The distribution of consecutive square-free numbers

Transcription

1 Ingela Mennema The distribution of consecutive suare-free numbers Master Thesis Supervisor: Dr E Sofos Co-supervisor: Dr J-H Evertse 6 May 207 Department of Mathematics Leiden University

2 Contents Introduction 2 2 Preliminaries 5 3 Proof of Theorem Proof of Theorem 22 5 Proof of Theorem 3 26 Introduction Dirichlet proved in 837 that for any two positive integers a and there are infinitely many primes p such that p a mod Let (x;, a) :=#{p apple x : p a mod } The Prime number theorem in arithmetic progressions states (x;, a) x () log x as x!, where is the Euler phi function Chebyshev observed in 853 that primes congruent to 3 modulo 4 are more freuent, in a specific uantitative sense, than primes congruent to modulo 4 Specifically, one would expect that the ineualities (x;4, ) > (x;4, 3) and (x;4, ) < (x;4, 3) each occur half of the time, since asymptotically half of the primes are of the form 4k + and half of the form 4k + 3 However, numerical evidence shows that the ineuality (x;4, ) < (x;4, 3) occurs much more freuently This phenomenon, which holds in greater generality, is known as Chebyshev s bias and has been studied extensively, see for example the work of Rubinstein and Sarnak [0] Let p n denote the n-th prime and for integers, a, b with gcd(, a) =gcd(, b) = let (x;, (a, b)) := # p n apple x : p n a mod, p n+ b mod This uantity is related to the Chebyshev s bias, because it considers the patterns of residues modulo among consecutive primes It was recently studied by Lemke Oliver and Soundararajan [6] They proved certain asymptotic estimates for it, conditionally on the, still open, Hardy Littlewood conjecture which regards prime values of collections of polynomials A very special case of their results is that for all large enough x one has (x;, (a, b)) > (x;, (a 0,b 0 )) 2

3 whenever a 6 b mod and a 0 b 0 mod Our aim in this thesis is to investigate similar phenomena for the seuence of suarefree integers Recall that a positive integer is called suare-free if it is not divisible by the suare of a prime This seuence is easier to handle and one might hope that such bias can be proved without resorting to an unproved hypothesis Let S(x) denote the number of suare-free integers up to x, then it is well-known that S(x) lim x! x = Y p p 2 = 6 2 Let a 2 Z, 2 N and let S(x;, a) be the number of suare-free numbers s apple x such that s a mod The proof of the following result is standard and can be found in Proposition 22, S(x;, a) = x Y p gcd(p 2,) a gcd (p 2,) + O( p x), p 2 where the definition of the Landau s O notation is given in the notation passage at the end of the introduction Denote by s n the n-th suare-free number and define for x 2 R and a, b, integers with >0 the uantity S (x;, (a, b)) := # {s n apple x : s n a mod, s n+ b mod } The main theorem in this thesis is about the probability that out of all suare-free integers s n apple x one has s n a mod and s n+ b mod, this probability will be denoted by S (x;, (a, b)) P x (;(a, b)) := x The proof of the following result is in 4 Theorem For any integers, a, b satisfying the following limit exists, 0 apple a< and a<bapple + a lim P x(;(a, b)) x!+ Furthermore, denoting its value by `(;(a, b)), the following rate of convergence holds: P x (;(a, b)) = `(;(a, b)) + O, log log x where the implied constant depends on, a and b 3

4 The size of gaps between consecutive suare-free integers has been a topic of intensive research; we shall draw upon work related to the entity M (x) := (s n+ s n ), ( 2 R 0 ) s n+ applex If one wishes to show that many of the gaps s n+ s n are small then upper bounds on M (x) for a single >0 and all large enough x are reuired It may be however that uite rarely there are very large gaps, these large gaps will make M (x) explode when is a large positive integer The first to study M (x) was Erdős [] who showed in 95 that for all 0 apple apple 2 there exists a constant C( ) such that M (x) lim x!+ x = C( ) () Several advances were made on the problem or proving () for larger values of later We give a table of all related developments here: 0 apple apple 2 Erdős [],95 0 apple apple 3 Hooley [4],973 0 apple apple 322 Filaseta [2],993 0 apple apple 3 =366 Huxley [5],995 To prove Theorem we shall partition the pairs (s n,s n+ ) in S(x;, (a, b)) according to the size of the gap s n+ s n Big gaps will be dealt with using the estimate of Erdős above, while small gaps will be dealt with by proving Theorem 2 below This theorem will be proven in 3 and forms the technical core of the present thesis We will not do so here, but we should mention that using the estimate of Huxley instead of the estimate of Erdős one can improve the error term in Theorem to O (log log x) for some > The Möbius function is defined by ( ) µ(n) :=, if n = p p t where p,,p t are distinct primes 0, if n is not suare-free Note that µ(n) 2 is an indicator function for suare-free numbers n For each small gap we will use the inclusion exclusion-principle For example if =4,a = and b =3we have that euals #{n apple z : n + is suare-free and the next suare-free is n +3} µ(4n +) 2 µ(4n +3) 2 napplez µ(4n +) 2 µ(4n +2) 2 µ(4n +3) 2 napplez The following theorem provides an expression for each summation that appears if one applies the inclusion-exclusion principle on every small gap 4

