On sparsity of representations of polynomials as linear combinations of exponential functions
Dragos Ghioca, Alina Ostafe, Sina Saleh, Igor E. Shparlinski
aa r X i v : . [ m a t h . N T ] F e b ON SPARSITY OF REPRESENTATIONS OFPOLYNOMIALS AS LINEAR COMBINATIONS OFEXPONENTIAL FUNCTIONS
DRAGOS GHIOCA, ALINA OSTAFE, SINA SALEH,AND IGOR E. SHPARLINSKI
Abstract.
Given an integer g and also some given integers m (sufficiently large) and c , . . . , c m , we show that the number of allnon-negative integers n M with the property that there existnon-negative integers k , . . . , k m such that n = m X i =1 c i g k i is o (cid:16) (log M ) m − / (cid:17) . We also obtain a similar bound when dealingwith more general inequalities (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Q ( n ) − m X i =1 c i λ k i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) B, where Q ∈ C [ X ] and also λ ∈ C (while B is a real number). Introduction
Set-up.
Motivated by applications to the dynamical Mordell-Langconjecture (for more details on this open problem in arithmetic dynam-ics, we refer the reader to [BGT16]), the authors [GOSS21] have re-cently considered the question about representations of values of poly-nomials Q ∈ Q [ X ] as fixed linear combinations of powers of a prime p . In particular, it is shown in [GOSS21] that for fixed coefficients c , . . . , c m ∈ Q and integral exponents a , . . . , a m the number of posi-tive integers n N for which Q ( n ) can be represented as Q ( n ) = m X i =1 c i p a i k i with some k , . . . , k m ∈ Z is bounded by O ((1 + log N ) m ) where theimplied constant depends only on the initial data. In fact it is easy Mathematics Subject Classification.
Key words and phrases.
Squares, to see that for Q ( n ) = n , this bound is tight. Furthermore, a similarresult is given in [GOSS21] for representations of the form(1.1) Q ( n ) = m X i =1 s X j =1 c i,j λ a i k i j , with algebraic integers λ , . . . , λ s , each one of them of absolute valueequal to q or √ q (where q is a given power of a prime number).Here we first consider representations of the form (1.1) with s = 1but arbitrary complex (rather than algebraic) parameters. We alsogeneralise this to approximations of polynomials rather than preciseequalities, that is, we consider inequalities of the form(1.2) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Q ( n ) − m X i =1 c i λ k i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) B, with Q ∈ C [ X ], c , . . . , c m , λ ∈ C and some B ∈ R .1.2. Notation.
We now recall that the notations A = O ( B ), A ≪ B and B > A are all equivalent to the inequality | A | cB with someconstant c . Throughout this work all implied constants may depend onthe polynomial Q and the parameters c i , i = 1 , . . . , m and λ in (1.2)and also on g in (1.4) below.For a finite set S we use S to denote its cardinality.1.3. New results.
We remark that the argument of [GOSS21] is basedon a result of Laurent [Lau84, Th´eor`eme 6], which required all parame-ters to be defined over a number field; furthermore, the result of [Lau84]refers to equalities, not inequalities. Hence here we use a different ap-proach to establish the following result.
Theorem 1.1.
Let c , . . . , c m , λ ∈ C , B ∈ R and let Q ∈ C [ X ] be anon-constant polynomial. Then for N > we have { n N : (1.2) holds for some k , . . . , k m ∈ Z } ≪ (log N ) m . We observe that the implied constant in Theorem 1.1 is effectivelycomputable in terms of the sizes of the initial data, while in the resultof [GOSS21] it is not.We also note (see Example 2.1) that if one considers inequalities ofthe form(1.3) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Q ( n ) − m X i =1 s X j =1 c i,j λ a i k i j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) B, for some arbitrary complex numbers λ j , then one cannot expect asimilar result as in Theorem 1.1. More precisely, there exists λ ∈ C PARSE REPRESENTATIONS OF SQUARES 3 such that for any ε > each sufficiently large integer n , thereexists some positive integer k n with the property that (cid:12)(cid:12)(cid:12)(cid:12) n − iπ · (cid:0) k n − λ k n (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ε, see Example 2.1 for more details.Furthermore, we consider the case of perfect squares and study rela-tions of the form(1.4) n = m X i =1 c i g k i , with non-zero integer coefficients c , . . . , c m and an integer basis g >
2. Using the square-sieve of Heath-Brown [H-B84] we improve theexponent m of log N .For m = 2, we write the equation (1.4) as n = g k (cid:0) c + c g k − k (cid:1) (with k > k ). Hence either c + c g k − k or c g + c g k − k +1 is aperfect square j for some j n N . Since the largest prime divisorof j + c for any c = 0 tends to infinity with j , see [Kea69], we see that k − k can take only finitely many values. Hence for m = 2 we have O (log N ) solution to (1.4) with n N . This bound is obviously thebest possible as the example of the numbers 2 k + 2 k , with k = k + 3and even k , shows. We also note that in [CGSZ, Theorem 5.1 (B)], itis established even more generally the precise set of all positive integers n for which u n is of the form c g k + c g k (for some given g , c , c ),where { u n } n > is an arbitrary linear recurrence sequence (the resultof [CGSZ, Theorem 5.1 (B)] is stated only when g = p is a primenumber, but as remarked in [CGSZ, Section 5], the method extendsverbatim to an arbitrary integer g ).So we are mostly interested in the case of m >
3; furthermore, wenote that for m = 3 (and in some cases, depending on g and the c i ,even for m = 4), more precise results are available in the literature(see [CoZa13]). However, when m >
5, it is very difficult to find aprecise description of all n ∈ N such that n is of the form (1.4) (forsome given integers g and c i ). Theorem 1.2.
Let m > and let c , . . . , c m and g > be integers.Then for N > we have { n N : (1.4) holds for some k , . . . , k m ∈ Z } (log N ) m − γ m + o (1) , where γ = 6771969 and γ m = 677 m m + 1354 for m > . D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI
We observe that γ m → / . . . . as m → ∞ . Thus forlarge m Theorem 1.2 saves more than 1 / γ m > / m > γ m .On the other hand defining s as the largest integer with s ( s + 1) / m and considering numbers (cid:0) g h + . . . + g h s (cid:1) with h i log ( N/s )2 log g , i = 1 , . . . , s, we see that for at least one choice c , . . . , c t > t m and c + . . . + c t s ( s + 1) / C (log N ) s times, for someconstant C >
0, which shows that the best possible exponent in anyresult of the type of Theorem 1.2 must grow with m (at least as about √ m for large m ).Note that cycling over all g m choices of( c , . . . , c m ) ∈ { , . . . , g − } m we obtain from Theorem 1.2 a result about the sparsity of the valuesof n for which n has at most m non-zero digits to base g . Variousfiniteness results on sparse digital representations of perfect powerscan be found in [BeBu14, BBM13, CoZa13, Mos21]. Note that as wehave just seen, in our setting of arbitrary m no finiteness result ispossible, and hence we can use Theorem 1.2 to provide a counting resultrelated to such representations. More precisely we have the followingstraightforward consequence: Corollary 1.3.
