30 HWASIN PARK, JOONGSOO PARK AND DAEYEOUL KIM Lemma 1.1. Let =2 k q +1, k 2 Z +. Then the set of rimitive roots modulo is the set of quadratic non-re

Transcription

1 J. KSIAM Vol.4, No.1, 29-38, 2000 A CRITERION ON PRIMITIVE ROOTS MODULO Hwasin Park, Joongsoo Park and Daeyeoul Kim Abstract. In this aer, we consider a criterion on rimitive roots modulo where is the rime of the form =2 k q +1, q odd rime. For such we also consider the least rimitive root modulo. Also, we deal with certain isomorhism classes of ellitic curves over nite elds. x0. Introduction In the famous book Disquisitiones Arithmeticae, C. F. Gauss had roved that the multilicative grou Z is cyclic and he had conjectured that 10 is a generator of Z for innitely many. We call a is a rimitive root modulo if a is a generator of Z. In 1927, E. Artin generalized Gauss' conjecture as: For a not equal to 1, -1, or a erfect square, do there exist innitely many rimes having a as a rimitive root. In 1986, Artin's conjecture was roved for almost all rimes but at most two rimes by assuming the generalized Riemann hyothesis ([2]). We note that Gauss' roof is not constructive, so that we have diculties to get rimitive roots modulo. In this aer, we restrict ourselves to be the rime of the form = 2q +1, 4q +1, 8q +1 where q is an odd rime. We consider criterions that for which rime, a = can be rimitive roots modulo. Also, we deal with isomorhism classes of ellitic curves over nite elds. These results are similar with the results of [8]. Let and q be odd rimes. x1. Primitive roots 1991 AMS Subject Classication: 11A07, 11A15, 11A41, 14H52. Key words and hrases: rimitive roots, ellitic curves over nite elds. 29 Tyeset by AMS-TEX

2 30 HWASIN PARK, JOONGSOO PARK AND DAEYEOUL KIM Lemma 1.1. Let =2 k q +1, k 2 Z +. Then the set of rimitive roots modulo is the set of quadratic non-residues modulo excet for a such that a s ;1 (mod ) s =2 k;1 : Proof. Let S be the set of rimitive roots modulo and let T be the subset of Z which are quadratic non-residues modulo. If a 2 S, then (a ) = 1 and a ;1 1 (mod ). Since ; 1 is the smallest, a ;1 2 ;1 (mod ). Then a 2 T. Thus S T. Also, j S j= (()) = (2 k q)=2 k;1 (q ; 1) and j T j= ;1 =2 k;1 q. Thus 2 j T j;js j= 2 k;1. Case I. k =1. We knowthat;1 2 T and ;1 =2 S, since( ;1 ;1 )=(;1) 2 =(;1) q = ;1 and (;1) 2 1 (mod ). In this case, j T j;js j= 1,anda = ;1 satises Lemma. Case II. k > 1. Then a 2 T, since ( a) a ;1 2 (mod ) = a 2k;1 q (;1) q (mod ) = ;1. But a =2 S, since a is a erfect square modulo if k > 1. In this case, j T j ; j S j= 2 k;1 and a 2k;1 ;1 (mod ) has 2 k;1 incongruent solutions. Thus Lemma holds. By Lemma 1.1, we can show the following: Theorem 1.2. Let =2q +1. (1) 2 is a rimitive root modulo if and only if q 1 (mod 4). In this case, 2 is the least rimitive root modulo. (2) 3 is a rimitive root modulo if and only if q =3. In this case, 3 is the least rimitive root modulo. (3) 5 is a rimitive root modulo if and only if q 1 3 (mod 5). In articular, 5 is the least rimitive root modulo if and only if q 3 11 (mod 20). (4) 6 is a rimitive root modulo if and only if q 5 (mod 12). (5) 7 is a rimitive root modulo if and only if q 5 11 (mod 14). In articular, 7 is the least rimitive root modulo if and only if q (mod 140). (6) 8 is a rimitive root modulo if and only if q 1 (mod 4). (7) 10 is a rimitive root modulo if and only if q (mod 20). (8) 11 is a rimitive root modulo if and only if q (mod 22), or q =11. In articular, 11 is the least rimitive root modulo if and only if q 79, 139, 279, 359, 419, 499, 519, 639, 799, 939, 1079, 1399 (mod 1540). (9) 12 is not a rimitive root for all. (10) 13 is a rimitive root modulo if and only if q (mod 13). In articular, 13 is the least rimitive rot modulo if and only if q 659, 699, 839, 919, 1219, 1359, 1539, 2239, 2319, 2459, 2759, 3039, 3299, 3779, 4139, 4299, 4579, 4839, 5179, 5319, 6119, 6159, 6379, 6819, 6939, 6999, 7079, 7519, 7699, 7919, 8479, 8759, 9239, 9799, 9939, 10019, 10299, 10459, 10779, 11339,