5 Theorem 2 Let 6= 0,a,,a k be integers with 0 apple a <a 2 < <a k and define the Euler product = Q p (!(p)p 2 ), where!(p) =# apple m apple p 2 : 9 apple i apple k with m + a i 0 mod p 2 Then we have for z a k, ky applenapplez i= µ(n + a i ) 2 = z + O z 2/3 e 2p a k (2k +logz) O(k) (2z) O where all implied constants depend at most on k log log z For the proof of this theorem we adopt the approach of Mirsky [7] Tsang [] and Reuss [8] also studied the sum in Theorem 2, however all previous work has focused on improving the dependence of the error term on z but does not provide uniformity of the error term with respect to k Making the dependence on k explicit is the new element of Theorem 2 This feature is important for us because in our proof of the asymptotic for S(x;, (a, b)) in Theorem we need to study suare-free values of many linear polynomials, the cardinality of which will be a function of x that tends to infinity The constant ` = `(;(a, b)) appearing in Theorem is unusually complicated, because it is an infinite sum of finite alternating sums of Euler products as is shown in (46) However, we can at least characterize all cases where `(;(a, b)) is non-zero, the proof of the following result is in 5 Theorem 3 The limit `(;(a, b)) in Theorem is strictly positive if and only if there exist suare-free integers s, t fulfilling s a mod and t b mod Notation Throughout this thesis we define N := Z >0 and (n, m) :=gcd(n, m) Furthermore p (n) will refer to the usual p-adic valuation and we write p k kn if k = p (n) We use the notation f(x) =O(g(x)) and the notation f(x) g(x) if for all x there exists a positive real C such that f(x) applecg(x) All the implied constants in Landau s O-notation and in Vinogradov s -notation may depend on the entities a, b and that appear in Theorem, but not on other variables 2 Preliminaries This sections is devoted to some auxiliary results that are well-known Definition 2 An arithmetic function is a function f : N! C!, 5

6 The associated Dirichlet series is defined as follows L f (s) := f(n)n s n= for every s 2 C for which the series converges Definition 22 A multiplicative function is an arithmetic function f : N! C such that f 6 0 and f(mn) =f(m)f(n) foreverym, n 2 N with (m, n) = A strongly multiplicative function is an arithmetic function f : N! C such that f 6 0 and f(mn) =f(m)f(n) foreverym, n 2 N Lemma 23 (Euler) Let f be a multiplicative function and let s 2 C be such that L f (s) converges absolutely Then L f (s) = Y! f p j p js p j=0 Corollary 24 Let f be a strongly multiplicative function and let s 2 C be such that L f (s) converges absolutely Then L f (s) = Y p f(p)p s Definition 25 The Riemann zeta function is given by (s) = for s 2 C with Re(s) > n s, n= Lemma 26 For every s 2 C with Re(s) > we have (s) = Y p p s Remark 27 The Möbius function is a multiplicative function Lemma 28 For every s 2 C with Re(s) > we have (s) = L µ(s) 6

7 Proof Note that L µ (s) converges absolutely for every s 2 C with Re(s) > Combined with Lemma 23 and Remark 27 we obtain L µ (s) = Y! µ(p j )p js = Y p s p p j=0 for every s 2 C with Re(s) > From Lemma 26 it follows that for every s 2 C with Re(s) > we have (s) = L µ(s) Lemma 29 Let n 2 N Then µ(d) =µ 2 (n) d 2 n Proof Write n = p p t t where p,,p t are distinct primes and,, t We have 0 0 d 2 n µ(d) =Y p m µ(p k ) C A = Y k2z p 2k n i= ( ) 0applekapple 2kapple i The inner sum euals if i = and 0 if i 2 Hence P d 2 n µ(d) euals if n is suare-free and 0 otherwise This implies that the sum is eual to µ(n) 2 We shall repeatedly make use of the following standard fact Lemma 20 Let a, b 2 Z and 2 N There exists n 2 N satisfying bn a mod if and only if (b, ) divides a Lemma 2 Let a, b 2 Z and 2 N and x 2 R with x # { apple n apple x : bn a mod } = where the implied constants are absolute x (b, )+O() O() Proof Let (b, ) - a, then from Lemma 20 it is clear that # { apple n apple x : bn a mod } = O() k Then C A if (b, ) a otherwise, Assume that (b, ) a By Lemma 20 there exists an integer n such that bn a mod b Euivalently, for this n the congruence n a mod b holds Since and (b,) (b,) (b,) (b,) (b,) are coprime, this congruence has a uniue solution mod,sayr Hence (b,) n o # { apple n apple x : bn a mod } =# apple n apple x : n r mod (b,) 7

8 Dividing the positive integers into blocks of length # apple n apple x : n r mod (b, ), we obtain (b,) = x (b, )+O() Proposition 22 Let a 2 Z and 2 N Then µ(n) 2 = x Y applenapplex n a mod p (p 2,) a (p 2,) + O( p x), p 2 where the implied constant is absolute Proof Using Lemma 29 we obtain µ(n) 2 = applenapplex n a mod d 2 n applenapplex n a mod Reordering the summation gives µ(d) = µ(d) dapple p x applenapplex n a mod applenapplex n a mod d 2 n µ(d) d 2 n = dapple p x µ(d) kapple x d 2 kd 2 a mod By Lemma 20 the congruence kd 2 a mod has a solution if and only if (d 2,) divides a Together with Lemma 2 it follows that the sum euals 0 x dapple p x (d 2,) a µ(d) d 2 Clearly, P dapple p x =O ( p x) Furthermore, (d 2,) a dapple p x (d 2,) a µ(d) d 2 (d2,)= d2n (d 2,) a d2n (d 2,) a µ(d) d 2 dapple p x (d 2,) a 0 C A d> p x (d 2,) a (d 2,) d 2 Note that (d 2,) is a multiplicative function with respect to d By Lemma 23 it follows that µ(d) d 2 (d2,)= Y (p 2,) p p 2 (p 2,) a 8 C A

