A SHARP RESULT ON m-covers. Hao Pan and Zhi-Wei Sun

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1 Proc. Amer. Math. Soc. 35(2007), no., A SHARP RESULT ON m-covers Hao Pan and Zhi-Wei Sun Abstract. Let A = a s + Z k s= be a finite system of arithmetic sequences which forms an m-cover of Z (i.e., every integer belongs at least to m members of A). In this paper we show the following sharp result: For any positive integers m,..., m k and θ [0, ), if there is I,..., k such that the fractional part of / is θ, then there are at least 2 uch subsets of,..., k. This extends an earlier result of M. Z. Zhang and an extension by Z. W. Sun. Also, we generalize the above result to m-covers of the integral ring of any algebraic number field with a power integral basis.. Introduction For an integer a and a positive integer n, we simply let a(n) represent the set a + nz = x Z : x a (mod n). Following Sun [S95, S96] we call a finite system A = a s ( ) k s= (.) of such sets an m-cover of Z (where m, 2, 3,... ) if every integer lies in at least m members of (.). We use the term cover (or covering system) instead of -cover. For problems and results in this area, the reader may consult [G04, pp ], [PS] and [S05]. P. Erdős [E97] once said: Perhaps my favorite problem of all concerns covering systems. Example.. For each integer m, there is an m-cover of Z which is not the union of two covers of Z. To wit, we let p,..., p r be distinct primes with r 2m, and set N = p p r. Clearly A = p s(n) I,...,r, I m does not cover any integer relatively prime to N. Let a,..., a n be the list of those integers in 0,,..., N not covered by A with each occurring exactly m times. If x Z is covered by A, then x 0(p s) for some I,..., r with I m. Therefore A 0 = 0(p ),..., 0(p r ), a (N),..., a n (N) 2000 Mathematics Subject Classifications: Primary B25; Secondary B75, D68, R04. The second author is responsible for communications, and supported by the National Science Fund for Distinguished Young Scholars (No ) in China.

2 2 HAO PAN AND ZHI-WEI SUN forms an m-cover of Z. Suppose that I I 2 =,..., r, J J 2 =,..., n and I I 2 = J J 2 =. For i =, 2 let A i be the system consisting of those 0(p s ) with s I i and those a t (N) with t J i. We claim that A or A 2 is not a cover of Z. Without loss of generality, let us assume that I I 2. Since 2 I 2 I + I 2 > 2(m ), we have I 2 m and hence 2 p s is covered by A. Therefore 2 p s n t= a t(n). Clearly 2 p s is not covered by 0(p s ) either. Thus A does not form a cover of Z. By means of the Riemann zeta function, in 989 M. Z. Zhang [Z89] proved that if (.) forms a cover of Z then / is a positive integer for some I,..., k. Let m,..., m k be any positive integers. If (.) is a cover of Z, then a s + ( / )Z k s= is also a cover of Z and hence Theorem 2 of [S95] indicates that for any J,..., k there is an I,..., k with I J such that / = s J /, where α denotes the fractional part of a real number α. When J = and m = = m k =, this yields Zhang s result. In 999 Z. W. Sun [S99] proved further that if (.) forms an m-cover of Z then for any J,..., k we have I,..., k : I J and = s J In this paper we will show the following sharp result. m. Theorem.. Let (.) be an m-cover of Z, and let m,..., m k be any integers. Then for any 0 θ < the set I A (θ) = I,..., k : = θ (.2) has at least 2 m elements if it is nonempty. Remark.. Clearly m copies of 0() form an m-cover of Z. This shows that the lower bound in Theorem. is best possible. Corollary.. Let (.) be an m-cover of Z, and let m,..., m k be any integers. Then S(A) 2 k m where S(A) = : I,..., k. (.3) Proof. As I A (θ) 2 m for all θ S(A), we have and hence S(A) 2 k m. S(A) 2 m I : I,..., k = 2 k Remark.2. Sun [S95, S96] showed that if m,..., m k are relatively prime to n,..., n k respectively then (.) forms an m-cover of Z whenever it covers S(A) consecutive integers at least m times.