3 A CRITERION ON PRIMITIVE ROOTS MODULO , 11619, 11839, 12139, 12279, 12539, 12979, 13159, 13239, 13379, 13679, 14419, 15219, 15399, 16099, 16179, 16239, 16619, 17599, 17859, 17999, 18439, 18699, 19039, 19139, (mod 20020). Proof. (1) By Lemma 1.1, 2 is a rimitive root modulo if and only if 2 is a quadratic non-residue modulo. By the quadratic recirocity law, must be congruent 3 (mod 8). Thus q 1 (mod 4). (2) By the quadratic recirocity law, ( 3 )=;1 ifand only if 5 (mod 12). Case. 5 (mod 12). Then q =6k + 2 for some k 2 Z. It is imossible. Case. ;5 (mod 12). Then q =6k +3. Thus q =3. (3) Note that ( 5 )=;1 if and only if 2 (mod 5). Case. q =5. Then =11and( 5 )=1. This case must be omitted. 11 Case. q = 5k +1. Then =2q +1=10k +3;2 (mod 5). In this case, we have ( 5)=;1. Case. q =5k +2. Then =2q +1=10k +5. It is imossible. Case. q = 5k +3. Then = 2q +1= 10k +7 2 (mod 5). In this case, we have ( 5)=;1. Case. q =5k +4. Then =2q +1=10k Then ( 5 )=1. This case must be omitted. In articular, 5 is the least rimitive root modulo if and only if q 3 (mod 4) q 6= 3, and q 1 3 (mod 5). By the chinese remainder theorem, q 3 11 (mod 20). (4) If q 5 (mod 12), then by (1), 2 is a rimitive root modulo. By Lemma 1.1, ( 2)=;1. Also, by (2), ( 3)=1. Thus ( 6 )=;1. That is 6 is a rimitive root modulo. Conversely, if6isarimitive root modulo. Then ( 6 )=;1 and 6= 7. We have two cases. Case I. ( 2)=1and ( 3 )=;1. Then by (1) and (2), q 3 (mod 4) and q =3. This case contradicts to 6= 7. Case II. ( 2)=;1 and ( 3 )=1. Then by (1) and the quadratic recirocity law, we get q 1 (mod 4) and 1 (mod 12). If 1 (mod 12), then we getacontradiction. Thus we have ;1 (mod 12) and then q 5 (mod 6). Thus we have q 5 (mod 12). (5) By the quadratic recirocity law, ( 7 ) = ;1 if and only if (mod 28). Case. =28k +5. Then q =14k +2. Case. =28k +23. Then q =14k Case. =28k +11. Then q =14k +5. Case. =28k +17. Then q =14k +8. Case. =28k +13. Then q =14k +6. Case. =28k +15. Then q =14k +7. Since q is odd rime, we have q 5 11 (mod 14).

4 32 HWASIN PARK, JOONGSOO PARK AND DAEYEOUL KIM In articular, 7 is the least rimitive root modulo if and only if 8 >< >: q 3 (mod 4) q 4 (mod 5) q 5 11 (mod 14): q 6= 3 By the chinese remainder theorem, q (mod 140). (6) Similar with (4). (7) By (1), (3) and the Chinese remainder theorem, we canshow (7). (8) By the quadratic recirocity law, 11 is a quadratic non-residue modulo if and only if (mod 44). Case. =44k +3. Then q =22k +1. Case. =44k +41. Then q =22k Case. =44k +13. Then q =22k +6. Case. =44k +31. Then q =22k Case. =44k +15. Then q =22k +7. Case. =44k +29. Then q =22k Case. =44k +17. Then q =22k +8. Case. =44k +27. Then q =22k Case. =44k +21. Then q =22k Case. =44k +23. Then q =22k Since q is odd rime, we have q (mod 22). In articular, 11 is the least rimitive root modulo if and only if 8 >< >: q 6 1 (mod 4) q (mod 5) q (mod 14) q 6= 3 q (mod 20) q (mod 22): That is, q 3 (mod 4) q 6= 3 (i) 8>< q 4 (mod 5) (ii) q (mod 14) or q =7 (iii) >: q (mod 20) q (mod 22): We donotneedq 19 (mod 20) in (iv) because of (i) and (ii). We also do not need (iv) because (i), (ii), (iii), (v) and (iv) has no simultaneous solution by the Chinese remainder theorem. Thus 11 is the least rimitive root modulo if and only if 8 >< >: q 3 (mod 4) q 4 (mod 5) q 6= 3 q (mod 14) or q =7 q (mod 22): (iv) (v)

