Bounded Gaps between Products of Special Primes

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1 Mathematics 2014, 2, 37-52; doi: /math Article OPEN ACCESS mathematics ISSN Bounded Gas between Products of Secial Primes Ping Ngai Chung 1, * and Shiyu Li 2, * 1 Massachusetts Institute of Technology, 305 Memorial Drive, Cambridge, MA 02139, USA 2 University of California, Berkeley, 1676 S. Blaney Ave. San Jose, CA 95129, USA * Authors to whom corresondence should be addressed; s: briancn@mit.edu P.N.C.); jjl2357@berkeley.edu S.L.); Tel: P.N.C.); S.L.). Received: 23 August 2013; in revised form: 18 February 2014 / Acceted: 25 February 2014 / Published: 3 March 2014 Abstract: In their breakthrough aer in 2006, Goldston, Graham, Pintz and Yıldırım roved several results about bounded gas between roducts of two distinct rimes. Frank Thorne exanded on this result, roving bounded gas in the set of square-free numbers with r rime factors for any r 2, all of which are in a given set of rimes. His results yield alications to the divisibility of class numbers and the triviality of ranks of ellitic curves. In this aer, we relax the condition on the number of rime factors and rove an analogous result using a modified aroach. We then revisit Thorne s alications and give a better bound in each case. Keywords: bounded rime gas; square-free numbers; modular ellitic curves 1. Introduction and Statement of Results The celebrated twin rime conjecture redicts that there are infinitely many airs of consecutive rimes. While a roof of the conjecture seems to be out of reach by current methods, there has been a sate of recent advances concerning the weaker conjecture: lim inf n n+1 n ) < In 2005, Goldston, Pintz and Yıldırım [1] roved that there exists infinitely many consecutive rimes, which are much closer than average, that is, lim inf n n+1 n log n = 0

2 Mathematics 2014, 2 38 Based on their methods, Zhang [2] showed that: lim inf n n+1 n ) Maynard [3] imroved the uer bound to 600 using a modified sieve method. The constant was subsequently imroved to 252 by an ongoing olymath roject [4]. In a similar vein, eole have investigated related roblems about almost rimes or numbers with few rime factors. Chen [5] roved that there are infinitely many rimes,, such that + 2 has at most two distinct rime factors. In [6], Goldston, Graham, Pintz and Yıldırım GGPY) considered the E 2 numbers, which are numbers with exactly two distinct rime factors, and showed that there are infinitely many airs of E 2 numbers that are at most six aart. Thorne [7] observed that the methods in [6] are highly adatable, and generalized the result to E r numbers, which are numbers with exactly r distinct rime factors. In [7], he showed that given an infinite set of rimes, P, satisfying certain conditions, and ositive integers ν and r with r 2, there exists an effectively comutable constant Cr, ν, P), such that: lim inf n q n+1 q n ) Cr, ν, P) where q n is the n-th E r number, whose rime factors are all in P. Using this theorem, Thorne roved several corollaries. With a result by Soundararajan [8], he showed that there are infinitely many airs of E 2 numbers, m and n, such that the class grous, ClQ m)) and ClQ n)), each contain elements of order four, with m n 64. As a second alication, he considered the quadratic twists of ellitic curves over Q without a Q-rational torsion oint of order two. Let E/Q be such an ellitic curve, LE, s) denote its Hasse Weil L-function, rke) := rke, Q) denote the rank of the grou of rational oints on E over Q and ED) denote the D-quadratic twist of E for a fundamental discriminant, D. Using the work of Ono [9], he showed that for good ellitic curve E/Q defined as in [10]), there are infinitely many airs of square-free numbers, m and n, such that LEm), 1) LEn), 1) 0, rkem)) = rken)) = 0 and m n C E hold simultaneously for some absolute constant, C E. For E = X 0 11), Thorne obtained a bound of C E 6,152,146. In this aer, we revisit Thorne s examles and obtain stronger bounds by relaxing the E r condition to instead consider bounded gas between square-free numbers with rime factors all in P. In this case, we can rove an analogous general theorem with a better bound on the gas. Theorem 1. Suose P is a set of rimes with ositive Frobenius density. Let ν be a ositive integer; and let q n denote the n-th square-free number, whose rime factors are all in P. Then: lim inf n q n+ν q n ) Cν, P) Remark 1. Theorem 1 also holds for arbitrary sets of rimes of ositive density satisfying a Siegel Walfisz-tye condition defined in Section 2). We observe that if we remove the restriction on the number of rime divisors in each of Thorne s examles, we can obtain better bounds. Relacing E 2 by a square-free number in his first examle, we obtain the following twin rime-tye result.

