})}\Big{)}\right)}{2\log N+O\left(\phi(k)\log N\exp\Big{(}-c\sqrt{\log(\frac{\phi(k)\log N}{T})}\Big{)}\right)-\log N}
=2||ϕ(k)T2logN+O(||ϕ(k)2logNT2exp(clog(ϕ(k)logNT)))1+O(ϕ(k)exp(clog(ϕ(k)logNT))).\displaystyle=\frac{\frac{2|\mathcal{I}|\phi(k)}{T^{2}}\log N+O\left(\frac{|\mathcal{I}|\phi(k)^{2}\log N}{T^{2}}\exp\Big{(}-c\sqrt{\log(\frac{\phi(k)\log N}{T})}\Big{)}\right)}{1+O\left(\phi(k)\exp\Big{(}-c\sqrt{\log(\frac{\phi(k)\log N}{T})}\Big{)}\right)}.

Recall that N=nkk2N=n^{\frac{k}{k-2}}. Thus, to obtain our desired result, we need to show

|A|(1+o(1))2||ϕ(k)T2logN1o(1),\displaystyle|A|\leq\frac{(1+o(1))\frac{2|\mathcal{I}|\phi(k)}{T^{2}}\log N}{1-o(1)},

and it suffices to show that

ϕ(k)exp(clog(ϕ(k)logNT))=o(1),\phi(k)\exp\Big{(}-c\sqrt{\log(\frac{\phi(k)\log N}{T})}\Big{)}=o(1),

as NN\to\infty, or equivalently,

logkclogϕ(k)T+loglogN,\log k-c\sqrt{\log\frac{\phi(k)}{T}+\log\log N}\to-\infty,

as NN\to\infty. We notice that ϕ(k)/T1/k\phi(k)/T\geq 1/\sqrt{k}. Let c=c/2c^{\prime}=c/2. Then the assumption logkcloglogn<cloglogN\log k\leq c^{\prime}\sqrt{\log\log n}<c^{\prime}\sqrt{\log\log N} implies

loglogN+logϕ(k)TloglogN12logk=(1o(1))loglogN,\log\log N+\log\frac{\phi(k)}{T}\geq\log\log N-\frac{1}{2}\log k=(1-o(1))\log\log N,

and

logkclogϕ(k)T+loglogN(co(1))loglogN,\log k-c\sqrt{\log\frac{\phi(k)}{T}+\log\log N}\leq-(c^{\prime}-o(1))\log\log N\to-\infty,

as required. ∎

Remark 5.1.

Note that when ={1}\mathcal{I}=\{1\}, that is to say, when we only consider primes pp such that p1(modk)p\equiv 1\pmod{k} for applying the Gallagher inequality, the condition p1(modk)p\equiv 1\pmod{k} guarantees that the kk-th powers are indeed kk-th powers modulo pp. We have T=T=kT=T_{\mathcal{I}}=\sqrt{k}, thus we trivially have ηk1k\eta_{k}\leq\frac{1}{k} in view of equation 5.1. In particular, if kk is fixed and nn\to\infty, Theorem 1.2 implies that

Mk(n)(2+o(1))ϕ(k)k2logn,M_{k}(n)\leq\frac{(2+o(1))\phi(k)}{k-2}\log n, (5.4)

which already provides a substantial improvement on the best-known upper bound Mk(n)(3ϕ(k)+o(1))lognM_{k}(n)\leq(3\phi(k)+o(1))\log n whenever k3k\geq 3 given in [7]. Moreover, note that ϕ(k)k\frac{\phi(k)}{k} can be as small as O(1loglogk)O(\frac{1}{\log\log k}) when kk is the product of distinct primes [28, Theorem 2.9]. Thus, in view of Theorem 1.2, the inequality 5.4 already shows there is k=k(n)k=k(n) such that logkloglogn\log k\asymp\sqrt{\log\log n} and

Mk(n)lognloglogklognlogloglogn.M_{k}(n)\ll\frac{\log n}{\log\log k}\ll\frac{\log n}{\log\log\log n}. (5.5)

Note that 5.5 already breaks the logn\log n barrier. On the other hand, we can still use other primes pp such that gcd(p1,k)>1\gcd(p-1,k)>1 for which kk-th powers are in fact gcd(k,p1)\gcd(k,p-1)-th powers modulo pp when we apply the Gallagher sieve inequality. We can take advantage of the improvement on the upper bound of Mk(n)M_{k}(n). In the next two subsections, we further provide a significant improvement on inequality 5.5.

Next, we define a strong Diophantine mm-tuple with property SDk(n)SD_{k}(n) to be a set {a1,,am}\{a_{1},\ldots,a_{m}\} of mm distinct positive integers such that aiaj+na_{i}a_{j}+n is a kk-th power for any choice of ii and jj. We have a stronger upper bound for the size of a strong Diophantine tuple with property SDk(n)SD_{k}(n). We define

MSk(n)=sup{|A|:A satisfies the property SDk(n)}.MS_{k}(n)=\sup\{|A|\colon A\subset{\mathbb{N}}\text{ satisfies the property }SD_{k}(n)\}.
Theorem 5.2.

There is a constant c>0c^{\prime}>0, such that as nn\to\infty,

MSk(n)(kk2+o(1))ηkϕ(k)logn,MS_{k}(n)\leq\bigg{(}{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\frac{k}{k-2}}+o(1)\bigg{)}\ \eta_{k}\phi(k)\log n,

holds uniformly for positive integers k,n3k,n\geq 3 such that logkcloglogn\log k\leq c^{\prime}\sqrt{\log\log n}. Moreover, if kk is even, under the same assumption (including the case k=2k=2), we have the stronger bound

MSk(n)min{(1+o(1))ηkϕ(k)logn,τ(n)},MS_{k}(n)\leq\min\{\big{(}1+o(1)\big{)}\eta_{k}\phi(k)\log n,\tau(n)\},

where τ(n)\tau(n) is the number of divisors of nn.

Proof.

The proof is very similar to the proof of Theorem 1.2 and we follow all the notations and steps as in the proof of Theorem 1.2, apart from the minor modifications stated below.

We prove the first part. For each p𝒫p\in\mathcal{P}, we have the stronger upper bound that |Ap|(p1)gcd(p1,k)+2|A_{p}|\leq\sqrt{\frac{(p-1)}{\gcd(p-1,k)}}+2 by Theorem 1.5 (2). To optimize the upper bound obtained from Gallagher’s larger sieve, we instead set Q=(ϕ(k)logNT)2Q=(\frac{\phi(k)\log N}{T})^{2}.

