On the Square-free Numbers in Shifted Primes Zerui Tan The High School Attached to The Hunan Normal University, China Advisor : Yongxing Cheng November 29, 204 Page - 504
On the Square-free Numbers in Shifted Primes Zerui Tan The High School Attached to The Hunan Normal University November 29, 204 Abstract For a fixed ositive integer k, let Q k x be the number of rimes not exceeding x and greater than k such that k is a square-free integer. In this aer, we rove that for any A > 2, we have Q k x = k + 2 lix + O x log x A as x +. Where the imlied constant deends only on A and k. 0 Keywords: shifted rimes, square-free numbers Page - 505
Introduction Number theory is a branch of ure mathematics which studies the roerties of integers. An integer is called square-free if it is not divisible by any erfect square excet. A ositive integer is called a rime number, if it has no more factors other than and itself, with the convention that is not a rime. In number theory, rime numbers and the roerties of objects closely related to integers or defined as generalizations of the integers are studied. Prime numbers lay a crucial role in number theory, and there are still a lot of unsolved roblems surrounding them. In [6], age 59, there is a table which resents all the rimes 2 < 5000, their least rimitive root and the factorization of. Part of the table is here unnecessary columns are removed. 3 2 37 2 2 3 2 5 2 2 4 2 3 5 7 2 3 43 2 3 7 2 5 47 2 23 3 2 2 3 53 2 2 3 7 2 4 59 2 29 9 2 3 2 6 2 2 3 5 23 2 67 2 3 29 2 2 7 7 2 5 7 3 2 3 5 73 2 3 3 2 By observation, we find that some of the rime numbers like 7,, 23, their corresonding factorization of only contains distinct rime factors, namely, their corresonding are square-free numbers. Some rimes do not have this roerty. So we are eager to know, How many such rimes are there in the interval [, x]? Do they have a ositive density among all rimes? And what about the case k for any given ositive integer k > 0? The book [6] did not mentioned these questions. Then we started searching these questions on the Internet including Baidu, Google, etc. in order to find out an answer. However, we did not find any of them. So we decided to study this question. Definition. For a fixed ositive integer k, we define Q k x to be the number of rimes not exceeding x and greater than k such that k is square-free. In this article, we rove that Theorem. For any A > 2, we have x Q k x = C k lix + O log x A x + 2 Page - 506
where C k is a constant deends on k, which can be written as C k = + 2 k and the imlied constant only deends on A and k. The function lix is the integral lix = x 2 dt log t and fx = Ogx denotes fx C gx for some ositive constant C. Notice that the constant C k and C l will have the same value if k has the same rime factors as l has, for examle we have C 6 = C 8. Let ϕn be the number of ositive integers not exceeding n that are relatively rime with n, we have that ϕn = n n where runs over all distinct rime factors of n. The Möbius function µn is defined as n = µn = 0 there is a rime that 2 n ωn otherwise where ωn = the number of distinct rime factors of n. To rove the theorem, we need the following basic ideas. µd = δn d n where { n = δn := 0 otherwise. Since any number n N can be written as the form n = a 2 b uniquely, where b is a square-free number, n is square-free if and only if a =, it follows that µ 2 n = µd = µd d a d 2 n where the sum d 2 n is taken over all ositive divisors d such that d2 n. A well-known theorem due to Bombieri and A.I.Vinogradov, the Mertens formula and the crude lower bound of the Euler function ϕd are also needed. 3 Page - 507
