On the size of quotient of two subsets of positive integers.

Transcription

1 arxv: v1 [math.nt] 13 Jun 2017 On the sze of quotent of two subsets of postve ntegers. Yur Shtenkov Abstract We obtan non-trval lower bound for the set A/A, where A s a subset of the nterval [1, Q]. 1 Introducton In a memory of Anatoly Alekseevch Karatsuba and Andrey Borsovch Shdlovsky Let A, B be subsets of ntegers of the nterval [1,Q], A wll denote the cardnalty of fnte set A. The sets AB and A/B are called the product and quotent of two sets A, B and are defned as a AB = ab : a A, b B, A/B = b : a A, b B, b 0. Recall that the multplcatve energy E(A, B) of two sets A, B s E(A, B) = a 1 b 1 = a 2 b 2 : a 1, a 2 A; b 1, b 2 B. When the sets are equal, A = B we wll smply wrte E(A) nstead of E(A, A). We note that usng good estmates of E(A, B) one can deduce non-trval lower estmates of the sze of AB but not vce versa the followng well-known nequalty whch s due to Shnrel man [10], whch can be also found n [9]. AB, A/B A 2 B 2 E(A, B). (1) Throughout the paper τ(n) (usual notaton) s the number of dvsors of n. Recall the well-known estmate of τ(n) whch can be found n the book [3], Theorem 5.2, Kaptel 1. (1+o(1)) log n τ(n) 2 log log n, n. (2) 1

2 Usng the above estmate t s easy to prove the followng result. For any fnte set A N such that a A a Q, we have the followng estmates AA A 2 exp ( 2 log 2 + o(1)) E(A) A 2 exp (2 log 2 + o(1)), Q ; (3) Q. (4) The constant 2 log 2 n (2),(3) can not be mproved, see t n the paper [1] for example. Usng (5) one can obtan that A/A A 2 exp ( 2 log 2 + o(1)) Q. (5) Ths bound cannot be mproved very much n general, except for the constant 2 log 2, see t for example n [2]. But there s queston that was posed n the paper [4] relatng to ths, we repeat the formulaton of t bellow. Queston. Is t possble to mprove the coeffcent 2 log 2 n (4)? The purpose of ths note s to gve postve answer to ths queston. So we formulate the man result of ths paper. Theorem 1. There s an absolute constant γ > 0, such that f A, B [1, Q] then we have the followng estmate A/B A B exp ( 2 log 2 + γ + o(1)) One can take γ = The notaton A B n ths paper denotes that A < B exp (o(1)) The paper s organzed as follows: n the next secton we formulate some prelmnary statements. In the thrd secton we gve the proof of Theorem 1. The last secton contans some fnal comments about ths result. 2 Preparatons and prelmnary results We need some defntons and prelmnary lemmas. We begn wth the smooth numbers. For postve nteger n let P + (n) denotes the maxmal prme dvsor of n, and P + (1) = 1. For x y 2 let 2

3 ψ(x, y) = n x : P + (n) y. We need some one upper estmate for ψ(x, y), whch can be found n [5], Theorem 1.4, whch s presented bellow. Lemma 2. Unformly for x y 2, we have where 1 log ψ(x, y) = Z 1 + O( log y + 1 log log x ), Z = Z(x, y) = log x log y log(1 + y log x ) + y log x log(1 + log y y ). Our second lemma gves some upper bound for the number of dvsors of postve nteger wth small redcal. Probably t was known before and we do not pretend on ths fact. Lemma 3. There exsts a functon C(ε) > 0, wth C(ε) 0, f ε 0 and wth the followng property. If n Q, rad(n) Q ε, then τ(n) exp (C(ε)) Proof. Let n = p α p αs s s the prme decomposton of n and p 1 < p 2 <... < p s. Consder the map on the set of dvsors of n π : p t p ls s p t 1 (1)... p ls (s), where p () - s the - ordered prme number, p (1) = 2, p (2) = 3, p (3) = 5,... By Prme Number Theorem f rad(n) Q ε then p (s) < (ε + o(1)) If d n then π(d) Q. So the number of such dvsors d does not exceed ψ(q, (ε + o(1) ). Usng Lemma 2 wth some easy computatons we get the desred property for the functon C(ε). Ths completes the proof of Lemma 3. Let τ(n, z) denotes the number of dvsors of n whch are less or equal to z. In other words τ(n, z) = d : d n, d z. The next proposton we present n the followng lemma. 3