Next, we assume that kk is even and prove the second part. Notice that for each xAx\in A, there is a positive integer yy, such that x2+n=y2x^{2}+n=y^{2}. Thus, |A||A| is bounded by the number of positive integral solutions to the equation x2+n=y2x^{2}+n=y^{2}, which is at most τ(n)\tau(n). On the other hand, this also implies that all the elements in AA are at most nn. Thus, we can set N=nN=n instead and obtain the stronger upper bound. ∎

5.2. Proof of Theorem 1.3

In this subsection, by finding a more refined upper bound on ηk\eta_{k} in equation 5.1, we show that the same approach significantly improves the upper bound of Mk(n)M_{k}(n) in inequality 5.5 when kk is the product of the first few distinct primes.

We label all the primes in increasing order so that 2=p1<p2<<p<2=p_{1}<p_{2}<\cdots<p_{\ell}<\cdots. Let P=i=1piP_{\ell}=\prod_{i=1}^{\ell}p_{i} be the product of first \ell primes. Let 1={1}\mathcal{I}_{1}=\{1\}. For 1\ell\geq 1, we define +1\mathcal{I}_{\ell+1} inductively:

+1={i+jP:i,0j<p+1,p+1(i+jP)}.\mathcal{I}_{\ell+1}=\{i+jP_{\ell}:i\in\mathcal{I}_{\ell},0\leq j<p_{\ell+1},p_{\ell+1}\nmid(i+jP_{\ell})\}. (5.6)

We note that +1\mathcal{I}_{\ell}\subset\mathcal{I}_{\ell+1} for any 1\ell\geq 1. Also, it is clear that

|+1|=||(p+11).\displaystyle|\mathcal{I}_{\ell+1}|=|\mathcal{I}_{\ell}|(p_{\ell+1}-1). (5.7)
Lemma 5.3.

Following the above definitions, we have

{1xP:gcd(x,P)=1,gcd(x1,P)>1}.\mathcal{I}_{\ell}\subset\{1\leq x\leq P_{\ell}:\gcd(x,P_{\ell})=1,\gcd(x-1,P_{\ell})>1\}. (5.8)
Proof.

We give an inductive proof. When =1\ell=1, the inclusion 5.8 holds. We assume that 5.8 holds for some 1\ell\geq 1. Let x=i+jP+1x=i+jP_{\ell}\in\mathcal{I}_{\ell+1}. By the assumption, we have gcd(i,P)=1\gcd(i,P_{\ell})=1, and it follows that gcd(x,P+1)=gcd(x,P)gcd(x,p+1)=gcd(i,P)gcd(x,p+1)=1.\gcd(x,P_{\ell+1})=\gcd(x,P_{\ell})\gcd(x,p_{\ell+1})=\gcd(i,P_{\ell})\gcd(x,p_{\ell+1})=1. This proves the claim. ∎

Furthermore, we introduce the following notation which is similar to the previously introduced on equation 5.2. For each 1\ell\geq 1, we let

T=ygcd(y1,P).T_{\ell}=\sum_{y\in\mathcal{I}_{\ell}}\sqrt{\gcd(y-1,P_{\ell})}.

Note that T1=2T_{1}=\sqrt{2}. We also establish a recurrence relation on the sequence.

Lemma 5.4.

The sequence (T)1(T_{\ell})_{\ell\geq 1} satisfies the recurrence relation

T+1=T(p+12+p+1).\displaystyle T_{\ell+1}=T_{\ell}(p_{\ell+1}-2+\sqrt{p_{\ell+1}}). (5.9)
Proof.

We have

T+1\displaystyle T_{\ell+1} =i0j<p+1p+1(i+jP)gcd(i+jP1,P+1)\displaystyle=\sum_{i\in\mathcal{I}_{\ell}}\sum_{\begin{subarray}{c}0\leq j<p_{\ell+1}\\ p_{\ell+1}\nmid(i+jP_{\ell})\end{subarray}}\sqrt{\gcd(i+jP_{\ell}-1,P_{\ell+1})}
=i0j<p+1p+1(i+jP)gcd(i+jP1,P)gcd(i+jP1,p+1)\displaystyle=\sum_{i\in\mathcal{I}_{\ell}}\sum_{\begin{subarray}{c}0\leq j<p_{\ell+1}\\ p_{\ell+1}\nmid(i+jP_{\ell})\end{subarray}}\sqrt{\gcd(i+jP_{\ell}-1,P_{\ell})}\sqrt{\gcd(i+jP_{\ell}-1,p_{\ell+1})}
=igcd(i1,P)(0j<p+1p+1(i+jP)gcd(i+jP1,p+1)).\displaystyle=\sum_{i\in\mathcal{I}_{\ell}}\sqrt{\gcd(i-1,P_{\ell})}\bigg{(}\sum_{\begin{subarray}{c}0\leq j<p_{\ell+1}\\ p_{\ell+1}\nmid(i+jP_{\ell})\end{subarray}}\sqrt{\gcd(i+jP_{\ell}-1,p_{\ell+1})}\bigg{)}.

It is easy to show that the inner sum consists of (p+12)(p_{\ell+1}-2) many 11 and a single p+1\sqrt{p_{\ell+1}}. It follows that

T+1=(p+12+p+1)igcd(i1,P)=T(p+12+p+1),T_{\ell+1}=(p_{\ell+1}-2+\sqrt{p_{\ell+1}})\sum_{i\in\mathcal{I}_{\ell}}\sqrt{\gcd(i-1,P_{\ell})}=T_{\ell}(p_{\ell+1}-2+\sqrt{p_{\ell+1}}),

proving the lemma. ∎

We are now ready to prove Theorem 1.3.

Proof of Theorem 1.3.

For each nn, we choose k=k(n)=Pk=k(n)=P_{\ell}, where =(n)\ell=\ell(n) is the largest integer such that logP<cloglogn\log P_{\ell}<c^{\prime}\sqrt{\log\log n}. It follows that logk=logPloglogn\log k=\log P_{\ell}\asymp\sqrt{\log\log n}. Thus, using equations 5.7 and 5.9, we have

ηkϕ(k)||ϕ(P)T2=pp(p1)2(p2+p)2.\eta_{k}\phi(k)\leq\frac{|\mathcal{I}_{\ell}|\phi(P_{\ell})}{T_{\ell}^{2}}=\prod_{p\leq p_{\ell}}\frac{(p-1)^{2}}{(p-2+\sqrt{p})^{2}}.

