Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 https://doi.org/10.1007/s12044-021-00604-z
Primes in Beatty sequence
C G KARTHICK BABU
Institute of Mathematical Science, HBNI C.I.T Campus, Taramani, Chennai 600 113, India E-mail: [email protected]
MS received 5 February 2020; revised 9 July 2020; accepted 28 July 2020
Abstract. For a polynomial g(x) of deg k ≥ 2 with integer coefficients and positive integer leading coefficient, we prove an upper bound for the least prime p such that g(p) is in non-homogeneous Beatty sequence {αn + β:n = 1, 2, 3,...},whereα, β ∈ R with α>1 is irrational and we prove an asymptotic formula for the number of primes p such that g(p) =αn + β. Next, we obtain an asymptotic formula for the number of primes p of the form p =αn + β which also satisfies p ≡ f (mod d),where f, d are integers with 1 ≤ f < d and ( f, d) = 1.
Keywords. Beatty sequence; prime number; estimates on exponential sums.
2010 Mathematics Subject Classification. 11B83, 11N13, 11L07.
1. Introduction Given a real number α>0 and a non-negative real β, the Beatty sequence associated with α, β is defined by B(α, β) ={nα + β:n ∈ N}, where x denotes the largest integer less than or equal to x. Beatty sequences appear in a variety of apparently unrelated mathematical settings and because of their versatility, the arithmetic properties of these sequences have been widely studied in the literature; see, for example, [1,5,9Ð11] and the references contained therein. If α is rational, then B(α, β) is a union of residue classes, hence we always assume that α is irrational. An irrational number θ is said to be of finite type t ≥ 1, if t is the infimum of ρ ∈ R such that for every ε>0, there exists q0(ε) > 0 such that p 1 θ − > q qρ+1+ε holds for all integers q > q0(ε). In 2016, Jörn Steuding and Marc Technau [13] proved that for every ε>0, there exists a computable positive integer l such that for every irrational α>1 the least prime p in the Beatty sequence B(α, β) satisfies the inequality ≤ 35−16εα2(1−ε) 1+ε , p L Bpm+l
© Indian Academy of Sciences 0123456789().: V,-vol 14 Page 2 of 19 Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 where B = max{1,β}, L = log(2αB), pn denotes the numerator of the n-th convergent to the regular continued fraction expansion of α =[a0, a1,...] and m is the unique integer such that
16 2 pm ≤ L α < pm+1.
The first result of this paper is the following.
k Theorem 1. Let g(x) = a0 + a1x +···+ak x , where a0,...,ak ∈ Z with ak ≥ 1 and − k ≥ 2.Put γ = 41 k. Then for any positive integer N ≥ 3, positive real number α with ak − a ≤ 1 ,(a, q) = and any ε> we have α q q2 1 0,
γ −( + )γ 1 ε −γ −γ/ 1 k 1 p = p+O N Nq +N 1 2 +q 1−γ N 1−γ . log α log p≤N p≤N g(p)∈B(α,β)
In particular, if ak/α is an irrational number of finite type t > 0, then we have
γ γ 1 1− k +ε 1− +ε p = p + O(N t+1 + N 2 ). log α log p≤N p≤N g(p)∈B(α,β)
k Theorem 2. Let g(x) = a0 + a1x +···+ak x , where a0,...,ak ∈ Z with ak ≥ 1 and k ≥ 2.Putγ = 41−k. For every ε>0, there exists a computable positive integer l such that for every irrational α>1, the least prime number p such that g(p) is in the Beatty sequence B(α, β) satisfies the inequality
− (γ + )ε − ε 1 ε 2k 1 − 1 k 1 − + γ ≤ α kγ γ k γ k , p B pm+l where B = max{1,β}, pn denotes the numerator of the n-th convergent to the regular continued fraction expansion of α and m is the unique integer such that ak
1+γ γ pm ≤ α B < pm+1. (1)
Remark 1. Instead of using Proposition 1 in (5) if we use Theorem 1 of [15], we may prove Theorem 1 with following error term:
37χ 1 1− 1 1− 2 ( k2 + ) p = p + O N 25k2 log(k+2) + N 100k log χ 4 , log α log p≤N p≤N g(p)∈B(α,β) (2) Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 Page 3 of 19 14
χ ak where χ>0 is a real number satisfying the condition N q and if α of finite type t > 0, 1 p = p + O(N 1−ρ), log α log (3) p≤N p≤N g(p)∈B(α,β) where 1 ρ = if 4k ≤ t + 1, 25k2 log(k + 2) 37 ρ = if 4k > t + 1. 100k(t + 1) log(k(t + 1) + 4)
It is easy to see that error terms obtained in (2) and (3) is better than the error term stated in Theorem 1 for large k.