Let m > and let K > and g > be integers. Thenthere are at most K m − γ m + o (1) integer squares with g -ary expansion oflength K and with at most m non-zero digits. Proof of Theorem 1.1
Counterexample to a possible extension to (1.3).
Beforeproceeding to the proof of Theorem 1.1, we provide the Example 2.1(mentioned in Section 1.3), which shows that one cannot expect to gen-eralise Theorem 1.1 to (1.3), that is, to the case when we approximate Q ( n ) with a sum of powers of different λ j . PARSE REPRESENTATIONS OF SQUARES 5
Example 2.1.
We consider the sequence of positive integers { b j } j > given by b = 2 and b j +1 = 2 b j + b j + 1 for j > . We let λ = 2 · e παi , where α = ∞ X j =2 j − b j . We let n be a positive integer and show that (2.1) n − iπ · (cid:16) bn − λ bn (cid:17) ≪ nb n +1 . Indeed, we first notice that λ bn = 2 bn · e πt n i , where t n = ∞ X j = n +1 j − b j − b n . Then bn − λ bn = 2 bn · ((1 − cos(2 πt n )) − i sin (2 πt n ))= 2 bn · πt n ) · (sin( πt n ) − i cos( πt n )) , and so, (2.2) iπ · (cid:16) bn − λ bn (cid:17) = 2 bn sin( πt n ) π · e πt n i . Now, by the definition of the rapidly increasing sequence { b j } j > , wehave that (2.3) 0 < t n − n b n +1 − b n < bn +1 ; also, clearly, t n → as n → ∞ . Furthermore, we know that when t is close to , then (2.4) | sin t − t | t . So, using the inequalities (2.3) and (2.4), along with the fact that b n +1 = 2 b n + b n + 1 , we get that (2.5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − bn sin ( πt n ) π (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < − b n ≪ /b n +1 D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI for all n sufficiently large. Also, for n large, using the inequality (2.3)we have that (2.6) (cid:12)(cid:12) − e πt n i (cid:12)(cid:12) < − b n ≪ /b n +1 . Recalling (2.2) and using (2.5) and (2.6), we derive (2.1).
Therefore, the conclusion of Theorem 1.1 cannot be generalised toinequalities (1.3) where we approximate a polynomial Q ( n ) with sumsof powers of different λ , . . . , λ s .Next we proceed to proving Theorem 1.1.2.2. Preliminaries.
We first note that if | λ | = 1, then the inequal-ity (1.2) yields that | Q ( n ) | is uniformly bounded above and therefore,we can only have finitely many n ∈ N satisfying such inequality since Q is a non-constant polynomial. Furthermore, since the exponents k i appearing in the inequality (1.2) are arbitrary integers, then withoutloss of generality, we may assume from now on that | λ | > k i , i = 1 , . . . , m , from (1.2) werenon-positive, then the absolute value of the corresponding terms c i λ k i is uniformly bounded above. So, at the expense of replacing B bya larger constant (but depending only on the absolute values of the c i ), we may assume from now on, that each exponent k i from (1.2) ispositive.Let Q ( X ) = a d X d + · · · + a X + a for complex numbers a , . . . , a d with a d = 0. There exists N > d and theabsolute values of the coefficients of Q ) such that(2.7) | Q ( n ) | (cid:12)(cid:12) a d n d (cid:12)(cid:12) N · n d for each n > N . Furthermore, at the expense of replacing N by a larger positive integer(but still depending only on d , the absolute values of the coefficientsof Q and also depending on B in this case), we may also assume that(2.8) | Q ( n + n ) − Q ( n ) | > B for each n , n > N . Induction.
We proceed to prove our desired result by inductionon m .We prove first the base case m = 1, which also constitutes the in-spiration for our proof for the general case in Theorem 1.1. So, wehave that | Q ( n ) | = O ( N d ) for each 0 n N (see also the inequal-ity (2.7)). Therefore, for n N satisfying the inequality(2.9) (cid:12)(cid:12) Q ( n ) − c λ k (cid:12)(cid:12) B one has | λ | k = O (cid:0) N d (cid:1) and thus, k = O (log N ) . On the other hand,for a given k , the inequality (2.9) is satisfied by O (1) non-negative PARSE REPRESENTATIONS OF SQUARES 7 integers n (see also (2.8)), thus proving the desired bound in the case m = 1.So, suppose that the result is true for m s and we prove thatTheorem 1.1 holds when m = s + 1; clearly we may assume each c i for i = 1 , . . . , s + 1 is nonzero. Since there are m powers of λ in theinequality (1.2), then in order to prove Theorem 1.1, it suffices to provethat the set S , consisting of all n ∈ N for which there exist integers(2.10) 1 k k · · · k s +1 such that(2.11) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Q ( n ) − s +1 X j =1 c j λ k j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) B satisfies(2.12) { n ∈ S : n N } ≪ (log N ) s +1 . Let ∆ ∈ N be sufficiently large (but depending only on | λ | , whichis larger than 1, and also depending on the absolute values of the c , . . . , c s +1 ) such that we have(2.13) | c s +1 | · | λ | k s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s +1 X j =1 c j λ k j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , for all integers k s +1 > · · · > k > k s +1 − k s > ∆.Now, we let U be the subset of S consisting of integers n ∈ N for which one can find integers k j satisfying (2.11) and in addition, k s +1 − k s < ∆. Then the existence of such a solution tuple ( k , . . . , k s +1 )for each n ∈ U means that (cid:12)(cid:12) Q ( n ) − (cid:0) c λ k + · · · + c s − λ k s − + (cid:0) c s + c s +1 λ k s +1 − k s (cid:1) λ k s (cid:1)(cid:12)(cid:12) B. Because k s +1 − k s ∈ { , . . . , ∆ − } , applying the induction hypothesisfor each of the possible ∆ values of k s +1 − k s , we obtain the desiredconclusion regarding the asymptotic growth given by (2.12) (further-more, we actually get that the exponent from the right-hand side ofthe inequality (2.12) is s = m − s + 1 = m ).On the other hand, for each n ∈ S \ U satisfying n > N , we knowthat there must exist some tuple of nonnegative integers ( k , . . . , k s +1 )satisfying (2.11) and in addition, k s +1 − k s > ∆. Then using both (2.7)and (2.13), we get | c s +1 | · | λ | k s +1 − B (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c + s +1 X j =1 c j λ k j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − B | Q ( n ) | | a d | · n d , D. GHIOCA, A. OSTAFE, S. SALEH, AND I. E. SHPARLINSKI which implies that(2.14) k s +1 b (1 + log n )for some positive real number b depending only on B , | λ | , d , | a d | and | c s +1 | .So, let N be an integer larger than N ; then for each integer N n N contained in S \ U , we know there exists an ( s + 1)-tupleof integers k i satisfying (2.10) and (2.11). Combining the fact that1 k i k s +1 with the inequality (2.14), we get that there are at most( b (1 + log N )) s +1 tuples ( k , . . . , k s +1 ) for which we could find some n ∈ S \ U satisfying the inequality N n N . However, since n > N , then the inequality (2.8) yields that for any such ( s + 1)-tupleof integers k i , there are at most N integers n ∈ ( S \ U ) ∩ [ N , N ]satisfying (2.11) with respect to the tuple ( k , . . . , k s +1 ). Hence, weget the inequality { n N : n ∈ S \ U } N · (cid:0) b · (1 + log N )) s +1 (cid:1) . for each positive integer N > N . This concludes our proof of Theo-rem 1.1.3. Construction and properties of the sieving set ofprimes
Multiplicative orders.