Note that for each prime pp, it is easy to verify that p1p2+p11p.\frac{p-1}{p-2+\sqrt{p}}\leq 1-\frac{1}{\sqrt{p}}. Recall that the inequality ex1+xe^{x}\geq 1+x holds for all real xx, and a standard application of partial summation gives

px1p=2xlogx+O(xlog2x).\sum\limits_{p\leq x}\frac{1}{\sqrt{p}}=\frac{2\sqrt{x}}{\log x}+O\left(\frac{\sqrt{x}}{\log^{2}x}\right).

Also, the prime number theorem implies that

logP=pplogp=θ(p)=(1+o(1))p\log P_{\ell}=\sum_{p\leq p_{\ell}}\log p=\theta(p_{\ell})=(1+o(1))p_{\ell}

and thus p=(1+o(1))logPp_{\ell}=(1+o(1))\log P_{\ell}. Putting the above estimates altogether, we have

ηkϕ(k)\displaystyle\eta_{k}\phi(k) pp(11p)2exp(2pp1p)\displaystyle\leq\prod_{p\leq p_{\ell}}\bigg{(}1-\frac{1}{\sqrt{p}}\bigg{)}^{2}\leq\exp\bigg{(}-2\sum_{p\leq p_{\ell}}\frac{1}{\sqrt{p}}\bigg{)}
=exp((4+o(1))plogp)=exp((4+o(1))logPloglogP)\displaystyle=\exp\bigg{(}-\frac{(4+o(1))\sqrt{p_{\ell}}}{\log p_{\ell}}\bigg{)}=\exp\bigg{(}-\frac{(4+o(1))\sqrt{\log P_{\ell}}}{\log\log P_{\ell}}\bigg{)}
exp(c′′(loglogn)1/4logloglogn).\displaystyle\leq\exp\bigg{(}-\frac{c^{\prime\prime}(\log\log n)^{1/4}}{\log\log\log n}\bigg{)}.

for some absolute constant c′′>0c^{\prime\prime}>0. It follows from Theorem 1.2 that

Mk(n)ηkϕ(k)lognexp(c′′(loglogn)1/4logloglogn)logn.M_{k}(n)\ll\eta_{k}\phi(k)\log n\ll\exp\bigg{(}-\frac{c^{\prime\prime}(\log\log n)^{1/4}}{\log\log\log n}\bigg{)}\log n.

5.3. An upper bound on ηk\eta_{k}

In this subsection, we deduce a simple upper bound of ηk\eta_{k}. It turns out that this upper bound well approximates ηk\eta_{k} empirically.

Theorem 5.5.

For any k2k\geq 2, we have

ηkμk,\eta_{k}\leq\mu_{k},

where μk=Rkmin{β(pα):pα||k}\mu_{k}=R_{k}\cdot\min\{\beta(p^{\alpha}):p^{\alpha}||k\} with

Rk=pαk(p1)pα1(pαpα1p(α1)/2+pα1/2)2,R_{k}=\prod_{p^{\alpha}\mid\mid k}\frac{(p-1)p^{\alpha-1}}{\big{(}p^{\alpha}-p^{\alpha-1}-p^{(\alpha-1)/2}+p^{\alpha-1/2}\big{)}^{2}},

and

β(pα)=(pαpα1p(α1)/2+pα1/2)2(p1)(p(α1)/2+pα1+pα1/2)2.\beta(p^{\alpha})=\frac{\big{(}p^{\alpha}-p^{\alpha-1}-p^{(\alpha-1)/2}+p^{\alpha-1/2}\big{)}^{2}}{(p-1)\big{(}-p^{(\alpha-1)/2}+p^{\alpha-1}+p^{\alpha-1/2}\big{)}^{2}}.
Proof.

We denote k=j=1pjαjk=\prod_{j=1}^{\ell}p_{j}^{\alpha_{j}}, where p1,p2,,pp_{1},p_{2},\ldots,p_{\ell} are distinct primes factors of kk such that

β(pα)=min{β(pα):pαk}.\beta(p_{\ell}^{\alpha_{\ell}})=\min\{\beta(p^{\alpha}):p^{\alpha}\mid\mid k\}.

Define

={1ik:gcd(k,i)=1,i1(modp)},T=igcd(i1,k).\mathcal{I}=\{1\leq i\leq k:\gcd(k,i)=1,i\equiv 1\pmod{p_{\ell}}\},\quad T_{\mathcal{I}}=\sum_{i\in\mathcal{I}}\sqrt{\gcd(i-1,k)}.

Then \mathcal{I} is obviously a subset of the set {1ik:gcd(i,k)=1,gcd(i1,k)>1}\{1\leq i\leq k:\gcd(i,k)=1,\gcd(i-1,k)>1\} consisting of residue classes that can be used in Gallagher’s larger sieve in the proof of Theorem 1.2. (In particular, when p=2p_{\ell}=2, \mathcal{I} consists of all the available residue classes with ||=ϕ(k)|\mathcal{I}|=\phi(k).) In view of the definition of ηk\eta_{k}, it suffices to show that

||T2=μk=Rkβ(pα).\frac{|\mathcal{I}|}{T_{\mathcal{I}}^{2}}=\mu_{k}=R_{k}\cdot\beta(p_{\ell}^{\alpha_{\ell}}).