For irrational α of finite type τ = τ(α), Banks and Yeager (Theorem 2, [3]) proved that 1 for any fixed ε>0, for all integers 1 ≤ c < d < N 4τ+2 with gcd (c, d) = 1, we have 1 1− 1 +ε (n) = (n) + O(x 4τ+2 ). α n≤x,n∈B(α,β) n≤x n≡c (mod d) n≡c (mod d) The following theorem improves the error term.
/ / Theorem 3. For any positive integers N ≥ 3, 1 ≤ f < d ≤ min{q1 2, N 1 6} such that ( f, d) = and positive real number α with 1 − a ≤ 1 with (a, q) = , for any ε> 1 α q q2 1 0, we have 1 p = p log α log p≤N p≤N p∈B(α,β) p≡ f (d) p≡ f (d) N N 4/5 + O N ε + N 1/2q1/2 + N 3/4d1/2 + . q1/2 d1/5
Furthermore, if α is an irrational number of finite type t > 0, then for all integers 1 ≤ 1 / < ≤ { 2(t+1) , 1 6} ( , ) = <ε< 1 , f d min N N with f d 1 and for any 0 4(t+1) we have 1 1− 1 + ε 3 +ε 1/2 4 +ε −1/5 p = p+O(N 2(1+t) +N 4 d +N 5 d ). log α log p≤N p≤N p∈B(α,β) p≡ f (d) p≡ f (d)
The proof of Theorem 3depends on the estimation of exponential sum of the type S(ϑ) = (n)e(lnϑ), (4) |l|≤L n≤N n≡ f (d) 14 Page 4 of 19 Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 where ϑ is irrational, L, N ≥ 1 and f < d,(f, d) = 1. We obtain an upper bound for S(ϑ) in Proposition 2 which is of independent interest.
Remark 2. Let d, f be natural numbers such that 1 ≤ f < d ≤ 500 and ( f, d) = 1. For every ε>0, there exists a computable positive integer l such that for every irrational α>1, the least prime number p ∈ B(α, β) such that p ≡ f (mod d) satisfies the inequality
− ε 1 ( − ε) − ε + ε ≤ α3 7 2 1 3 3 10 1 3 , p B d pm+l where B = max{1,β} and pn denotes the numerator of the n-th convergent to the regular continued fraction expansion of α and m is the unique integer such that
7/3 1/2 10/3 pm ≤ α B d < pm+1.
This fact can be proved in a similar way as Theorem 2 using Theorem 3 and Corollary 1.6 of [4].
2. Notation Throughout this paper, the implied constants in the symbols O and are either absolute or depend only on α and ε. We recall that the notation f = O(g) and f g are equivalent to the assertion that the inequality | f |≤cg holds for some constant c > 0. The notation f ≈ g means that f g and f g. It is important to note that our bounds are uniform with respect to all of the involved parameters other than α, ε and degree of the polynomial k; in particular, our bounds are uniform with respect to β. The letters a, d, f, q always denote non-negative integers and m, n, l, u,vand t denote integers. We use x and {x} to denote the greatest integer less than or equal to x and the fractional part of x respectively. Finally, recall that the discrepancy D(M) of a sequence , ..., ∈[ , ) of (not necessarily distinct) real numbers a1 a2 aM 0 1 is defined by V (I, M) D(M) = sup −|I |, I ⊂[0,1) M where the supremum is taken over all sub-intervals I of [0, 1), V (I, M) is the number of positive integers m ≤ M such that am ∈ I and |I | is the length of I .