Let τ ℓ ( g ) denote the multiplicative orderof an integer g > ℓ , that is, the smallest positiveinteger τ for which g τ ≡ ℓ .Let α be a fixed real number such that(3.1) { ℓ z : ℓ is prime and P ( ℓ − > ℓ α } ≫ z log z for all sufficiently large z , where P ( k ) denotes the largest prime divisorof an integer k >
2, and the implied constant depends only on α .We recall the following well known result which follows from thedivisibility τ ℓ ( g ) | ℓ − g, ℓ ) = 1) and the bound (cid:8) ℓ z : ℓ is prime, P ( ℓ − > ℓ / (cid:9) = (1 + o (1)) z log z as z → ∞ , which easily follows from a stronger result of Erd˝os andMurty [ErMu96, Theorem 3]. Details can be found in the work ofKurlberg and Pomerance [KuPo05, Lemma 20]. Lemma 3.1.
For any fixed α > / satisfying (3.1) and any fixedinteger g > we have { ℓ z : ℓ is prime, τ ℓ ( g ) > ℓ α } ≫ z log z PARSE REPRESENTATIONS OF SQUARES 9 as z → ∞ . For an integer s > ν ( s ) the 2-adic order of s , thatis, the largest power ν such that 2 ν | s . Lemma 3.2.
For any fixed α > / satisfying (3.1) and any fixedinteger g > there are some absolute constants C , C > , such thatfor every sufficiently large real number z > , there exist some integer u and a set L z ⊆ [ z, C z ] of primes of cardinality L z > C z log z log log z such that for every ℓ ∈ L z we have P ( ℓ − > z α , P ( ℓ − | τ ℓ ( g ) , ν ( τ ℓ ( g )) = u . Proof.
Lemma 3.1 obviously implies that for some absolute constants C , C there are at least C z/ log z primes ℓ ∈ [ z, C z ] satisfying onlythe first two conditions, see also [LuSh09, Lemma 5.1]. Let L z be thisset. Trivially, there are at at most z/ v primes ℓ with ν ( τ ℓ ( g )) > v .Hence taking a sufficiently large C , and v = ⌊ C log log z ⌋ , we seethat if we remove these primes from L z we obtain the set of f L z ⊆ L z of cardinality f L z > L z − z/ v > . L z > . C z/ log z. Since obviously ν ( τ ℓ ( g )) ν ( ℓ − v , making a majority decisionwe can find a set of L z of cardinality L z > f L z v > . C zv log z with ν ( τ ℓ ( g )) = u for some fixed u v for every ℓ ∈ L z . Taking C = 0 . C /C we conclude the proof. (cid:3) We note that the
Brun-Titchmarsh theorem (see [IwKo04, Theo-rem 6.6]) can be used to remove log log z in the bound on L z ofLemma 3.2. However in our final result we do not try to optimise termsof this order, so we ignore this and similar potential improvements.3.2. Sieving set L z . We see that using a result of Baker and Har-man [BaHa98] one can take(3.2) α = 0 . , in Lemmas 3.1 and 3.2.From now on, for any positive real number z , we fix a set L z satis-fying the conclusion of Lemma 3.2 with α given by (3.2). Bounds of some arithmetic sums.
For an integer K we con-sider the set(3.3) K = K m ( K )where(3.4) K m ( K ) = { , . . . , K } m , and for k = ( k , . . . , k m ) ∈ K we define(3.5) F ( k ) = m X i =1 c i g k i . For a real z > ω z ( n ) be the number of distinct prime factors ℓ ∈ L z of n . Lemma 3.3.
Let an integer K and a real z be sufficiently large. For K and F ( k ) as in (3.3) and (3.5), respectively, we have X k ∈ K ω z ( F ( k )) ≪ (cid:0) K m z − α + K m − (cid:1) L z . Proof.
We have X k ∈ K ω z ( F ( k )) ≪ X k ∈ K X ℓ ∈ L z ℓ | F ( k ) X ℓ ∈ L z X k ∈ K ℓ | F ( k ) . Clearly the last sum can be estimated as X k ∈ K ℓ | F ( k ) ( K + 1) m − (cid:18) K + 1 τ ℓ ( g ) + 1 (cid:19) ≪ K m ℓ − α + K m − K m z − α + K m − , and the result follows. (cid:3) Remark 3.4.
The proof of Lemma 3.3 appeals to essentially trivialbound O ( K m − ( K/τ ℓ ( g ) + 1)) on the number of solution to the con-gruence F ( k ) ≡ ℓ ) , k ∈ K . Using bounds of exponentialsums one can obtain a better bound, which however does not improveour final result (see also our Appendix). For a real κ we define the sums D κ = X ℓ,r ∈ L z P ( ℓ − = P ( r − gcd ( ℓ − , r − κ . PARSE REPRESENTATIONS OF SQUARES 11
Lemma 3.5.
Let a real z be sufficiently large. Then for κ > wehave D κ z κ + α − ακ +1+ o (1) . Proof.