We first compute the size of \mathcal{I}. Equivalently, we can write

={1ik:i0(modpj) for each 1j<, and i1(modp)},\mathcal{I}=\{1\leq i\leq k:i\not\equiv 0\pmod{p_{j}}\text{ for each }1\leq j<\ell,\text{ and }i\equiv 1\pmod{p_{\ell}}\},

and hence, we deduce ||=j=11(pj1)pjαj1pα1|\mathcal{I}|=\prod_{j=1}^{\ell-1}(p_{j}-1)p_{j}^{\alpha_{j}-1}\cdot p_{\ell}^{\alpha_{\ell}-1}. In order to compute TT_{\mathcal{I}}, we first count the number of solutions to vpj(i1)=sv_{p_{j}}(i-1)=s over 1ipjαj1\leq i\leq p_{j}^{\alpha_{j}} such that pjip_{j}\nmid i for 0sαj0\leq s\leq\alpha_{j} separately, and then use the Chinese remainder theorem. Set

Cj,s\displaystyle C_{j,s} ={1ipjαj:i0(modpj),vpj(i1)=s},for 0sαjj<;\displaystyle=\{1\leq i\leq p_{j}^{\alpha_{j}}:i\not\equiv 0\pmod{p_{j}},~{}v_{p_{j}}(i-1)=s\},\quad\text{for $0\leq s\leq\alpha_{j}$, $j<\ell$;}
C,s\displaystyle C_{\ell,s} ={1ipα:i1(modp),vp(i1)=s},for 0sα.\displaystyle=\{1\leq i\leq p_{\ell}^{\alpha_{\ell}}:i\equiv 1\pmod{p_{\ell}},~{}v_{p_{\ell}}(i-1)=s\},\quad\text{for $0\leq s\leq\alpha_{\ell}$}.

Note that

|Cj,s|\displaystyle|C_{j,s}| =ϕ(pjαjs),for 0<sαjj<;\displaystyle=\phi(p_{j}^{\alpha_{j}-s}),\quad\quad\quad\quad\text{for $0<s\leq\alpha_{j}$, $j<\ell$;}
|Cj,0|\displaystyle|C_{j,0}| =ϕ(pjαj)pjαj1,for j<;\displaystyle=\phi(p_{j}^{\alpha_{j}})-p_{j}^{\alpha_{j}-1},\quad\;\;\text{for $j<\ell$;}
|C,s|\displaystyle|C_{\ell,s}| =ϕ(pαs),for 0<sα,\displaystyle=\phi(p_{\ell}^{\alpha_{\ell}-s}),\quad\quad\quad\quad\text{for $0<s\leq\alpha_{\ell}$},

and |C,0|=0|C_{\ell,0}|=0. It follows that

T=igcd(i1,k)=dkdi,gcd(i1,k)=d1=j=1(s=0αjpjs|Cj,s|).\displaystyle T_{\mathcal{I}}=\sum_{i\in\mathcal{I}}\sqrt{\gcd(i-1,k)}=\sum_{d\mid k}\sqrt{d}\sum_{\begin{subarray}{c}i\in\mathcal{I},\\ \gcd(i-1,k)=d\end{subarray}}1=\prod_{j=1}^{\ell}\left(\sum_{s=0}^{\alpha_{j}}\sqrt{p_{j}^{s}}|C_{j,s}|\right).

For each 1j11\leq j\leq\ell-1, we calculate

s=0αjpjs|Cj,s|=ϕ(pjαj)pjαj1+s=1αjpjsϕ(pjαjs)=pjαjpjαj1pj(αj1)/2+pjαj1/2.\displaystyle\sum_{s=0}^{\alpha_{j}}\sqrt{p_{j}^{s}}|C_{j,s}|=\phi(p_{j}^{\alpha_{j}})-p_{j}^{\alpha_{j}-1}+\sum_{s=1}^{\alpha_{j}}\sqrt{p_{j}^{s}}\phi(p_{j}^{\alpha_{j}-s})=p_{j}^{\alpha_{j}}-p_{j}^{\alpha_{j}-1}-p_{j}^{(\alpha_{j}-1)/2}+p_{j}^{\alpha_{j}-1/2}.

Similarly, we have

s=0αps|C,s|\displaystyle\sum_{s=0}^{\alpha_{\ell}}\sqrt{p_{\ell}^{s}}|C_{\ell,s}| =s=1αpsϕ(pαs)=p(α1)/2+pα1+pα1/2.\displaystyle=\sum_{s=1}^{\alpha_{\ell}}\sqrt{p_{\ell}^{s}}\phi(p_{\ell}^{\alpha_{\ell}-s})=-p_{\ell}^{(\alpha_{\ell}-1)/2}+p_{\ell}^{\alpha_{\ell}-1}+p_{\ell}^{\alpha_{\ell}-1/2}.

Putting these all together, we compute

T\displaystyle T_{\mathcal{I}} =j=11[pjαjpjαj1pj(αj1)/2+pjαj1/2](p(α1)/2+pα1+pα1/2).\displaystyle=\prod_{j=1}^{\ell-1}\bigg{[}p_{j}^{\alpha_{j}}-p_{j}^{\alpha_{j}-1}-p_{j}^{(\alpha_{j}-1)/2}+p_{j}^{\alpha_{j}-1/2}\bigg{]}\cdot\bigg{(}-p_{\ell}^{(\alpha_{\ell}-1)/2}+p_{\ell}^{\alpha_{\ell}-1}+p_{\ell}^{\alpha_{\ell}-1/2}\bigg{)}.

Hence,

||T2\displaystyle\frac{|\mathcal{I}|}{T_{\mathcal{I}}^{2}} =j=11(pj1)pjαj1(pjαjpjαj1pj(αj1)/2+pjαj1/2)2pα1(p(α1)/2+pα1+pα1/2)2.\displaystyle=\prod_{j=1}^{\ell-1}\frac{(p_{j}-1)p_{j}^{\alpha_{j}-1}}{\big{(}p_{j}^{\alpha_{j}}-p_{j}^{\alpha_{j}-1}-p_{j}^{(\alpha_{j}-1)/2}+p_{j}^{\alpha_{j}-1/2}\big{)}^{2}}\cdot\frac{p_{\ell}^{\alpha_{\ell}-1}}{\big{(}-p_{\ell}^{(\alpha_{\ell}-1)/2}+p_{\ell}^{\alpha_{\ell}-1}+p_{\ell}^{\alpha_{\ell}-1/2}\big{)}^{2}}.

Therefore, Theorem 1.2 implies

Corollary 5.6.

There is a constant c>0c^{\prime}>0, such that as nn\to\infty,

Mk(n)(2kk2+o(1))μkϕ(k)logn,M_{k}(n)\leq\bigg{(}{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\frac{2k}{k-2}}+o(1)\bigg{)}\ \mu_{k}\phi(k)\log n,

holds uniformly for positive integers k,n3k,n\geq 3 such that logkcloglogn\log k\leq c^{\prime}\sqrt{\log\log n}.

Remark 5.7.