3. Preliminaries 3.1 Case of polynomial values of prime m β−1 β Note that an integer m ∈ B(α, β) if and only if α ∈ α , α (mod 1) and m >α+β−1. This is equivalent to m 1 − 2β 1 + < . α 2α 2α Hence m 1 − 2β #{m ≤ N : m ∈ B(α, β)}= χ 1 + , 2α α 2α m≤N Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 Page 5 of 19 14
χ δ> where δ for 0 is defined by 1if θ <δ, χδ(θ) = 0 otherwise, k for θ ∈ R.Letg(x) = a0 + a1x +···+ak x , where a0,...,ak ∈ Z with ak ≥ 1, k ≥ 2. Therefore, g(p) 1 − 2β #{p ≤ N : g(p) ∈ B(α, β)}= χ 1 + . 2α α 2α p≤N
∈ N ± Lemma 1([8], Lemma 2.1). For any L , there are coefficients Cl such that 1 − 1 + 2δ − + C e(lθ) ≤ χδ(θ) ≤ 2δ + + C e(lθ), L + 1 l L + 1 l 1≤l≤L 1≤l≤L | ±|≤ δ + 1 , 3 . with Cl min 2 L+1 2l
Using Lemma 1, we get g(n) 1 − 2β 1 N (n)χ 1 + = (n) + O 2α α α α + ≤ 2 ≤ L 1 n N n N lg(n) + O |C | (n)e , l α 1≤|l|≤L n≤N (5) | |≤ 1 + 1 , 3 . where Cl min α L+1 2l To estimate the exponential sum, we use the following proposition.
PROPOSITION 1 (Equation (22) of [7])
Suppose ε>0 is given. Let f (x) be a real-valued polynomial in x of degree k ≥ 2. Put γ = 41−k. Suppose α is the leading coefficient of f and there are integers a, q with (a, q) = 1 such that
|qα − a| < q−1.
Then we have +ε − − / −k − γ (n)e(lf(n)) (NL)1 (q 1 + N 1 2 + qN L 1) . l≤L n≤N
3.2 Case of primes in arithmetic progression Now we are interested in prime numbers p of the form p ≡ f mod d, which is in B(α, β), where ( f, d) = 1 and f < d. As we discussed above, in order to find a prime number p ∈ B(α, β), we need to show that 14 Page 6 of 19 Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 p 1 − 2β 1 + < . α 2α 2α
By Lemma 1,wehave n 1 − 2β 1 N (n)χ 1 + = (n) + O 2α α 2α α Lϕ(d) n≤N n≤N ≡ ( ) ≡ ( ) n f d n f d + O |Cl | (n)e(ln/α) , 1≤|l|≤L n≤N n≡ f (d) (6) | |≤ 1 + 1 , 3 . where Cl min α L+1 2l Now we want to estimate the exponential sum of the form (4). To estimate the exponential sum, we use the following proposition.
PROPOSITION 2 (ϑ) ϑ − a ≤ −2, Let S be defined by 4 with q q where a and q are positive integers satisfying (a, q) = 1. Then for any real number ε>0, we have 4/5 ε NL 1/2 1/2 1/2 3/4 1/2 LN S(ϑ) ε (NL) + L N q + LN d + . q1/2 d1/5
We will give the proof of Proposition 2 in Section 6. In the end of this section, we give some lemmas require for the proof. The following lemma gives explicit bound for the average of von Mangoldt function.
Lemma 2[12]. For any N ∈ N, (n) ≤ c0 N, n≤N for some constant c0, where one may take c0 = 1.04.
1 Lemma 3. Let α be of finite type t if and only if α is of finite type t.
Proof. Let α be an irrational number of finite type t.Let pm be the m-th convergent to the qm continued fraction expansion of α. We denote μ(α) = {ρ ∈ R : ≤ ρ ∈ N}. inf qm+1 qm holds for all but finitely many m
We will first show that μ(α) = t. Since μ(α) is the infimum, for every ε>0, qm+1 ≥ μ(α)−ε qm holds for infinitely many m ∈ N. By using Theorem 171 of [6], we have α − pm < 1 ≤ 1 . μ(α)+ −ε + 1 qm qmqm 1 qm Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 Page 7 of 19 14
Therefore μ(α) − ε
To prove reverse inequality, we use the definition of finite type to obtain 1 ≤ α − pm ≤ 1 t+1−ε 2qmqm+1 qm qm which holds for every ε>0 and for all but finitely many m ∈ N. Hence, we have ε t− t−ε ≤ 2 ≤ 2 ≤ qm 2qm+1, which implies qm ε/2 qm+1 qm+1 holds for all but finitely many qm ∈ N. − ε <μ(α) ε> ≤ μ(α). m Therefore, t 2 for every 0 from which it follows that t Thus we have μ(α) = t. Since the convergents of 1 are reciprocals of the convergents pn to the continued fraction α qn expansion of α, and pn satisfies same recurrence relation that is satisfied by qn, we have
1 ρ μ( ) = {ρ ∈ R : p + p m ∈ N}=t. α inf m 1 m holds for all but finitely many
This completes the proof of Lemma 3.