3 Mathematics 2014, 2 39 Corollary 1. There are infinitely many square-free numbers, n, such that the class grous, ClQ n)) and ClQ n 8)), each contain elements of order four. In the second examle, the bound, C E, can be imroved analogously. We give an exlicit bound in the case when E = X 0 11). Corollary 2. Let E := X 0 11). Then, there are infinitely many airs of square-free numbers, m and n, for which the following hold simultaneously: i) LEm), 1) LEn), 1) 0, ii) rkem)) = rken)) = 0, iii) m n 48. Remark 2. There are more general alications of Theorem 1. Thorne [7] described an alication to the nonvanishing of Fourier coefficients of weight one newforms, where Theorem 1 can also be alied. If fz) = n=1 an)qn is a newform of integer weight, then the set of integers, n, such that an) is nonzero modulo l has zero density for all rime l by the theory of Deligne and Serre [11]. Nonetheless, our result shows that there are bounded gas between such n for almost all l, yielding better bounds than [7]. Our result also alies to the quadratic twists of ellitic curves over Q that have a given 2-Selmer F 2 -rank. If K is a number field, E is an ellitic curve over K and r is a suitable nonnegative integer, Mazur and Rubin [12] conjecture that for a ositive roortion of quadratic extensions, F/K, the quadratic twist, E F, of E by F/K has the 2-Selmer rank r. Using [12] Proosition 4.2)), our result shows that for K = Q, an ellitic curve, E/Q, with no two-torsion oints, and a given integer, r 0, either no quadratic twists have 2-Selmer rank r or there are bounded gas between the square-free numbers, d, such that Ed) has 2-Selmer rank r. 2. Main Result We borrow our notation from [6], using k to denote an integer greater than one, L = {L 1,..., L k } to denote an admissible k-tule of linear forms defined in Section 2.2) with L i n) := a i n + b i for some a i, b i Z, a i > 0, and P to denote a set of rimes with ositive density α. The constants imlied by O and may deend on k, L and P. Let τ k n) denote the number of ways of writing n as a roduct of k factors and ωn) denote the number of distinct rime factors of n. φn) and µn) are the usual Euler and Möbius functions. N and R will denote real numbers regarded as tending to infinity, and we will always assume R N 1/2. Given a set of rimes, P, with density α, we call a square-free number with rime factors only in P an E P number. Let PN) be the set of rimes in P greater than ex log N) and ξ P be the characteristic function of all E PN) numbers. Given a ositive integer, M, to be chosen later, let δ m be the density of