3 A SHARP RESULT ON m-covers 3 Corollary.2. Suppose that (.) forms an m-cover of Z but a s ( ) k s= does not. If the covering function w A (x) = s k : x a s ( ) is periodic modulo n k, then for any r = 0,..., n k we have I,..., k : = r 2 m. (.4) n k Proof. By Theorem of Sun [S06], : I,..., k and = r m. n k In particular, / = r/n k for some I,..., k, and hence (.4) holds in the case m =. For A k = a s ( ) k s=, clearly w Ak (x) m for all x Z. In the case m >, we obtain (.4) by applying Theorem. to A k with m = = m k = and θ = r/n k. Remark.3. When n k is divisible by all the moduli n,..., n k, Corollary.2 was stated by the second author in [S03, Theorem 2.5]. When w A (x) = m for all x Z, the following result stronger than (.4) was proved in [S97]: I,..., k : for every n = 0,..., m. = n + r n k ( ) m For an algebraic number field K, let O K be the ring of algebraic integers in K. For α, β O K, we set α + βo K = α + βω : ω O K and call it a residue class in O K. For a finite system A = α s + O K k s= (.5) of such residue classes, if s k : x α s + O K m for all x O K (where m, 2, 3,... ), then we call A an m-cover of O K. Covers of the ring Z[ 2] = O Q( 2) were investigated by J. H. Jordan [J68]. An algebraic number field K of degree n is said to have a power integral basis if there is γ O K such that, γ,..., γ n form a basis of O K over Z. It is well known that all quadratic fields and cyclotomic fields have power integral bases. Here is a generalization of Theorem.. n

4 4 HAO PAN AND ZHI-WEI SUN Theorem.2. Let K be an algebraic number field with a power integral basis. Suppose that (.5) forms an m-cover of O K, and let ω,..., ω k O K. Then, for any µ K, the set I,..., k : is empty or it has at least 2 m elements. ω s µ + O K Remark.4. We conjecture that the requirement in Theorem.2 that K has a power integral basis can be cancelled. 2. Proof of Theorem. Lemma 2.. Let (.) be an m-cover of Z. Let m,..., m k Z and define S(A) as in (.3). Then, for any given θ S(A), there exists t,..., k such that both θ and θ m t /n t lie in S(A t ), where A t = a s ( ) s k, s t. Proof. Choose a maximal J,..., k such that s J / = θ. As (.) is a cover of Z, by [S95, Theorem 2] or [S99, Theorem (i)] there exists an I,..., k for which I J and / = θ. Note that J I and hence t J \ I for some t k. Clearly θ = / S(A t ) and also θ m t /n t = s J\t / S(A t ). This concludes the proof. Proof of Theorem.. We use induction on m. The m = case, as mentioned above, has been handled in [S95, S99]. Now let m > and assume that Theorem. holds for smaller positive integers. Let θ S(A). In light of Lemma 2., there is a t,..., k such that both θ and θ = θ m t /n t lie in S(A t ). As A t forms a (m )-cover of Z, by the induction hypothesis we have I At (θ) 2 m and I At (θ ) 2 m. Observe that Therefore I A (θ) = I At (θ) I t : I I At (θ ). I A (θ) = I At (θ) + I At (θ ) 2 m + 2 m = 2 m. We are done. 3. Proof of Theorem.2 At first we give a lemma on algebraic number fields with power integral bases.

5 A SHARP RESULT ON m-covers 5 Lemma 3.. Let K be an algebraic number field with a power integral basis, γ,..., γ n. For any µ = n r=0 µ rγ r K with µ 0,..., µ n Q, we have µ O K ψ(µ), ψ(µγ),..., ψ(µγ n ) Z, where ψ(µ) denotes the last coordinate µ n. Proof. If µ O K, then µ, µγ,..., µγ n O K and hence ψ(µγ j ) Z for every j = 0,..., n. Now assume that ψ(µγ j ) Z for all j = 0,..., n. We want to show that µ O K (i.e., µ 0,..., µ n Z). Clearly µ n = ψ(µγ 0 ) Z. If 0 r < n and µ r+,..., µ n Z, then µ r =ψ(µ 0 γ n r + µ γ n r + + µ r γ n ) =ψ(µγ n r ) ψ(µ r+ γ n + + µ n γ 2n 2 r ) and hence µ r Z since µ r+ γ n + + µ n γ 2n 2 r O K. So, by induction, µ r Z for all r = 0,..., n. We are done. Proof of Theorem.2. In the spirit of the proof of Theorem., it suffices to handle the case m =. That is, we should prove that for any J,..., k there is I,..., k with I J such that ω s/ s J ω s/ O K. Let, γ,..., γ n be a power integral basis of K, and define ψ as in Lemma 3.. Let x 0,..., x n Z and x = n r=0 x rγ r. Since x O K is covered by A = α s + O K k s=, we have 0 = k s= = ( e 2πiψ(ω s(x+α s )/ ) ) = I,...,k( ) I = I,...,k = θ 0 S 0 e 2πix0θ0 I,...,k( ) I e 2πi(ψ(ω sα s / )+ n r=0 x rψ(ω s γ r / )) ( ) I e 2πiψ( ω sα s / ) θ S e 2πixθ n r=0 e 2πix rψ( ω sγ r / ) e 2πiψ(ω s(x+α s )/ ) θ n S n e 2πixn θn f(θ 0,..., θ n ), where S r = ( ψ ω s γ r ) : I,..., k