4. Proof of Theorem 1 and Theorem 2 In the previous section, we stated essential results to prove Theorem 1 and Theorem 2.In this section, we give proof of these theorems.
Proof of Theorem 1. It follows from Proposition 1 and partial summation formula that g(n) 1 − 2β Cl (n)e l + α 2α 1≤|l|≤L n≤N 1+ε ε γ −γ/2 γ −kγ −γ ε N L (q + N + q N L ) +ε γ − γ + N 1 q N k . (7)
By (5) and (7), we have g(n) 1 − 2β 1 N (n)χ 1 + = (n) + O 2α α 2α α L + 1 n≤N n≤N + O(N 1+ε Lε(q−γ + N −γ/2) + N 1+εqγ N −kγ ).
Choosing L = qγ ,wehave g(n) 1 − 2β 1 (n)χ 1 + = (n) 2α α 2α α n≤N n≤N 14 Page 8 of 19 Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 γ −( + )γ ε −γ −γ/ 1 k 1 + O N Nq + N 1 2 + q 1−γ N 1−γ .
This leads to 1 p + p = p + ξ(N, q), log log α log (8) pν ≤N pν ≤α+β−1 pν ≤N g(pν )∈B(α,β) g(pν )∈B(α,β−α+β) where γ 1−(k+1)γ ε −γ 1−γ/2 |ξ(N, q)|≤ε N Nq + N + q 1−γ N 1−γ . (9)
The number of prime powers pν ≤ N with ν ≥ 2isO(π(N 1/2)), thus we have 1 N 1/2 log p = log p + O α α log N p≤N p≤N ( )∈B(α,β) g p γ −( + )γ ε −γ −γ/ 1 k 1 + O N Nq + N 1 2 + q 1−γ N 1−γ . (10)
ak Suppose we assume α is an irrational number of finite type t. For any positive ε, there is a positive constant c such that ak p c − ≥ , (11) α q qt+1+ε holds for all integers p, q with q > 0. On the other hand, by using Dirichlet’s approximation kt kt theorem with Q = N t+1 , we obtain a rational p/q with 1 ≤ q ≤ N t+1 such that a p 1 k − ≤ . (12) α kt q qN t+1
Suppose we assume that there exists a convergent to the simple continued fraction expan- ak sion of α which satisfies (12). Then by (11), its denominator satisfies
k +ε kt N t+1 q ≤ N t+1 . (13)
Therefore, by (10) and (13), we obtain
γ γ 1 1− k +ε 1− +ε p = p + O(N t+1 + N 2 ). log α log p≤N p≤N g(p)∈B(α,β)
To complete the proof, we want to show that there exists a convergent to the continued ak p fraction of α which satisfies (12). Let q be a rational number which satisfies (12). By using Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 Page 9 of 19 14 recurrence relation satisfied by the denominator of convergent to the continued fraction of ak , ∈ N ≤ ≤ . α we obtain unique n such that qn q qn+1 It follows from Theorem 171 of [6] kt 1 < ak − pn > t+1 that kt α and qn+1 N both cannot hold simultaneously. Therefore, + qn qn N t 1 by Theorem 181 of [6], we get that either pn or pn+1 satisfies (12). This completes the qn qn+1 proof of Theorem 1.
