Lenny Jones Department of Mathematics, Shippensburg University, Shippensburg, Pennsylvania Daniel White

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1 #A10 INTEGERS 1A (01): John Selfrige Memorial Issue SIERPIŃSKI NUMBERS IN IMAGINARY QUADRATIC FIELDS Lenny Jones Deartment of Mathematics, Shiensburg University, Shiensburg, Pennsylvania Daniel White Deartment of Mathematics, Shiensburg University, Shiensburg, Pennsylvania Receive: 5/30/11, Revise: 4/18/1, Accete: 6/30/1, Publishe: 10/1/1 Abstract In 1960, Sieriński rove that there exist infinitely many o ositive rational integers k such that k n + 1 is comosite in Z for all n 1. Any such integer k is now known as a Sieriński number, an the smallest value of k rouce by Sieriński s roof is k = In 196, John Selfrige showe that k = is also a Sieriński number, an he conjecture that this value of k is the smallest Sieriński number. This conjecture, however, is still unresolve toay. In this article, we investigate the analogous roblem in the ring of integers O of each imaginary fiel Q( ) having class number one. More recisely, for each O, with < 0, that has unique factorization, we etermine all α O, with minimal o norm larger than 1, such that α n + 1 is comosite in O for all n 1. We call these numbers Selfrige numbers in honor of John Selfrige. 1. Introuction In 1960, using a covering of the integers (a concet originally ue to Erős [1]; see Definition 17), Sieriński [6] rove that there exist infinitely many o ositive rational integers k such that k n + 1 is comosite for all n 1. Such values of k are the classical Sieriński numbers. The smallest value of k rouce by the covering argument in Sieriński s roof is k = This value of k, however, is not the smallest Sieriński number. In 196, John Selfrige use a ifferent covering to show that k = is also a Sieriński number. It is still unknown whether k = is the smallest Sieriński number, although currently only six smaller numbers remain as viable caniates for the smallest (see Section 6.). There is an aitional roblem, known as the rime Sieriński roblem, of trying to establish that 7119 is the smallest rational rime such that n +1 is comosite in Z for all integers n 1. There are currently eleven caniates below The intereste reaer shoul visit [].

2 INTEGERS: 1A (01) In this article, we investigate the analogous roblems of etermining the smallest Sieriński number an Sieriński rime in imaginary quaratic fiel extensions of the rational numbers with class number one. To make this iea recise, we give the following efinitions. For a square-free integer, we let O enote the ring of integers in the quaratic fiel Q( ). It is well-known that Q( ), < 0, has class number one recisely when D = { 1,, 3, 7, 11, 19, 43, 67, 163}, an O has unique factorization exactly for these negative values of. Throughout this article, we assume that D. Recall that the norm of an element α O is N(α) = αα, where α is the algebraic conjugate of α in O. Definition 1. We call α O a Sieriński number if N(α) > 1 is o, an α n +1 is comosite in O for all n 1. A Sieriński number σ O is calle a Selfrige number if N(σ) = min N(α) α is a Sieriński number in O. A Sieriński number π O is calle a Selfrige rime if N(π) = min N(α) α is a rime Sieriński number in O. Remarks. (i) The conition that N(α) be o in Definition 1 is analogous to the requirement that k be o in Sieriński s original theorem. (ii) It is immeiate from Definition 1 that α is a Sieriński number in O if an only if α is a Sieriński number in O. We rove the following theorems in this article: Theorem 3. For each D, all Selfrige numbers σ O are given in Table 7. Theorem 4. For each { 1,, 3, 7, 11, 19}, all Selfrige rimes in O are given in Table 7. Theorem 5. Let S be the set of all o ositive rational integers k such that, for all n 1, the integers k n + 1 are simultaneously comosite in O for all D. Then S roerly contains the set of classical Sieriński numbers. We rovie the reaer with an overview of the aroach use in this aer. The roofs of Theorem 3 an Theorem 4 rely on Theorem 15, Corollary 16 an quaratic recirocity. Theorem 15 is well-known an gives a escrition of the rimes in the ring of integers of a quaratic fiel in which unique factorization hols. Corollary 16, which follows from Theorem 15, gives criteria to etermine whether certain elements of O are rime in O. To fin caniates α O that might be Sieriński numbers, we first use a comuter search to eliminate values of α with small norm. Theorem 15 an Corollary 16 rovie us with the necessary tools