Our computations indicate that when 2k100,0002\leq k\leq 100{,}000, the inequality μk2ηk\mu_{k}\leq 2\eta_{k} holds for all but 501501 of them. This numerical evidence suggests that μk\mu_{k} provides a good approximation for ηk\eta_{k} for a generic kk. Note that the computational complexity for computing μk\mu_{k} is the same as that of the prime factorization of kk: a naive algorithm takes O(k)O(\sqrt{k}) time. The best theoretical algorithm has running time O(exp((logk)1/3+o(1)))O\big{(}\exp((\log k)^{1/3+o(1)})\big{)} using the general number field sieve [5]. On the other hand, computing ηk\eta_{k} requires O(klogk)O(k\log k) time; we refer to Appendix A for an algorithm and some computational results.

6. Multiplicative decompositions of shifted multiplicative subgroups

In this section, we present our contributions to 1.9. In particular, we make significant progress towards Sárközy’s conjecture (1.8). We recall Sd=Sd(𝔽q)={xd:x𝔽q}S_{d}=S_{d}({\mathbb{F}}_{q})=\{x^{d}:x\in{\mathbb{F}}_{q}^{*}\}.

6.1. Applications to Sárközy’s conjecture

In this subsection, we show that for almost all primes p1(modd)p\equiv 1\pmod{d}, the set (Sd(𝔽p)1){0}(S_{d}({\mathbb{F}}_{p})-1)\setminus\{0\} cannot be decomposed as the product of two sets non-trivially. This confirms the truth of Sárközy’s conjecture (1.8) as well as the truth of its generalization in the generic case (1.9) when the shift of the subgroup is given by λ=1\lambda=1.

Theorem 6.1.

Let d2d\geq 2 be fixed. As xx\to\infty, the number of primes pxp\leq x such that p1(modd)p\equiv 1\pmod{d} and (Sd(𝔽p)1){0}(S_{d}({\mathbb{F}}_{p})-1)\setminus\{0\} can be decomposed as the product of two sets non-trivially (that is, it can be written as the product of two subsets of 𝔽p{\mathbb{F}}_{p}^{*} with size at least 2) is o(π(x))o(\pi(x)).

Proof.

Let 𝒫d\mathcal{P}_{d} be the set of primes pp such that p1(modd)p\equiv 1\pmod{d} and (Sd(𝔽p)1){0}(S_{d}({\mathbb{F}}_{p})-1)\setminus\{0\} admits a non-trivial multiplicative decomposition. By the prime number theorem for arithmetic progressions, it suffices to show that |𝒫d[0,x]|=o(x/logx)|\mathcal{P}_{d}\cap[0,x]|=o(x/\log x).

Let p𝒫dp\in\mathcal{P}_{d}. Then we can write (Sd1){0}(S_{d}-1)\setminus\{0\} as the product of two sets A,B𝔽pA,B\subset{\mathbb{F}}_{p}^{*} such that |A|,|B|2|A|,|B|\geq 2. Then Corollary 1.10 implies that |A||B|=p1d1,|A||B|=\frac{p-1}{d}-1, that is,

d|A||B|=p(d+1).\displaystyle d|A||B|=p-(d+1). (6.1)

On the other hand, Proposition 3.4 implies that we can find an absolute constant Cd(0,1)C_{d}\in(0,1) such that

Cdp<min{|A|,|B|}<p.C_{d}\sqrt{p}<\min\{|A|,|B|\}<\sqrt{p}.

It follows that p(d+1)p-(d+1) has a divisor in the interval (Cdp,p)(C_{d}\sqrt{p},\sqrt{p}). To summarize, if p𝒫dp\in\mathcal{P}_{d}, then we have τ(p(d+1);Cdp,p)1,\tau(p-(d+1);C_{d}\sqrt{p},\sqrt{p})\geq 1, where τ(n;y,z)\tau(n;y,z) denotes the number of divisors of nn in the interval (y,z](y,z]. Now, we use results by Ford [14, Theorem 6] on the distribution of shift primes with a divisor in a given interval. Denote

H(x,y,z)\displaystyle H(x,y,z) =#{1nx:τ(n;y,z)1};\displaystyle=\#\{1\leq n\leq x:\tau(n;y,z)\geq 1\}; (6.2)
Pd(x,y,z)\displaystyle P_{d}(x,y,z) =#{px:τ(p(d+1);y,z)1}.\displaystyle=\#\{p\leq x:\tau(p-(d+1);y,z)\geq 1\}. (6.3)

Setting y=Cdx/2y=C_{d}\sqrt{x}/2 and z=xz=\sqrt{x}, [14, Theorem 6] and [14, Theorem 1, third case of (v)] imply that

Pd(x,y,z)H(x,y,z)logxxlogxuδ(log2u)3/2P_{d}(x,y,z)\ll\frac{H(x,y,z)}{\log x}\ll\frac{x}{\log x}u^{\delta}\bigg{(}\log\frac{2}{u}\bigg{)}^{-3/2}

where δ=11+loglog2log2\delta=1-\frac{1+\log\log 2}{\log 2} and u=log(Cd/2)/logyu=\log(C_{d}/2)/\log y. It follows that as xx\to\infty, we have Pd(x,y,z)=o(x/logx)P_{d}(x,y,z)=o(x/\log x). Therefore, we have

#{p𝒫d:x/2px}#{x/2px:τ(p(d+1);Cdx/2,x)1}=o(x/logx).\#\{p\in\mathcal{P}_{d}:x/2\leq p\leq x\}\leq\#\{x/2\leq p\leq x:\tau(p-(d+1);C_{d}\sqrt{x}/2,\sqrt{x})\geq 1\}=o(x/\log x).

We conclude that as xx\to\infty,

|𝒫d[0,x]|\displaystyle|\mathcal{P}_{d}\cap[0,x]| =O(x)+#{p𝒫d:xpx}\displaystyle=O(\sqrt{x})+\#\{p\in\mathcal{P}_{d}:\sqrt{x}\leq p\leq x\}
=O(x)+0j(log2x)/2o(x/2jlog(x/2j))\displaystyle=O(\sqrt{x})+\sum_{0\leq j\leq(\log_{2}x)/2}o\bigg{(}\frac{x/2^{j}}{\log(x/2^{j})}\bigg{)}
=O(x)+(0j(log2x)/212j)o(xlogx)=o(xlogx).\displaystyle=O(\sqrt{x})+\bigg{(}\sum_{0\leq j\leq(\log_{2}x)/2}\frac{1}{2^{j}}\bigg{)}o\bigg{(}\frac{x}{\log x}\bigg{)}=o\bigg{(}\frac{x}{\log x}\bigg{)}.