Proof of Theorem 2. By (8) and (9), we have
1 p + p = p + ξ(N, q), log log α log (14) pν ≤N pν ≤α+β−1 pν ≤N g(pν )∈B(α,β) g(pν )∈B(α,β−α+β) where
γ 1−(k+1)γ ε −γ 1−γ/2 |ξ(N, q)|≤ε N (Nq + N + q 1−γ N 1−γ ). (15)
By Lemma 2, the second sum on the left-hand side of (14)is< 1.04 (α + β − 1). Therefore, we have 1 p ≥ p + ξ(N, q) − . (α + β − ) log α log 1 04 1 p≤N p≤N ( )∈B(α,β) g p 1 + − p. α 1 log pν ≤N ν≥2
Notice that the last term is negative, it is obviously bounded by 1 1 − p < − π(N 1/2) N. 1 α log 1 α log pν ≤N ν≥2
We will use inequality (3.2) of Rosser and Schoenfeld [12]forπ(x).Wehave 1 3 / 1 − log p < 2 1 + N 1 2. α log N pν ≤N ν≥2
We also use inequality (3.16) of Rosser and Schoenfeld which is
N log p > N − , log N p≤N for N ≥ 41. Therefore, we obtain N 1 log p ≥ 1 − α log N p≤N g(p)∈B(α,β) 14 Page 10 of 19 Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 3 / + ξ(N, q) − 1.04(α + β − 1) − 1 + N 1 2. log N
We thus find a prime p ≤ N such that g(p) ∈ B(α, β) if we show that the following inequality: N 1 3 / 1 − >ξ(N, q) + 1.04(α + β − 1) + 1 + N 1 2, α log N log N may also be replaced by
N . > . (α + β − ) + . N 1/2 + ξ(N, q). 0 73 α 1 04 1 1 81
By (15), we have
α α ε −γ −γ/ 0.73 > 1.04 (α + β − 1) + 1.81 + C(ε)N α(q + N 2 N N 1/2 γ −kγ + q 1−γ N 1−γ ) and appropriate absolute constant C(ε) depending only on ε but not on α. Obviously N need to be larger than Max{α2/γ , B} and q larger than α1/γ . We shall take both N and q somewhat larger so that the above inequality holds. Now choose
2k−1 k−1 1 ε γ +1 N = α kγ B k q k η γ , q = α γ Bη with some large parameter η to be specified later and B = max{1,β}. Then the latter inequality can be rewritten as
( − )γ − k 1 2k − − 1 − ε 0.73 >1.04(α + β − 1)(α kγ B 1η k γ ) ( − )γ − 2k 1 2k − / − 1 − ε + 1.81(α 2kγ B 1 2η 2k 2γ ) ( +γ)ε 2 −γ + 2k −γ +ε −γ + ε + ε + C α kγ B η k γ
γ ( +γ)ε γ 2 − + 2k − γ +ε − + (2−k)ε + ε + α 2k kγ B 2 η 2k 2k γ
2(1−k)−γ + (2k+γ)ε (1−k)γ +ε (1−k2−γ)ε+ ε2 + α 1−γ kγ B 1−γ η 1−γ γ .
≥ γ = 1−k ε< γ 2 α, Since k 2 and 4 assuming 2(2k+γ), as we may, all exponents of B and η are negative. Therefore, the above inequality is satisfied for all sufficiently large η,say η ≥ η0. Since η is intertwined with q, a little care needs to be taken. Since α is irrational a there are infinitely many rational q satisfying α − a < 1 2 ak q q Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 Page 11 of 19 14
α Let qn be the n-th convergent to the continued fraction expansion of . We shall choose pn ak p + l such that η ≤ m l , where m is defined by (1), for then the choice q = p + will yield 0 pm m l an η ≥ η0. The choice of η is as in (12) of [13]. Once η is chosen, l is also determined as it depends only on η.