4 Mathematics 2014, 2 40 integers, n, congruent to m mod M in the set of E P numbers and the minimum density δ of a set of linear forms, L, be the minimum of δ bj, 1 j k. We define: P,b N; q, a) := n amod q) n bmod M) ξ P n) 1 φq) n,q)=1 n bmod M) ξ P n) Following [7], we say that P satisfies a Siegel-Walfisz condition SW M) if for each b corime to M and for any ositive C, φq) A N log C N N< 2N, P N< 2N, P bmod M) amod q) bmod M) holds uniformly for all q with q, Ma) = 1. We also recall that a set of rimes, P, has Frobenius densityα, 0 < α < 1 cf. [13]), if there is a Galois extension, K/Q, and a union of conjugacy classes, H, in G = GalK/Q), such that for all rimes,, sufficiently large, Frob H if and only if P, and #H/#G = α. Analogous to the aroach in [6,7], Theorem 1 follows from the following main result. Theorem 2. Let P be an infinite set of rimes with ositive Frobenius density α < 1 that satisfies SW M). Let L i x)1 i k) be an M-admissible defined in Section 2.2) k-tule of linear forms with minimum density δ. There are at least ν + 1 forms among them that infinitely, often, simultaneously reresent square-free numbers with rime factors all in P, rovided that: where: bk) := k > ν 41 α b1) Γα)Γ2 α) δφm) bk) Γ1 α)γk1 α) + 1) Γk + 1)1 α) + 1)) = B1 α, k1 α) + 1) Remark 3. Our method is not directly alicable in the case when α = 1. Nonetheless, if we take the limit, α 1, on the right-hand side of the inequality, we get k > ν/δϕm)). In ractice, one can take a subset of P with Frobenius density close to one that satisfies SW M), so that the same k still satisfies the inequality in Theorem 2. In the case when k = 2 and ν = 1, we have the following twin rime-tye result. Corollary 3. Let P be an infinite set of rimes with ositive Frobenius density α < 1 that satisfies SW M). For any even number, d, let δ = max m minδ m, δ m+d ). Assume that: δ φm) > 2 1 2α b1) Γα)Γ2 α) b2) Then, there are infinitely many n for which n and n + d are simultaneously square-free numbers with rime factors all in P.

5 Mathematics 2014, 2 41 We have lotted the right-hand side of the inequality against α to illustrate the conditions under which one can obtain a twin rime-tye result Figure 1). Figure 1. The right-hand side of Corollary 3 vs. α As another samle alication of our theorem, we consider the roblem of reresenting square-free integers by translates of tules. This roblem was actually answered by Hall [14], who even obtained an asymtotic exression for the number of such reresentations, but, by taking the limit, α 1, in Theorem 2.1, we easily obtain the following corollary. Corollary 4. Let {b 1, b 2,..., b k } be an admissible k-tule. Then, there are infinitely many n, such that all of the n + b i are simultaneously square-free. Here, we recall from [1] that a k-tule of integers is admissible if for all rime,, they do not cover all the residue classes modulo. We remark that admissibility is not a necessary condition for this corollary to hold, but it is a natural limit of our method. Α 2.1. The Level of Distribution of E P Numbers In this section, we will rove a Bombieri Vinogradov-tye result for E P numbers if P satisfies SW M), generalizing a result of Orr [15]. More recisely, we shall show that the E P numbers have a level of distribution ϑ = 1/2. We remark that a set, P, with ositive Frobenius density satisfies SW M) for some M as a consequence of Lemma 3.1 in [7]. Lemma 1. Suose that P satisfies SW M) for some M. Then, for each b corime to M and for any C, there exists some B = BC) > 0, such that: q N 1/2 log B N q,m)=1 max a P,b N; q, a) C N log C N a,q)=1