6 6 HAO PAN AND ZHI-WEI SUN and f(θ 0,..., θ n ) = I,...,k ψ( ω sγ r / )=θ r for all r=0,...,n ( ) I e 2πiψ( ω sα s / ). For each r = 0,..., n, if θ r S r e 2πix rθ r F (θ r ) = 0 for all x r = 0,..., S r, then F (θ r ) = 0 for every θ r S r, because the Vandermonde determinant det(e 2πix rθ r ) 0 xr < S r, θ r S r does not vanish. So, by the above, we have f(θ 0,..., θ n ) = 0 for all θ 0 S 0,..., θ n S n. Now suppose that µ K and s J ω s/ µ+o K for a unique subset J of,..., k. We want to deduce a contradiction. Set θ r = ψ(µγ r ) for r = 0,..., n. For any I,..., k we have ( ψ (( ψ ω s γ r ) = θ r for all r = 0,..., n ) )γ r ω s µ Z for all r = 0,..., n ω s µ + O K (by Lemma 3.) I = J. Thus the expression of f(θ 0,..., θ n ) only contains one summand, and therefore 0 = f(θ 0,..., θ n ) = ( ) J e 2πiψ( s J ω sα s /) 0 which is a contradiction. The proof of Theorem.2 is now complete. Acknowledgment. The authors are indebted to the referee for his/her valuable suggestions. References [E97] P. Erdős, Some of my favorite problems and results, in: The mathematics of Paul Erdős, I, 47 67, Algorithms Combin., 3, Springer, Berlin, 997. [G04] R. K. Guy, Unsolved Problems in Number Theory, 3rd edition, Springer, New York, [J68] J. H. Jordan, A covering class of residues with odd moduli, Acta Arith. 3 (968), [PS] Š. Porubský and J. Schönheim, Covering systems of Paul Erdös: past, present and future, in: Paul Erdös and his Mathematics. I (edited by G. Halász, L. Lovász, M. Simonvits, V. T. Sós), Bolyai Soc. Math. Studies, Budapest, 2002, pp

7 A SHARP RESULT ON m-covers 7 [S95] Z. W. Sun, Covering the integers by arithmetic sequences, Acta Arith. 72 (995), [S96] Z. W. Sun, Covering the integers by arithmetic sequences. II, Trans. Amer. Math. Soc. 348 (996), [S97] Z. W. Sun, Exact m-covers and the linear form k s= x s/, Acta Arith. 8 (997), [S99] Z. W. Sun, On covering multiplicity, Proc. Amer. Math. Soc. 27 (999), [S03] Z. W. Sun, Unification of zero-sum problems, subset sums and covers of Z, Electron. Res. Annnounc. Amer. Math. Soc. 9 (2003), [S05] Z. W. Sun, On the range of a covering function, J. Number Theory (2005), [S06] Z. W. Sun, A connection between covers of the integers and unit fractions, Adv. in Appl. Math., in press. [Z89] M. Z. Zhang, A note on covering systems of residue classes, Sichuan Daxue Xuebao (Nat. Sci. Ed.) 26 (989), Special Issue, Department of Mathematics, Nanjing University, Nanjing 20093, People s Republic of China addresses: haopan79@yahoo.com.cn, zwsun@nju.edu.cn