5. Proof of Theorem 3 The present section is devoted to a proof of Theorem 3.
Proof of Theorem 3. It follows from Proposition 2 and partial summation formula that Cl (n)e(ln/α) 1≤|l|≤L n≤N n≡ f (d) 1/2 1/2 4/5 ε N 1/2 1/2 N q 3/4 1/2 N ε (NL) + N q + + N d + . (16) q1/2 L1/2 d1/5
By (6) and (16), we obtain n 1 − 2β 1 N (n)χ + = (n) + O α 2α α L(ϕ(d)) n≤N n≤N n≡ f (d) n≡ f (d) N N 4/5 + O (NL)ε + N 1/2q1/2 + N 3/4d1/2 + . q1/2 d1/5
Choose
q1/2 L = . ϕ(d)
Therefore, we obtain an estimate n 1 − 2β 1 (n)χ + = (n) α 2α α n≤N n≤N n≡ f (d) n≡ f (d) N N 4/5 + O N ε + N 1/2q1/2 + N 3/4d1/2 + . q1/2 d1/5
We rewrite the above equality as
1 p + p = p log log α log pν ≤N pν ≤α+β−1 pν ≤N pν ∈B(α,β) pν ∈B(α,β−α+β) pν ≡ f (d) pν ≡ f (d) pν ≡ f (d) N N 4/5 + O N ε + N 1/2q1/2 + N 3/4d1/2 + . q1/2 d1/5 14 Page 12 of 19 Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14
By using prime number theorem, we have √ log p N. pν ≤N pν ≡ f (d) ν≥2
Thus we have 1 N 1/2 p = p + O log α log α p≤N p≤N p∈B(α,β) p≡ f (d) p≡ f (d) 4/5 ε N / / / / N + O N + N 1 2q1 2 + N 3 4d1 2 + . (17) q1/2 d1/5
Suppose we assume that α is an irrational number of finite type t. Note that by Lemma 3, 1 α and α are of same type. Hence, for any positive ε, there is a positive constant c such that 1 p c − ≥ (18) α q qt+1+ε holds for all integers p, q with q > 0. As discussed in Theorem 1, by using Dirichlet’s t approximation theorem with Q = N 1+t , we obtain a convergent to the simple continued t 1 p ≤ ≤ 1+t fraction expansion of α ,say q , with 1 q N such that 1 p 1 − ≤ . (19) α t q qN t+1
Then by (19) and (18), there exists a convergent to the simple continued fraction expansion 1 of α whose denominator satisfies
1 +ε t N 1+t q ≤ N 1+t . (20)
Therefore, by (17) and (20), we obtain 1 ε 1− 1 / / / − / p = p+O(N (N 2(1+t) +N 3 4d1 2 +N 4 5d 1 5)). log α log p≤N p≤N p∈B(α,β) p≡ f (d) p≡ f (d)
This completes the proof of Theorem 3. If N is a prime number, Fermat’s little theorem asserts that aN ≡ a (mod N) for all a ∈ Z. A Carmichael number is a composite number N with the same property. Banks and Yeager (Theorem 1, [3]) proved that for α>1 of finite type irrational, there are infinitely many Carmichael numbers composed solely of primes from the Beatty sequence B(α, β). They also proved the quantitative version of the above result (Theorem 3of[3]). As an application of Theorem 3, we deduce the following corollary which extends Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 Page 13 of 19 14 the range of B in Theorem 3 of [3] for type t > 1. To state the corollary, we require the following notations of [3]: Let π(x) be the number of primes p ≤ x, and let π(x, y) be the number of those primes for which p − 1 is free of prime factors exceeding y. We denote by E the set of numbers E in the range 0 < E < 1 for which there exists a number x0(E), γ (E)>0 such that
1−E π(x, x ) ≥ γ(E)π(x) for all x ≥ x0(E).
COROLLARY 1
∈ E, ∈ ( , { 1 , 1 }) ε> , = For each E B 0 min 2t+2 6 and 0 there is a number x0 EB−ε x0(α, β, E, B,ε)such that for any x ≥ x0, there are at least x Carmichael numbers up to x composed solely of primes from B(α, β).
The proof of Corollary 1 follows from Lemma 5, Lemma 6 of [3] by using Theorem 3 in place of Theorem 2 of [3].