6 Mathematics 2014, 2 42 Proof. The result is a variant of Motohashi [16]. Similar results are also treated by Bombieri, Friedlander and Iwaniec [17]. We will use the modern treatment given by [18] Theorem 17.4)), following the aroach of Thorne [7] Lemma 3.2)). Let ξ P,b n) be the characteristic function of E PN) numbers congruent to b modulo M and χ P,b n) be the characteristic function of rimes in PN) congruent to b modulo M. We further define ξ P,b n) = ωn) 1 ξ P,b n). Borrowing the notation from [18], given an arithmetic function, fn), define: D f N; q, a) = n N n amod q) fn) 1 φq) n N n,q)=1 Finally, we let f I denote the restriction of fn) to the interval, I, i.e., f I n) = fn) if n I, f I n) = 0 otherwise. We first remark that: ξ P,b = ξ P,i χ P,j, P,b N; q, a) = D ξp,b [N,2N] 2N; q, a) ij bmod M) fn) Hence, for Q = N 1/2 log B N, where B = BC) will be chosen later, max P,bN; q, a) = max D ξ P,b [N,2N] 2N; q, a) a,q)=1 a,q)=1 q Q q Q max D ξ P,i a,q)=1 χ P,j [N,2N] 2N; q, a) ij bmod M) q Q Fixing ε > 0, we now slit the interval [x, 2N/x] into intervals of the form [t, 1 + ε)t), where x = ex log N). The number of such intervals is log N. Then, we remark that ε t ξ P,i [t,1+ε)t) χ P,j [N/t, 2N 1+ε)t ] closely aroximates ξ P,i χ P,j [N,2N]. Both functions are suorted in [N, 2N] and identical on [N1+ε), 2N/1+ε)]. The differences on the intervals, [N, N1+ε)) and 2N/1+ε), 2N], contribute εn/φq) to each D ξ P,i χ P,j [N,2N] 2N; q, a). Summing over all q Q and all airs i, j), such that ij bmod M), the total contribution of the error is: εn q Q 1 φq) εn log N On the other hand, we may aly Theorem 17.4 in [18] with α = ξ P,i [t,1+ε)t) and β = χ P,j [N/t, 2N ]. 1+ε)t Note that Condition 17.13) on β is satisfied for some U log U N, since P satisfies the SW M) condition, and: ) 1/2 εt α1 2ε)N/t ξ P,i [t,1+ε)t) log 1 α)/2 εt), χ P,j 2N [N/t, 1+ε)t] log1 2ε)N/t) where: ) 1/2 f := fn) 2 Hence, the theorem gives note that x log U N for any U > 0): max D ξ P,i a,q)=1 [t,1+ε)t) χ P,j [N/t, 2N ]2N; q, a) ε 1/2 N log B+2.5 N 1+ε)t t q Q n

7 Mathematics 2014, 2 43 We take ε = log 2B/3+1 N, and sum over all the airs i, j), such that ij bmod M) to conclude that: max P,bN; q, a) N log 2B/3+2 N a,q)=1 q Q From there, taking B = 3C/2 + 3 gives the desired result Linear Forms and Admissibility Following [7] and [6], we will rove our results for k-tules of linear forms: L i x) := a i x + b i 1 i k), a i, b i Z, a i > 0 We will rove that for any admissible k-tule with k sufficiently large, there are infinitely many x for which several L i x) simultaneously reresent square-free numbers with all rime factors in P. The basic setu is the same as [7]. We shall recall only the imortant notions and hyotheses in this section and refer our readers to [7] Section 2.2]) and [6] Section 3)) for a detailed exosition. As in [7] and [6], we define the quantities: k P L n) := i=1 L i n), A := lcm i a i ), SL) := A We recall from [7] the following admissibility constraint. 1 1 ) k A 1 k ) 1 1 ) k Definition 1. Given a ositive integer, M, a k-tule of linear forms L = {L 1,..., L k } is M-admissible if the following conditions hold simultaneously. n i) For every rime,, there exists an integer, x, such that a i x + b i ); ii) for each i, M divides a i ; iii) for each i, M is corime to a i /M. A k-tule of linear forms, L, is called admissible if it satisfies only i). The above stronger admissibility constraint is introduced to incororate the fact that P may fail to be well-distributed modulo M. We will rimarily consider the case when a 1 = = a k = M. Given a set of linear forms L = {L 1,..., L k } with L i n) = a i n + b i, remove finitely many rimes from P, so that A, ) = 1 for all P. Throughout the aer, we shall use for a sum over all the values relatively rime to A and any rime PN) this is different from [7], which only requires the values to be relatively rime to A). As in [7] and [6], we may assume without loss of generality that M-admissibility can be relaced by a stronger condition, which we label Hyothesis AM). The justification for this hyothesis aears in [7]. Hyothesis AM). L = {L 1,..., L k } is an M-admissible k-tule of linear forms. The functions L i n) = a i n + b i 1 i k) have integer coefficients with a i > 0. Each of the coefficients, a i, is divisible by the same set of rimes, none of which divides any of the b i. If i j, then any rime factor of a i b j a j b i divides each of the a i. i=1

