NNO LATTICE TILING OF Z n BY LEE SPHERE OFRADIUS 2
KA HIN LEUNG AND YUE ZHOU † Abstract.
We prove the nonexistence of lattice tilings of Z n byLee spheres of radius 2 for all dimensions n ≥
3. This impliesthat the Golomb-Welch conjecture is true when the common radiusof the Lee spheres equals 2 and 2 n + 2 n + 1 is a prime. As adirect consequence, we also answer an open question in the degree-diameter problem of graph theory: the order of any abelian Cayleygraph of diameter 2 and degree larger than 5 cannot meet theabelian Cayley Moore bound. Introduction
The
Lee distance (also known as (cid:96) l -norm , taxicab metric , rectilineardistance or Manhattan distance ) between two vectors x = ( x , x , · · · , x n )and y = ( y , y , · · · , y n ) ∈ Z n is defined by d L ( x, y ) = n (cid:88) i =1 | x i − y i | . Let S ( n, r ) denote the Lee sphere of radius r centered at the origin in Z n , i.e. S ( n, r ) = (cid:40) ( x , · · · , x n ) ∈ Z n : n (cid:88) i =1 | x i | ≤ r (cid:41) . If there exists a subset C ∈ Z n such that T = { S ( n, r ) + c : c ∈ C } forms a partition of Z n , then we say that T is a tiling of Z n by S ( n, r ).If C is further a lattice, then we call T a lattice tiling . Department of Mathematics, National University of Singapore,119076 Singapore College of Liberal Arts and Sciences, National University of De-fense Technology, 410073 Changsha, China † Corresponding author
E-mail addresses : [email protected], [email protected] .1991 Mathematics Subject Classification.
Key words and phrases.
Golomb-Welch conjecture; Lattice tiling; Algebraictiling; Degree-diameter problem; Perfect Lee code. a r X i v : . [ m a t h . C O ] O c t K. H. LEUNG AND Y. ZHOU
One may get a geometric interpretation of tilings of Z n by Leespheres in the following way. Let R denote the set of real numbersand C ( x , · · · , x n ) = { ( y , · · · , y n ) : | y i − x i | ≤ / } which is the n -cube centered at ( x , · · · , x n ) ∈ R n . Let L ( n, r ) be the union of n -cubescentered at each point in S ( n, r ). Figure 1 depicts L ( n, r ) for n = 2 , r = 1 , Figure 1.
Figures of L (2 , L (2 , L (3 ,
1) and L (3 , Z n by S ( n, r ) exits if and only if atiling of R n by L ( n, r ) exists. Figure 2 shows a (lattice) tiling of R by L (2 , R n by L ( n, r ) for n = 1 , r always exist and lattice tilings of R n by L ( n,
1) also exist forany n ; see [6]. Figure 2.
Tiling of R by L (2 , R n is quite important, because L ( n, r )is close to a cross-polytope when r is large enough. It follows that a tiling of R n with L ( n, r ) induces a dense packing of R n by cross-polytopes. One can use the cross-polytope packing density or the lin-ear programming method which is originally applied on the Euclideansphere packing density in [3] to show the following type of results. Result 1.1.
For any n ≥ , there exists r n such that for r > r n , R n cannot be tiled by L ( n, r ) . Result 1.1 was first obtained by Golomb and Welch who showed in [6]only the existence of r n . However, the value of r n three is unspecified.Later, several lower bounds on r n for the periodic case were obtainedby Post [15] and Lepist¨o [13]. In [10] the very first lower bound on r n is stated.In the same seminal paper [6], Golomb and Welch proposed the fol-lowing conjecture originally given in the language of perfect Lee codes. Conjecture 1.
