LARGE GAPS BETWEEN CONSECUTIVE PRIME NUMBERS CONTAINING SQUARE-FREE NUMBERS AND PERFECT POWERS OF PRIME NUMBERS

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1 LARGE GAPS BETWEEN CONSECUTIVE PRIME NUMBERS CONTAINING SQUARE-FREE NUMBERS AND PERFECT POWERS OF PRIME NUMBERS HELMUT MAIER AND MICHAEL TH. RASSIAS Abstract. We rove a modification as well as an imrovement of a result of K. Ford, D. R. Heath-Brown and S. Konyagin [] concerning rime avoidance of square-free numbers and erfect owers of rime numbers. 010 Mathematics Subject Classification: 11P3. 1. Introduction In their aer [], K. Ford, D. R. Heath-Brown and S. Konyagin rove the eistence of infinitely many rime-avoiding erfect k-th owers for any ositive integer k. They give the following definition of rime avoidance: an integer m is called rime avoiding with constant c, if m + u is comosite for all integers u satisfying 1 u c log m log m log 4 m log 3 m). In this aer, we rove the following two theorems: Theorem 1.1. There is a constant c > 0 such that there are infinitely many rimeavoiding square-free numbers with constant c. Theorem 1.. For any ositive integer k, there are a constant c = ck) > 0 and infinitely many erfect k-th owers of rime numbers which are rime-avoiding with constant c. We largely follow the roof of [].. Proof of the Theorem 1.1 Lemma.1. For large and z log 3 /10 log ), we have {n : P + n) z} log ) 5, where P + n) denotes the largest rime factor of a ositive integer n. Proof. This is Lemma.1 of [] see also [8]). Date: October 3, We denote by log = log log, log 3 = log log log, and so on. 1

2 Lemma.. Let R denote any set of rimes and let a Z \ {0}. Then, for large, we have { : amodr) r R)} 1 1 ). log R Note. Here and in the sequel will always denote a rime number. Proof. This is Lemma. of [] see also [4]). Lemma.3. Let N =. Then there is m 0 Z, such that for all m m 0 mod N) we have: m + u is comosite for u [ y, y]. Proof. The argument for the roof aears in [8] Proof of Theorem 1.1. We now consider the arithmetic rogression *) m = kn + m 0, k N. By elementary methods see Heath-Brown [6] for references) the arithmetic rogression *) contains a square-free number 1) m N 3/+ε, where ε > 0 is arbitrarily small. By the rime number theorem, we have ) N e +o). We know that m + u is a comosite number for u [ y, y] see [8]). estimates 1) and ), we obtain y c log m log m log 4 m log 3 m) for a constant c > 0, which roves Theorem Primes in arithmetic rogressions The following definition is borrowed from [7]. By the Definition 3.1. Let us call an integer q > 1 a good modulus, if Ls, χ) 0 for all characters χ mod q and all s = σ + it with σ > 1 C 1 log [q t + 1)]. This definition deends on the size of C 1 > 0. Lemma 3.. There is a constant C 1 > 0 such that, in terms of C 1, there eist arbitrarily large values of for which the modulus is good. P ) = < Proof. This is Lemma 1 of [7]

3 Lemma 3.3. Let q be a good modulus. Then π; q, a) φq) log, uniformly for a, q) = 1 and q D. Here the constant D deends only on the value of C 1 in Lemma 3.. Proof. This result, which is due to Gallagher [3], is Lemma from [7]. 4. Congruence conditions for the rime-avoiding number Let be a large ositive number and y, z be defined as in Definition??. Set P ) =. We will give a system of congruences that has a single solution m 0, with 0 m 0 P ) 1 having the roerty that the interval [m k 0 y, m k 0 + y] contains only few rime numbers. Definition 4.1. We set Lemma 4.. We have P 1 = { : log or z < /40k}, P = { : log < z}, U 1 = {u [ y, y], u Z, u for at least one P 1 }, U = {u [ y, y] : u U 1 }, U 3 = {u [ y, y] : u is rime}, U 4 = {u [ y, y] : P + u ) z}, U 5 = {u U 3 : u + k 1 for P } U = U 3 U 4. Proof. Assume that u U \ U 4. Then by Definition 4.1 there is a rime number 0 P with 0 u. Since u U 1, we have 0 > /4. Thus, there is no rime 1 u 0, since otherwise u 0 1 > 4 log > y, a contradiction. Thus u = 0 and therefore u U 3. Lemma 4.3. We have Proof. This follows from Lemma.1. U 4 log ) 4. A trivial consequence of Lemma.. is the following Lemma: Lemma 4.4. We can choose the constants c 1, c such that U 5 30k log. 3

4 For the net definitions and results we follow the aer []. For the convenience of the reader we reeat the elanations of []. Let k be odd. For each u U associate with u a different rime u 40k, ] such that u 1, k) = 1 e.g. one can take u mod k), if k 3). Then every residue modulo u is a k-th ower residue. Let k be even. There do not eist rimes for which every residue modulo is a k-th ower residue. We maimize the density of k-th ower residues by choosing rimes such that 1, k) =, e.g. taking 3mod 4k). For such rimes every quadratic residue is a k-th ower residue. Definition 4.5. Let { { : 40k P 3 = <, mod k)}, if k is odd { : 40k <, 3 mod 4k)}, if k is even, We now define the ecetional set U 6 as follows: For k odd we set U 6 =. For k even and δ > 0, we set { ) u U 6 = u [ y, y] : = 1 for at most Lemma 4.6. if δ is sufficiently small. U 6 ε 1/+ε, Proof. Each u may be written uniquely in the form u = s a u 1u, } δ log rimes P 3. where s = ±1, a {0, 1} and u is odd and squarefree. From 3mod4k) it follows by the law of quadratic recirocity, that ) ) 1 = 1, = 1. Therefore *) We consider the sum ) u = s 1) u 1 S = u U P 3 ) ) a. u Given u, there are at most y/u y choices for u 1. Each of the eight ossibilities ) for the choices s { 1, 1}, a {0, 1}, u 1 or 3 mod 4) leads to a coefficient of u on the right hand side of *) that is indeendent of. Thus, we have S y ) 1/ ε 5/+ε u u y P 3 4 ) u