4 Lemma 4. Let n Q, µ(n) 0, z Q δ and δ (0, 1/2]. Then we have the followng estmate τ(n, z) exp (δ log( 1 δ ) + (1 δ) log( 1 1 δ ) + o(1)) Proof. Consder any dvsor d of n and ts prme decomposton: d = p 1... p s. It s easy to see, that s The number n has at most (1+o(1)) (δ+o(1)), Q dfferent prme dvsors. Dong some computatons together wth the asymptotc expresson for bnomal coeffcent ( ) m exp (δ log( 1 δm δ ) + (1 δ) log( 1 1 δ ) + o(1))m, m we get the desred bound. where Let z = Q δ. In the notatons above we n fact have shown that ( ) m τ(n, z) δ m m = [ ], δ = mn(δ, 1 2 ). In fact the condton µ(n) 0 n the prevous lemma can be removed. Lemma 5. Let n Q, z Q δ. Then we have the followng estmate τ(n, z) exp (δ log( 1 δ ) + (1 1 δ ) log( 1 δ )), Q, where δ = mn(δ, 1 2 ). Proof. We may assume that δ < 1, as n the opposte stuaton ths Lemma easly 2 follows from the general estmate for τ(n). Let ε > 0 be small fxed real number and K be fxed large nteger. The proof conssts of several steps and we begn wth the frst one. Step 1. We show that there exsts a presentaton of n n the followng form n = n 1... n s m, where µ(n ) 0, n > Q ε, rad(m) Q ε (We allow the stuaton wth s = 0, where there are no n n ths presentaton.) The argument of the proof s a sort of an algorthm. If rad(n) Q ε then we are done wth n = m. If rad(n) > Q ε then 4

5 n n n = rad(n) and we proceed ths procedure wth nstead of n. It s easy to rad(n) rad(n) see that the algorthm wll be fnshed and we get the desred representaton. Step 2. We can easly get an upper estmate for the number of dvsors of m. Indeed m Q, rad(n) Q ε. We use Lemma 3 and see that τ(n) exp (C(ε)), Q, where C(ε) 0 f ε 0. Step 3. In ths step we ntroduce some defntons. Let the quanttes δ be defned from the denttes n = Q δ. Now we defne Ks ntervals Ω,j 1 s, 1 j K by settng Ω,j = [Q δ j 1 j K, Q δ K ]. Step 4. Any dvsor of n 1... n s can be presented as d 1... d s, d n. Suppose that for any 1 s we fx the nterval Ω,j. Now we wll obtan upper estmate for the number of vectors (d 1,..., d s ), d n and d Ω,j. Each d s a dvsor of n, µ(n ) 0, d Q δ j K. So the number of such d by Lemma 4 does not exceed ( m ) δ,j m where m = [ δ ] and δ,j = mn( 1, j ). Therefore the number of such 2 K vectors (d 1... d s ) s bounded by the product ( ) ( m m ). δ,j m δ,j m 1 s It s easy to see that m. Next we are gong to estmate δ,j m. We see that δ,j 1 m δ δ,j ( ) Now we estmate each term n the last sum. We have j 1 log d [δ K, δ j ] K and δ,j j K. So we can wrte δ,j j δ δ K log d + δ. K Insertng ths nequalty to the expresson (*) we obtan δ,j m δ + 5 δ K

6 Fnally we have that the number of such vectors (d 1,..., d s ) such that d n, d Ω,j (the sets Ω,j are fxed) s bounded by ( ) M, αm where M =, α = mn(1 2, δ(1 + 1 K )). Step 5. Now we obtan an upper bound for the number of dfferent choces of the ntervals Ω,j. Ths number does not exceed K s and s some bounded constant, (whch does not depends on Q). Our Lemma now follows f one uses statements of Steps 2,4,5 and takes suffcently large constant K and suffcently small ε. Lemma 6. For any nteger n > 1 we have log τ(n2 ) log τ(n) log 3 log 2. Proof. Let n = p β p β l l, then log τ(n 2 ) log τ(n) = log(1 + 2β 1) log(1 + 2β l ) log(1 + β 1 ) log(1 + β l ). The last expresson s always less than log 3. Wth that we fnsh the proof of ths log 2 lemma. Next, we ntroduce some notatons. Let n N and let l(n) denotes the maxmal postve nteger m such that m 2 n. We are gong to prove the followng lemma. Lemma 7. Let n be postve nteger, n Q 2 and the quantty c s defned from the equaton τ(n) = exp (2 log 2 c). Then there we have where l(n) Q δ(c)+o(1), Q, δ(c) c 2 log 2 log 3. 6