3 INTEGERS: 1A (01) 3 to ecie if a caniate α O is a Sieriński number by calculating either the norm N(α n +1) or the Legenre symbol α n. Aitionally, Theorem an Corollary 16 allow us to etermine if a Sieriński number α is in fact a rime in O by checking the rimality of the norm N(α) or comuting the Legenre symbol. For the values of α not eliminate by the initial norm search, it is often useful α to use a covering system (see Definition 17) to verify that α is inee a Sieriński number. All of these ieas are escribe in greater etail in Section. Comuter calculations were one using MAPLE TM an PARI/GP.. Preliminaries In this section we rovie the necessary backgroun material neee to establish Theorem 15 an Corollary 16, the main tools in the roofs of Theorem 3 an Theorem 4. We begin by remining the reaer of some basic ieas concerning quaratic recirocity an quaratic fiel extensions of the rational numbers [5]. We let enote the Legenre symbol. Theorem 6. Let be an o rime. Then 1 = ( 1) ( 1)/ an = ( 1) ( 1)/8. Theorem 7. (Quaratic Recirocity) Let an q be istinct o rimes. Then q = ( 1) ( 1)(q 1)/4. q 3 Corollary 8. Let be an o rime. Then = 1 if an only if 1 or 11 (mo 1). Although all of the following ieas aly to ositive values of as well, our focus here is on values of < 0. Proosition 9. The ring of integers in Q( ) is given by a + b a, b Z if, 3 (mo 4) O = a + b a, b Z, a b (mo ) if 1 (mo 4).

4 INTEGERS: 1A (01) 4 Definition 10. The iscriminant δ of Q( ) is efine as 4 if, 3 (mo 4) δ = if 1 (mo 4). Definition 11. An element u O is calle a unit in O if N(u) = ±1. Two elements α an β in O are calle associates if α = uβ, for some unit u O. Proosition 1. Let U be the set of units in Q( ). Then ±1, ± 1 if = 1 U = ±1, ±1 ± 3 if = 3 {±1} otherwise. Proosition 13. Let α O. If N(α) = ±, where is a rational rime, then α is rime in O. Remark 14. Note that in our situation we have < 0 so that N(α) 0 for all α O. Theorem 15. Suose that Q( ) has class number one. Then 1. Any rational rime is either a rime π in O, or a rouct π 1 π of two rimes π 1 an π, not necessarily istinct, in O.. The totality of rimes π, π 1, π obtaine by alying Part 1. to all rational rimes, together with their associates, constitutes the set of all rimes in O. 3. The behavior of the rational rime in O is given below. (a) If δ 1 (mo ) an 1 (mo 8), then = π 1 π, where π 1 = π are rimes in O. (b) If δ 1 (mo ) an 5 (mo 8), then remains rime in O. (c) If δ 0 (mo ), then = uπ, where π O is rime, an u O is a unit. 4. The behavior of an o rational rime in O is given below. (a) If δ 0 (mo ) an = 1, then = π 1 π, where π 1 = π are rimes in O. Furthermore, π 1 an π are not associates, but π 1 an π are, an π an π 1 are. (b) If δ 0 (mo ) an = 1, then remains rime in O. (c) If δ 0 (mo ), then = uπ, where π O is rime, an u O is a unit.