6. Proof of Proposition 2 The proof of Proposition 2 is based on the work of Balog and Perelli [2].
6.1 Some lemmas Here we list several lemmas required for the proof.
Lemma 4. x − e(mθ) min + 1, θd 1 . d x Lemma 5[16]. Suppose that X, Y ≥ 1 are positive integers. Also suppose that |α −a/q| < q−2, where α is a real number, a and q integers satisfying (a, q) = 1. Then − XY min(Y, αx 1) + (X + q) log 2q, q x≤X XY − XY min , αx 1 + (X + q) log(2XYq). x q x≤X Lemma 6[14]. For any real number ϑ and natural numbers N, l and 1 ≤ f < dsuch that ( f, d) = 1, we have 1/2 (n)e(ln ϑ) = O(N ) + S1 − S2 − S3, n≤N n≡ f (d) 14 Page 14 of 19 Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 where S1 = μ(m)(log n)e(lmnϑ), m≤U n≤N/m ≡ ( ) mn f d S2 = φ1(m)e(lmnϑ), m≤U 2 n≤N/m ≡ ( ) mnf d S3 = φ2(m) (n)e(lmnϑ), U φ1(m) log m,φ2(m) d2(m). Here U is an arbitrary parameter to be chosen later satisfying 1 ≤ U ≤ N 1/2. Lemma 7. Suppose that ε>0 and that φ(u), ψ(v) are real-valued functions such that |φ(u)|T, |ψ(v)|F. Suppose that |ϑ − a/q| < q−2, where ϑ is a real number, a and q are integers satisfying (a, q) = 1. For positive integers N, W, X and L, write S = φ(u)ψ(v)e(luvϑ). (21) |l|≤L X Then LWX1/2 LXW S TF +(LXd)ε + LXW1/2d1/2 1/2 1/2 d q +L1/2q1/2 X 1/2W 1/2 . Proof. For the moment, we shall ignore the condition uv ≤ N in (21). Consider S = φ(u)ψ(v)e(luvϑ). (22) |l|≤L X By using CauchyÐSchwarz inequality, we obtain 2 | |2 ≤ φ( )ψ(v) ( vϑ) R f1, f2 L u e lu |l|≤L u≤W X We may write R1 in the form R1 = ψ(v1)ψ(v2)e(lukϑ) |l|≤L u≤W |k|≤X X F2 X ζ (k) = ψ(v )ψ(v ) . 1 1 2 d X Applying Lemma 4 for the innermost sum of (25), we get 2 F X W − |R |≤ min + 1, lkdϑ 1 . 1 d d |l|≤L |k|≤X k≡0(d) Let r = lkd so that 1 ≤|r|≤2LXd and r will run through all the integers in the interval above, also the number of representations of r is not more than d2(r). Therefore we have 2 F X ε W − |R |≤ (LXd) min + 1, rϑ 1 . 1 d d |r|≤2LXd 14 Page 16 of 19 Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 Then by using Lemma 5, we obtain 2 ε LX W qX R F2(LXd) + LX2 + . (26) 1 qd d By (24) and (26), we have 1/2 LWX ε R , TF + (LXd) f1 f2 d3/2 LXW LXW1/2 L1/2q1/2 X 1/2W 1/2 + + . (27) q1/2d d1/2 d Thus lemma follows from (23) and (27). Lemma 8. Suppose we have the hypotheses and notations of Lemma 7 with either φ(x) = 1 or φ(x) = log x for all x. Then ε − S F(LXd) (LXWq 1 + LXd + q). (28) Proof. The log x factor may easily be removed by partial summation formula, so we presume that φ(x) ≡ 1. Again we may ignore the condition uv ≤ N. Therefore, we need to estimate S = ψ(v)e(luvϑ) |l|≤L X Let r = lvd so that 1 ≤|r|≤2LXd and r will run through all the integers in the interval above. Also, the number of representations of r is not more than d2(r). Therefore, we have ε W − S ≤ F(LXd) min + 1, rϑ 1 . (29) d |r|≤2LXd Thus (28) follows from (29) and Lemma 5. 6.2 Proof of the Proposition 2 We may assume that − N ≥ max(qd2 L 1, d6), q ≥ d2 (30) Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 Page 17 of 19 14 otherwise (16) is a consequence of the trivial bound, LN S(ϑ) ≤ . d Using Lemma 6, we have the following sums to estimate: = μ( )( ) ( ϑ), S1 m log n e lmn |l|≤L m≤U n≤N/m ≡ ( ) mn f d = φ ( ) ( ϑ), S2 1 m e lmn |l|≤L m≤U 2 n≤N/m ≡ ( ) mn f d = φ ( ) ( ) ( ϑ). S3 2 m n e lmn |l|≤L U By dyadic division, we write log U log 2 ≤ , S1 S1t t=0 where S1t = μ(m)(log n)e(lmnϑ). l≤L 2t By dividing S31 dyadically, we obtain R = , S31 S31t t=0 14 Page 18 of 19 Proc. Indian Acad. Sci. (Math. Sci.) (2021) 131:14 where S31t = μ(m)(log n)e(lmnϑ) l≤L U2t Then (16) follows from (31) with the choice of N 2/5 U = , d3/5 and the observation q ≤ L1/2 N 1/2q1/2. Acknowledgements The author would like to express his sincere thanks to his thesis supervisor, Anirban Mukhopadhyay, for his valuable and constructive suggestions during the planning and development of this paper. 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