9 Further, where we have used This proves the proposition x d> p x (d 2,) a d> p x (d 2,) d 2 apple x d> p x d = O px 2 d 2, The following result was proved in a slightly stronger form by Tsang [, Th 3] Theorem 23 (Tsang) Let f,,f r 2 Z[x] such that f i is irreducible over Z and deg f i apple 3 for all i Then applenapplex i= ry µ(f i (n)) 2 = cx + O(x 3/4 ), where c = Q p (! f(p)p 2 ) with! f (p) :=# apple n apple p 2 : 9 apple i apple r such that f i (n) 0 mod p 2 The implied constant depends on r as well as on the coe cients of each f i We will use the following upper bound for the number of primes up to x, a proof of which can be found in [9, Cor, E(36)] Lemma 24 For all x> we have (x) < x log x Definition 25 Let s 2 N Define s (d) to be the number of vectors (d,,d s ) 2 N s with d d s = d We shall henceforth use the following notational convention (d) := 2 (d) = d n Lemma 26 Let s 2 Z Then s+ (d) apple (d) s 9

10 Proof The proof is by induction For s =, it is clear Let s all k<sthe theorem holds Consider s+ (d) = d s d 2 d 2 d and assume that for Using d d 2 d for all d 2 d, the induction hypothesis gives s+ (d) apple s (d) d 2 d = (d) s (d) apple (d) s Lemma 27 For every >0 there exists a constant C > 0 such that (n) apple C n for all n Proof Clearly, p k = k + for any prime p and integer k 0, hence (n) n = Y p kn + p We divide the product into two parts, one part with the primes p<2 / and the other part with the primes p 2 / First, let p 2 /, so that p 2 Furthermore using the binomial formula we find 2 + Hence Next, let p<2 / We have Hence + p apple + 2 apple p 2 = e log 2 log 2 + p Combining all of the above gives = p + apple p log 2 + (n) n = Y p kn apple p<2 / Y p<2 / + p Y p kn p 2 / + log p

11 Hence with C = Q p<2 +,weget / log 2 (n) apple C n Lemma 28 For every n, s 2 N we have s (n) apple n O constant is absolute s log log 3n, where the implied Proof Applying the ineuality + x apple e x for x 2 R and using the proof of Lemma 27 we get C apple Y log 2, euivalently By Lemma 24 it follows that With = p<2 / e log(c ) apple log 2 papple2 / log 2 apple 2 (log 2) 2 2/ papple2 / 2log2 log log 3n We note that 2/ =(log3n) /2, therefore log(c ) apple 2 (log 2) 2 (log 3n)/2 This yields the following ineuality C apple n hence by Lemma 26 and 27 we obtain 2 (log 2) 2 (log 3n) /2, s (n) apple (n) s apple n O s log log 3n Lemma 29 There exists C > such that for all r 2 N 2 and for all x 2 R we have µ(n) 2 r (n) apple Cx(log r x) r napplex

12 Proof Letting f(n) =µ(n) 2 r (n) one easily sees that 8 < i =0 f(p i )= r i = : 0 i> Thus we infer and Lemma 24 reveals that Furthermore f(p)logp = r log p apple r (y)logy pappley pappley f(p)logp apple 2ry pappley p i> By [3, Th 0] with A =2r and B = 0, we find Note that napplex f(p i ) p i log p i =0 x f(n) apple (2r +) log x napplex f(n) n apple Y papplex + r apple p napplex Y papplex f(n) n! r +, p where the second ineuality follows from the binomial formula Note that Y + = Y Y p p p 2 papplex papplex papplex Mertens third theorem gives us that Y p = O(log x) papplex Furthermore by Lemma 26 it follows that Y lim x! papplex p 2 = (2) = 6 2 The euality (2) = 6 is a very well-known theorem by Euler Hence there exists C 2 > such that Y + apple C log x p 3 papplex 2

13 We now have f(n) apple Cx(log r x) r, napplex since (2r +) C 3 r apple C r for every r 2 Lemma 220 There exists C 2 > such that for all r 2 N 2 and for all x 2 R we have µ(d) 2 r (d) apple Cr 2(2(r ) + log x) r d 2 x d>x Proof We partition the interval (x, ) as follows therefore d>x (x, ) = µ(d) 2 r (d) d 2 = [ (e i x, e i+ x], i=0 i=0 e i x<dapplee i+ x µ(d) 2 r (d) d 2 By Lemma 29 there exists a C > such that for all r 2 N 2 and for all x 2 R we have Let z := 2(r i=0 e i x<dapplee i+ x ) + log x Then µ(d) 2 r (d) apple Cr d 2 x i=0 e i (i ++logx)r (i ++logx) r =(i ++z 2(r )) r apple (i + z) r = z r + i z r This implies i=0 e i (i ++logx)r apple z r i=0 + i r e i z Note that z 2(r ), euivalently + i z r apple + i 2(r ) r Furthermore + i 2(r ) apple e i 2(r ), hence r i + apple i r e e i 2(r ) e i 2(r ) =e i/2 3