Clearly for each pair of primes ( ℓ, r ) in the sum D κ we havegcd ( ℓ − , r − ( ℓ − /P ( ℓ − H for some integer H ≪ z − α .Hence D κ H X h =1 h κ X ℓ,r ∈ L z P ( ℓ − = P ( r − ℓ − ,r − h . We estimate the inner sum trivially as O (( z/h ) ) and derive D κ z H X h =1 h κ − z o (1) H κ − z κ − − α )+ o (1) , and the desired result follows. (cid:3) Bounds of character sums
Complete character sums with diagonal forms over finitefields.
Let q be an odd prime power and let F q be the finite field of q elements. We note that for the purpose of proving Theorem 1.2,we only need to estimate the sums of this section over a prime finitefield. However, since our proofs work over arbitrary finite fields, wepresent them in this more general setting with the hope they would beof independent interest.We let m > d > d coprime with q .Let X denote the set of multiplicative characters of F ∗ q and let X ∗ = X \ { χ } be the set of non-principal characters, we referto [IwKo04, Chapter 3] for a background on characters. We also denoteby η ∈ X ∗ the quadratic characters (that is η = χ ).We recall that the implied constant may depend on m (but not on d , q and other parameters).We start with ‘pure’ bounds of sums of quadratic characters.We note that in our next result we have an additional condition of d being an even integer. Lemma 4.1.
Assume that the integer d > satisfies gcd( d, q ) = 1 and is even. Let a , . . . , a m ∈ F ∗ q . Then for S = X x ,...,x m ∈ F q η (cid:0) a x d + · · · + a m x dm (cid:1) we have | S | d m − ( q − q ( m − / . Proof.
The proof follows by induction on m . For m = 1, since d iseven, the sum becomes (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X x ∈ F q η (cid:0) a x d (cid:1)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = q − . We assume the bound true for m − m . We have S = X x m ∈ F ∗ q X x ,...,x m − ∈ F q η (cid:0) a x d + · · · + a m x dm (cid:1) + X x ,...,x m − ∈ F q η (cid:0) a x d + · · · + a m − x dm − (cid:1) . By the induction hypothesis, the second sum in the above is boundedby d m − ( q − q ( m − / . Hence, we have(4.1) | S | | S ∗ | + d m − ( q − q ( m − / , where S ∗ = X x m ∈ F ∗ q X x ,...,x m − ∈ F q η (cid:0) a x d + · · · + a m x dm (cid:1) , to which we apply [Kat02, Theorem 2.1]. Indeed, since x m = 0 in S ∗ ,we make the transformation x i → x i x m , i = 1 , . . . , m −
1, which doesnot change the sum. Moreover, since again d is even and η ( x dm ) = 1,we obtain S ∗ = X x m ∈ F ∗ q X x ,...,x m − ∈ F q η (cid:0) a x d + · · · + a m − x dm − + a m (cid:1) = ( q − X x ,...,x m − ∈ F q η (cid:0) a x d + · · · + a m − x dm − + a m (cid:1) . (4.2)Let now F ( X , . . . , X m − ) = a X d + · · · + a m − X dm − + a m ∈ F q [ X , . . . , X m − ] . We note that the equation F ( X , . . . , X m − ) = 0 defines a smoothhypersurface in the affine space A m − ( F q ). Indeed, considering thepartial derivatives of F with respect to each variable X i , we ob-tain that the only possible singular point would be (0 , . . . , a m = 0, this point does not belong to the hypersurface F ( X , . . . , X m − ) = 0. PARSE REPRESENTATIONS OF SQUARES 13
Similarly, the equation given by the leading homogenous part of F , a X d + · · · + a m − X dm − = 0, defines a smooth hypersurface in theprojective space P m − ( F q ).Applying now [Kat02, Theorem 2.1], we conclude from (4.2) that(4.3) | S ∗ | ( d − m − ( q − q ( m − / . Substituting (4.3) in (4.1), we obtain | S | ( d − m − ( q − q ( m − / + d m − ( q − q ( m − / d m − ( q − q ( m − / (cid:16) ( d − q + 1 (cid:17) . Since ( d − q + 1 < dq / , we conclude the proof. (cid:3) Next we need the following bound on multidimensional sum of qua-dratic characters, twisted by arbitrary characters. In the next resultwe do not use that d is even. Lemma 4.2.
Assume that the integer d > satisfies gcd( d, q ) = 1 .Let a , . . . , a m ∈ F ∗ q . Then for any χ , . . . , χ m ∈ X we have X x ,...,x m ∈ F q η (cid:0) a x d + · · · + a m x dm (cid:1) χ ( x ) . . . χ m ( x m ) ≪ d m q ( m +1) / . Proof.
First we note that if each χ i is equal to the principal character,then the result follows from Lemma 4.1. So, from now on, we assumethat not all of the characters χ i are equal to the principal character.We have(4.4) X x ,...,x m ∈ F q η (cid:0) a x d + · · · + a m x dm (cid:1) χ ( x ) . . . χ m ( x m ) = S − S , where S = X x ,...,x m ∈ F q χ ( x ) . . . χ m ( x m )and S = X y ∈ F ∗ q X x ,...,x m ∈ F q a x d + ··· + a m x dm = y χ ( x ) . . . χ m ( x m ) . Indeed, we observe that each vector ( x , . . . , x m ) ∈ F mq contributes2 χ ( x ) . . . χ m ( x m ) to the sum S . It is also easy to see that S vanishesunless χ = . . . = χ m = χ ; therefore, due to our assumption fromabove, we get that S = 0. We now fix a nontrivial additive character ψ of F q . By the orthogonality relation,1 q X λ ∈ F q ψ ( λu ) = ( , if u = 0 , , if u ∈ F ∗ q , see [IwKo04, Section 3.1]. Hence we write S = X x ,...,x m ∈ F q X y ∈ F ∗ q q X λ ∈ F q ψ (cid:0) λ (cid:0) a x d + · · · + a m x dm − y (cid:1)(cid:1) χ ( x ) . . . χ m ( x m )= 1 q X λ ∈ F q X y ∈ F ∗ q ψ (cid:0) − λy (cid:1)X x ,...,x m ∈ F q ψ (cid:0) λ (cid:0) a x d + · · · + a m x dm (cid:1)(cid:1) χ ( x ) . . . χ m ( x m )= 1 q X λ ∈ F q X y ∈ F ∗ q ψ (cid:0) − λy (cid:1) m Y i =1 X x i ∈ F q ψ (cid:0) λa i x di (cid:1) χ i ( x i ) . The contribution from the terms corresponding to λ = 0 is obviouslyequal to q − q · S = 0 since S = 0 (because not all of the characters χ i are equal to the principal character). Hence(4.5) S = W, where W = 1 q X λ ∈ F ∗ q X y ∈ F ∗ q ψ (cid:0) − λy (cid:1) m Y i =1 X x i ∈ F q ψ (cid:0) λa i x di (cid:1) χ i ( x i ) . Now the sum over y differs from the classical Gauss sums by only oneterm corresponding to y = 0, and so we have(4.6) X y ∈ F ∗ q ψ (cid:0) − λy (cid:1) ≪ q / , see [IwKo04, Theorem 3.4]. For the remaining sums, using that λa i ∈ F ∗ q we apply the Weil bound [Wei74, Appendix 5, Example 12] of mixedsums of additive and multiplicative characters which implies(4.7) X x i ∈ F q ψ (cid:0) λa i x di (cid:1) χ i ( x i ) ≪ dq / , PARSE REPRESENTATIONS OF SQUARES 15 see also [Li96, Chapter 6, Theorem 3]. Therefore, the bounds (4.6)and (4.7), combined together yield W ≪ d m q ( m +1) / . and together with (4.4) and (4.5) we conclude the proof. (cid:3) We remark that some, or all, of the characters χ , . . . , χ m ∈ X can be principal and that the implied constant from the conclusion ofLemma 4.2 depends only on m .4.2. Incomplete character sums with exponential functions.