5 A CRITERION ON PRIMITIVE ROOTS MODULO 33 By the chinese remainder theorem, we have q 79, 139, 279, 359, 419, 499, 519, 639, 799, 939, 1079, 1399 (mod 1540). (9) By Lemma 1.1, 12 is a rimitive root modulo if and only if ( 12 )=;1. That is, ( 3 )=;1. Then by (2), must be 7. But 12 is not a rimitive root modulo 7. (10) 13 is a quadratic non-residue modulo if and only if (mod 13). Case. q =13. Then = 27. Case. q =13k +1. Then =26k +3 3 (mod 13). Case. q =13k +2. Then =26k +5 5 (mod 13). Case. q =13k +3. Then =26k +7 7 (mod 13). Case. q =13k +4. Then =26k +9 9 (mod 13). Case. q =13k +5. Then =26k (mod 13). Case. q =13k +6. Then =26k Case. q =13k +7. Then =26k (mod 13). Case. q =13k +8. Then =26k (mod 13). Case. q =13k +9. Then =26k (mod 13). Case. q =13k +10. Then =26k (mod 13). Case. q =13k +11. Then =26k (mod 13). Case. q =13k +12. Then =26k (mod 13). Since is the rime of the form (mod 13), q (mod 13). In articular, 13 is the least rimitive root modulo if and only if 8 >< >: q 3 (mod 4) q 4 (mod 5) q 6= 3 q (mod 14) or q =7 q (mod 22) q (mod 13): By the chinese remainder theorem, we have q 659, 699, 839, 919, 1219, 1359, 1539, 2239, 2319, 2459, 2759, 3039, 3299, 3779, 4139, 4299, 4579, 4839, 5179, 5319, 6119, 6159, 6379, 6819, 6939, 6999, 7079, 7519, 7699, 7919, 8479, 8759, 9239, 9799, 9939, 10019, 10299, 10459, 10779, 11339, 11559, 11619, 11839, 12139, 12279, 12539, 12979, 13159, 13239, 13379, 13679, 14419, 15219, 15399, 16099, 16179, 16239, 16619, 17599, 17859, 17999, 18439, 18699, 19039, 19139, (mod 20020). Corollary 1.3. ([3, 4, 5]) Let =2q and 8 are rimitive roots modulo if and only if so is 2. In articular, 2, 6, and 8 are not rimitive roots modulo if and only if q 3 (mod 4). Proof. By Theorem 1.2 (1), 2 is a rimitiverootmodulo if and only if q 1 (mod 4). Then q 5 (mod 12). For q 1 (mod 12) or q 9 (mod 12) contradict to and q are rimes. By Theorem 1.2 (4), 6 must be a rimitive root modulo. By Theorem 1.2 (1) and (6), 2 and 8 occur simultaneously as a rimitive rootmodulo. Remark 1.4. (1) We use Mathematica 3.0 to solve the Chinese remainder theorem and to get the following (2).

6 34 HWASIN PARK, JOONGSOO PARK AND DAEYEOUL KIM (2) Let =2q +1. The least rimitive root modulo are relatively small. If () denotes the least rimitive root modulo. Then for , () = 2 takes lace aroximately 50%, and () = 5 haens aroximately 33%. If we denote that h(0) is the total number of rimes, , and h(a) isthe number of rimes which hasa as the least rimitive root modulo. Then we have the following table: a h(a) a h(a) a h(a) a h(a) a Theorem 1.5. Let =4q +1. Then 2 is the least rimitive root modulo for all. Proof. Since q is odd, = 4(2k +1)+1=8k +5. By the quadratic recirocity law, 2 is a quadratic non-residue modulo for all. The smallest is 13, so ;1 (mod ) for all. By Lemma 1.1, 2 is a rimitive root modulo for all. In articular, 2 is the least rimitive root modulo for all. Theorem 1.6. Let =8q +1. (1) 2 is not a rimitive root modulo for all. (2) 3 is the least rimitive root modulo for all excet for =41. Proof. (1) 2 is not a rimitive rootmodulo, since ( 2)=1. (2) Case. q =3. Then =25. It is imossible. Case. q =3k +1. Then = 3(8k + 3). It is imossible. Case. q = 3k +2. Then = 24k +17. Then 5 (mod 12). By the quadratic recirocity law, ( 3 )=;1 forall. Thus by Lemma 1.1, 3 is a rimitive rootmodulo for all excet for = 41, since 3 4 ;1 (mod ) has only one, =41. Actually, 7 is the least rimitive root modulo 41. Remark 1.7. (1) Similarly, we get the following: If =2 n q +1 n 4 q 6= 3,>3 2n;1, then 3 is the least rimitive root modulo for all. In articular, 3 is the least rimitive root modulo for all =16q +1,32q +1,64q +1,. (2) We get a result that is similar to Theorem 1.2 for =4q +1,8q +1.

