Non-existence of partial difference sets of order 8p^3 in Abelian groups
aa r X i v : . [ m a t h . C O ] J u l Non-existence of partial difference sets of order 8 p inAbelian groups Stefaan De Winter ∗ †
Zeying Wang ∗‡ Abstract
In this paper we prove non-existence of nontrivial partial difference sets inAbelian groups of order 8 p , where p ≥ Keywords—
Partial difference set
Let G be a finite Abelian group of order v , and let D ⊆ G be a subset of size k .We say D is a ( v, k, λ, µ )- partial difference set (PDS) in G if the expressions gh − , g ∈ D , h ∈ D , g = h , represent each non-identity element in D exactly λ times,and each non-identity element of G not in D exactly µ times. Further assume that D ( − = D (where D ( s ) = { g s : g ∈ D } ) and e / ∈ D , where e is the identity of G ,then D is called a regular partial difference set. A regular PDS is called trivial if D ∪ { e } or G \ D is a subgroup of G . The condition that D be regular is not a veryrestrictive one, as D ( − = D is automatically fulfilled whenever λ = µ , and D isa PDS if and only if D ∪ { e } is a PDS. The importance of regular PDS lies in thefact that they are equivalent to strongly regular Cayley graphs. A detailed studyof PDS was started by Ma in [8]. By now there is a rich literature on PDS, bothwith a focus on existence conditions and classification (see for example [1], [4], [7],[10],. . . ), as well as with a focus on constructions (see for example [2], [6], [11], [12],. . . ).When the group G is Abelian only a limited number of examples of regular PDSare known: (negative) Latin square type PDS, reversible difference sets, Paley type ∗ Michigan Technological University † [email protected] ‡ [email protected] DS, PCP type PDS, and projective two-weight sets. However, all these exampleshave many connections to other areas in combinatorics, and so it is natural to lookfor other examples. On the other hand, not that many classification results areknown, leaving an abundance of parameter sets for which no PDS is known, butfor which existence has not been excluded. In recent work the authors proved non-existence of PDS in 18 specific cases [3, 5], finalizing the classification of parametersfor which there exists a regular PDS with k ≤
100 in an Abelian group. Thisclassification was started by Ma [9], and had been open for almost 20 years. Buildingon some of the techniques developed in [3] the authors obtained in [4] a completeclassification of regular PDS in Abelian groups of order 4 p , p ≥ p , p ≥ p . The main reason for this is that in the4 p case the number of hypothetical parameter sets for each p grows rapidly as p grows, whereas in the 8 p case this is not the case. When searching for parametersthat survive the basic integrality and divisibility conditions we noticed that for mostprimes p only two parameter sets appeared, and for the remaining primes exactly4 sets appeared. This explains why we use an approach focussed on restrictingparameters in this paper (Section 2), rather than the constructive approach takenin [4]. Equally surprising to the authors was the fact that some of the seemingly adhoc methods used to prove nonexistence of PDS in Abelian groups of order 2 · remain largely valid in the general case (Section 3). The results of this paper seemto further confirm that PDS in Abelian groups are actually rare objects.We end this section by listing a few useful results which we will need in the restof the paper. The first three results can be found in Ma ([9], [10]), and the fourthis a recent local multiplier theorem from [3]. Given a ( v, k, λ, µ )-PDS, one definesthe following two parameters: β := λ − µ , and ∆ := ( λ − µ ) + 4( k − µ ). Proposition 1.1