8 Mathematics 2014, Preliminary Lemmas In this section, we shall rovide the setu of the roof of the main theorem and rove a few key lemmas. We first recall a lemma from [6], which we shall use frequently in this section. Lemma 2. [6] Lemma 4)) Suose that γ is a multilicative function, and suose that there are ositive real numbers κ, A 1, A 2, L, such that: 0 γ) 1 1 A 1 and: L w <z γ) log if 2 w z. Let g be the multilicative function defined by: gd) = d κ log z w A 2 γ) γ) Let: c γ := 1 γ) ) ) κ Assume that F : [0, 1] R is a iecewise differentiable function. Then: ) log z/d µ 2 log z) κ 1 d)gd)f = c γ F 1 x)x κ 1 dx + Oc γ LMF )log z) κ 1 ) log z Γκ) d<z 0 where: MF ) = su{ F x) + F x) ) : 0 x 1}. The constant imlied by O may deend on A 1, A 2 and κ, but it is indeendent of L and F. Following the aroach in [7] and [6], we shall consider the sum: S := k ) ξ P L i n)) ν i=1 d P L n) where λ d are real numbers to be described later. As in [7] and [6], Theorem 2 will follow from the ositivity of the sum, S. Note that there is a key distinction between our definition of S and the definition in [7] and [6]. Here, we sum u λ d over only the square-free numbers, d, that are relatively rime to all the rimes in PN). Intuitively, this gives a bigger sieve weight to the values, n, where P L n) has many rime factors in PN); hence, the ositivity of S can be satisfied for smaller k. As we can see in the roof of Lemma 4, this change also simlifies the calculation of S. Define: µ 2 r)sl) if r < R, r, A) = 1 and r, ) = 1 for all PN) y r = 0 otherwise λ d 2

9 Mathematics 2014, 2 45 For each square-free number, d, let: fd) := d τ k d) = d k, f 1 := f µ The sieve weights, λ d, are related to the quantities, y r, by: Then, by Möbius inversion, we have: λ d = µd)fd) y r = µr)f 1 r) r d y rd f 1 rd) λ dr fdr) Since the sum of λ d is taken over all the square-free numbers relatively rime to the rimes in PN), we take y r to be suorted only on integers corime to PN), which imlies that the λ d are also suorted on integers corime to PN). To determine S, we break the sum into arts and evaluate each of them individually. Let: S 1,j := ξ P L j n)) 2 and S 0 := 2 Then: d P L n) λ d S = k S 1,j νs 0 We shall now estimate S 0 and S 1,j in the following two lemmas. j=1 d P L n) Lemma 3. Suose that L is a set of linear forms satisfying Hyothesis AM). There is a constant, C, such that if R N 1/2 log N) C, then: S 0 = SL) 1 1 ) kα 1 k ) Nlog R) k1 α) Γk1 α) + 1) + ONlog N)k1 α) 1 ) PN) Proof. From the definition of S 0, we have: S 0 = [] P L n) 1 = N f[d, e]) + O where for each square-free d with d, A) = 1 and d, ) = 1 for all P, r d := d P L n) 1 N fd) r [] As in the roof of Theorem 7 in [6], note that the error term is ON) if R N 1/2 log N) 3k. For the main term, ) λ d f[d, e]) = fd)fe) r ) f 1 r) = f 1 r) r d λ dr fdr) ) 2 = r µ 2 r)y 2 r f 1 r) = µ 2 r)sl) 2 f 1 r) r