For n ≥ r ≥
2, there is no perfect r -error-correcting Lee code in Z n , i.e. Z n cannot be tiled by Lee spheres ofradius r .Conjecture 1 is still far from being solved, though various approacheshave been applied on it. We refer the reader to the recent survey [10]and the references therein.In [10] Horak and Kim suggest that r = 2 appears to be the mostdifficult case of Conjecture 1 for two reasons. First it is the thresholdcase, because Z n can always be tiled by S ( n, ≤ n ≤ r ≥ Z n by S ( n,
2) for the given n .In this direction, there are several recent advances. In [9], Conjecture1 is proved for n ≤
12 and r = 2. In [12], Kim presents a methodbased on symmetric polynomials to show that Conjecture 1 is true for r = 2 and a certain class of n satisfying that | S ( n, | is a prime. Thisapproach has been further applied to the lattice tilings of Z n by S ( n, r )with larger r in [16] and [18]. In [19], Zhang and the second authortranslated the lattice tilings of Z n by S ( n,
2) or S ( n,
3) into group ringequations. By applying group characters and algebraic number theory,they have obtained more nonexistence results for infinitely many n with r = 2 and 3.In this paper, we completely solve the lattice tiling cases of Conjec-ture 1 for r = 2 and any n . Theorem 1.1.
For any integer n ≥ , there is no lattice tiling of Z n by S ( n, . It is worth noting that, in contrast to Result 1.1 which is proved forfixed dimension n , Theorem 1.1 is for fixed radius r and arbitrary n . K. H. LEUNG AND Y. ZHOU
It is straightforward to show that | S ( n, | = 2 n + 2 n + 1. Accordingto [17, Theorem 28] (see [11, Exampe 2] for an alternative proof), when2 n + 2 n + 1 is a prime, a tiling of Z n by S ( n,
2) must be a lattice tiling.Thus Theorem 1.1 implies the following result.
Corollary 1.2.
For r = 2 and n ≥ satisfying that n + 2 n + 1 isprime, the Golomb-Welch conjecture is true. Our result also answers an important question in graph theory. The degree-diameter problem is to determine the largest graph of givenmaximum degree d and diameter k . For the general case, the famous Moore bound is an upper bound for the orders of such graphs. Exceptfor k = 1 or d ≤
2, graphs achieving the Moore bound are only possiblefor d = 3 , ,
57 and k = 2; see [1] [4] and [7].Let G be a multiplicative group with the identity element e and S ⊆ G such that S − = S and e (cid:54)∈ S . Here S − = { s − : s ∈ S } . The(undirected) Cayley graph Γ( G, S ) has a vertex set G , and two distinctvertices g, h are adjacent if and only if g − h ∈ S . In particular, when G is abelian, we call Γ( G, S ) an abelian Cayley graph .Let AC ( d, k ) denote the largest order of abelian Cayley graphs ofdegree d and diameter k . In [5], an upper bound for AC (2 n, r ) isobtained which actually equals | S ( n, r ) | = min { n,r } (cid:88) i =0 i (cid:18) ni (cid:19)(cid:18) ri (cid:19) . This value is often called the abelian Cayley Moore bound . An impor-tant open question in graph theory is whether there exists an abeliangraph whose order meets this bound. For more details about thedegree-diameter problems, we refer to the survey [14].By checking the proof of the upper bound for AC (2 n, r ) in [5], itis not difficult to see that an abelian Cayley graph of degree 2 n anddiameter r achieves this upper bound if and only if there is a latticetiling of Z n by S ( n, r ); see [19, Section 2.1] for the detail. This linkis also pointed out in [2]. Hence, Theorem 1.1 is equivalent to thefollowing statement. Corollary 1.3.
The number of vertices in any abelian Cayley graphof diameter and even degree d ≥ is strictly less than the abelianCayley Moore bound. The rest of this paper is organized as follows: In Section 2, we in-troduce the group ring conditions for the existence of a lattice tiling of Z n by S ( n, Preliminaries
Let Z [ G ] denote the set of formal sums (cid:80) g ∈ G a g g , where a g ∈ Z and G is any (not necessarily abelian) group which we write here multiplica-tively. The addition of elements in Z [ G ] is defined componentwise, i.e. (cid:88) g ∈ G a g g + (cid:88) g ∈ G b g g := (cid:88) g ∈ G ( a g + b g ) g. The multiplication is defined by( (cid:88) g ∈ G a g g ) · ( (cid:88) g ∈ G b g g ) := (cid:88) g ∈ G ( (cid:88) h ∈ G a h b h − g ) · g. Moreover, λ · ( (cid:88) g ∈ G a g g ) := (cid:88) g ∈ G ( λa g ) g for λ ∈ Z . For A = (cid:80) g ∈ G a g g and t ∈ Z , we define A ( t ) := (cid:88) g ∈ G a g g t . For any set A whose elements belong to G ( A may be a multiset), wecan identify A with the group ring element (cid:80) g ∈ G a g g where a g is themultiplicity of g appearing in A . Moreover, we use | A | to denote thenumber of distinct elements in A , rather than the counting of elementswith multiplicity.The existence of a lattice tiling of Z n by S ( n,
2) can be equivalentlygiven by a collection of group ring equations.