No non-trivial PDS exists in • an Abelian group G with a cyclic Sylow- p -subgroup and o ( G ) = p ; • an Abelian group G with a Sylow- p -subgroup isomorphic to Z p s × Z p t where s = t . Proposition 1.2
Let D be a nontrivial regular ( v, k, λ, µ ) -PDS in an Abeliangroup G. Suppose ∆ = ( λ − µ ) + 4( k − µ ) is a perfect square. If N is a sub-group of G such that gcd( | N | , | G | / | N | ) = 1 and | G | / | N | is odd, then D = D ∩ N is a (not necessarily non-trivial) regular ( v , k , λ , µ ) -PDS with | D | = 12 (cid:20) | N | + β ± q ( | N | + β ) − (∆ − β )( | N | − (cid:21) . ere ∆ = π with π = gcd( | N | , √ ∆) and β = β − θπ where β = λ − µ and θ isthe integer satisfying (2 θ − π ≤ β < (2 θ + 1) π . Proposition 1.3
Suppose there exists a regular ( v , k , λ , µ )-PDS D in a group G . (a) If D = ∅ and D = G \ { e } , then ≤ λ ≤ k − and ≤ µ ≤ k − . (b) The parameters β and ∆ have the same parity. (c) The PDS D is nontrivial if and only if −√ ∆ < β < √ ∆ − . Also, if D = G \ { e } , then D is nontrivial if and only if ≤ µ ≤ k − . (d) If ∆ is not a square, then ( v , k , λ , µ )=( t + 1 , t , t − , t ) for some positiveinteger t ; furthermore, if G is Abelian, then v = p s +1 for some prime p ≡ . (e) If G is Abelian and D = ∅ and D = G \ { e } , then v ≡ (2 k − β ) ≡ ; furthermore, if D is nontrivial, then v , ∆ , and v / ∆ have the sameprime divisors. (f) The set ( G \ D ) \ { e } is a PDS with parameters ( v ′ , k ′ , λ ′ , µ ′ ) = ( v, v − k − , v − k − µ, v − k + λ ) called the complement of D . (g) If D is nontrivial, then there exists a nontrivial regular ( v, k + , λ + , µ + ) -PDS D + (in an Abelian group of order v ) with ∆ + = ( λ + − µ + ) + 4( k + − µ + ) = v / ∆ . (The PDS D + is called the dual of D .) Proposition 1.4 [LMT]
Let D be a regular ( v, k, λ, µ ) -PDS in an Abelian group G . Furthermore assume ∆ = ( λ − µ ) + 4( k − µ ) is a perfect square. Then g ∈ G belongs to D if and only if g s belongs to D for all s coprime with o ( g ) , the order of g . Throughout the rest of the paper we will assume that p is a prime with p ≥ p = 3 already was dealt with in [5], andsome of our arguments are only valid if p ≥
5. Furthermore, we will always assumethat D is a nontrivial regular PDS. k By part (f) of Proposition 1.3 we may assume that k ≤ v/
2. By part (g) of Propo-sition 1.3, we may assume that ∆ ≤ v . Before proving a better upper bound on k we will show that under our assumptions ∆ can only take two values. emma 2.1 If D is a regular ( v , k , λ , µ )-PDS in an Abelian group with v = 8 p , k ≤ v/ and ∆ ≤ v then either ∆ = 4 p or ∆ = 16 p . Proof.
By Ma’s Proposition 1.3 (d) and (e), ∆ must be a square, and v , ∆, v / ∆have the same prime divisors. It follows that ∆ = 4 p or ∆ = 16 p . (cid:3) Since the Cayley graph of a ( v, k, λ, µ )–PDS is a ( v, k, λ, µ )-strongly regulargraph, we have k ( k − λ −
1) = µ ( v − k − . (1)By substituting k ( k − λ − v − k − for µ into ∆ = ( λ − µ ) + 4( k − µ ), we get∆ = (cid:18) λ − k ( k − λ − v − k − (cid:19) + 4 (cid:18) k − k ( k − λ − v − k − (cid:19) . (2) Lemma 2.2
If a non-trivial ( v, k, λ, µ ) -PDS exists in an Abelian group with v =8 p , p ≥ a prime, ∆ = 16 p and k ≤ v , then we have k ≤ p + 3 p − . Proof.