10 Mathematics 2014, 2 46 We use Lemma 2 with: k if A and / PN) γ) = 0 otherwise and F x) = 1. Then, gd) = f 1 d) 1. It can verified as in [6] Lemma 7)) that the conditions in Lemma 2 are satisfied with κ = k1 α), using the fact that the Frobenius density of PN) is α < 1. Then, the main term becomes: SL) 1 1 ) kα 1 k ) Nlog R) k1 α) 1 x k1 α) 1 dx Γk1 α)) PN) 0 with the desired error term. The result follows by evaluating the integral. Remark 4. This argument breaks down when α = 1, since κ needs to be ositive in order to aly Lemma 2. A similar henomenon occurs in the roof of Lemma 4. Lemma 4. Let L be a set of linear forms satisfying Hyothesis AM). There is a constant, C, such that if R = N 1/4 log N) C, then: where: S 1,j D := SL)δ b j φm)c P Γ21 α) + 1) Γ2 α) 2 Γk + 1)1 α) + 1) DNlog R)k+1)1 α) log 1 α N 1 1 ) αk+1) 1 1 ) 1 k ) PN) and c P = c P N) > 0 satisfies: ξ P n) = c P N log 1 α N) Remark 5. The ratio, c P N), aroaches a ositive constant as N tends to infinity by Theorem 2.4 in [13]. Proof. From the definition of S 1,j, we have: S 1,j = ξ P L j n)) d P L n) λ d 2 = [] P L n)/l j n) ξ P L j n)) We remark that in the second equality, we have used the condition that [d, e] is not divisible by any rime in PN), and hence, the condition [d, e] P L n) in the sum imlies that [d, e] P L n)/l j n) for nonzero ξ P L j n)). This contributes to the much simler estimate of S 1,j than that in [7]. Let Ω x) denote the set of residue classes, a, modulo x, such that x P L a)/l j a). Note that Ω x) = τ k 1 x) by Hyothesis AM) cf. 4.4) in [7]). Then: S 1,j = ξ P L j n)) a Ω []) n amod [])

11 Mathematics 2014, 2 47 Write n = L j n) = a j n + b j, then a j N + b j < n 2a j N + b j, with n aa j + b j mod [d, e]) and n b j mod a j ). By assumtion, the values, [d, e], a j /M, M, are all corime. Let u = [d, e]a j /M. Then, we may use the Chinese Remainder Theorem to combine the congruence conditions modulo [d, e] and a j /M into a single condition on u. Let Ω 1x) denote the set of all ossible residue classes of n modulo x. Then: S 1,j = ξ P n ) Now, we decomose the inner sum: a j N+b j <n 2a j N+b j n a mod u ) n b j mod M) ξ P n ) = 1 φu ) a Ω 1 u ) a j N<n 2a j N n b j mod M) a j N+b j <n 2a j N+b j n a mod u ) n b j mod M) ξ P n ) + P,bj a j N; u, a ) + O bj 1) Accordingly, we can decomose S 1,j into its main term and error term S 1,j = M 1,j + E 1,j. Let P,bj X; u ) := max a,u )=1 P,bj X; u, a). The error term can be estimated using Lemma 1 and Cauchy s inequality as in the roof of Lemma 4.1 in [7]. Let v = [d, e], then u = a j v/m. Note that Ω 1u ) = Ω [d, e]) = τ k 1 [d, e]). Moreover, by Hyothesis AM), we have a, u ) = 1 for all a Ω 1u ). Hence: E 1,j = a Ω 1 u ) P,bj a j N; u, a ) + O bj 1)) τ k 1 [d, e]) P,bj a j N; u ) + O1)) log 2k N v R 2 3k 3) ωv) P,bj U log 2k N)a j N) log U a j N) N log 2k U N a j N; a jv M for any U. To obtain the third line, we use the fact that λ d log k R log k N by 4.3) of [6]. For the main term of S 1,j, M 1,j = τ k 1 [d, e]) φu ) φm) δ bj φa j ) a j N<n 2a j N a j N<n 2a j N n b j mod M) ξ P n) ) ξ P n) 1) τ k 1 [d, e]) φ[d, e]) where δ bj is the density of the elements congruent to b j mod M in the set of E PN) numbers.