We now extend the definition of τ ℓ ( g ) to orders modulo any compositemoduli q with gcd( g, q ) = 1. We also use ( u/q ) to denote the Jacobisymbol modulo an odd q .Here we need to obtain multidimensional analogues of the result oncharacter sums from [BaSh17, Section 3]. Although this does not re-quire new ideas and can be achieved at the cost of merely typographicalchanges we present some short proofs of these results.As usual, we write e ( t ) = exp(2 πit ) for all t ∈ R .We use the following variant of the result of [BaSh17, Lemma 3.1],which in turn is based on some ideas of Korobov [Kor70, Theorem 3]. Lemma 4.3.
Let a , b . . . , a m , b m ∈ Z and let ϑ ∈ Z with ϑ > . Let ℓ and r be distinct primes with t ℓ = τ ℓ ( ϑ ) , t r = τ r ( ϑ ) , t = τ ℓr ( ϑ ) and such that gcd ( ℓr, a . . . a m ϑ ) = gcd ( t ℓ , t r ) = 1 . We define integers b i,ℓ and b i,r by the conditions b i,ℓ t r + b i,r t ℓ ≡ b i (mod t ) , b i,ℓ < t ℓ , b i,r < t r , for i = 1 , . . . , m . Then, for S = t X k ,...,k m =1 (cid:18) a ϑ k + . . . + a m ϑ k m ℓr (cid:19) e (cid:18) b k + . . . + b m k m t (cid:19) we have S = S ℓ S r where S ℓ = t ℓ X x ,...,x m =1 (cid:18) a ϑ x + . . . + a m ϑ x m ℓ (cid:19) e (cid:18) b ,ℓ x + . . . + b m,ℓ x m t ℓ (cid:19) ,S r = t r X y ,...,y m =1 (cid:18) a ϑ y + . . . + a m ϑ y m r (cid:19) e (cid:18) b ,r y + . . . + b m,r y m t r (cid:19) . Proof.
As in the proof of [BaSh17, Lemma 3.1], using that gcd( t ℓ , t r ) =1, we see that the integers xt r + yt ℓ , x < t ℓ , y < t r , run through the complete residue system modulo t = t ℓ t r . Moreover,(4.8) ϑ xt r + yt ℓ ≡ ϑ xt r (mod ℓ ) , ϑ xt r + yt ℓ ≡ ϑ yt ℓ (mod r ) , and(4.9) e ( b ( xt r + yt ℓ ) /t ) = e ( bx/t ℓ ) e ( by/t r ) . Hence, S = t ℓ X x ,...,x m =1 t r X y ,...,y m =1 (cid:18) a ϑ x t r + y t ℓ + . . . + a m ϑ x m t r + y m t ℓ ℓr (cid:19) e (cid:18) b ( x t r + y t ℓ ) + . . . + b m ( x m t r + y m t ℓ ) t (cid:19) . (4.10)Using the multiplicativity of the Jacobi symbol, and recalling thecongruences (4.8), we derive (cid:18) a ϑ x t r + y t ℓ + . . . + a m ϑ x m t r + y m t ℓ ℓr (cid:19) = (cid:18) a ϑ x t r + y t ℓ + . . . + a m ϑ x m t r + y m t ℓ ℓ (cid:19)(cid:18) a ϑ x t r + y t ℓ + . . . + a m ϑ x m t r + y m t ℓ r (cid:19) = (cid:18) a ϑ x t r + . . . + a m ϑ x m t r ℓ (cid:19)(cid:18) a ϑ y t ℓ + . . . + a m ϑ y m t ℓ r (cid:19) . (4.11) PARSE REPRESENTATIONS OF SQUARES 17
Furthermore, by (4.9) we have e (cid:18) b ( x t r + y t ℓ ) + . . . + b m ( x m t r + y m t ℓ ) t (cid:19) = e (cid:18) b x + . . . + b m x m t ℓ (cid:19) e (cid:18) b y + . . . + b m t m t r (cid:19) . (4.12)Using (4.11) and (4.12) in (4.10), we see that the sum S can be de-composed into a product of two sums as follows S = t ℓ X x ,...,x m =1 (cid:18) a ϑ x t r + . . . + a m ϑ x m t r ℓ (cid:19) e (cid:18) b x + . . . + b m x m t ℓ (cid:19) t r X y ,...,y m =1 (cid:18) a ϑ y t ℓ + . . . + a m ϑ y m t ℓ r (cid:19) e (cid:18) b y + . . . + b m t m t r (cid:19) . We now replace x i with x i t − r (mod t ℓ ) and y i with y i t − ℓ (mod t r ),and take into account that b i t − r ≡ b i,ℓ (mod t ℓ ) and b i t − ℓ ≡ b i,r (mod t r ) , for i = 1 , . . . , m . This concludes the proof. (cid:3) Next we estimate the sums S ℓ and S r which appear in Lemma 4.3.Namely we now establish an analogue of [BaSh17, Lemma 3.2]. Lemma 4.4.
Let a , b . . . , a m , b m ∈ Z and let ϑ ∈ Z with ϑ > . Let ℓ be a prime with t ℓ = τ ℓ ( ϑ ) and such that gcd ( ℓ, a . . . a m ϑ ) = 1 and gcd( t ℓ ,
2) = 1 . Then for S ℓ = t ℓ X x ,...,x m =1 (cid:18) a ϑ x + . . . + a m ϑ x m ℓ (cid:19) e (cid:18) b x + . . . + b m x m t ℓ (cid:19) we have S ℓ ≪ ( ℓ ( m +1) / , for arbitrary b , . . . , b m ,t ℓ ℓ ( m − / , for b = . . . = b m = 0 . Proof.