5 INTEGERS: 1A (01) 5 A main tool use in the roofs of Theorem 3 an Theorem 4 is the following artial converse of Proosition 13. Recall Remark 14. Corollary 16. Let α O. Suose that α is not a rational integer an N(α) is not rime in Z. 1. If { 1, 3}, then α is not rime in O.. If = 1 or = 3, then either (i) α is not rime in O, or (ii) α = u, where is a rational rime with = 1, an u is a unit in O. Proof. Assume that α is rime in O. By Theorem 15 (Part.), we have that there exists a rational rime such that either = vαβ or = vα, where v is a unit an β O is rime. If = vαβ, then = N(α) N(β). Since N(α) is not rime an α is not a unit, we have that N(α) =. But then N(β) = 1, which contraicts the fact that β is rime in O. Therefore, = vα or equivalently, α = u, where u = v 1 is a unit. Since α is not a rational integer, it must be that u is not a rational integer as well. But this is imossible if { 1, 3}, which roves Part 1. To finish the roof of Part., if = 1, then = π 1 π, where π 1 an π are (not necessarily istinct) rimes in O. But this contraicts the assumtion that α is rime in O. The following concet is originally ue to Erős [1]. Definition 17. A (finite) covering system, or simly a covering, of the rational integers is a system of congruences n r i (mo m i ), with 1 i t, such that every integer n satisfies at least one of the congruences. Using Definition 17, we can escribe a covering (or a artial covering) C as a set of orere airs (r, m), where the congruence n r (mo m) is a congruence in the covering. In this article, however, we escribe C as a set of orere triles (r, m, ), where the congruence n r (mo m) is a congruence in the covering, an is a rational rime associate to the articular congruence. This association tyically means that is a rime ivisor of m 1, an this will be the case throughout the aer, unless state otherwise. 3. The Proof of Theorem 3 The general strategy here is to first use a comuter to search for caniate Sieriński numbers α in O. For any fixe ositive rational integer M, there are only finitely many elements α O such that N(α) = M, an they are straightforwar to

6 INTEGERS: 1A (01) 6 calculate. So, we set uer limits for n an M, an fin all α O with N(α) o an 1 < N(α) M. Then we check that α n + 1 is comosite in O using Theorem 15 an Corollary 16. Once a nonemty set T of caniates is foun, then each α T is examine in etail, starting with elements of smallest norm, to etermine if α can be roven to be a Sieriński number in O, or whether a value of n can be foun such that α n + 1 is rime in O. If no element of T is foun to be a Sieriński number in O, then a new comuter search is conucte increasing the limits of M in the search. Since many of the roofs are similar, we give etails only for certain values of. When roving that a articular σ O from Table 7 is a Selfrige number, or whether a articular π O from Table 7 is a Selfrige rime, it will be convenient to write s n for σ n + 1 or π n + 1, resectively = 1 For the urose of reaability, we let i = 1 in this section. In a ersonal communication with E. Weisstein at MathWorl, E Pegg Jr. note that i, 5 + 3i, an i are all Sieriński numbers in O [7]. This fact is easy to see. For examle, if α = i, we have that N(α n + 1) = 109 n + 0 n 0 (mo 3) if n 1 (mo ) (mo 5) if n 0 (mo ). Since N(α n + 1) > 5 for all n 1, it is clear that α n + 1 = u, for some rational rime an some unit u. Thus, we can conclue from Corollary 16 that α n + 1 is comosite in O for all n 1. We show, however, that i is not a Selfrige number in O. With a comuter we are able to eliminate as caniates for Sieriński numbers in O all α with o norm smaller than 5. We show that σ = 4 + 3i is a Sieriński number in O, an hence a Selfrige number in O, since N(σ) = 5. There aears to be no simle attern to the rime ivisors of N(s n ) as in the case of N((10 + 3i) n + 1), an so we use a ifferent metho to establish that is a Sieriński number. For even n, we have For o n, we have s n = ( 4 + 3i) n + 1 s n = ( 4 + 3i) n + 1 n (mo 3). = i (1 i) i (1 + i) n + 1 (i (1 i)(1 + i) = n ) + 1 if n 1 (mo 4) ((1 i)(1 + i) n ) + 1 if n 3 (mo 4) ((1 i)(1 + i) = n + 1) ((1 i)(1 + i) n 1) if n 1 (mo 4) ((1 i)(1 + i) n + i) ((1 i)(1 + i) n i) if n 3 (mo 4),