Using a similar argument, we can prove Theorem 1.11:

Proof of Theorem 1.11.

Consider the family of primes pp such that p1(modd)p\equiv 1\pmod{d} and nn is a dd-th power modulo pp. By a standard application of the Chebotarev density theorem, the density of such primes is given by 1[(e2πi/d,n1/d):]\frac{1}{[\mathbb{Q}(e^{2\pi i/d},n^{1/d}):\mathbb{Q}]}. Among the family of such primes pp, we can repeat the same argument as in the proof of Theorem 6.1 to show that if (Sd(𝔽p)n){0}(S_{d}({\mathbb{F}}_{p})-n)\setminus\{0\} admits a non-trivial multiplicative decomposition, then p(d+1)p-(d+1) necessarily has a divisor which is “close to” p\sqrt{p}. We remark that it is important to assume that nn is a dd-th power modulo pp, so that we can take advantage of Corollary 1.10. Similar to the proof of Theorem 6.1, we can show that among the family of primes p1(modd)p\equiv 1\pmod{d}, the property that p(d+1)p-(d+1) has a divisor with the desired magnitude fails to hold for almost all pp. This finishes the proof of the theorem. ∎

Remark 6.2.

When nn is a fixed negative integer, one can obtain a similar result to Theorem 1.11 following the idea of the above proof.

Remark 6.3.

Theorem 1.11 essentially states if dd is fixed, p1(modd)p\equiv 1\pmod{d} is a prime, and λSd(𝔽p)\lambda\in S_{d}({\mathbb{F}}_{p}), then it is very unlikely that we can decompose (Sd(𝔽p)λ){0}(S_{d}({\mathbb{F}}_{p})-\lambda)\setminus\{0\} as the product of two subsets of 𝔽p{\mathbb{F}}_{p}^{*} non-trivially. On the other hand, when λSd(𝔽p)\lambda\notin S_{d}({\mathbb{F}}_{p}), the above technique does not apply. Nevertheless, when λSd\lambda\notin S_{d} and we have two sets A,B𝔽pA,B\subset{\mathbb{F}}_{p}^{*} such that AB=(Sdλ){0}=SdλAB=(S_{d}-\lambda)\setminus\{0\}=S_{d}-\lambda, Theorem 1.1 implies that

|Sd||A||B||Sd|+min{|A|,|B|}1.|S_{d}|\leq|A||B|\leq|S_{d}|+\min\{|A|,|B|\}-1.

In particular, we get the following non-trivial fact: if |A||A| is fixed, then |B||B| is also uniquely fixed.

6.2. Applications to special multiplicative decompositions

In this subsection, we verify the ternary version of 1.9 in a strong sense, which generalizes [31, Theorem 2].

Shkredov [33, Theorem 3] showed if GG is a multiplicative subgroup of 𝔽p{\mathbb{F}}_{p} with 1ϵ|G|p6/7ϵ1\ll_{\epsilon}|G|\leq p^{6/7-\epsilon}, then there is no A𝔽pA\subset{\mathbb{F}}_{p} and ξ𝔽p\xi\in{\mathbb{F}}_{p}^{*} such that A/A=ξG+1A/A=\xi G+1. In fact, due to the analytic nature of the proof, he pointed out that his proof can be slightly modified to show something stronger, namely A/A(ξG+1)CA/A\neq(\xi G+1)\cup C, as long as CC is small(see also [33, Remark 15]). The following corollary of Theorem 1.1 is of a similar flavor.

Corollary 6.4.

Let pp be a prime. If GG is a proper multiplicative subgroup of 𝔽p{\mathbb{F}}_{p} with |G|8|G|\geq 8, and λ,ξ𝔽p\lambda,\xi\in{\mathbb{F}}_{p}^{*}, then there is no A𝔽pA\subset{\mathbb{F}}_{p}^{*} such that AA=(ξGλ){0}AA=(\xi G-\lambda)\setminus\{0\}.

Proof.

We assume, otherwise, that AA=(ξGλ){0}AA=(\xi G-\lambda)\setminus\{0\} for some A𝔽pA\subset\mathbb{F}_{p}^{*}. Then we observe that aa=aaaa^{\prime}=a^{\prime}a for each a,aAa,a^{\prime}\in A, it follows that

|G|1|AA||A|2+|A|2.|G|-1\leq|AA|\leq\frac{|A|^{2}+|A|}{2}.

Since |G|8|G|\geq 8, it follows that |A|4|A|\geq 4. Let B=A/ξB=A/\xi, and λ=λ/ξ\lambda^{\prime}=\lambda/\xi. Then we have AB=(Gλ){0}AB=(G-\lambda^{\prime})\setminus\{0\} and thus Theorem 1.1 implies that |A|2=|A||B||G|+|A|1|A|^{2}=|A||B|\leq|G|+|A|-1. Comparing the above two inequalities, we obtain that

|A|2|A||G|1|A|2+|A|2,|A|^{2}-|A|\leq|G|-1\leq\frac{|A|^{2}+|A|}{2},

which implies that |A|3|A|\leq 3, contradicting the assumption that |A|4|A|\geq 4. ∎

Lemma 6.5.

Let A,B,CA,B,C be nonempty subsets of 𝔽q{\mathbb{F}}_{q} and let λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}. Then |ABC+λ|2|AB+λ||BC+λ||CA+λ||ABC+\lambda|^{2}\leq|AB+\lambda||BC+\lambda||CA+\lambda|.

Proof.

It suffices to show |ABC|2|AB||BC||CA||ABC|^{2}\leq|AB||BC||CA|, which is a special case of [29, Theorem 5.1] due to Ruzsa. ∎

The following two theorems generalize Sárközy [31, Theorem 2] and confirm the ternary version of 1.9 in a strong form.

Theorem 6.6.

There exists an absolute constant M>0M>0, such that whenever pp is a prime, GG is a proper multiplicative subgroup of 𝔽p{\mathbb{F}}_{p} with |G|>M|G|>M, and λ𝔽p\lambda\in{\mathbb{F}}_{p}^{*}, there is no ternary multiplicative decomposition ABC=(Gλ){0}ABC=(G-\lambda)\setminus\{0\} with A,B,C𝔽pA,B,C\subset{\mathbb{F}}_{p}^{*} and |A|,|B|,|C|2|A|,|B|,|C|\geq 2.