Lemma 2.1 ([19]) . Let n ≥ . There exists a lattice tiling of Z n by S ( n, if and only if there exists a finite abelian group G of order n + 2 n + 1 and a subset T of size n + 1 viewed as an element in Z [ G ] satisfying(a) the identity element e belongs to T ,(b) T = T ( − ,(c) T = 2 G − T (2) + 2 n . We also need the following nonexistence results summarized in [19].
Lemma 2.2.
For ≤ n ≤ , there is no lattice tiling of Z n by S ( n, except possibly for n = 16 , , , , , , , . K. H. LEUNG AND Y. ZHOU Proof of the main result
Our objective is to show the nonexistence of T satisfying Conditions(a)–(c) in Lemma 2.1. To do so, we do assume such T exists and tryto deduce some necessary consequences.The outline of our proof of Theorem 1.1 is as follows: we first in-vestigate T (2) T (mod 3), which provides us some strong restrictions onthe multiplicities of elements in T (2) T . In particular, it leads to a proofof Theorem 1.1 when n ≡ T (4) T (mod 5). For each of the rest 10 possible valueof n modulo 15, we can get a contradiction.First, by using Condition (c), we immediately obtain the following: Lemma 3.1.
For any g ∈ G \ { e } , |{ ( t , t ) ∈ T × T : t t = g }| = (cid:40) , if g ∈ T (2) ;2 , if g ∈ G \ T (2) . In particular, T ∩ T (2) = { e } . Moreover, if t, t , t ∈ T , then t t = t if and only if t = t = t . Observe that by (c), T = 2(2 n + 1) G − T (2) T + 2 nT which means(1) T (2) T = 2(2 n + 1) G − T + 2 nT. Our strategy is to exploit the above equation. For convenience, we keepthe following notation through this section. We write T (2) T = N (cid:88) i =0 iX i where { X i : i = 0 , , . . . , N } forms a partition of G . It is easy to deducethe following:(2) 2 n + 2 n + 1 = N (cid:88) i =0 | X i | , and(3) (2 n + 1) = N (cid:88) i =1 i | X i | . Note that | G | is odd and | T (2) | = 2 n + 1. Moreover, as T (2) ∩ T = { e } ,it follows that e ∈ X . Besides the above two equations on | X i | ’s, wederive another equation based on the inclusion-exclusion principle asfollows: Lemma 3.2. (4) N (cid:88) i =1 | X i | = 4 n + 1 + N (cid:88) s =3 ( s − s − | X s | . Proof.
By (a) and (b), we can write T (2) = (cid:80) ni =0 a i with a = e and a − i = a n +1 − i for i = 1 , , . . . , n . Clearly all the a i ’s are distinct fromeach other and(5) T (2) T = n (cid:88) i =0 a i T. First, we prove the following claim.
Claim 1.