Theorem A. Bombieri-Vinogradov Let A > 0, There is some constant B = BA such that max liy a,q= πy; q, a ϕq x x 2 log x A q x /2 /log x B max y x where the imlied constant deends only on A. Where f g means there is a ositive constant C such that f C g, the function πx; q, l is defined as πx; q, l = For the roof of the theorem, see [4]. x x lmod q Theorem B. Mertens formula For x 2, = e γ log x where γ is the Euler-Mascheroni constant. + O. log x The Mertens formula is a classical result, and the roof can be found in []. Lemma.. If d > 5 is a square-free number, then we have where c 2 > 0 is some constant. ϕd d log log d Proof. Let n be the n-th rime, by Mertens formula we have that ϕd = d d d ωd = de γ + O log ωd log ω d since log N log N and ωd log d, it follows that log ωd log log d. Therefore we can get ϕd d log log d. 4 Page - 508
2 Proof of Theorem Now we have that By the formula Q k x = µ 2 k. k< x µ 2 n = d 2 n µd and by changing the order of summation we obtain Q k x = k< x d 2 k = d< x k< x kmod d 2 µd = d< x µd + O k< x kmod d 2 x log x. µd + O x k<d< x k< x kmod d 2 We can change the summation condition k < x to x, the error we get is O x, so Q k x = µdπx; d 2, k + O x. d< x For the sake of simlicity, let L = log x. We divide the sum into two ieces by the intervals [, x α L B0 [x α L B0, x /2, where 0 < α < /2 and B 0 > 0 are constants, their value will be determined later. We write Q k x = S α + S α,/2 + O x where and S α = S α,/2 = µdπx; d 2, k x α L B 0 d<x /2 µdπx; d 2, k. For the first sum, we can write πx; d 2, k as the form πx; d 2, k = ρ kd ϕd 2 lix + πx; d 2, k ρ kd ϕd 2 lix Where ρ k n := δn, k is the characteristic function of numbers n that are relatively rime with k. Namely, it equals to if and only if n is relatively rime with k, otherwise it equals to 0. Inserting the above formula into S α we can get S α = lix ρ k dµd dϕd + 5 µd. πx; d 2, k ρ kdlix ϕd 2. Page - 509
Notice that µd = 0 if d is not square-free, we can assume d is square-free, in this case we have ϕd 2 = dϕd. By Lemma., it is easy to check that the sum ρ k dµd dϕd d= converges absolutely. Thus ρ k dµd S α = lix + µd πx; d 2, k ρ kdlix dϕd ϕd 2 ρ k dµd ρ k dµd = lix + O lix dϕd ϕd 2 d= d x α L B 0 + µd πx; d 2, k ρ kd ϕd 2 lix = C k lix + OS + R α. Where and ρ k dµd C k = ϕd 2 d= = + ρ kµ ϕ 2 = + 2 k R α = S = lix µd πx; d 2, k ρ kd ϕd 2 lix d x α L B 0 ρ k dµd ϕd 2. 2. Uer bound for S α,/2, S and R α Now, we are going to find the uer bound for S α,/2 = µdπx; d 2, k. x α L B 0 d<x /2 We only need to consider the trivial bound πx; d 2, k x d 2. 6 Page - 50
So we obtain S α,/2 = µdπx; d 2, k x α L B 0 d<x /2 x x d 2 x α L B 0 d<x /2 x /2 x α L B 0 dt t 2 x α log x B0 x /2 rovided x sufficiently large. Recall that /2 > α > 0, namely S α,/2 x α log x B0. As for the sum S = lix n x α L B 0 ρ k nµn ϕn 2 it is not difficult to get an uer bound, since ϕd 2 is aroximately equals to d 2 for a square-free number d. Lemma 2.. For sufficiently large x, we have lix n x α L B 0 ρ k nµn ϕn 2 x α L B0. Proof. Because µn = 0 if n is not square-free, we can assume n is square-free. We have ϕn 2 = nϕn and n x and by Lemma. we have ρ k nµn ϕn 2 n x α L B 0 n 2 dt x t 2 x n x α L B 0 nϕn log log x x α log x. B0 By the fact lix x/ log x, we have lix ρ k nµn ϕn 2 x α log log xlog x B0 n x α L B 0 which comletes the roof of the lemma. 7 Page - 5
Now we are going to consider the R α. R α = µd πx; d 2, k ρ kd ϕd 2 lix where the constant B 0 = BA/2, BA is the constant in the Bombieri- Vinogradov Theorem. Now we let α = /4, by the Bombieri-Vinogradov Theorem we obtain lix R = 4 µd πx; d 2, k ρ kd ϕd 2 d<x /4 /log x B 0 πx; d 2, k lix ϕd 2 + πx; d 2, k d 2 <x /2 L 2B 0 ρ k d= x log x A. d 2 <x /2 L 2B 0 ρ k d=0 Where the last inequality is true since the function πx; d 2, k in the second sum is bounded due to the fact d, k >. 2.2 Comletion Combining the results we got, for a given ositive integer k > 0, any A > 2, and sufficiently large x, we have Q k x = C k lix + O x + OS + S 4, + R 2 4 = C k lix + O x /2 + x 3/4 log x B0 + x 3/4 x log x B0 + log x A x = C k lix + O log x A. Inserting the exression of C k, we can get the formula in Theorem Q k x = lix + 2 + O k This is the result we desired. As a consequence, we have that Q k x lim = C k > 0 x lix x log x A Which means the rimes such that k is square-free have a ositive density among all rimes.. 8 Page - 52