Proof.

Assume that there are sets A,B,C𝔽pA,B,C\subset{\mathbb{F}}_{p}^{*} with |A|,|B|,|C|2|A|,|B|,|C|\geq 2, such that ABC=(Gλ){0}ABC=(G-\lambda)\setminus\{0\} for some proper multiplicative subgroup GG of 𝔽p{\mathbb{F}}_{p} and some λ𝔽p\lambda\in{\mathbb{F}}_{p}^{*}.

Then we can write (Gλ){0}(G-\lambda)\setminus\{0\} in three different ways: A(BC),B(CA),C(AB)A(BC),B(CA),C(AB), so that we can apply the results in previous sections to each of them. Note that Lemma 3.6 implies that

|A|,|B|,|C||G|1/2+o(1).|A|,|B|,|C|\geq|G|^{1/2+o(1)}.

On the other hand, Theorem 1.1 implies that

|AB||C|,|BC||A|,|CA||B||G|.|AB||C|,|BC||A|,|CA||B|\ll|G|.

Therefore, from Lemma 6.5 and the fact |ABC|{|G|,|G|1}|ABC|\in\{|G|,|G|-1\}, we have

|G|2|A||B||C||ABC|2|A||B||C|(|AB||C|)(|BC||A|)(|CA||B|)|G|3.|G|^{2}|A||B||C|\ll|ABC|^{2}|A||B||C|\ll(|AB||C|)(|BC||A|)(|CA||B|)\ll|G|^{3}.

It follows that

|G|3/2+o(1)|A||B||C||G|,|G|^{3/2+o(1)}\ll|A||B||C|\ll|G|,

that is, |G|1|G|\ll 1, where the implicit constant is absolute. This completes the proof of the theorem. ∎

Theorem 6.7.

Let ϵ>0\epsilon>0. There is a constant Q=Q(ϵ)Q=Q(\epsilon), such that for each prime power q>Qq>Q and a divisor dd of q1q-1 with 2dq1/10ϵ2\leq d\leq q^{1/10-\epsilon}, there is no ternary multiplicative decomposition ABC=(Sd(𝔽q)λ){0}ABC=(S_{d}({\mathbb{F}}_{q})-\lambda)\setminus\{0\} with A,B,C𝔽qA,B,C\subset{\mathbb{F}}_{q}^{*}, |A|,|B|,|C|2|A|,|B|,|C|\geq 2, and λ𝔽q\lambda\in{\mathbb{F}}_{q}^{*}.

Proof.

The proof is similar to the proof of Theorem 6.6. While Lemma 3.6 does not hold in the new setting (see Remark 3.8), we can instead use Proposition 3.4. If ABC=(Sd(𝔽q)λ){0}ABC=(S_{d}({\mathbb{F}}_{q})-\lambda)\setminus\{0\}, then Proposition 3.4 implies that

|A|,|B|,|C|qd,|A||BC|,|BC||A|,|CA||B|q.|A|,|B|,|C|\gg\frac{\sqrt{q}}{d},\quad|A||BC|,|BC||A|,|CA||B|\ll q.

A similar computation leads to dq1/10d\gg q^{1/10}, which implies that qϵ1q\ll_{\epsilon}1 since we assume that dq1/10ϵd\leq q^{1/10-\epsilon}. ∎

Acknowledgements

The authors thank Andrej Dujella, Greg Martin, and József Solymosi for helpful discussions. The third author was supported by the KIAS Individual Grant (CG082701) at the Korea Institute for Advanced Study and the Institute for Basic Science (IBS-R029-C1).

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Appendix A Algorithm and Computations

We continue our discussion from the introduction on the following constant

γk=lim supnMk(n)logn.\gamma_{k}=\limsup_{n\to\infty}\frac{M_{k}(n)}{\log n}.

It is implicit in [7] that γk3ϕ(k)\gamma_{k}\leq 3\phi(k). We also write νk=2kk2ηkϕ(k).\nu_{k}=\frac{2k}{k-2}\eta_{k}\phi(k).

Our main result, Theorem 1.2, implies that γkνk\gamma_{k}\leq\nu_{k}. In particular, in view of Remark 5.1, it follows that γk6\gamma_{k}\leq 6 for all k2k\geq 2 and γk2+o(1)\gamma_{k}\leq 2+o(1) when kk\to\infty. In Figure A.1, we pictorially compare our new bound νk\nu_{k} with the bound 3ϕ(k)3\phi(k) when 2k10002\leq k\leq 1000.

Refer to caption
Figure A.1. Comparison between the new bound νk\nu_{k} and the bound 3ϕ(k)3\phi(k) in [7] when 2k10002\leq k\leq 1000. The black dots denote 3ϕ(k)3\phi(k), and the blue dots denote νk\nu_{k}.

Recall that for k2k\geq 2, we defined the constant ηk=min||/T2,\eta_{k}=\underset{\mathcal{I}}{\min}|\mathcal{I}|/T_{\mathcal{I}}^{2}, where the minimum is taken over all nonempty subsets \mathcal{I} of {1ik:gcd(i,k)=1,gcd(i1,k)>1},\{1\leq i\leq k:\gcd(i,k)=1,\gcd(i-1,k)>1\}, and T=igcd(i1,k).T_{\mathcal{I}}=\sum_{i\in\mathcal{I}}\sqrt{\gcd(i-1,k)}.

To compute ηk\eta_{k}, we use the following simple greedy algorithm with running time O(klogk)O(k\log k). The observation is as follows. If |||\mathcal{I}| is fixed, our goal is to minimize ||/T2|\mathcal{I}|/T_{\mathcal{I}}^{2}. Thus, we should choose those residue classes i(modk)i\pmod{k} with gcd(i1,k)\gcd(i-1,k) as large as possible to maximize TT_{\mathcal{I}}. Then, we can sort these gcds in decreasing order, and when |||\mathcal{I}| is fixed, we pick those residue classes corresponding to the largest |||\mathcal{I}| gcds. The following is a precise description of the algorithm:

Algorithm A.1.

Let k2k\geq 2. We follow the notations defined in Subsection 5.2.