7 Proof. Let the quantty δ s defned from the equalty l(n) = Q δ. We also may assume that log log l(n) = (1 + o(1)), as n the opposte stuaton the Lemma 7 s true. Usng Lemma 6 and upper estmate for τ(l(n)) we conclude that τ(l 2 (n)) exp (log 3 + o(1)) δ We see that l(n) 2 n and we can wrte exp (2 log 2 c) = τ(n) τ(l(n) 2 n )τ( l(n) ). 2 It s easy to see that the last expresson does not exceed (δ ) exp log 3 + (2 2δ) log 2 + o(1). Comparng ths quantty wth the left-sde expresson n the last nequalty and dong some easy computatons we obtan the desred estmate for δ. Wth that we fnsh the proof of Lemma 7. Now we are ready to prove Theorem 1 and we are gong to the next secton. 3 The proof of Theorem 1 Proof. Let the quantty c be defned from the equalty From the nequalty 1 we see that E(A, B) = A B exp (2 log 2 c) A/B A B exp ( 2 log 2 + c) Our next step s to fnd another lower bound for A/B, ths nequalty wll work well n the case of small c. Let us denote the quantty L from the dentty E(A, B) = A B L, and let r A,B (z) = (a 1, b 1 ) A B : a 1 b 1 = z... Defne the set M 1 = z AB : r A,B (z) L/2 7

8 and We see that and so For nteger 0 let M 2 = AB \ M 1. ra,b(z) 2 A B L/2; z M 1 ra,b(z) 2 A B L/2. z M 2 We see that M 2, = z AB : r A,B (z) (2 1 L, 2 L]. M 2 = 0 M 2,. Hence by pgeonhole prncple there exsts 0 such that r 2 A,B(z) A B exp (2 log 2 c + o(1)) z M 2, Let us fx such and let the quantty c be defned from the dentty 2 L = exp (2 log 2 c ) It s easy to see that c [o(1), c]. Next we wll show that r A,B (z) A B exp (c c + o(1)) z M 2, Indeed,.. A B exp (2 log 2 c+o(1)) r A,B(z) 2 max r A,B (z) r A,B (z). z M 2, z M 2, z M 2, The quantty max z M2, r A,B (z) s less than exp (2 log 2 c ). So nsertng ths bound to the prevous nequalty we get the desred estmate. Next we consder the set G: G = (a 1, b 1 ) A B : a 1 b 1 M 2,. From the prevous estmate G > A B exp 8 (c c + o(1))

9 Next we consder the followng set a1 W = : (a 1, b 1 ) G, b 1 and wll show that W s large. For every element z M 2, we use Lemma 7 and see that where l(z) = Q δ(c )+o(1), Q, δ(c ) c 2 log 2 log 3. Ths means that for every par (a 1, b 1 ) G gcd(a 1, b 1 ) Q δ(c )+o(1). Defne r A/B,G (z) = (a 1, b 1 ) G : a 1 = z. b 1 We can wrte A B exp (c c + o(1)) = G = r A/B,G (z) z W 1/2 Our am s to obtan good upper estmate for z r 2 A/B,G(z) 1/2. σ = z r 2 A/B,G(z). where The σ does not exceed the number of soluton to the equaton We may wrte a 1 b 1 = a 2 b 2, a A; b B gcd(a 1, b 1 ), gcd(a 2, b 2 ) Q δ(c )+o(1). a 1 = tu, b 1 = tv, a 2 = su, b 2 = sv; gcd(u, v) = 1 and t, s Q δ(c )+o(1). Let us fx a 1 and b 2. If for these a 1 and b 2 we choose t and s we then dentfy a 2 and a 3. For any fxed a 1, b 2 the parameters t, s are the dvsors of a 1, b 2 respectvely. These t, s do not exceed Q δ(c )+o(1). Usng Lemma 5 we see that the number of dfferent pars t, s do not exceed exp (2δ(c 1 ) log( δ(c ) ) + 2(1 1 δ(c )) log( 1 δ(c ) ) + o(1)) 9