Denoting d = ( ℓ − /t ℓ , we can write ϑ = ρ d with some prim-itive root ρ modulo ℓ . Then, S ℓ = 1 d m ℓ − X x ,...,x m =1 (cid:18) a ρ dx + . . . + a m ρ dx m ℓ (cid:19) e (cid:18) d ( b x + . . . + b m x m ) ℓ − (cid:19) = 1 d m ℓ − X w ,...,w m =1 (cid:18) a w d + . . . + a m w dm ℓ (cid:19) χ ( w ) . . . χ m ( w m ) , (4.13)where for w ∈ F ℓ we define χ i by χ i ( w ) = e ( b i dx/ ( ℓ − , i = 1 , . . . , m, where x is any integer for which w ≡ ρ x (mod ℓ ).As in the proof of [BaSh17, Lemma 3.2] we observe χ i is a multi-plicative character of F ℓ for each i = 1 , . . . , m . Recalling Lemma 4.2,we derive from (4.13) S ℓ ≪ d m d m ℓ ( m +1) / = ℓ ( m +1) / , which establishes the desired bound for arbitrary b , . . . , b m ∈ Z .For b = . . . = b m = 0 we observe that since t ℓ is odd, d is evenand hence we can use Lemma 4.1 instead of Lemma 4.2. Thus in thiscase (4.13) implies S ℓ ≪ d m d m − ℓ ( m +1) / = 1 d ℓ ( m +1) / ≪ t ℓ ℓ ( m − / , which concludes the proof. (cid:3) Lemmas 4.3 and 4.4 combined together imply the following bound.
Corollary 4.5.
Let a , b . . . , a m , b m ∈ Z and let ϑ ∈ Z with ϑ > .Let ℓ and r be distinct primes with t ℓ = τ ℓ ( ϑ ) , t r = τ r ( ϑ ) , t = τ ℓr ( ϑ ) and such that gcd ( ℓr, a . . . a m ϑ ) = gcd ( t ℓ , t r ) = gcd( t ℓ t r ,
2) = 1 . Then, for S = t X k ,...,k m =1 (cid:18) a ϑ k + . . . + a m ϑ k m ℓr (cid:19) e (cid:18) b k + . . . + b m k m t (cid:19) PARSE REPRESENTATIONS OF SQUARES 19 we have S ≪ ( ( ℓr ) ( m +1) / , for arbitrary b , . . . , b m ,t ( ℓr ) ( m − / , for b = . . . = b m = 0 . Clearly in Lemma 4.4 and Corollary 4.5 the parity condition on mul-tiplicative orders is important only in the case where b = . . . = b m = 0,as only these parts appeal to Lemma 4.1 (which required d to be even).Combining Corollary 4.5 with the completing method, see [IwKo04,Section 12.2], we derive an analogue of [BaSh17, Lemma 3.4], which isour main technical tool. Lemma 4.6.
Let a , . . . , a m ∈ Z and let ϑ ∈ Z with ϑ > . Let ℓ and r be distinct primes with gcd ( ℓr, a . . . a m ϑ ) = gcd ( τ ℓ ( ϑ ) , τ r ( ϑ )) = gcd( τ ℓ ( ϑ ) τ r ( ϑ ) ,
2) = 1 . Then, for any integers L , . . . L m > , we have L X k =1 . . . L m X k m =1 (cid:18) a ϑ k + . . . + a m ϑ k m ℓr (cid:19) ≪ L . . . L m t − m +1 ( ℓr ) ( m − / + (cid:0) L m − t − m +1 + 1 (cid:1) ( ℓr ) ( m +1) / (log( ℓr )) m , where L = max { L , . . . L m } and t = τ ℓr ( ϑ ) , and the implied constant is absolute.Proof. Clearly we can split the above sum into ⌊ L /t ⌋ · · · · · ⌊ L m /t ⌋ complete sums, where each variable runs over the complete residuesystem modulo t and into at most O (( L/t ) m − + 1) incomplete sumsover a complete residue system modulo ℓr .By Corollary 4.5 each of these complete sums can be estimated as O (cid:0) t ( ℓr ) ( m − / (cid:1) , so they contribute O (cid:0) L . . . L m t − m +1 ( ℓr ) ( m − / (cid:1) intotal.By the standard completing techniques, see, for example, [IwKo04,Section 12.2], we derive from Corollary 4.5 that each incomplete sumcan be estimated as O (cid:0) ( ℓr ) ( m +1) / (log( ℓr )) m (cid:1) . Therefore, in total theycontribute O (cid:0) ( L m − t − m +1 + 1) ( ℓr ) ( m +1) / (log( ℓr )) m (cid:1) .Combining both contributions together, we conclude the proof. (cid:3) Proof of Theorem 1.2
Preliminary transformations.
We can always assume that k m > . . . > k . We note that there is an integer constant h depending only on theinitial data such that if k m > k m − + h then n = m X i =1 c i g k i > . g k m and hence for some(5.1) K ≪ log N we have(5.2) k m = max { k , . . . , k m } ≪ K. On the other hand, for k m < k m − + h , writing k m = k m − + h , h = 0 , . . . , h − n = m X i =1 c i g k i = m − X i =1 c i g k i + ( c m − + c m g h ) g k m − by Theorem 1.1 we obtain at most O ((log N ) m − ) solutions n N .Hence we now estimate the number of solutions to (1.4) with (5.2)(for K as in (5.1)).We recall the notation (3.3) and (3.5), and we write (1.4) as(5.3) n = F ( k ) . We also recall the definitions of the set L z in Section 3.2 and of ω z ( n )from Section 3.3.To simplify the exposition everywhere below we replace logarithmic,and double logarithmic factors of z with z o (1) (implicitly assuming that z → ∞ ). In particular we simply write(5.4) L z = z o (1) . Since z o (1) also absorbs all implied constants, we use instead of ≪ in the corresponding bounds.5.2. Sieving.