14 There exists C 2 > such that C r x zr i=0 e i/2 apple Cr 2 x zr, since the series converges, this concludes the proof Erdős proved the following lemma in 95 [] Its meaning is that big gaps between consecutive suare-free numbers are rare Lemma 22 (Erdős) Let t, x 2 R >0 Then where the implied constant is absolute #{s n+ apple x : s n+ s n >t} x t 2 (log t) 2, Proof Let s n+ s n = r>tand A = s n <m<s n+ : 9p > t log t such that p 2 m 00 r We want to show that A for every r> 6 Assume t<6, then we have t log t < This implies that A = r, since every integer 00 m 2 N \ (s n,s n+ ) is divisible by the suare of a prime Note that r > r for r>, 6 r which implies A Assume that t 6, then it follows that r>6 We first look at 6 the number of integers between s n and s n+ that are divisible by the suare of a prime less than t log t Let 00 B = sn <m<sn+ 9papple t log t : 00 p2 m Note that B apple Calculations show that P p papple t log t 00 p 2 apple r p 2 +apple r p < 046 Hence p 2 + t log t 00 r p p 2 apple 2 r We want to show that t log t t log t < 2, hence =0apple 3t Assume t>50 By Lemma 24 we obtain t log t function t log t 50 log(t log t/00) f(t) = 3 t 8 t log t is smaller than 3 t for t 6 For t apple 50 we have apple t log t Consider the 50 log(t log t/00) 4

15 We want to show that f(t) apple Note that f(t) = 8 log t 3 50(log t +loglogt log 00) 8 Clearly, we have f(t) apple if and only if 50 log t +50loglogt 50 log 00 log t We 3 infer that f() apple if and only if t 7 75 log t 00 The latter holds for t = 50 and hence for t>50, since the function t 7! t 7 75 log t is increasing for t>50 Hence B apple 7r Furthermore A + B r, since every integer m 2 N \ (s 8 n,s n+ ) 7 is divisible by the suare of a prime This implies A r r = r > r, since we have assumed that r > 6 It follows that # <m<x 9p> t log t 00 such that p 2 m On the other hand # <m<x 9p> t log t 00 such that p 2 m apple t 6 #{s n+ apple x : s n+ s n t} p> t log t 00 apple x p 2 apple x p> t log t 00 p 2 Using Lemma 24 and the fact that the function /p 2 is decreasing we obtain x p> t log t 00 x p 2 t log t 00 log( t log t 00 ) Observe that log t log t 00 =logt +loglogt 00 and furthermore that log log t = O(log t) and also log 00 = O(log t) This implies and concludes the proof # <m<x 9p> t log t 00 : p2 m x t(log t) 2 3 Proof of Theorem 2 We would like to write each term of the summation in Theorem 2 as one Möbius function For this to happen we need all n + a i to be coprime in pairs For a lot of choices of a,,a k this does not happen We introduce the two following functions and study their properties 5

16 Definition 3 Let W 2 N and define µ W and eµ by 0 µ W (n) Y p W p p(n) A and 0 eµ(n) Y p-w p p(n) A Lemma 32 Let n, W 2 N Then µ(n) =µ W (n)eµ(n) Proof Obvious Lemma 33 Let n, m 2 N be such that n m mod W and assume that W is an integer suare Then µ W (n) =µ W (m) Proof Note that p W if and only if p 2 W Letµ W (n) = 0 Then there exists a p W such that n 0 mod p 2 It follows that m 0 mod p 2, since n m mod p 2 Hence µ W (n) = 0 if and only if µ W (m) =0, since the relation congruence modulo W is symmetric Let p W Then the congruence n m mod p implies that p n if and only if p m This implies # {p prime : p W, p n} =#{p prime : p W, p m} Let p be a prime such that p W and p n, but p 2 - n The congruence n m mod p 2 implies p 2 - m Hence p prime : p W and p n implies p 2 - n = p prime : p W and p m implies p 2 - m This concludes the proof Lemma 34 Let n, W 2 N Then eµ(n) 2 = d 2 n (d,w )= µ(d) Proof Let n = p p t t p t+ t+ p s s where p i W for all apple i apple t and p j - W for all t<japple s Define n = p p t t and n 2 = p t+ t+ p s s Clearly (n,n 2 ) = and n n 2 = n Furthermore eµ(n) 2 = µ(n ) 2 and by Lemma 29 we find µ(n ) 2 = P d 2 n µ(d) Note that {d 2 N : d 2 n } = {d 2 N : d 2 n, (d, W )=} 6

17 Hence it follows that eµ(n) 2 = d 2 n (d,w )= µ(d) For the rest of this thesis we will have W := 2 Y p 2 applea k a p 2 Lemma 35 Let 6= 0,n,a,,a k be integers with 0 apple a < a 2 < < a k If d 2 i n + a i, (d i,w)=, d 2 j n + a j and (d j,w)=, then (d i,d j )=for all j 6= i Proof Assume that d i and d j are not coprime Then there exists a prime p such that p d i and p d j So p 2 d 2 i and p 2 d 2 j Therefore p 2 n + a i and p 2 n + a j We infer that p 2 a j a i, since p 2 divides the di erence of (n + a j ) and (n + a i ) This implies p 2 apple a j a i Furthermore (d i,w) = and (d j,w) = imply p 2 >a k a Hence (d i,d j ) =, since a k a a j a i for all j 6= i Define et m (z) = applenapplez n m mod W ky eµ(n + a i ) 2 By Lemma 32 and Lemma 33 it follows that the sum in Theorem 2 euals Define M = ( applemapplew i= apple m apple W : Hence the sum in Theorem 2 euals i= ky µ W (m + a i ) 2 Tm e (z) m2m ) ky µ W (m + a i ) 2 = i= since Q k i= µ W (m + a i ) 2 is either zero or one et m (z), (3) 7