Setting ∆ = 16 p and v = 8 p in Equation (2), and solving the obtainedquadratic equation for λ , we get λ = − k (8 p −
1) + k (8 p + 1) ± p (1 + k − p ) (4 p (8 p − + k (1 + k − p )8 p )(8 p − . As λ is an integer, the discriminant must be nonnegative, hence4 p (8 p − + k (1 + k − p )8 p ≥ . (3)Solving the quadratic equation 4 p (8 p − + k (1 + k − p )8 p = 0 for k yields k = − p + 8 p ± p p − p − p + 32 p + p − p p . As the largest root is greater than 4 p , and k ≤ v/
2, in order for Inequality (3)to hold, we have k ≤ − p + 8 p − p p − p − p + 32 p + p − p p . We estimate 64 p − p − p + 32 p + p − p = (8 p − p − p ) + 32 p − p − p + p − p ≥ (8 p − p − p ) for p ≥
5. Hence k ≤ − p + 8 p − p (8 p − p − p ) p = 4 p + 3 p − . (cid:3) emma 2.3 If a non-trivial ( v, k, λ, µ ) -PDS exists in an Abelian group with v =8 p , p ≥ a prime, ∆ = 4 p and k ≤ v , then we have k ≤ p + p − . Proof.
This proof is very similar to that of the preceding lemma. Setting ∆ = 4 p and v = 8 p in Equation (2), and solving the obtained quadratic equation for λ , wenow obtain p (8 p − + k (1 + k − p )8 p ≥ . (4)Solving the quadratic equation p (8 p − + k (1 + k − p )8 p = 0 for k yields k = − p + 16 p ± p p − p − p + 32 p + 4 p − p p . Again as the largest root is greater than 4 p , and k ≤ v/ k ≤ − p + 16 p − p p − p − p + 32 p + 4 p − p p . We estimate 256 p − p − p + 32 p + 4 p − p = (16 p − p − p ) + 16 p − p + 31 p + 4 p − p ≥ (16 p − p − p ) for p ≥
5. Hence k ≤ − p + 16 p − p (16 p − p − p ) p = p + p − . (cid:3) µ From Lemma 2.1 we know that we may assume that ∆ equals either 4 p or 16 p . Inthis subsection we will show that ∆ = 4 p cannot occur, and that either µ = 2 p + 2or µ = 2 p − p . Proposition 2.4
The case ∆ = 4 p cannot occur. Proof.
If ∆ = 4 p , by Proposition 1.3-(e), we have (2 k − β ) ≡ p ). Thuswe can write k = px + β with x an integer.By substituting v = 8 p , λ = µ + β , and k = px + β , Equation (1) becomes p x − β − px − β − p µ + µ = 0 . (5)As ∆ = 4 p = ( λ − µ ) + 4( k − µ )= β + 4( px + β − µ ), it follows that β + 2 β = 4 p − px + 4 µ. (6) ombining Equations (5) and (6), we have obtained µ = x − p . (7)Since µ is an integer, x + 1 and x − p is a primenumber ≥
5, it follows that 2 p | x + 1 or 2 p | x −
1. So we have x = 2 tp − x = 2 tp + 1, where t is an integer. Thus k = px + β = 2 tp ± p + β . Since k ≤ p + p − by Lemma 2.3 and −√ ∆ < β < √ ∆ − t = 0. But when t = 0, we have x = ± µ = 0, implying a trivial PDS. (cid:3) Lemma 2.5 If ∆ = 16 p , then either µ = 2 p + 2 or µ = 2 p − . Proof.