For 0 ≤ i < j ≤ n , | a i T ∩ a j T | = (cid:40) , i < j ;2 , < i < j. Observe that a i t = a j t (cid:48) ∈ a i T ∩ a j T if and only if a i a − j = t − t (cid:48) forsome t, t (cid:48) ∈ T . Recall that T (2) also satisfies Condition (c) in Lemma2.1. Hence, a i a − j / ∈ T (2) if and only if i (cid:54) = 0.If a i a − j ∈ G \ T (2) , then by Lemma 3.1, there exist two distinct t , t ∈ T such that ( t − , t (cid:48) ) = ( t , t ) or ( t , t ). Hence, t = t − or t − . Consequently, | a i T ∩ a j T | = 2. On the other hand, if i = 0, a i a − j = ea − j = s ∈ T (2) for some s ∈ T . By Lemma 3.1, t − = s = t (cid:48) .Hence, | a i T ∩ a j T | = 1.By the inclusion–exclusion principle and (5), we count the distinctelements in T (2) T ,(6) | T (2) T | = n (cid:88) i =0 | a i T |− (cid:88) i 3. It meansthat g ∈ X s for some s ≥ 3. Then the contribution for g in the sum K. H. LEUNG AND Y. ZHOU (cid:80) r ≥ ( − r − | a i T ∩ a i T ∩ · · · ∩ a i r T | is( − − (cid:18) s (cid:19) +( − − (cid:18) s (cid:19) + · · · = (cid:18) s (cid:19) − (cid:18) s (cid:19) + (cid:18) s (cid:19) = ( s − s − . Therefore, (cid:88) r ≥ ( − r − | a i T ∩ a i T ∩ · · · ∩ a i r T | = (cid:88) s ≥ | X s | ( s − s − . Plugging the above equation, (7) and (8) into (6), we obtain (4). (cid:3) Our strategy is to derive a contradiction using Equation (2), (3)and (4). We need to further exploit (1). It is natural to consider (1)modulus 3 as T ≡ T (3) (mod 3). We then have(9) T (2) T = 2(2 n + 1) G − T (3) + 2 nT (mod 3) . Note that | G | = 2 n + 2 n + 1 is not divisible by 3. Therefore, | T (3) | =2 n + 1. We first investigate the case when n ≡ Proposition 3.3. Theorem 1.1 is true for n ≡ .Proof. Now (9) becomes T (2) T ≡ G − T (3) (mod 3) . Since all coefficients are non-negative, | T (3) | = 2 n +1 and all coefficientsof T (3) is 1, we conclude that(10) (cid:88) i =0 | X i +1 | = 2 n + 1 , (cid:88) i =0 | X i +2 | = 2 n and (cid:88) i =0 | X i | = 0 . By (3) and (10),(11) 2 n = (cid:88) i =1 i ( | X i +1 | + | X i +2 | ) . We recall that N is the largest integer with | X N | (cid:54) = 0. By (2) and (4),we have(12) 2 n − n = N (cid:88) s =4 ( s − s − | X s | ≤ N − N (cid:88) s =4 ( s − | X s | . As T ( − = T and T ( − = T (2) , it is clear that X N = X ( − N . Recallthat e ∈ X . It then follows that | X N | ≥ 2. From (11), we derive that2( N − ≤ n .By (11), N ≡ , (cid:80) i =0 | X i | = 0, we conclude (cid:80) Ns =4 ( s − | X s | ≤ n . Therefore, we obtain from (12) that 2 n − n ≤ ( n + 1) n . This is possible only if n ≤ 3. As for n = 3, it is alreadyknown in Lemma 2.2 that Theorem 1.1 is true. (cid:3) Unfortunately, using the above argument in case n ≡ ± Lemma 3.4. Suppose n ≡ . Then, | X i | = 0 for all i ≥ .Moreover, we have (13) | X | = 4 n ( n − , | X | = 2 n ( n − , | X | = 4 n and | X | = 1 . Proof. Now (9) becomes(14) T (2) T ≡ T − T (3) (mod 3) . We first show that T ∩ T (3) = { e } . Suppose that t and t ∈ T satisfythat t = t ∈ T . Then tt − = t ∈ T (2) which means t = t − = t