18 Proposition 36 Let 6= 0,a,,a k be integers with 0 apple a <a 2 < <a k and let m be an integer with apple m apple W Then there exists a constant := (, a,,a k,w,m) such that for all z a k, et m (z) = z + O z 2/3 (2k +logz) O(k) (2z) O k log log z!, where the implied constants depend at most on Remark 37 Before we start the proof, we describe its main idea First we use Lemma 34 which provides the euality 0 ky et m (z) = µ(d i ) C A (32) applenapplez n m mod W i= d 2 i n+a i (d i,w )= Next we partition the summation domain into specific parts as follows Let A,,A k be positive reals to be chosen later Assume that d d k >A for some integers d,,d k By demanding d apple appled k we get an extra factor k k in the error term, since the number of ways to order d,,d k is eual to k! apple k k For all 2 apple m apple k define D m := (d,,d k ) 2 N k d : apple appled k, (d i,w) = for all i d m d k apple A m, d m d k >A m We have z a k, therefore d 2 i apple n + a i apple z + a i apple 2z for all i, hence d k apple p 2z Take A k := p 2z Let d d k >A Ifd 2 d k apple A 2, then d 2 D 2 Letd 2 d k >A 2 Ifd 3 d k apple A 3, then d 2 D 3 Let d 3 d k > A 3 Continue this way until d k d k > A k Since d k apple p 2z = A k, we obtain d 2 D k This proves that whenever d d k >A then there exists an integer m such that d 2 D m Proof of Proposition 36 We begin by studying the contribution of d i in (32) with d d k apple A This contribution will provide our main term It is eual to eb m (z) := applenapplez n m mod W ky i= d 2 i n+a i (d i,w )= d d k applea µ(d i )= d,,d k 2N (d i,w )=, d d k applea ky µ(d i ) i= applenapplez n+a i 0modd 2 i n m mod W By Lemma 35 we find Q k i= µ(d i)=µ(d d k ) Note that (d 2 i,) =, since (d i,w)=, therefore has a multiplicative inverse modulo d 2 i Hence n+a i 0 mod d 2 i is euivalent 8

19 a i to n mod d2 i By the Chinese remainder theorem there exists an integer n 0 such that n n 0 mod Wd 2 d 2 k Thus, we get eb m (z) = µ(d d k ) (33) d,,d k 2N (d i,w )=, d d k applea = d,,d k 2N (d i,w )=, d d k applea d,,d k 2N (d i,w )=, d d k applea applenapplez n n 0 mod Wd 2 d2 k z µ(d d k ) W (d d k + O() (34) ) 2 Let d = d d k Clearly (d, W ) =, since (d i,w) = for all i Recalling Definition 25 we have z z µ(d d k ) W (d d k ) + O() µ(d) 2 Wd + O() 2 k (d) Completing the series gives d2n (d,w )=, dapplea µ(d) k (d) d 2 = d= (d,w )= = d2n (d,w )=, dapplea = z W µ(d) k (d) d 2 d2n (d,w )=, dapplea + O µ(d) k (d) d 2 0 d2n (d,w )=, d>a 0 + O µ(d) 2 k (d) C A d2n (d,w )=, dapplea µ(d) 2 k (d) d 2 Since µ(d) k (d) is a multiplicative function, we obtain by Lemma 23 that By Lemma 220 we have d2n (d,w )=, d>a d= (d,w )= µ(d) k (d) d 2 = Y p-w k p 2 µ(d) 2 k (d) d 2 apple Ck 2 (2(k ) + log A ) k A C A 9

20 for some C 2 > Furthermore, by Lemma 29 we obtain µ(d) 2 k (d) apple C k A (log A ) k for some C > Define d2n (d,w )=, dapplea := (, a,,a k,w,m)= W d= (d,w )= µ(d) k (d) d 2 It follows that B e m (z) euals z C2 k (2(k ) + log A ) k z + O + O C k A (log A ) k W A Define C := max{c,c 2 } The choice A = p z makes the error terms to be O p zc k (2k +logz) k Let d =(d,,d k ) 2 N k with d d k >A and let m be an integer with 2 apple m apple k such that d 2 D m We have shown in Remark 37 that such m exists for every (d,,d k ) 2 N k with d d k >A Then d m d k apple A m and d m d k >A m Furthermore d m d k d j apple A m for all j with m apple j apple k, since d m apple d j for all such j Multiplying all these ineualities implies (d m d k ) k m+2 d m d k apple A k m m+2 Hence d m d k apple A k m+2 k m+ m Define m := k m+2 k m+ The contribution of d 2 D m is given by ky µ(d i ) applenapplez i= d 2 i n+a i (d i,w )= A m <d m d k applea m m apple d m,,d k 2N (d i,w )=, A m <d m d k applea m m µ(d m d k ) my 2 applenapplez i= d 2 m n+a m d 2 k n+a k d i 2N d 2 i n+a i Clearly, d i 2N d 2 i n+a i apple (n + a i ) Note that n + a i apple 2z for every i By Lemma 28 it follows that (n + a i ) apple (2z) O 20 log log 6z