If ∆ = 16 p , by Proposition 1.3-(e), we have (2 k − β ) ≡ p ).Thus we can write k = 2 px + β with x an integer.By substituting v = 8 p , λ = µ + β , and k = 2 px + β , Equation (1) becomes4 p x − β − px − β − p µ + µ = 0 . (8)As ∆ = 16 p = ( λ − µ ) + 4( k − µ )= β + 4(2 px + β − µ ), it follows that β + 2 β = 16 p − px + 4 µ. (9)Combining Equations (8) and (9), we have obtained µ = x − p . (10)Since µ is an integer, x + 1 and x − p is a primenumber ≥
5, it follows that 2 p | x + 1 or 2 p | x −
1. So we can assume that x = 2 tp − x = 2 tp + 1, where t is an integer. If t = 0 then x = ± µ = 0, and we obtaina trivial PDS.Thus k = 2 px + β = 4 tp ± p + β . Since k ≤ p + 3 p − by Lemma 2.2 and −√ ∆ < β < √ ∆ − t = 1. Hence either x = 2 p + 1and µ = 2 p + 2; or x = 2 p − µ = 2 p − (cid:3) p with µ = 2 p + 2 When x = 2 p + 1, we have µ = x − p = 2 p + 2 and k = 2 px + β = 4 p + 2 p + β .Also Equation (9) becomes β + 2 β − hus β = − β = 2. If β = −
4, then k = 4 p + 2 p − µ = 2 p + 2, and λ = 2 p − β = 2, then k = 4 p + 2 p + 1, µ = 2 p + 2, and λ = 2 p + 4. Theorem 3.1
There does not exist a regular ( p , p +2 p − , p − , p +2 )–PDSin an Abelian group, where p ≥ is a prime. Proof.
Let G be an Abelian group of order 8 p . Assume by way of contradictionthat D is a (8 p , 4 p + 2 p −
2, 2 p −
2, 2 p + 2)–PDS in G . By Proposition 1.1, weknow that G ∼ = Z × Z p . Let g , g , · · · , g p − be all elements of order p in G , andlet B g i = { ag i | o ( a ) = 1 or 2 , ag i ∈ D } , and B i = |B g i | , i = 1 , , · · · , p −
1. That is, B i equals the number of elements in D whose ( p + 1)-th power equals g i . We nextprove the simple yet important observation that the LMT implies that |B g i | = |B g si | whenever 1 ≤ s ≤ p − Observation (O)
It holds that |B g i | = |B g si | for ≤ s ≤ p − . Proof. If ag i ∈ B g i , that is, if ag i ∈ D with o ( a ) = 1 or 2, then, by the LMT,( ag i ) s = ag si ∈ D if s is odd; or ( ag i ) s + p = ag si ∈ D if s is even. Thus |B g i | ≤ |B g si | .On the other hand, since gcd( s, p ) = 1, we can find integers r and t such that rs + tp = 1. It is clear that gcd( r, p )=gcd( r + p, p )=1. If ag si ∈ B g si , that is, if ag si ∈ D with o ( a ) = 1 or 2, then, by the LMT, ( ag si ) r = ag i ∈ D if r is odd, and( ag si ) r + p = ag i ∈ D if r is even. Thus |B g si | ≤ |B g i | .Hence |B g i | = |B g si | for 1 ≤ s ≤ p − (cid:3) Let N be the Sylow 2-subgroup of G . Using Proposition 1.2, we know that D contains either 0 or 4 elements of order 2. First assume that D contains no elementsof order 2. We see that Σ i B i = 4 p + 2 p − i B i ( B i −
1) = 14 p + 14, as all 7elements of order 2 are not in D , thus each of them has exactly µ = 2 p + 2 differencerepresentations.According to observation (O), we can assume, by relabeling the g i if neces-sary, that C j := B ( j − p − = B ( j − p − = · · · = B ( j − p − p − for j = 1 , , · · · , p + p + 1, and C ≥ C ≥ · · · ≥ C p + p +1 . We now obtainΣ j C j = 4 p + 2 p − p − p + 6) + 4 p − , (12)Σ j C j = 4 p + 16 p + 12 p − . (13)Since p − is an integer only when p − p =2, 3, or 5,Equation (12) has no integer solutions when p ≥ p is a prime number.When p = 5, Equations (12) and (13) become Σ j C j = 27 and Σ j C j = 48. ThusΣ j C j ( C j −
1) = 21, which contradicts with the fact that C j ( C j −
1) is always even. econdly assume that D contains 4 elements of order 2. It follows that Σ i B i +4 =4 p + 2 p −
2. By counting the number of difference representations of elements oforder 2 in DD ( − , we obtain that Σ i B i ( B i −
1) + 4 · p −