by(c). Hence t = t = e because | G | is odd.By comparing the coefficients in (14), we see that except for thoseelements in T ∪ T (3) , all are congruent to 0 mod 3. Since T ∩ T (3) = { e } and e ∈ X , the coefficients of all the elements in T ∪ T (3) \{ e } arecongruent to 2 mod 3. Therefore, we get | X | = 1 , | X i +1 | = 0 for i ≥ (cid:88) i =0 | X i +2 | = 4 n. Plugging them into (4), we get1 + | X | + N (cid:88) i =3 | X i | = (cid:88) i =0 | X i +2 | + 1 + N (cid:88) s =3 ( s − s − | X s | , from which we conclude that2 | X | + 4 | X | + 9 | X | + · · · ≤ . This implies that | X i | = 0 for i ≥ | X | = 4 n . Hence, by (2) and(3), (cid:26) n + 2 n + 1 = | X | + 1 + 4 n + | X | , n + 4 n + 1 = 1 + 2 · n + 3 | X | . Solving the above equations, we get the desired result. (cid:3) Next, we consider the case when n ≡ Lemma 3.5. Suppose n ≡ . Then, | X i | = 0 for all i ≥ .Moreover, we have (15) | X | = 4 n − n + 33 , | X | = | X | = 2 n, | X | = 2 n − n and | X | = 0 . Proof. Now (9) becomes T (2) T ≡ G − T (3) + T (mod 3) . As shown before, we have T ∩ T (3) = { e } . Thus, X = ∅ and T (3) \{ e } = (cid:91) i =0 X i , T \{ e } = (cid:91) i =0 X i +2 and G \ ( T ∪ T (3) ) ∪{ e } = (cid:91) i =0 X i +1 . Hence(16) (cid:88) i =0 | X i +1 | = 2 n − n + 1 , (17) (cid:88) i =0 | X i +2 | = 2 n and (cid:88) i =0 | X i | = 2 n. By (4), we have2 n + 2 n + 1 ≥ n + 1 + N (cid:88) s =3 ( s − s − | X s | . Summing up the above equation and (16), we have4 n − n + 1 ≥ | X | + | X | + 4 | X | + 6 | X | + 10 | X | + 16 | X | + · · · . On the other hand, plugging (3), (17) and X = ∅ into (cid:80) i =1 i | X i | − (cid:80) i =0 | X i | − (cid:80) i =0 | X i +2 | , we have4 n − n + 1 = | X | + | X | + 4 | X | + 3 | X | + 4 | X | + 7 | X | + · · · . Thus | X | = | X | = · · · = 0. It follows from (17) that | X | = 2 n . Bysolving n − n + 1 = | X | + | X | + 4 | X | , n + 4 n + 1 = | X | + 2 | X | + 3 | X | + 4 | X | , n − n + 1 = | X | + | X | . we get our desired result. (cid:3) In view of the above results, we see that by just considering modulus3, it doesn’t rule out the case for n ≡ , T = (8 n + 8 n + 2)(2 n + 1) G − T (4) T − nT (2) T + (4 n + 2 n ) T which implies(18) T (4) T = (8 n + 8 n + 2)(2 n + 1) G − T − nT (2) T + (4 n + 2 n ) T. As before, we write T (4) T = M (cid:88) i =0 iY i where { Y i : i = 0 , , . . . , M } forms a partition of G . Since | T (4) | =2 n + 1, we have : 2 n + 2 n + 1 = M (cid:88) i =0 | Y i | , and(19) (2 n + 1) = M (cid:88) i =1 i | Y i | . However, the situation is slightly different now. Lemma 3.6. There exists an integer ∆ ∈ [ − n, such that (20) M (cid:88) i =1 | Y i | = 4 n + 1 + ∆ + M (cid:88) s =3 ( s − s − | Y s | . Moreover, we have (21) 2 | Y | + 3 | Y | + 3 | Y | + 2 | Y | ≥ n + 6 n + 2 . Proof. The proof is quite similar to the one for Lemma 3.2. The onlydifferent part is Claim 1.By (a), we may write T (4) = (cid:80) ni =0 a i with a = e and a − i = a n +1 − i for i = 1 , , . . . , n . Hence T (4) T = n (cid:88) i =0 a