21 We infer that my 2 i= d i 2N d 2 i n+a i apple (2z) O k log log z since m 2 apple k for every m Following the same steps as in (33) and (34) we obtain z µ(d m d k ) µ(d m d k ) + d 2 m d 2 k d m,,d k 2N (d i,w )=, A m <d m d k applea m m applenapplez d 2 m n+a m d 2 k n+a k d,,d k 2N (d i,w )=, A m <d m d k applea m m, This euals z d2n (d,w )=, A m <dapplea m m µ(d) 2 k m+2 (d) + d 2 d2n A m (d,w )=, <dapplea m m µ(d) 2 k By Lemma 29 and Lemma 220 this is less than or eual to m+2 (d) We take We note that m Furthermore z A m = z ) k z Ck 2 (2k +loga m + C k A m m (log(a m m )) k A m A m := z + m = z k m+ 2k 2m+3 for m apple k (35), therefore A m apple z /2 Observe that k m+ k m+2 2k 2m+3 = z 2k 2m+3 for every m with 2 apple m<k k m +2 2k 2m +3 apple 2 3 for every m with 2 apple m apple k Using our previous choice A k = p 2z we get for each 2 apple m apple k and every d 2 D m that ky applenapplez i= d 2 i n+a i (d i,w )= A m <d m d k applea m m Finally using C k k k apple 2k O(k), we obtain et m (z) = z + O µ(d i ) z 2/3 C k k k (2k +logz) k (2z) O z 2/3 (2k +logz) O(k) (2z) O k log log z k log log z! 2

22 Proof of Theorem 2 By Proposition 36 there exists et m (z) = m z + O z 2/3 (2k +logz) O(k) (2z) O m := (, a,,a k ) such that! k log log z, where the implied constant depends at most on Define := (, a,,a k )= m2m m By (3) it follows that the sum in Theorem 2 euals z + O Wz 2/3 (2k +logz) O(k) (2z) O k log log z! To obtain an upper bound for W, observe that log W apple 2log +2 p< p a k a log p, By Lemma 24 we find p ak log p apple 2 p< p log p log ( p a k ) a k a k a We infer that W 2 e 2p a k From Theorem 23 it follows that applenapplex i= ry µ(n + a i ) 2 = cx + O x 3/4, where c = Q p (!(p)p 2 ) and!(p) :=# apple n apple p 2 : 9 apple i apple r such that n + a i 0 mod p 2 This concludes the proof 4 Proof of Theorem Let x be a positive real and a, b, positive integers and define S(x;, (a, b)) := # {s n apple x : s n a mod, s n+ b mod } (4) 22

23 From now on a, b, are fixed integers such that 0 apple a<and a<bapple + a Observe that S(x;, (a, b)) = S h (x;, (a, b)), (42) where S h (x;, (a, b)) = # n apple x a : h= n + a is suare-free and the next suare-free is (n + h ) + b (43) The main idea of the proof of Theorem is to partition the summation (42) into the two ranges apple h apple H and h>h, where H = log log x for some positive that will be optimized later Proof of Theorem By the inclusion-exclusion principle S h (x; ;(a, b)) = l h c=0 where l h = b + (h ) a and T (z; i,,i c,h)= napplez ( ) c µ(n + a) 2 a<i < <i c<(h )+b T ( x a; i,,i c,h), (44)! cy µ(n + i j ) 2 µ((n + h ) + b) 2 j= Let a = a, a 2 = i,, a k = i c and a k = b + (h ) By Theorem 2 there exists such that! T ( x a; i,,i c,h)= x a + O x 2/3 e 2p b+(h ) (2c +logx) O(c) (2(x a)) O c log log x Define h := (, a, b, W, h) = a l h ( ) c c=0 a<i < <i c<(h )+b Observe that P l h c=0 ( ) P c a<i < <i c<(h )+b is eual to the number of possible seuences of length l h where each element is either zero or one We obtain that l h ( ) c c=0 a<i < <i c<(h )+b =2 b+(h ) a apple 2 O(h) 23

24 The last ineuality holds, since a, b and are fixed Note that c apple b + (h ) a for every c This implies c h, since a, b, are fixed Furthermore (x a) apple x and apple h It follows that h log log x S h (x;, (a, b)) = h x + O Let H be a function of x to be chosen later We have S(x;, (a, b)) = Note that furthermore for any H By Lemma 22 we get Define (, a, b, H) := applehappleh x2/3 e O(p h) (2h +logx) O(h) x O applehappleh S h (x;, (a, b)) + h>h h log log x S h (x;, (a, b)) S h (x;, (a, b)) apple #{s n apple x : s n+ s n = h + b a}, 2 and for every h>h we have h + b a>h >H S h (x;, (a, b)) h>h P applehappleh H 2 x 2 log H 2 H 2 2 x H 2 (, a, b, W, h) It follows that S h (x;, (a, b)) = (, a, b, H)x + O x 2/3 He O(p h) (2H +logx) O(H) x O Clearly, He O(p h) apple (2H +logz) O(h) Note that the implied constant for x O depends at most on, sayc > 0 Let H := log log x where c 2 (0, ) Define 3 and note that L := (2H +logx) O(H) log L = O (log log x) 2 Hence log L apple log x for x su ciently 4 x 2/3 Lx c apple x 2/3+/4+c The choice of with c 2 (0, 2! H log log x! H log log x large It follows that L apple x/4 This implies ) provides us with S(x;, (a, b)) = (;(a, b),h)x + O 24 x (log log x) 2