7 INTEGERS: 1A (01) 7 so that s n factors nontrivially in O. 3.. = an = 11 Since the aroach is the same in each of these cases, we give etails only for =. It is easy to check by comuter that no element of O with o norm smaller than 11 is a Sieriński number in O. Let σ = 3 +. Then, since N(s n ) = 11 n + 6 n (mo 3), it follows from Corollary 16 that σ is a Selfrige number in O. The roof is ientical for the other values of σ in Table 7 when = = 3 an = 7 Since the techniques use are similar for each value of, we give etails only for = 7. Let σ = 7. First note that s n is comosite when n = 1. Then, if s n is rime for some n, we have by Theorem 7 that sn 1 = = = 1, 7 7 s n which establishes that σ is a Sieriński number in O by Theorem 15 (Part 4.). To rove that σ is a Selfrige number, we examine the set T of all elements in O other than σ having o norm N with 3 < N 49 = N(σ): T = { 7, ±5, ±3, 7, ±1+ 7, ±3+ 7, ±+ 7, ±4+ 7, ±6+ 7}. Fortunately, it is easy to show that no element of T is a Sieriński number in O. For examle, if α = 5, then 7 α 3 = = 1, an so α = 5 is not a Sieriński number in O by Theorem 15 (Part 4.). As another examle, let α = Then N(α 3 + 1) = = 1873, an since 1873 is rime in Z, we have that α is rime in O by Proosition 13. Hence, α = is not a Sieriński number in O = 19 Let σ = We use techniques similar to those use for = 7 to verify that there are no Sieriński numbers in O with o norm smaller than N(σ) = 49. Then we observe that N(s n ) = 49 n + 5 n 0 (mo 3) if n 1 (mo ) (mo 5) if n 0 (mo ), which imlies, by Corollary 16, that σ is a Selfrige number in O. The result is ientical for the other value of σ in Table 7.

8 INTEGERS: 1A (01) = 43, = 67 an = 163 The techniques for these cases are similar so we resent the etails only for = 43. To verify that there are no Sieriński numbers in O with o norm smaller than 169, we use PARI/GP to obtain certificates of rimality in certain cases. For examle, let α = Then N(α) = 17. To show that α is not a Sieriński number in O, we examine N(α n + 1) = 17 n + 11 n + 1, which is easily seen to be comosite for all n 39. We then use PARI/GP to rove that N(α ) is rime, an conclue by Proosition 13 that α is not a Sieriński number in O. Now, let σ = Observe that if n 1 (mo ), then For n 0 (mo ), we can write N(s n ) = 169 n + 17 n (mo 3). N(s n ) = (13 n + 1) (3 n/ ) = 13 n n/ 13 n n/. Thus, by Corollary 16, σ is a Selfrige number in O. The roof for the other value of σ is ientical. 4. The Proof of Theorem 4 The strategy here is the same as the strategy emloye in the roof of Theorem 3 with an aitional ste. The aitional ste require here is that our Sieriński caniates α must be rime in O. Primality of α in O can be verifie by simly checking that N(α) is rime in Z when α Z (Proosition 13), or that α = 1 when α is a rime in Z (Theorem 15 (Part 4.)). Since many of the roofs are similar for various values of, we give etails only in selecte cases. For convenience, we efine s n := π n + 1, where π is a Selfrige rime from Table = 1 an = Using the techniques use to rove Theorem 3, it is straightforwar to establish that the Selfrige rimes for = are π = ±3 ±, an that the only Selfrige rime for = 1 is π = 7.