  1. Step 1.

    Let 𝒜={1ik:gcd(i,k)=1,gcd(i1,k)>1}\mathcal{A}=\{1\leq i\leq k:\gcd(i,k)=1,\gcd(i-1,k)>1\}. We list the elements of 𝒜\mathcal{A} by {a1,a2,}\{a_{1},a_{2},\ldots\} such that gcd(aj1,k)\gcd(a_{j}-1,k) is decreasing by using a sorting algorithm.

  2. Step 2.

    Set Ir={a1,,ar}I_{r}=\{a_{1},\ldots,a_{r}\}, TIr=iIrgcd(i1,k)T_{I_{r}}=\sum_{i\in I_{r}}\sqrt{\gcd(i-1,k)}, and ξIr=|Ir|/TIr2\xi_{I_{r}}=|I_{r}|/T_{I_{r}}^{2}.

  3. Step 3.

    Return ηk=minrξIr\eta_{k}=\min_{r}\xi_{I_{r}} and terminate the algorithm.

Note that the running time of the above algorithm is O(klogk)O(k\log k): sorting takes O(klogk)O(k\log k) time, while other steps take linear time.

Next, we also consider the minimum value mkm_{k} of the upper bounds {νi:2ik}\{\nu_{i}\colon 2\leq i\leq k\} for each k2k\geq 2. Table A.1 shows the values of mkm_{k} for 2k1,200,0002\leq k\leq 1{,}200{,}000 when they are changed.

kk mkm_{k}
2 2.6071
4 1.37258
6 0.80385
8 0.72776
12 0.44134
24 0.31910
36 0.29027
48 0.25836
60 0.21636
120 0.16570
kk mkm_{k}
180 0.15191
240 0.13876
360 0.11708
720 0.09693
840 0.09266
1260 0.08465
1440 0.08445
1680 0.07624
2520 0.06465
5040 0.05317
kk mkm_{k}
7560 0.05171
10080 0.04592
15120 0.04252
20160 0.04111
25200 0.03887
27720 0.03665
30240 0.03647
50400 0.03343
55440 0.02997
83160 0.02877
kk mkm_{k}
110880 0.02574
166320 0.02343
221760 0.02280
277200 0.02138
332640 0.02008
498960 0.01985
554400 0.01827
665280 0.01774
720720 0.01654
1081080 0.01587
Table A.1. The minimum mkm_{k} of the upper bounds {νi:1ik}\{\nu_{i}\colon 1\leq i\leq k\} for 2k1,200,0002\leq k\leq 1{,}200{,}000.

We also report our computations on νk\nu_{k} for 2k2012\leq k\leq 201 in the following table.

kk νk\nu_{k}
2 2.6071
3 4.0000
4 1.3726
5 2.6667
6 0.8038
7 2.4000
8 0.7278
9 1.1077
10 0.7295
11 2.2222
12 0.4413
13 2.1818
14 0.7185
15 0.7222
16 0.5383
17 2.1333
18 0.4522
19 2.1176
20 0.4450
21 0.7355
22 0.7251
23 2.0952
24 0.3191
25 1.1180
26 0.7313
27 0.7508
28 0.4552
29 2.0741
30 0.3351
31 2.0690
32 0.4555
33 0.7709
34 0.7438
35 0.7311
36 0.2903
37 2.0571
38 0.7497
39 0.7873
40 0.3353
41 2.0513
kk νk\nu_{k}
42 0.3465
43 2.0488
44 0.4739
45 0.4581
46 0.7604
47 2.0444
48 0.2584
49 1.1716
50 0.4974
51 0.8163
52 0.4818
53 2.0392
54 0.3596
55 0.7678
56 0.3486
57 0.8290
58 0.7742
59 2.0351
60 0.2164
61 2.0339
62 0.7783
63 0.4746
64 0.4124
65 0.7860
66 0.3643
67 2.0308
68 0.4950
69 0.8515
70 0.3548
71 2.0290
72 0.2171
73 2.0282
74 0.7892
75 0.5005
76 0.5006
77 0.7644
78 0.3713
79 2.0260
80 0.2730
81 0.6359
kk νk\nu_{k}
82 0.7956
83 2.0247
84 0.2263
85 0.8195
86 0.7985
87 0.8796
88 0.3682
89 2.0230
90 0.2239
91 0.7794
92 0.5103
93 0.8877
94 0.8040
95 0.8346
96 0.2261
97 2.0211
98 0.5376
99 0.5038
100 0.3344
101 2.0202
102 0.3827
103 2.0198
104 0.3757
105 0.3626
106 0.8114
107 2.0190
108 0.2355
109 2.0187
110 0.3768
111 0.9094
112 0.2864
113 2.0180
114 0.3874
115 0.8621
116 0.5220
117 0.5160
118 0.8179
119 0.8087
120 0.1657
121 1.2615
kk νk\nu_{k}
122 0.8200
123 0.9220
124 0.5254
125 0.8820
126 0.2335
127 2.0160
128 0.3877
129 0.9278
130 0.3869
131 2.0155
132 0.2413
133 0.8226
134 0.8256
135 0.3709
136 0.3878
137 2.0148
138 0.3955
139 2.0146
140 0.2400
141 0.9387
142 0.8291
143 0.7812
144 0.1809
145 0.8976
146 0.8307
147 0.5506
148 0.5342
149 2.0136
150 0.2468
151 2.0134
152 0.3928
153 0.5366
154 0.3772
155 0.9081
156 0.2471
157 2.0129
158 0.8353
159 0.9532
160 0.2385
161 0.8485
kk νk\nu_{k}
162 0.3140
163 2.0124
164 0.5392
165 0.3857
166 0.8382
167 2.0121
168 0.1737
169 1.2965
170 0.4049
171 0.5453
172 0.5416
173 2.0117
174 0.4053
175 0.5104
176 0.3077
177 0.9660
178 0.8422
179 2.0113
180 0.1519
181 2.0112
182 0.3854
183 0.9700
184 0.4014
185 0.9365
186 0.4080
187 0.8001
188 0.5459
189 0.3843
190 0.4129
191 2.0106
192 0.2075
193 2.0105
194 0.8471
195 0.3948
196 0.3652
197 2.0103
198 0.2493
199 2.0102
200 0.2517
201 0.9811
Table A.2. The upper bound νk\nu_{k} of γk\gamma_{k} when 2k2012\leq k\leq 201