5 by Lemma.3 of []. If u U 6, then clearly P 3 ) u η log with η = ηk) > 0. It follows that S U 6 / log ), and consequently that Definition 4.7. We set Lemma 4.8. We have U 6 ε 1/+ε. U 7 = U 4 U 5. U 7 0k log. Proof. This follows from Definition 4.7 and Lemmas 4.3, 4.4 We now introduce the congruence conditions, which determine the integer m 0 uniquely mod P )). Definition 4.9. C 1 ) m 0 1 mod ), for P 1, C ) m 0 mod ), for P. For the introduction of the congruence conditions C 3 ) we make use of Lemma 4.8. Since P 3 U 7, there is an injective maing Φ : U 7 /U 6 P 3, u P u. We set P 3 = ΦU 7 /U 6 ). Every residue modulo u is a k-th ower residue and we take m u such that m k u u 1) mod u ) The set C 3 ) of congruences is then defined by C 3 ) m 0 m u mod u ), u P 3. Let P 4 = { [0, ) : P 1 P P 3 }. The set of congruences is then defined by C 4 ) m 0 1 mod ), P 4. Lemma The congruence systems C 1 ) C 4 ) and the condition 1 m 0 P ) 1 determine m 0 uniquely. We have m 0, P )) = 1. Proof. The uniqueness follows from the Chinese Remainder Theorem. The corimality follows, since by the definition of C 1 ) C 4 ) m 0 is corime to all, with 0 <. 5

6 Lemma Let m m 0 mod P )). Then m, P )) = 1 and the number is comosite for all u [ y, y] \ U 6. m k + u 1) Proof. For u U 1, there is P 1 with u. Therefore, since by Definition 4.9, the system C 1 ) imlies that m 0 1 mod ), we have i.e. m k + u 1) m k 0 + u 1) 1 + u 1 u 0 mod ), m k + u 1). For u U 3, u U 5, there is P with u + k 1. Since by C ) m 0 mod ), we have i.e. m k 0 + u 1) k k 0 mod ), m k + u 1). There is only one remaining case, namely u U 7 /U 6, and one uses C 3 ). 5. Conclusion of the roof of Theorem 1. Let now be such that P ) is a good modulus in the sense of Definition 3.1. By Lemma 3., there are arbitrarily large such elements. Let D be a sufficiently large ositive integer. Let M be the matri with P ) D 1 rows and U = y + 1 columns, with the r, u element being a r,u = m 0 + rp )) k + u 1, where 1 r P ) D 1 and y u y. Let N 0, k) be the number of erfect k-th owers of rimes in the column C 1 = {a r,1 : 1 r P ) D 1 }. Since P ) is a good modulus, we have by Lemma 3. that 5.1) N 0, k) C 0 k) P ) D 1 logp ) D 1 ). Let R 1 be the set of rows R 1, in which these owers of rimes aear. We now give an uer bound for the number N 1 of rows R r R 1, which contain rimes. We observe that for all other rows R r R 1, the element a r,1 = m 0 + rp )) k is a rime avoiding k-th ower of the rime m 0 + rp ). Lemma 5.1. For sufficiently small c, we have Proof. For all v with v 1 U 6, let N 1 1 N 0, k). T v) = {r : 1 r P ) D 1, m 0 + rp ) and m 0 + rp )) k + v 1 are rimes}. We have 5.) N 1 v U 6 T v). 6

7 A standard alication of sieves gives 5.3) T v) P ) D 1 < P ) < P ) By Lemma 3.1 of [], we have 1 ρ) ) 1 1 ) 1 ρ) ). < P ) k,ε v ε log log P ). Lemma 5.1 now follows from 5.), 5.3) and the bound for U 6. This comletes the roof of Theorem 1.. Acknowledgements. We would like to thank the referee for his very valuable comments which imroved the resentation of the aer. References [1] A. C. Cojocaru, M. Ram Murty, An Introduction to Sieve Methods and their Alications, Cambridge Univ. Press, 006. [] K. Ford, D. R. Heath-Brown and S. Konyagin, Large gas between consecutive rime numbers containing erfect owers, In: Analytic Number Theory. In honor of Helmut Maier s 60th birthday, Sringer, New York, 015 to aear). [3] P. X. Gallagher, A large sieve density estimate near σ = 1, Invent. Math., ), [4] H. Halberstam and H. -E. Richert, Sieve Methods, Academic Press, London, [5] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th edition, Oford Univ. Press, [6] D. R. Heath-Brown, The least square-free number in an arithmetic rogression, J. Reine Angew. Math., 33198), [7] H. Maier, Chains of large gas between consecutive rimes, Adv. in Math., ), [8] R. A. Rankin, The difference between consecutive rime numbers, J. London Math. Soc., ), Deartment of Mathematics, University of Ulm, Helmholtzstrasse 18, 8901 Ulm, Germany. address: helmut.maier@uni-ulm.de Deartment of Mathematics, ETH-Zürich, Rämistrasse 101, 809 Zürich, Switzerland & Deartment of Mathematics, Princeton University, Fine Hall, Washington Road, Princeton, NJ , USA address: michail.rassias@math.ethz.ch, michailrassias@math.rinceton.edu 7