9 INTEGERS: 1A (01) = 7 Using a comuter to check u to n = 400 an norm 09, the only caniate for a Selfrige rime is π = 47. To verify that π = 47 is inee a Selfrige rime, we must show that s n is comosite in O for all n 1. To accomlish this task, we use the artial covering C = {(0,, 3), (1, 4, 5), (3, 1, 13), (11, 1, 7)}. It is then straightforwar to show that 47 n +1 0 (mo ), when n r (mo m) for each (r, m, ) C. The only hole in our artial covering C is n 7 (mo 1). But for these values of n we have that 47 n (mo 7). Hence, if = 47 n + 1 is 7 rime in Z, it follows that = 1, so that is comosite in O. Therefore, we have establishe that 47 is the Selfrige rime in O by Theorem 15 (Part 4.) = 11 In this case, each Selfrige number ±3 ± 11 has norm 5, an therefore each Selfrige number is also a Selfrige rime by Proosition = 19 Let π = Note that N (s n ) = 199 n 5 n (mo 3) if n 1 (mo ) 0 (mo 5) if n 0 (mo ), which roves, by Corollary 16, that π is a Sieriński number in O. Since N(π) = 199 is rime, we have by Proosition 13 that π is rime in O. Using a comuter, it is straightforwar to show that π is a Sieriński number in O with smallest o rime norm, which imlies that π is a Selfrige rime. The roof that is a Selfrige rime is ientical. 5. Proof of Theorem 5 Recall that D = { 1,, 3, 7, 11, 19, 43, 67, 163}, an that S is the set of all o ositive rational integers k such that, for all n 1, the integers k n + 1 are simultaneously comosite in O for all D. Clearly, all classical Sieriński numbers are containe in S. To see that this containment is roer, let z 111 (mo 130) be a ositive integer, an let k = z. Then k + 1 an k + 1 are both comosite in Z since k (mo 5) an k (mo 13).

10 INTEGERS: 1A (01) 10 Let D, assume that n 3, an suose that = k n + 1 is rime in Z. Then, 1 (mo 8) an, by Theorem 6 an Theorem 7, we have that = 1. Hence, k S. The smallest value of k rouce by this rocess is k = , an we note that this value of k is not a classical Sieriński number since = is rime in Z. 6. Comments, Conjectures an Conclusions 6.1. Extening Theorem 4 While we were not able to fin the Selfrige rimes for { 43, 67, 163}, we give in Table 7 a list of conjecture Selfrige rimes for = 43 an = 67, an some justification below for these conjecture values = 43 Since π = has norm 4649, which is rime, we know by Proosition 13 that π is rime in O. Now consier the covering C = {(0,, 3), (1, 4, 5), (3, 4, 17)}. Note that here the rime 17 associate to the thir congruence in C is not a rime ivisor of 4 1. However, 17 oes ivie N(s n ) for values of n 3 (mo 4). Aly C to the exonent n in N(s n ) = N(π n + 1) to get that 0 (mo 3) if n 0 (mo ) N (s n ) = 4649 n + 89 n (mo 5) if n 1 (mo 4), 0 (mo 17) if n 3 (mo 4). Hence, π is a Sieriński number in O. To rove that π is a Selfrige rime, we must rule out all rimes in O with norm less than Unfortunately, there are several caniates which we have been unable to eliminate. For examle, for the rime α = , with norm 3541, the number c n = α n + 1 is comosite for all n Because there aears to be no nice attern to the rime ivisors of N(c n ), we believe that there exists m > such that N(c m ) is rime. Thus, c m woul be rime in O, an α woul not be a Selfrige rime = 67 The covering C = {(0,, 3), (1, 4, 5), (3, 4, 17)} use in can be use in this case to verify that the rime π = in O is a Sieriński number. But we are face with the same ifficulties as in the case of = 43 in showing that π is a Selfrige rime.