i T. Claim 1. | T ∩ a i T | = 2 for i > n − n ≤ (cid:88) i > 0, as shown before, | a i T ∩ a j T | = (cid:40) , a − i a j ∈ T (2) , , otherwise.To find the number of pairs ( i, j ) with 0 < i < j when | a i T ∩ a j T | = 1,we need to find for each s ∈ T (2) , the number of pairs of ( i, j ) with i < j and a − i a j = s . Since a − i , a j ∈ T (4) and T (4) also satisfies Condition(c), it follows that the number of pairs is at most 1. Therefore,0 ≤ δ = |{ ( i, j ) : 0 < i < j, a − i a j ∈ T (2) }| ≤ n. Thus (cid:88) i Proposition 3.7. Theorem 1.1 is true for n ≡ .Proof. By (18), we obtain(22) T (4) T ≡ G − T (5) (mod 5) . In this case, 5 doesn’t divide | G | . Therefore, | T (5) | = 2 n + 1. Conse-quently, (cid:91) i =0 Y i +1 = T (5) , (cid:91) i =1 Y i +2 = G \ T (5) . Therefore, | Y i | = 0 for i (cid:54)≡ , (cid:88) i =0 | Y i +1 | = 2 n + 1 , (cid:88) i =0 | Y i +2 | = 2 n . Hence(24) (cid:88) i =0 | Y i +1 | + (cid:88) i =0 | Y i +2 | = 4 n + 2 n + 1 . On the other hand, | Y | + 2 | Y | + 6 | Y | + 7 | Y | + · · · = M (cid:88) i =1 i | Y i | = 4 n + 4 n + 1 . Together with (24), we get(25) 5 (cid:88) i =1 i ( | Y i +1 | + | Y i +2 | ) = 2 n. Recall that M = max { i : Y i (cid:54) = ∅} . By (20) and (23),(26) 2 n − n − ∆ = M (cid:88) s =3 ( s − s − | Y s | ≤ M − M (cid:88) s =3 ( s − | Y s | . As | Y i | = 0 for i (cid:54)≡ , M (cid:88) s =3 ( s − | Y s | ≤ n. Case (i) If | Y M | ≥ 2, then as in the proof of Proposition 3.3, weobtain 2( M − ≤ n and M − ≤ n + 1. Plugging them into (26) and(27), we get 2 n − n − ∆ ≤ ( n + 1) n and n ≤ 3. This is impossible. Case (ii) | Y M | = 1. Note that Y M = Y − M . Hence, Y M = { e } .If M < n + 1, then M − < n . Then, in view of (25), there exists j (cid:54) = M such that | Y j | ≥ 1. Suppose j = 5 i + c where i ≥ c = 1or 2. Again, Y j = Y − j implies, M − ≤ n − 10. Consequently,2 n − n − ∆ ≤ ( M − M (cid:88) s =3 ( s − | Y s | ≤ (2 n − n. This is impossible as ∆ ≤ M = 2 n + 1 and Y M = { e } . This is possible onlywhen T = T (4) . In that case, T (4) T = T = 2 G − T (2) + 2 n . It followsfrom (22) that T (5) = T (2) . For any t ∈ T , there exists s ∈ T suchthat t = s . As T (4) = T , t ∈ T . Hence, t t = s and s ∈ T (2) . ByLemma 3.1, this is possible only when t = t = s . But it then followsthat t = e . Hence | T | ≤ (cid:3) Proposition 3.8. Theorem 1.1 is true if n ≡ .Proof. By Proposition 3.7, we only have to consider the 4 cases when n (cid:54)≡ (i) n ≡ T (4) T ≡ G + T (2) T + T − T (5) (mod 5) ≡ G + X + 2 X + 3 X + T − T (5) (mod 5) . As e ∈ T, T (5) , X , G and e (cid:54)∈ X , X , the identity element e appears in Y i for some i ≥ 1. In view of the above equation, we deduce that e / ∈ X \ ( T ∪ T (5) ) ⊂ (cid:91) i =1 Y i +2 ,e / ∈ X \ ( T ∪ T (5) ) ⊂ (cid:91) i =1 Y i +4 , and e / ∈ X \ ( T ∪ T (5) ) ⊂ (cid:91) i =1 Y i +1 . In view of (19), we get(2 n + 1) ≥ (cid:88) i =0 i | Y i | + (cid:88) i =0 | Y i +2 | + (cid:88) i =0 | Y i +4 | + (cid:88) i =0 | Y i +1 |≥ (cid:88) i =0 | Y i +2 | + (cid:88) i =0 | Y i +4 | + (cid:88) i =0 | Y i +1 |≥ | X \ ( T ∪ T (5) ) | + 4 | X \ ( T ∪ T (5) ) | + | X \ ( T ∪ T (5) ) |≥ | X | + 4 | X | + | X | − | ( T ∪ T (5) ) \ { e }|≥ n − n T ∪ T (5) in the disjoint union of X , X and X is at mostthe size of ( T ∪ T (5) ) \ { e } .This means 4 n − n + 12 ≤ n ≤ 15. However, accordingto Lemma 2.2, this is impossible. (ii) n ≡ T (4) T ≡ T (2) T − T (5) (mod 5) ≡ X + 2 X + 4 X + X − T (5) (mod 5) . Recall that X , X , X and X form a partition of G and all nonzerocoefficients in T (5) are 1. Therefore, Y \ T (5) ⊂ X , Y \ T (5) ⊂ X , Y \ T (5) = ∅ and Y \ T (5) ⊂ X . It follows that | Y | ≤ | X | + x , | Y | ≤ | X | + x , | Y | ≤ x and | Y | ≤ | X | + x where x + x + x + x ≤ | T (5) | = 2 n + 1. Hence, from(13) we can derive(28)2 | Y | +3 | Y | +3 | Y | +2 | Y | ≤ n ( n − n +3(2 n +1) = 83 n + 343 n +6 . On the other hand, it follows from (21) and (28), we have83 n + 343 n + 6 ≥ n + 6 n + 2 , which means n ≤ 6. Hence n = 2. However, it contradicts the assump-tion that n ≡ (iii) n ≡ T (4) T ≡ G − T (2) T + 2 T − T (5) (mod 5) ≡ G + 0 X + 3 X + X + 4 X + 2 T − T (5) (mod 5) ≡ X + 4 X + 2 X + 0 X + 2 T − T (5) (mod 5) . In this case, 5 divides | G | and it is not necessarily true that all nonzerocoefficients in T (5) are 1. One may write T (5) = (cid:80) ki =1 iZ i where (cid:83) ki =1 Z i = T (5) and (cid:80) ki =1 i | Z i | = 2 n + 1. However, as we will seebelow, we can still get some contradiction by checking the bounds for | Y | , | Y | , | Y | and | Y | as before. Observe that Y \ ( T ∪ T (5) ) ⊂ X , Y \ ( T ∪ T (5) ) ⊂ X ,Y \ ( T ∪ T (5) ) = ∅ and Y \ ( T ∪ T (5) ) = ∅ . Note that the last equation is true because X = { e } .2 | Y | + 3 | Y | + 3 | Y | + 2 | Y |≤ | X | + 3 | X | + 3 | T ∪ T (5) |≤ n ( n − n + 12 n + 3 (by Lemma 3.4)= 4 n n . By (21), we have 4 n n ≥ n + 6 n + 2 , which implies n ≤ 6. Taking account of the values of n modulo 3 and5, we see that n = 13. However, according to Lemma 2.2, there is nolattice tiling of Z by S (13 , (iv) n ≡ T (4) T ≡ G − T (2) T + 2 T − T (5) (mod 5) ≡ G + 0 X + 4 X + 3 X + 2 X + 2 T − T (5) (mod 5) ≡ X + 2 X + X + 0 X + 2 T − T (5) (mod 5) . As before, we obtain Y \ ( T ∪ T (5) ) ⊂ X , Y \ ( T ∪ T (5) ) = ∅ ,Y \ ( T ∪ T (5) ) ⊂ X and Y \ ( T ∪ T (5) ) = ∅ . Together with (13) we get2 | Y | + 3 | Y | + 3 | Y | + 2 | Y |≤ | X | + 3 | X | + 3 | X | + 3 | T ∪ T (5) |≤ n + 3 2 n ( n − n + 3=2 n + 18 n + 3 . By (21), 4 n + 6 n + 2 ≤ n + 18 n + 3 . Hence n ≤ n = 4. However, this value has been alreadyexcluded by Lemma 2.2. (cid:3) Proposition 3.9. Theorem 1.1 is true for n ≡ .Proof. By Proposition 3.7, we only have to investigate the 4 cases when n (cid:54)≡ (i) n ≡ T (4) T ≡ G + T (2) T + T − T (5) (mod 5) ≡ G + X + 2 X + 3 X + 4 X + T − T (5) (mod 5) . Thus, Y \ ( T ∪ T ) ⊂ X , Y \ ( T ∪ T ) ⊂ X , Y \ ( T ∪ T ) ⊂ X , Y \ ( T ∪ T ) = ∅ . Hence, 2 | Y | + 3 | Y | + 3 | Y | + 2 | Y |≤ | X | + 3 | X | + 3 | X | + 3 | T ∪ T (5) |≤ n + (2 n − n ) + 12 n + 3 (By Lemma 3.5)=2 n + 18 