2) + 3(2 p + 2). Usingsimilar labelling as above, we now obtainΣ p + p +1 j =1 C j = 4 p + 6 and Σ p + p +1 j =1 C j = 4 p + 20 . (14)Since C j is a non-negative integer, and 0 − − − − − − p + p +1 j =1 ( C j − C j ) = 14, the system of Equations (14)only has the following nonnegative integer solutions, listed as (decreasing) p + p + 1tuples:(4 , , , , · · · , | {z } p , , , · · · , , , , , , · · · , | {z } p − , , , · · · , , , , , , , , · · · , | {z } p − , , , · · · , , , , , , , , , , · · · , | {z } p − , , , · · · , N is the unique subgroup isomorphic to Z in G . Let P , . . . , P p + p +1 be the p + p + 1 subgroups of G isomorphic to Z p , and let L , . . . , L p + p +1 be the p + p + 1 subgroups of G isomorphic to Z p . Now consider the incidence structure P with points the subgroups P i × N , i = 1 , . . . , p + p + 1, of G , with blocks thesubgroups L i × N , i = 1 , . . . , p + p + 1, of G , and with containment as incidence.Then it is easily seen that P is a 2 − ( p + p + 1 , p + 1 ,
1) design, or equivalently,the unique projective plane of order p . We next assign a weight to each point of P in the following way: if point p corresponds to subgroup P i × N then the weight of p is p − | (( P i × N ) \ N ) ∩ D | . In this way the weights of the p + p + 1 points of P correspond to the p + p + 1 values C , C , . . . , C p + p +1 , that is, 1 / ( p −
1) times thenumber of elements of order p or 2 p from D in the subgroup underlying the givenpoint. Without loss of generality we may assume the labeling is such that point P i × N has weight C i . The weight of a block will simply be the sum of the weightsof the points in that block.We next count how many elements of order p or 2 p from D a specific subgroupof the form L i × N can contain. Assume that | ( L i × N ) ∩ D | = m . Let ag and bh be two distinct elements from D , with a = b = g p = h p = e . Then agh − b − belongs to L i × N if and only if gh − ∈ L i . It is easy to see that if g ∈ L i thereare m − bh such that gh − ∈ L i . When g L i there are several ases to discuss (recall that, by the LMT, when an element bh belongs to D , so doall elements bh l for 1 ≤ l ≤ p −
1) : • if bh ∈ L i × N then obviously agh − b − does not belong to L i × N ; • if bh = ag l for some l ∈ { , . . . , p − } then clearly agh − b − cannot be anonidentity element of L i × N ; • if bh = bg l for some l ∈ { , . . . , p − } and b = a then agh − b − will be in L i × N if and only if l = 1; • if h = g l for any l ∈ { , . . . , p − } and h / ∈ L i , then it is easy to see there is aunique r ∈ { , . . . , p − } such that gh − r ∈ L i , and hence such that agh − r b − belongs to L i × N .Combining the above observations yields that when g / ∈ L i there are | D |− m − ( p − p − possibilities for bh such that agh − b − ∈ L i × N .Counting the number of differences of elements of D that are in L i × N in twoways, we obtain m ( m −
1) + ( k − m )( k − m − ( p − p − λm + µ (8 p − − m ) , (15)where ( k, λ, µ ) = (4 p + 2 p − , p − , p + 2). This yields that m = 2( p + 1) or2(3 p − m ′ := p − | (( L i × N ) \ N ) ∩ D | . We obtain m ′ = p +1) − p − = 2 or m ′ = p − − p − = 6 since D contains 4 elements of order 2.We now note that the values m ′ must be the weights of the blocks of P , andthat in both cases these weights are even. We first show that no value C i can beodd. Assume by way of contradiction that C i is odd for some i . Let the weight ofthe p + 1 blocks that contain P i × N be n , . . . , n p +1 respectively. Then p + p +1 X j =1 C j = C i + p +1 X t =1 ( n t − C i ) . As n t is even for all t (the n t are m ′ values), this implies that P p + p +1 j =1 C j isodd. This contradicts with the fact that P p + p +1 j =1 C j = 4 p + 6.Since all the solutions to the system of Equations (14) contain at least one odd C j , it follows that no nontrivial (8 p , 4 p + 2 p −
2, 2 p −
2, 2 p + 2)–PDS exists in anAbelian group. (cid:3) heorem 3.2 There does not exist a ( p , p + 2 p + 1 , p + 4 , p + 2 )–PDS inan Abelian group, where p ≥ is a prime. Proof.