10 We wll just wrte δ c nstead of δ(c ). And so we conclude that σ < A B exp (2δ c log( 1 1 ) + 2(1 δ c ) log( ) + o(1)) δ c 1 δ c So we can obtan the lower bound for W : W A B exp (2c 2c 2δ c log( 1 δ c 1 ) 2(1 δ c ) log( )+o(1)) 1 δ c Recall that c [o(1), c]. We may assume c It s easy to see that the expresson 2c 2δ c log( 1 δ c takes the smallest value f c = c. So, we can rewrte the last estmate 1 ) 2(1 δ c ) log( ) 1 δ c W A B exp ( 2δ c log( 1 1 ) 2(1 δ c ) log( ) + o(1)), Q, δ c 1 δ c where δ c = δ(c). As t was noted before there s trval estmate A/B A B exp ( 2 log 2 + c). We have these two estmates, one work well wth small c, another work well wth large c. It s easy to see that the explct absolute constant γ > 0 can be taken as the soluton of the followng equaton 2 log 2 + c = 2δ c log( 1 1 ) 2(1 δ c ) log( ), δ c 1 δ c c where δ c =. 2 log 2 log 3 Computer calculatons show that the soluton s equal to , so one can ths value for the γ. Wth that we fnsh the proof of Theorem 1. 4 Fnal remarks One can easly deduce the followng corollary, whch follows from the proof of Theorem 1. 10

11 Corollary 8. Let A, B [1, Q] and E(A, B) = A B exp (2 log 2 + o(1)). Then we have A/B = A B exp (o(1)) In partcular f AB = A B exp ( 2 log 2 + o(1)), Q, then A/B = A B exp (o(1)) Indeed, the the condton AB = A B exp ( 2 log 2 + o(1)) mply E(A, B) = A B exp (2 log 2 + o(1)). It seems that usng more precse arguments for fndng pars (a, b) A B for the set G wth smaller gcd(a, b) may lead to a better coeffcent nstead of Acknowledgements The work s supported by the Russan Scence Foundaton under grant I wsh to thank Serge Konyagn for valuable comments, advces and attenton to ths work. References [1] Shtenkov Yu., On the product sets of ratonal numbers, Proceedngs of the Steklov Insttute of Mathematcs, Vol. 296, 2017, preprnt. [2] J. Clleruelo, D.S.Ramana and O.Ramare, Quotents and product sets of thn subsets of the postve ntegers, Proceedngs of the Steklov Insttute of Mathematcs, Vol. 296, 2017, preprnt. [3] Prachar K. Prmzahlvertelung // Sprnger Verlag Berln Gőttngen Hedelberg, [4] Shtenkov Yu. N. Addendum to the paper Quotents and product sets of thn subsets of the postve ntegers by J. Clleruelo, D.S. Ramana and O. Ramare. // Proceedngs of the Steklov Insttute of Mathematcs, 296, 2017,... [5] Adolf Hldebrand, Gerald Tenenbaum. Integers wthout large prme factors // Journal de Theore des Nombres de Bordeaux 5, (1993), [6] Bourgan J., Konyagn S.V., Shparlnsk I.E. Product sets of ratonals, multplcatve translates of subgroups n resdue rngs and fxed ponts of the dscrete logarthm // Int. Math Research Notces rnn 090, P

12 [7] Clleruelo J. A note on product sets of ratonals // Internatonal Journal of Number Theory, Vol. 12, No. 05, pp (2016) [8] Clleruelo J., Garaev M. Congruences nvolvng product of ntervals and sets wth small multplcatve doublng modulo a prme and applcatons // Math. Proc. Cambrdge Phl. Soc., Vol. 160, Issue 03, pp , May [9] Tao T., Vu V. Addtve combnatorcs // Cambrdge Unversty Press 2006, P [10] Shnrel man L.G. Uber addtve Egenschaften von Zahlen // Mathematsche Annalen, V. 107 (1933), P Steklov Insttute of Mathematcs, Russan Academy of Scence, yursht@yandex.ru 12