3 Related Questions Now we will discuss some interesting questions that aeared directly in the research of Q k x. Question. Is there any ositive integer k, such that for sufficiently large x, we have Q k x < C k lix? See aendix for more details. It seems that the bound for R is too crude that it lost some information. 4 We used the Bombieri-Vinogradov Theorem to get this result. This theorem is quite imortant in number theory since it is essentially of the same strength with the Generalized Riemann Hyothesis usually written as GRH. on average, which hyothesis has a consequence that lix πx; q, k ϕq x log x rovided q, k =. So, if we assume GRH and α 4 we can get πx; d 2, k lix ϕd 2 x log x x /2+α L B0 therefore k,d= k,d= all remainder terms x /2+α L B0 + x α L B0. we know that we should choose α = 4 and B 0 = 2 to get an otimal remainder term all remainder terms x 3/4 log x. If 4 < α 2, we can just do as what we did before, divide the sum into two ieces by the intervals [, x /4 L B [x /4 L B, x α L B0, where B is some constant. Then use GRH to estimate the first sum and use an integral to bound the second, one gets πx; d 2, k lix x 3/4 L B + x 3/4 L B k,d= ϕd 2 which is exactly the same as the case α /4, also notice that the above uer bound is indeendent with B 0 and α. The above arguments suggests the following two questions. 9 Page - 53
Question 2. The question is, can we rove that for any A > 0, there is a constant B = BA that π y; d 2, k liy ϕ d 2 x/2+α L A d<x α L B max y x max k,d= k<d 2 for some 0 < α /4 esecially for the case α = /4. without assuming GRH? Question 3. Is there an absolute constant B such that for sufficiently large x, we have max y x π y; d 2, k liy ϕ d 2 x3/4 log x B? d max k,d= k<d 2 It is not difficult to check that the above sum converges for every x > 0. Remark. The above two questions can be regarded as the analogy of the Bombieri-Vinogradov Theorem. As for Question, in fact, we guess that it might be wrong. To answer these questions, we still need more research. 0 Page - 54
Aendix Tables of Q k x For k =, 2, 3 We made a C++ rogram on our comuter to calculate some numerical value of Q k x for k =, 2, 3 and x 0 7. The exact value of Q k x, C k lix and their ratio Q k x : C k lix are resented as follows. x Q x C lix Q x : C lix 0 3.948.5667 50 8 6.556.2278 00 3 0.875.954 500 40 37.676.067 000 68 66.027.0299 5000 255 255.50 0.99804 0 4 467 465.6.0030 5 0 4 943 93.7.0058 0 5 3599 3600.7 0.99953 5 0 5 5602 5559.0028 0 6 29397 29403 0.99980 5 0 6 3039 30375.000 0 7 24858 248650 0.99947 x Q 2 x C 2 lix Q 2 x : C 2 lix 50 3.032 0.844 00 20 2.750 0.995 500 74 75.352 0.982 000 27 32.05 0.968 5000 506 5.00 0.99022 0 4 925 93.22 0.99332 5 0 4 384 3863.4 0.99420 0 5 775 720.4 0.99639 5 0 5 3020 38 0.99685 0 6 58653 58806 0.99740 5 0 6 26038 260750 0.99858 0 7 496848 497300 0.99909 x Q 3 x C 3 lix Q 3 x : C 3 lix 50 6 7.887 0.76739 00 0 3.050 0.76628 500 4 45.2 0.90686 000 74 79.232 0.93386 5000 295 306.60 0.9627 0 4 548 558.73 0.98079 5 0 4 2280 238.0 0.98359 0 5 4292 4320.8 0.99333 5 0 5 8603 867 0.99637 0 6 3553 35284 0.99630 5 0 6 56249 56450 0.99872 0 7 298075 298380 0.99898 Page - 55
References [] G. Tenenbaum, Introduction to Analytic and Probabilistic Number Theory translated version, Higher Education Press, 20 [2] Melvyn B. Nathanson, Elementary Methods in Number Theory, Sringer- Verlag, 999 [3] Kenneth Ireland and Michael Rosen, A Classical Introduction to Modern Number Theory Second Edition, Sringer, 990 [4] Dimitris Koukoulooulos, Sieve Methods, Prerint, 202 [5] Karatsuba, Basic Analytic Number Theory translated version, Harbin Institute of Technology Press, 202 [6] Luogeng Hua, Introduction to Number Theory in Chinese, Science Press, 957 2 Page - 56