This case is dealt with in a very similar way. We will only provide a sketchof the proof. Assume by way of contradiction D is a (8 p , 4 p + 2 p + 1, 2 p + 4,2 p + 2)-PDS in an Abelian group G .As before G ∼ = Z × Z p , and using Proposition 1.2 we obtain that D containseither 3 or 7 elements of order 2. If D contains 3 elements of order 2 we obtainΣ j C j = 4 p + 2 p − p − p + 6) + 4 p − , (16)Σ j C j = 4 p + 16 p + 12 p − . (17)which is the same as the system of Equations (12), (13). Hence this case cannotoccur.If D contains 7 elements of order 2 we obtainΣ j C j = 4 p + 6 and Σ j C j = 4 p + 20 , (18)which is the same as the system of equations in (14), and thus has the same set ofsolutions.With similar notation as in the previous theorem, and using the same countingargument for ( k, λ, µ )=(4 p + 2 p + 1, 2 p + 4, 2 p + 2), one obtains m = 2 p + 5 or m = 6 p + 1.Now define m ′ := p − | (( L i × N ) \ N ) ∩ D | . We obtain m ′ = p +5 − p − = 2 or m ′ = p +1 − p − = 6 since D contains 7 elements of order 2.As before the weights of all blocks of P must be even, and the proof can befinished in the same way as in the (8 p , 4 p + 2 p −
2, 2 p −
2, 2 p + 2) case. (cid:3) p with µ = 2 p − When x = 2 p −
1, we have µ = x − p = 2 p − k = 2 px + β = 4 p − p + β .Also Equation (9) becomes β + 2 β − p −
1) = 0 (19)Thus β = − ± √ p −
7. It is easy to observe that when √ p − If √ p − y + 1, we have p = 4 y + y + , contradicting with the assump-tion that p is an integer; • If √ p − y + 3, we have p = 4 y + 3 y + 1; • If √ p − y + 5, we have p = 4 y + 5 y + 2; • If √ p − y + 7, we have p = 4 y + 7 y + , contradicting with theassumption that p is an integer. p = 4 y + 3 y + 1 In this subsection, we let p = 4 y + 3 y + 1 and p ≥ y ≥
2, and β = − ± (8 y + 3)= − y − y + 2 by Equation (19). Theorem 4.1
Let p = 4 y + 3 y + 1 and p ≥ be a prime number. Then no non-trivial regular partial difference sets exist with µ = 2 p − , β = − y − or y + 2 ,and k = 4 p − p + β in Abelian groups of order p . Proof.