12 Mathematics 2014, 2 48 Our next ste is to evaluate the last sum. Let f n) = φn)/τ k 1 n) and f 1 = f µ. Then: τ k 1 [d, e]) φ[d, e]) by an analogue of Lemma 6 in [6], where: Thus: We use Lemma 2 with: = f [d, e]) = f d)f e) f d, e)) = f f d)f 1 r) e) r ) = f1 r) f d) r = r r d µ 2 r) f 1 r) y r) 2 y r := µ2 r)r φr) y r = µ2 r)rsl) φr) m λ d y mr φm) m<r/r m,ra)=1 µ 2 m) φm) 1 if ra and / PN) γ) = 0 otherwise and F x) = 1. Again, the conditions are verified as in [6] Lemma 7)) with κ = 1 α, giving: c γ = φra) 1 1 ) α 1 1 ) ra PN) Hence for square-free r with r, A) = 1 and r, ) = 1 for all PN), yr φa)sl) 1 1 ) α 1 1 ) log R/r) 1 α A Γ1 α) Thus: r = φa)sl) A µ 2 r) f1 r) y r) 2 φa)2 SL) 2 A 2 PN) 1 1 ) α PN) 1 1 ) log R) 1 α Γ2 α) x α dx log R/r log R ) 1 α 1 1 ) 2α 1 1 ) 2 log R) 21 α) µ 2 r) Γ2 α) 2 f PN) r<r 1 r) We evaluate the last sum using Lemma 2, taking F x) = x 21 α) and: k 1) if ra and / PN) γ) = 1 0 otherwise ) 21 α) log R/r log R

13 Mathematics 2014, 2 49 The conditions are satisfied when κ = k 1)1 α), as verified in [6] [Lemma 8]). Then: c γ = 1 A 1 1 ) αk 1) 1 1 ) 1 1 k ) SL) φa) Hence: µ 2 r) f1 r) y r) 2 r φa)sl) A PN) 1 1 ) αk+1) 1 1 ) 1 k ) log R) k+1)1 α) Γ21 α) + 1) Γ2 α) 2 Γk + 1)1 α) + 1) PN) Plugging the definition of c P and this identity into 1), we have the desired result. Note that here we have used the fact that a j /φa j ) = A/φA) by Hyothesis AM) Proof of Theorems 1 and 2 Proof of Theorem 2. Let R = N 1/4 log N) C, where C is the constant in Lemma 2.6. Noting that 4 log R log N, we have, as N, S SL) 1 1 ) kα 1 k ) Nlog R) k1 α) kdk) ν) Γk1 α) + 1) where: PN) Dk) := c PδφM) Γk1 α) + 1)Γ21 α) + 1) 4 1 α Γ2 α) 2 Γk + 1)1 α) + 1) Hence, S is ositive for all large enough N if: where: and: bk) := 4 1 α kdk) ν k > ν b1) bk) c P δφm) Π := 1 1 ) α 1 1 ) PN) Γ2 α) Π 1 1 ) α 1 1 ) PN) Γ1 α)γk1 α) + 1) Γk + 1)1 α) + 1)) = B1 α, k1 α) + 1) is the beta function. Finally, by a variant of the Tauberian theorem [19] Theorem 2.4.1)), we deduce that: 1 c P Π ζ PN) 2)Γα) = ) Γα) 2 PN) where, as we recall from Lemma 4, c P = c P N) > 0 is the function that satisfies: ξ P n) = c P N log 1 α N) For N large enough, the infinite roduct aroaches one. Now, the result follows from Equation 2). 2)