Note that if F ( k ) is a perfect square, then we alwayshave X ℓ ∈ L z (cid:18) F ( k ) ℓ (cid:19) = L z − ω z ( F ( k )) . Hence L z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ℓ ∈ L z (cid:18) F ( k ) ℓ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + ω z ( F ( k )) . PARSE REPRESENTATIONS OF SQUARES 21
Denote by M the set of values of k ∈ K satisfying (5.3) and let M = M be its cardinality. Invoking Lemma 3.3, we obtain M L z X k ∈ M (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ℓ ∈ L z (cid:18) F ( k ) ℓ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + X k ∈ M ω z ( F ( k )) X k ∈ M (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ℓ ∈ L z (cid:18) F ( k ) ℓ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + X k ∈ K ω z ( F ( k )) ≪ X k ∈ M (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ℓ ∈ L z (cid:18) F ( k ) ℓ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:0) K m z − α + K m − (cid:1) L z . Therefore either(5.5) M L z ≪ X k ∈ M (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ℓ ∈ L z (cid:18) F ( k ) ℓ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) or(5.6) M ≪ K m z − α + K m − . Assuming that (5.5) holds, by the Cauchy inequality( M L z ) M X k ∈ M (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ℓ ∈ L z (cid:18) F ( k ) ℓ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) and extending summation back to all k ∈ K and using (5.4), weobtain(5.7) M z − o (1) W, where W = X k ∈ K (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ℓ ∈ L z (cid:18) F ( k ) ℓ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = X k ∈ K X ℓ,r ∈ L z (cid:18) F ( k ) ℓr (cid:19) . Combining (5.6) and (5.7), we see that in any case we have(5.8) M (cid:0) K m z − α + K m − + z − W (cid:1) z o (1) . We further split the sum W into two sums as W = U + V , where U = X ℓ,r ∈ L z P ( ℓ − P ( r − X k ∈ K (cid:18) F ( k ) ℓr (cid:19) ,V = X ℓ,r ∈ L z P ( ℓ − = P ( r − X k ∈ K (cid:18) F ( k ) ℓr (cid:19) . (5.9) To estimate U (which also includes the diagonal case ℓ = r ), we usethe trivial bound ( K + 1) m on each inner sum, deriving that U ( K + 1) m X ℓ,r ∈ L z P ( ℓ − P ( r − ( K + 1) m X d > z α X ℓ,r ∈ L z ℓ ≡ r ≡ d ≪ K m X d > z α z d ≪ K m z − α . Hence we can write (5.8) as(5.10) M (cid:0) K m z − α + K m − + z − V (cid:1) z o (1) . Bounds of character sums.
To estimate V , we first observethat for every ℓ ∈ L z the inequality τ ℓ ( g ) > P ( ℓ − > ℓ α implies(since α > ) that P ( ℓ − | τ ℓ ( g ).Fix a pair ( ℓ, r ) ∈ L z × L z with P ( ℓ − = P ( r −
1) and define h = gcd ( τ ℓ ( g ) , τ r ( g )) and ϑ = g h . We observe that τ ℓ ( ϑ ) = τ ℓ ( g ) /h and τ r ( ϑ ) = τ r ( g ) /h. Furthermore, due to our choice of the set L z in Section 3.2 we have ν ( τ ℓ ( g )) = ν ( τ r ( g )) = ν ( h )and hence both τ ℓ ( ϑ ) and τ r ( ϑ ) are odd.We now write(5.11) X k ∈ K (cid:18) F ( k ) ℓr (cid:19) = h X j ,...,j m =1 T ℓ,r ( j , . . . , j m ) , where T ℓ,r ( j , . . . , j m ) = X k ( K − j ) /h . . . X k m ( K − j m ) /h (cid:18) c g k h + j + . . . + c m g k m h + j m ℓr (cid:19) = X k ( K − j ) /h . . . X k m ( K − j m ) /h (cid:18) c g j ϑ k + . . . + c m g j m ϑ k m ℓr (cid:19) . PARSE REPRESENTATIONS OF SQUARES 23
We can certainly assume that z is large enough so thatgcd( c · · · c m , ℓr ) = 1for ℓ, r ∈ L z . Therefore Lemma 4.6 applies to T ℓ,r ( j , . . . , j m ) andimplies T ℓ,r ( j , . . . , j m ) ≪ ( K/h ) m ( τ ℓ ( ϑ ) τ r ( ϑ )) − m +1 ( ℓr ) ( m − / + (cid:0) ( K/h ) m − ( τ ℓ ( ϑ ) τ r ( ϑ )) − m +1 + 1 (cid:1) ( ℓr ) ( m +1) / (log( ℓr )) m . Using τ ℓ ( ϑ ) τ r ( ϑ ) ≫ z α and ℓr ≪ z we see that T ℓ,r ( j , . . . , j m ) (cid:0) ( K/h ) m z ( m − − α ) + (cid:0) ( K/h ) m − z ( m − − α )+2 + z m +1 (cid:1)(cid:1) z o (1) . Therefore, after the substitution in (5.11) we obtain (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X k ∈ K (cid:18) F ( k ) ℓr (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:0) K m z ( m − − α ) + K m − hz ( m − − α )+2 + h m z m +1 (cid:1) z o (1) . Since obviously h gcd( ℓ − , r − V in (5.9)and using (5.4), we now derive V (cid:0) K m z ( m − − α ) + D K m − z ( m − − α ) + D m z m − (cid:1) z o (1) , where D and D m are as in Lemma 3.5 and thus we get D z o (1) and D m z m + α − αm +1+ o (1) . Therefore V (cid:0) K m z ( m − − α ) + K m − z ( m − − α )+2 + z m + α − αm (cid:1) z o (1) . which after the substitution in (5.10) yields M ( K m z − α + K m − + K m z ( m − − α ) + K m − z ( m − − α )+2 + z m + α − αm ) z o (1) . Optimisation.
Clearly we can assume that(5.12) z K / (2 − α ) as otherwise the last term z m + α − αm exceeds the trivial bound K m .Furthermore, for m > α as in (3.2), we have(5.13) ( m − α − > α and hence (using that z ( m − − α ) < z − α due to the inequality (5.13)),we can simplify the above bound as follows: M (cid:0) K m z − α + K m − + K m − z − ( m − α − + z m + α − αm (cid:1) z o (1) . Moreover, since we have (5.12) and α <
1, we see that K m z − α > K m − which means that(5.14) M (cid:0) K m z − α + K m − z − ( m − α − + z m + α − αm (cid:1) z o (1) . First we note that for m >
5, with our choice of α = 0 .