Let G be an Abelian group of order 8 p . We will prove this theorem in twoparts based on the β values:(i) Let β = − y −
4. Assume on the contrary that D is a non-trivial regular(8 p , p − p − y − , p − y − , p − G . Assume that | D ∩ Z | = a ,where 0 ≤ a ≤
7. Using the notation from Section 3, we have X B i = k − a = 4 p − p − y − − a. It follows that X C i = 4 p − p − y − − ap − p + 2 − y + ap − . Since y ≥ ≤ a ≤
7, and p = 4 y + 3 y + 1, it is easy to check that0 < y + a ≤ y + 7 < p − y + 3 y, thus y + ap − is not an integer, and hence no such D exists.(ii) Let β = 8 y + 2. Assume on the contrary that D is a non-trivial regular(8 p , p − p + 4 y + 1 , p + 8 y, p −
2) PDS in G . Assume that | D ∩ Z | = a ,where 0 ≤ a ≤
7. Using the notation from Section 3, we have X B i = 4 p − p + 4 y + 1 − a. t follows that X C i = 4 p − p + 4 y + 1 − ap − p + 2 + 4 y + 3 − ap − . Since y ≥ ≤ a ≤
7, and p = 4 y + 3 y + 1, it is easy to check that0 < y + 3 − a ≤ y + 3 < p − y + 3 y, thus y +3 − ap − is not an integer, and hence no such D exists. (cid:3) p = 4 y + 5 y + 2 In this subsection, we let p = 4 y + 5 y + 2 ≥ β = − ± (8 y + 5)= − y − y + 4 by Equation (19). We note that the firstpart of the following theorem is slightly more subtle than the proof of Theorem 4.1as one needs to invoke Proposition 1.2 in order to exclude the case of 5 elements oforder 2 when p = 11. Theorem 4.2
Let p = 4 y + 5 y + 2 and p ≥ be a prime number. Then no non-trivial regular partial difference sets exist with µ = 2 p − , β = − y − or y + 4 ,and k = 4 p − p + β in Abelian groups of order p . Proof.
Let G be an Abelian group of order 8 p . We will prove this theorem in twoparts based on the β values:(i) Let β = − y −
6. Assume on the contrary that D is a non-trivial regular(8 p , p − p − y − , p − y − , p −
2) regular PDS in G .Let N = Z be the Sylow 2-group of G . By Proposition 1.2, we know that D contains either 3 or 7 elements of order 2. Thus X B i = 4 p − p − y − X B i = 4 p − p − y − . Hence X C i = 4 p + 2 − y + 4 p − X C i = 4 p + 2 − y + 8 p − . Clearly 0 < y + 4 < p − y + 5 y + 1 when y ≥
1, so y +4 p − is not an integer.Also 0 < y + 8 < p − y + 5 y + 1 when y ≥
2, and y +84 y +5 y +1 = when y = 1, so y +84 y +5 y +1 is not an integer for any y ≥
1. Hence no such D exists. ii) Let β = 8 y + 4. Assume on the contrary that D is a non-trivial regular(8 p , p − p + 4 y + 2 , p + 8 y + 2 , p −
2) regular PDS in G . Assume that | D ∩ Z | = a , where 0 ≤ a ≤
7. Then X B i = 4 p − p + 4 y + 2 − a. It follows that X C i = 4 p − p + 4 y + 2 − ap − p + 2 + 4 y + 4 − ap − . Since y ≥ ≤ a ≤
7, and p = 4 y + 5 y + 2, it is easy to check that0 < y + 4 − a ≤ y + 4 < p − y + 5 y, thus y +4 − ap − is not an integer. Hence no such D exists. (cid:3) Combining all results from this paper and [5] we have obtained the following theo-rem.
Theorem 5.1
All regular PDS in Abelian groups of order p , p an odd prime,are trivial. Proof.
When p = 3 this is the main result of [5]. When p ≥ (cid:3) As mentioned in the introduction, this nonexistence result seems to provide fur-ther evidence that regular PDS in Abelian groups are rare. Nevertheless, classifyingor completely characterizing regular PDS in Abelian groups seems to be completelyout of reach at this point.
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