14 Mathematics 2014, 2 50 Proof of Theorem 1. To derive Theorem 1.1 from our main theorem, we consider for given k and m a set of k rimes, b 1,..., b k, with the aroriate residue classes as needed. Then, {Mx + b i } forms an M-admissible k-tule, which can be normalized to fit Hyothesis AM). This gives us the value, b k b 1, for the constant, Cν, P). 3. Discussion of Examles In this section, we shall revisit two of the examles in [7]. We observe that both Theorem 1.2 and Corollary 1.3 in [7] rely on theorems that are not just alicable to E r numbers, but any square-free numbers whose rime factors satisfy a Chebotarëv condition. Therefore, Theorem 2 alies in these two examles, and we can get better bounds in both cases Examle 1: Ideal Class Grous with Order Four Elements We first recall a result of Soundararajan [8]. Proosition 1 and 2 in [8] show that for any ositive square-free number d 1 mod 8), whose rime factors are all congruent to ±1 mod 8), the class grou ClQ d)) contains an element of order four. Alying Theorem 2 with ν = 1, α = 1/2, δ = 1/2 and M = 8, the right-hand side of Theorem 2 is < 2 when k = 2. Hence, we may take k = 2. Considering the eight-admissible two-tule {8n + 17, 8n + 25}, we obtain Corollary 1, which is an imrovement on the bound in [7] Examle 2: Rank Zero Quadratic Twists of Modular Ellitic Curves As in Section 6 of [7], we shall focus on the ellitic curve E = X 0 11), recalling the setting from [9]. We take the cubic model of X 0 11) to be: y 2 = fx) = x 3 4x 2 160x 1264 The Galois reresentation: ρ f : GalQ/Q) GL 2 Z/2Z) induced by the natural action of GalQ/Q) on the two-torsion oints of E 11) has the roerty that: trρ f Frob )) a) mod 2) for all, excet finitely many, rimes,. Here, a) = + 1 #EF ), and Frob is the Frobenius element at in GalQ[fx)]/Q). Let S be the set of rimes,, such that trρ f Frob )) 1 mod 2). We remark that S is also the set of all rimes, such that fx) is irreducible mod. Equation 1.4) and Theorem 1.1 of Boxer-Diao [10] establish that for any ositive square-free number, d, with rime factors all in S, we have: LE d), 1) 0 and rke d), Q) = 0 We note that GalQ[fx)]/Q) = S 3 ; hence, the Chebotarëv density theorem shows that S has density α = 1/3. Murty and Murty s theorem [20] now imlies that S satisfies SW 11), as analyzed in [7]. Furthermore, since a) is odd only if = 11 or is a quadratic residue mod 11 cf. the roof of

15 Mathematics 2014, 2 51 Corollary 1 in [9]) and these values distribute uniformly among the five quadratic residue classes, one may take δ = 0.2. Given ν = 1, α = 1/3, δ = 0.2 and M = 11, the right-hand side of Theorem 2 is < 8 when k = 8. Hence, we may take k = 8. One may check that the eight-tule: {1, 3, 9, 15, 25, 31, 45, 49} contains only quadratic residues mod 11 and, hence, forms an 11-admissible eight-tule {11n+b j }, such that δ 0.2. As a result, there are infinitely many airs of square-free m and n with: LE m), 1) LE n), 1) 0, rke 11m)) = rke 11n)) = 0 and: m n 48 Acknowledgments The authors are grateful to their advisors, Ken Ono and Robert Lemke Oliver, and the suort from the National Science Foundation of the Research Exerience for Undergraduates at Emory University. Conflicts of Interest The authors declare no conflict of interest. References 1. Goldston, D.; Pintz, J.; Yıldırım, C. Primes in tules. I. Ann. Math. 2009, 2, Zhang, Y. Bounded gas between rimes. Ann. Math. to aear. 3. Maynard, J. Small gas between rimes. Available online: htt://arxiv.org/abs/ accessed on 28 February 2014). 4. Polymath, Bounded gas between rimes. Available online: htt://michaelnielsen.org/ olymath1/index.h?title=bounded gas between rimes accessed on 28 February 2014). 5. Chen, J.-R. On the reresentation of a large even integer as the sum of a rime and a roduct of at most two rimes. Sci. Sinica 1973, 16, Goldston, D.; Graham, S.; Pintz, J.; Yıldırım, C. Small gas between roducts of two rimes. Proc. Lond. Math. Soc. 2009, 3, Thorne, F. Bounded gas between roducts of rimes with alications to ideal class grous and ellitic curves. Int. Math. Res. Not. IMRN 2008, 5, doi: /imrn/rnm Soundararajan, K. Divisibility of class numbers of imaginary quadratic fields. J. Lond. Math. Soc. 2000, 61, Ono, K. Twists of ellitic curves. Comos. Math. 1997, 106, Boxer, G.; Diao, P. 2-Selmer grous of quadratic twists of ellitic curves. Proc. Am. Math. Soc. 2010, 6, Deligne, P.; Serre, J.-P. Formes modulaires de oids 1. Ann. Sci. École Norm. Su. 1974, 7,

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