n + 3 . Therefore, 4 n + 6 n + 2 ≤ n + 18 n + 3 which means n ≤ 6. By Lemma2.2, this is impossible.Alternatively, as 5 | n + 2 n + 1, 8 n − (cid:54) = 5 k for some k ∈ Z , and8 n + 1 is not a square, one may also use [19, Corollary 3.9 (1)] to provethis case. (ii) n ≡ T (4) T ≡ T (2) T − T (5) (mod 5) ≡ X + 4 X + X + 3 X − T (5) (mod 5) . It is easy to see that X \ T (5) ⊂ (cid:91) i =0 Y i +2 and X \ T (5) ⊂ (cid:91) i =0 Y i +3 . It follows that (cid:88) i =0 (5 i + 2) | Y i +2 | ≥ | X | − x and (cid:88) i =0 (5 i + 3) | Y i +3 | ≥ | X | − y with x + y ≤ n + 1. Thus, by (15)(2 n + 1) = M (cid:88) i =1 i | Y i | ≥ | X | + 3 | X | − n − n − n − . By calculation, we get 2 n − n − ≤ n ≤ n is congruent to 2 mod 5, n (cid:54) = 16 , 21. Therefore by Lemma 2.2,we have a contradiction. (iii) n ≡ T (4) T ≡ G − T (2) T + 2 T − T (5) (mod 5) ≡ G + 3 X + X + 4 X + 2 X + 2 T − T (5) (mod 5) ≡ X + 2 X + 0 X + 3 X + 2 T − T (5) (mod 5) . This implies that X \ ( T ∪ T (5) ) ⊂ (cid:91) i =0 Y i +4 and X \ ( T ∪ T (5) ) ⊂ (cid:91) i =0 Y i +3 . Therefore, (cid:88) i =0 | Y i +4 | ≥ | X | − x, and (cid:88) i =0 | Y i +3 | ≥ | X | − y where 0 ≤ x + y ≤ n + 1. Hence, by (15)(2 n + 1) = M (cid:88) i =1 i | Y i | ≥ | X | + 3 | X | − n − n − n , which implies that 10 n − n − ≤ n ≤ 7. According toLemma 2.2, there is no such a lattice tiling of Z n by S ( n, (iv) n ≡ T (4) T ≡ G − T (2) T + 2 T − T (5) (mod 5) ≡ G + 4 X + 3 X + 2 X + X + 2 T − T (5) (mod 5) ≡ X + X + 0 X + 4 X + 2 T − T (5) (mod 5) . As before, we obtain (cid:88) i =0 | Y i +2 | ≥ | X | − x, and (cid:88) i =0 | Y i +4 | ≥ | X | − y where 0 ≤ x + y ≤ n + 1. Hence, by (15)(2 n + 1) = M (cid:88) i =1 i | Y i | ≥ | X | + 4 | X | − n − n − n − , which implies that 4 n − n − ≤ n ≤ 18. As n is congruentto 4 mod 5, n (cid:54) = 16. Thus by Lemma 2.2, this is impossible. (cid:3) Proof of Theorem 1.1. Propositions 3.3, 3.8 and 3.9 together form acomplete proof of Theorem 1.1. (cid:3) Remark 1. Although our main result shows that there is no latticetiling of Z n by S ( n, 2) for all n ≥ 3; we still do not know whetherGolomb-Welch conjecture has been proved in the case of r = 2 for in-finitely many values of n . The reason is that we do not know whether f ( n ) = 2 n + 2 n + 1 is a prime for infinitely many values of n . Apositive answer to this question would solve a special case of the fa-mous conjecture of Bunyakovsky (1857) that asks whether there existsan irreducible quadratic polynomial attaining a prime number valueinfinitely many times. Acknowledgment The authors thank the referees for their helpful comments and sug-gestions, which improve the presentation of this paper. The first authoris supported by grantR-146-000-158-112, Ministry of Education, Singa-pore. The second author is supported by the National Natural ScienceFoundation of China (No. 11771451) and Natural Science Foundationof Hunan Province (No. 2019RS2031). References [1] E. Bannai and T. Ito. 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