677 in (3.2)along with our assumption (5.12), we have that K m z − α > K m − z − ( m − α − and thus the second term in (5.14) never dominates and we choose z = K m/ (2 m +2 α − αm ) to balance the first and the third terms. Hence for if m >
5, we obtain:(5.15) M K m − mα/ (2 m +2 α − αm )+ o (1) . For m = 4 and with (3.2), direct calculations show that as for m > m = 4.Finally, for m = 3 one checks that balancing the first and the secondterms in (5.14) with z = K / (4 − α ) leads to an optimal result M (cid:0) K − α/ (4 − α ) + K (6 − α ) / (4 − α ) (cid:1) K o (1) K − α/ (4 − α )+ o (1) , which concludes the proof (see also (5.1)).6. Comments
The proof of Theorem 1.2, depends on the bound X k ∈ K ω z ( F ( k )) X ℓ ∈ L z T m ( K, ℓ ) , where T m ( K, ℓ ) is the number of solutions to the congruence F ( k ) ≡ ℓ ) , k ∈ K m ( K ) , PARSE REPRESENTATIONS OF SQUARES 25 where K m ( K ) is given by (3.4), see the proof of Lemma 3.3. In fact,in the proof of Lemma 3.3 we use the trivial bound(6.1) T m ( K, ℓ ) ( K + 1) m − (cid:18) K + 1 τ ℓ ( g ) + 1 (cid:19) ≪ K m z − α + K m − , which holds for any ℓ ∈ L z and is the best possible for m = 2. For m > W in (5.7)become available (or maybe with some other modifications of the argu-ment) we present such a better bound in Appendix A, see Lemma A.2.The method of the proof of Theorem 1.2 also works for relations ofthe form n = m X i =1 c i g k i i , with integer coefficients c , . . . , c m of the same sign and arbitrary in-teger bases g , . . . , g m >
2. Indeed, in this case we still have a bound O (log N ) on the exponents k , . . . , k m , which is important for ourmethod. It is an interesting open question to establish such a boundfor arbitrary c , . . . , c m . Similarly, our method can also be used toestimate the number of n N which can be represented as n = u + . . . + u m for some S -units u , . . . , u m , that is, as a sum of m integers whichhave all their prime factors from a prescribed finite set of primes S .Again, if negative values of u , . . . , u m are allowed then some additionalarguments are needed to bound the powers of primes in each S -unit.Furthermore, as in [BaSh17] we observe that under the GeneralisedRiemann Hypothesis we can obtain a slightly large value of γ m . Wenow recall that p is called a Sophie Germain prime if p and 2 p + 1are both prime. Under the assumption of the existence of the expectednumber of Sophie Germain primes in intervals, or in fact of just z o (1) such primes up to z , we can choose a set L z in the argument of theproof of Theorem 1.2 with any α < γ m = m/ ( m + 2) for m > n ν for a fixed ν = 3 , , . . . ,can be investigated by our method. However, one needs a version of aresult of Baker and Harman [BaHa98] for primes ℓ in the arithmeticprogression ℓ ≡ ν ), so that there are multiplicative charactersmodulo ℓ of order ν . Appendix A. Congruences with exponential functions
First we recall the following special case of a classical result of Ko-robov [Kor72, Lemma 2].
Lemma A.1.
Let a ∈ Z and let ϑ ∈ Z with ϑ > . Let ℓ be a primewith t = τ ℓ ( ϑ ) , and such that gcd ( ℓ, aϑ ) = 1 . Then, we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t X k =1 e (cid:0) aϑ k /t (cid:1)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ℓ / . We now have a bound on T m ( K, ℓ ) which improves (6.1) in someranges.
Lemma A.2.
Let m > . Then for K > z and ℓ ∈ L z , where L z isas in Section 3.2, we have T m ( K, ℓ ) ≪ K m z − + K m z m/ − α ( m − − . Proof.
Let t = τ ℓ ( g ); then since ℓ ∈ L z , we know that t > z α . Wealso let T m ( ℓ ) = T m ( t − , ℓ ) . First we observe that K > z ≫ t and thus(A.1) T m ( K, ℓ ) (cid:18) K + 1 t + 1 (cid:19) m T m ( ℓ ) ≪ K m t − m T m ( ℓ ) . Now, using the orthogonality of exponential functions, we write T m ( ℓ ) = 1 ℓ X k ∈ K m ( t − ℓ − X a =0 e ( aF ( k ) /ℓ ) , where K m ( t −
1) consists of all m -tuples ( k , . . . , k m ) of non-negativeintegers k i < t . Now changing the order of summation, we obtain T m ( ℓ ) = 1 ℓ ℓ − X a =0 X k ∈ K m ( t − e ( aF ( k ) /ℓ ) = 1 ℓ ℓ − X a =0 m Y i =1 t − X k i =0 e (cid:0) ac i g k i /ℓ (cid:1) . The term corresponding to a = 0 is equal to t m /ℓ . We can as-sume that z is large enough (as otherwise the bound is trivial) so PARSE REPRESENTATIONS OF SQUARES 27 that gcd( c · · · c m , ℓ ) = 1 for ℓ ∈ L z . We apply now the bound ofLemma A.1 to m − k , . . . , k m and derive(A.2) T m ( ℓ ) t m /ℓ + ℓ ( m − / ℓ R, where R = ℓ − X a =0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t − X k =0 e (cid:0) ac g k /ℓ (cid:1)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t − X k =0 e (cid:0) ac g k /ℓ (cid:1)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (note that after an application of Lemma A.1 we have added the termcorresponding to a = 0 back to the sum). By the Cauchy inequality R ℓ − X a =0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t − X k =0 e (cid:0) ac g k /ℓ (cid:1)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) · ℓ − X a =0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t − X k =0 e (cid:0) ac g k /ℓ (cid:1)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Using the orthogonality of exponential functions again, we derive ℓ − X a =0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t − X k =0 e (cid:0) ac g k /ℓ (cid:1)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ℓt and similarly for the sum over k . Hence R ℓt , and after substitutionin (A.2) we derive T m ( ℓ ) t m /ℓ + ℓ ( m − / t, which together with the inequality (A.1) and the fact that t > z α concludes the proof. (cid:3) Acknowledgements
D. G. and S. S. were partially supported by a Discovery Grant fromNSERC, A. O. by ARC Grants DP180100201 and DP200100355, andI. S. by an ARC Grant DP200100355.
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Dragos Ghioca, Department of Mathematics, University of BritishColumbia, Vancouver, BC V6T 1Z2, Canada
Email address : [email protected] PARSE REPRESENTATIONS OF SQUARES 29
Alina Ostafe, School of Mathematics and Statistics, University ofNew South Wales, Sydney NSW 2052, Australia
Email address : [email protected] Sina Saleh, Department of Mathematics, University of British Co-lumbia, Vancouver, BC V6T 1Z2, Canada
Email address : [email protected] Igor E. Shparlinski, School of Mathematics and Statistics, Univer-sity of New South Wales, Sydney NSW 2052, Australia
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