25 Define `(;(a, b)) := h= S h (x;, (a, b)) (, a, b, W, h) and c h (, a, b) := lim x! x Reformulating the definition of S h (x;, (a, b)) gives S h (x;, (a, b)) = #{s n apple x : s n a mod and s n+ = s n a + b + (h )} Clearly this is less than or eual to By Lemma 22 we obtain for every h> Observe that #{s n apple x : s n+ s n >b a + (h ) } S h (x;, (a, b)) x (;(a, b),h)=`(;(a, b)) + O h 2 h>h c h (;(a, b))! (45) It follows that Hence c h (;(a, b)) h 2 H h>h h>h S(x;, (a, b)) = `(;(a, b))x + O x log log x We now proceed to provide an explicit expression for `(;(a, b)) Combining (42) and (44) we obtain S(x;, (a, b)) = l h ( ) c h= c=0 a<i < <i c<(h )+b T ( x a ; i,,i c,h) where l h = b + (h ) a Recall that T ( x a ; i,,i c,h)= napple x a µ(n + a) 2! cy µ(n + i j ) 2 µ((n + h ) + b) 2 j= By Theorem 23, T ( x a ; i,,i c,h) euals Y! i,,i c (p)p 2 x + O(x 3/4 ), p 25

26 where! i,,i c (p) is the number of residue classes n modulo p 2 such that at least one of the uantities n + a, n + i,,n+ i c,n+ (h ) + n is divisible by p 2 Hence `(;(a, b)) = l h ( ) c h= c=0 a<i < <i c<(h )+b Y p! i,,i c (p)p 2 (46) 5 Proof of Theorem 3 For 2 N we define G := a 2 Z \ [,]:(a, ) = suare-free Remark 5 Note that 2 G for every Furthermore G =[,] if is suare-free Proposition 52 Let a, 2 Z Then the following statements are euivalent a 2 G 2 The progression a mod contains infinitely many suare-free numbers 3 The progression a mod contains at least one suare-free number Proof ( ) 2) Assume a 2 G By Proposition 22 it is su following absolutely convergent product Y (p 2,) p 2 p (p 2,) a cient to prove that the does not vanish Indeed, if it vanishes then there exists a prime p such that (p 2,)=p 2 and (p 2,) a There two conditions however imply that p 2 (, a), thus contradicting our assumption that a 2 G (2 ) 3) Obvious (3 ) ) Assume that there exists a suare-free integer s such that s a mod Hence, there exists m such that s = m + a Writing s =(a, ) (a, ) m + a (a, ) we deduce that the factor (a, ) must also be suare-free Lemma 53 Let a, b 2 Z and define N p (a, b) :=# apple x apple p 2 : ax + b 0 mod p 2 If p 2 (a, b) then N p (a, b) =p 2 and in the opposite case we have N p (a, b) apple p 26

27 Proof If p 2 a and p 2 b then the congruence ax + b 0 mod p 2 holds for every integer x Hence N p (a, b) =p 2 Assume that p 2 - a or p 2 - b If p - a then the euation ax b mod p 2 is euivalent to x b/a mod p 2, thus N p (a, b) = The two remaining cases are pka and p 2 a Let pka Ifp - b then by Lemma 20 we obtain N p =0 Ifp b, then there exists a 0 and b 0 with (p, a 0 ) = such that a 0 x+b 0 0 mod p 2 This implies that there exists a uniue solution mod p, since (p, a 0 ) b 0 It follows that N p apple p Let p 2 a Then N p (a, b) is either 0 or p 2, according to if p 2 - b or p 2 b respectively The assumption p 2 - (a, b) allows us to rule our the second case, thus leaving us with the desired bound Corollary 54 Let a,b,a 2,b 2 2 Z and define M p (a,b ; a 2,b 2 ):=# apple x apple p 2 : a x + b 0 mod p 2 or a 2 x + b 2 0 mod p 2 Then M p (a,b ; a 2,b 2 ) apple 2p except if p 2 (a,b ) or p 2 (a 2,b 2 ) Proof Clearly, M p (a,b ; a 2,b 2 ) apple N p (a,b )+N p (a 2,b 2 ) The proof then follows directly from Lemma 53 Lemma 55 Let a,b,a 2,b 2 2 Z and assume that 4 - (a,b ) and 4 - (a 2,b 2 ) If we have M 2 (a,b ; a 2,b 2 )=4then 2 (a,b,a 2,b 2 ) Proof Using Lemma 53 for p = 2 shows that we have which, in addition to N 2 (a,b ) apple 2 and N 2 (a 2,b 2 ) apple 2, M 2 (a,b ; a 2,b 2 ) apple N 2 (a,b )+N 2 (a 2,b 2 ), shows that if N 2 (a,b ) < 2orN 2 (a 2,b 2 ) < 2 then we have a contradiction Therefore we can safely deduce that N 2 (a,b ) = 2 and N 2 (a 2,b 2 )=2 This shows that 2 (a,a 2 ), since otherwise at least one of the two last uantities would be Of course, if 2 - b then N 2 (a,b ) = 0, and similarly, we must have 2 b 2 Lemma 56 Let a, be integers such that (a, ) is suare-free positive integer k 0 such that (k 0, (a, )) = and k 0 a (a, ) mod (a, ) Then there exists a 27