11 INTEGERS: 1A (01) = 163 We teste, u to n = 5000, all elements α O such that N(α) is rime an N(α) is less than the 50000th rime, an we foun no Selfrige rime caniates. 6.. Comments on Theorem 5 Of course the classical Sieriński number k = foun by Selfrige is containe in S, but is it the smallest element of S? It can be shown, with little effort, that is the smallest rational integer that is a Sieriński number in O for each { 3, 7, 11, 19, 43, 67, 163}, although to rule out smaller rational integers requires checking u to some large exonents. For examle, to show that 47 is not a Sieriński number in the case of = 67, we have to check u to n = 6115 since this is the smallest value of n for which 47 n + 1 is rime in O. It is also straightforwar to show that the smallest rational Sieriński numbers in O for = 1 an = are 7 an 5, resectively. For examle, to show that 5 is the smallest rational Sieriński number in O when =, we first observe that = 7 is rime in O. This rules out 3 as a Sieriński number in O. Next, we see that while 5 n + 1 is rime in Z for n = 1 an n = 3, these rimes (11 an 41) are both comosite in O. Also, = 1 is obviously comosite. Then, for n 3, if = 5 n + 1 is rime in Z, we have that 1 (mo 8) an so = ( 1) ( 1)/ ( 1) ( 1)/8 = 1, which roves that 5 is a Sieriński number in O, by Theorem 15 (Part 4.). Piecing together this information is what is one in the roof of Theorem 5 to get a value of k that is a Sieriński number in O for all D. However, the smallest value of k rouce by these methos is much larger than the Sieriński number k = foun by Selfrige. Comuter calculations show that all of the values of k < for which a value of n is known such that k n + 1 is rime in the rational integers, also fail to be Sieriński in O for at least one D. The remaining six rational Sieriński caniates [] are the elements of B = {103, 1181, 699, 4737, 55459, 67607}. If = k n + 1 is a 3 rational rime for any k B, then 5 (mo 1) an thus = 1, so that remains rime for = 3, by Theorem 15 (Part 4.). In other wors, if k B is not a rational Sieriński number, then k S. Thus we have the following: Theorem 18. The smallest element of S is either k = or the smallest Sieriński number in B Two Avenues for Future Investigation For { 1, 3, 7}, the rational rime oes not remain rime in O. So one coul conuct an investigation similar to the toics in this aer to fin α O

12 INTEGERS: 1A (01) 1 such that α π n +1 is comosite in O for all n 1, where π O is a rime ivisor of. A ossible secon area of investigation woul be to focus on real quaratic fiels. This situation might be comlicate by the fact that there coul be infinitely many solutions to the equation x y = N, for a nonzero integer N, an fining these solutions can be somewhat involve [5, 3, 4]. In aition, another comlicating factor is that it is conjecture that there are infinitely many such fiels with class number one. 7. Tables σ N(σ) 1 4 ± ±3 ± ±3 ± ± ± ± ± Table 1: Selfrige Numbers σ in O π N(π) ±3 ± ±3 ± ± Table : Selfrige Primes π in O

13 INTEGERS: 1A (01) 13 π N(π) ?? Table 3: Conjecture Selfrige Primes π in O Acknowlegements The authors thank the referee for the many valuable suggestions. References [1] P. Erős, On integers of the form k + an some relate roblems, Summa Brasil. Math., (1950), [] L.Helm, D. Norris et al., Seventeen or Bust: A Distribute Attack on the Sieriński Problem, available online at htt:// [3] K. Matthews, The Diohantine equation x Dy = N, D > 0. Exo. Math. 18 (000), 33-?331. [4] K. Matthews, Thue s theorem an the iohantine equation x Dy = ±N, Math. Com. 71 (00), [5] I. Niven, H. S. Zuckerman, an H. L. Montgomery, An Introuction to the Theory of Numbers, 5th e., John Wiley, New York, [6] W. Sieriński, Sur un roblème concernant les nombres k n +1, Elem. Math. 15 (1960) [7] E. Weisstein, Sieriński Number of the Secon Kin, from MathWorl A Wolfram Web Resource, htt://mathworl.wolfram.com/sierinskinumberoftheseconkin.html.