28 Proof By the Chinese remainder theorem our lemma is euivalent to the following: for each prime p that divides there exists k (a,) p 2 Z such that k p,p min{ p(a), p()} = and k p a (a, ) mod p p() min{ p(a), p()} (5) Clearly, this holds if min{ p (a), p ()} =0 Ifmin{ p (a), p ()} 6= 0, then the minimum must eual, since (a, ) is suare-free There are two cases to consider: p () = and p () 2 If p () =, then (5) is euivalent only to p - k p ; such a k p can obviously always be found If p () 2, then our assumptions on a, guarantee that p (a) = Hence (5) is euivalent to The integer k p = a (a,) p - k p and k p a (a, ) mod p+( p() 2) is admissible here, because its p-adic valuation euals p (a) min{ p (a), p ()} =0 Proof of Theorem 3 One direction is obvious: if `(;(a, b)) > 0 then the progressions a mod and b mod contain at least one suare-free number For the other direction assume that the progressions a mod and b mod contain at least one suare-free number, thus Proposition 52 provides us with µ 2 ((a, )) = = µ 2 ((b, )) Let r := b a and choose r distinct primes p,,p r with the property that 2 <p <p 2 <<p r (52) Recalling euation (4) allows us to observe the following simple ineuality, S(x;, (a, b)) µ(m) 2 µ(m + b a) 2 mapplex,m a mod m+i 0 modp 2 i (appleiappler) The congruence conditions imposed on m may imply that no such m exist, in this case the sum above is taken to be zero However, we shall now see that there are in fact plenty of m satisfying all congruences Let m = k(a, ) for some k 2 N Note that m is suare-free if and only if µ 2 (k) = and gcd(k, gcd(a, )) =, thus getting S(x;, (a, b)) µ(k) 2 µ (k(a, )+b a) 2 kapple x (a,),(k,(a,))= k a (a,) mod (a,) k(a,)+i 0 modp 2 i 8i 28

29 By the Chinese remainder theorem, the fact that each p i is coprime to and Lemma 56, there exists an integer k 0 that satisfies Letting (k 0, (a, )) =, k 0 a (a, ) mod (a, ),k 0(a, )+i 0 mod p 2 i 8 apple i apple r (53) D := (a, ) and writing k = k 0 + nd we are therefore led to S(x;, (a, b)) T (x) := µ(k 0 + nd) 2 µ ((k 0 + nd)(a, )+b a) 2 k 0D applenapple x D(a,) ry i= p 2 i We may now invoke Theorem 23 to obtain where!(p) is given by T (x) lim x!+ x = Y p!(p), p 2 # apple n apple p 2 : k 0 + nd 0 mod p 2 or nd(a, )+k 0 (a, )+b a 0 mod p 2 It will therefore be su cient to show that the limit above is non-zero If p is coprime to D D(a, ), then!(p) apple 2, therefore the last infinite product converges absolutely Hence it is enough to show that each individual term is non-vanishing and we dedicate the rest of the proof solely to this task We shall now prove that for each prime p one has! ry p 2 - (k 0,D) and p 2 - p 2 i,k 0 (a, )+b a (54) i= By Corollary 54 this shows that!(p) apple 2p, hence the reuired ineuality!(p) 6= p 2 holds as long as p 6= 2; we shall prove that!(2) 6= 4 later We prove (54) via contradiction First, assume that there exists p such that p 2 (k 0,D) If p = p i for some apple i apple r, then p 2 i k 0, hence k 0 (a, )+i 0 mod p 2 i implies p 2 i divides i This is a contradiction, since i apple r = b a <b a apple 29

30 and we have chosen all p i such that p i > 2 If p 6= p i, then p 2 D, hence p 2 must divide /(a, ) and by (53) combined with p 2 k 0, we deduce that p 2 a/(a, ) This is a a contradiction, since the integers, are coprime (a,) (a,) It remains to prove that for each prime p one cannot have! ry p 2 p 2 i,k 0 (a, )+b a If p = p i for some apple i apple r, then (53) provides us with i= b a i 0 mod p 2 i This is clearly not possible because of (52) and b a i<ifp6= p i, then p 2 divides (, k 0 (a, )+b a) The congruence k 0 (a, ) a mod is implied by (53) and it shows that p 2 b This is a contradiction, since by assumption b 2 G The last task in our proof is to prove that!(2) 6= 4 If we had!(2) = 4 then Lemma 55 and (54) would imply that! ry 2 D, k 0, p 2 i,k 0 (a, )+b a In light of (52) we know that 2 - Q i p i, hence we may thus infer 2 (a, ),k 0,b a i= However, (53) supplies us with k 0 a mod, thus a/(a, ) must be even This (a,) (a,) is a contradiction since /(a, ) and a/(a, ) are always coprime but here we found that they are both even This remark concludes our proof References [] P Erdős Some problems and results in elementary number theory Publ Math Debrecen, 2:03 09, 95 [2] M Filaseta On the distribution of gaps between suarefree numbers Mathematika, 40:88 0, 993 [3] R R Hall and G Tenenbaum Divisors, volume 90 of Cambridge Tracts in Mathematics Cambridge University Press, Cambridge, 988 [4] C Hooley On the distribution of suare-free numbers Can J Math, 25:26 223,

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