7 A CRITERION ON PRIMITIVE ROOTS MODULO 35 x2. Some other cases Lemma 2.1. Let =2 n +1 be a rime with n 1. Then the set S of rimitive root modulo is the set T of quadratic non-residues modulo. Proof. In the roof of Lemma 1.1, we have S T. Also, j S j= (()) = 2 n;1 and j T j= ;1 2 =2 n;1. Thus we have S = T. Proosition 2.2. Let =2 n +1 be a rime with n 1. (1) 2 is the least rimitive root modulo when n =1 2. (2) For n 3, 3 is the least rimitive root modulo for all. Proof. (1) By comutation, it is clear. (2) Since =2 n +1andn 3, 1 (mod 4). Also, 2 (mod 3). For if 1 (mod 3), then 2 n +1 1 (mod 3), get a contradiction. >From 1 (mod 4) and 2 (mod 3), we have 5 (mod 12). By the quadratic recirocity law, ( 3)=;1. By Lemma 2.1, 3 is a rimitive root modulo for all. Note that 1 (mod 8), since n 3. By the quadratic recirocity law, ( 2 )=1. Thus 2 is not a rimitive root modulo for all. Corollary 2.3. Let be a Fermat's rime. Then 3 is the least rimitive root modulo. Remark 2.4. Erdos ([1]) asks if is large enough, is there always a rime r so that r is a rimitive rootmodulo? If =2 n + 1 is a rime with n 1. Then this is true. For if b is a quadratic nonresidue modulo and b = e 1 1 e 2 2 e t t, then ( 1 1 ( )e t )e t = ;1. Then ( i )=;1 for some i. Then by Lemma 2.1, i is a rimitive root modulo for large enough. x3 Alication to ellitic curves over nite eld F Let and q be odd rimes and let K be a eld with char(k) > 3. Proosition 3.1. ([6], [7]) Two ellitic curves E b : y 2 = x 3 + ax + b and E b0 : a a 0 y2 = x 3 + a 0 x + b 0 dened over K are isomorhic over K if and only if there exists u 2 K such that u 4 a 0 = a and u 6 b 0 = b. If E b a = E b0 a 0 over K, then the isomorhism is given by : E b a! E b0 0 :(x y) 7! a (u;2 x u ;3 y) or equivalently : E b0 a 0! Eb a :(x y) 7! (u2 x u 3 y): Theorem 3.2. Let E 0 a : y 2 = x 3 + ax, E 0 ag 2i : y 2 = x 3 + ag 2i x and E 0 ag 4i : y 2 = x 3 + ag 4i x be ellitic curves dened over F and let g be a rimitive root modulo. (1) If 1 (mod 4), then E 0 a is isomorhic to E 0 ag 4i where 1 i ;1 4. (2) If 3 (mod 4), then E 0 a is isomorhic to E0 ag 2i where 1 i ;1 2.

8 36 HWASIN PARK, JOONGSOO PARK AND DAEYEOUL KIM Proof. (1) For each i =1 2 ;1,take 4 u4 = g ;1;4i. Then u 4 ag 4i = g (;1);4i ag 4i = ag ;1 = a. Also, u = g ;1;4i 4 = g ;1 4 g ;i 2 F, since 1 (mod 4). This u satises the conditon of Proosition 3.1. Thus E 0 a = E 0 for i = 1 2 ;1. That is, ag 4i 4 E 0 a = E 0 ag 4 = E 0 ag 8 = = E 0 ag ;1. (2) By the same way with (1), E 0 a = E 0 ag 4 = = E 0. ag ;1 = E 0 ag 2i for i =1 2 ;1 2. That is, E0 a = E 0 ag 2 Corollary 3.3. Let T be the set of ellitic curves of the form y 2 = x 3 + ax dened over F. We denote [E 0 a] be the isomorhism class containing E 0 a. (1) If 1 (mod 4), then the number of isomorhism classes of ellitic curves in T is 4: [E 0 1] 3 y 2 = x 3 + x [E 0 g] 3 y 2 = x 3 + gx [E 0 g 2] 3 y2 = x 3 + g 2 x [E 0 g 3] 3 y2 = x 3 + g 3 x where g is a rimitive root modulo. (2) If 3 (mod 4), then the number of isomorhism classes of ellitic curves in T is 2: [E 0 1] 3 y 2 = x 3 + x [E 0 g] 3 y 2 = x 3 + gx where g is a rimitive root modulo. Proof. (1) We have four isomorhism classes: E 0 1 = E 0 g 4 = E 0 g 8 = E 0 g = E 0 g 5 = E 0 g 9 = E 0 g 2 = E 0 g 6 = E 0 g 10 = E 0 g 3 = E 0 g 7 = E 0 g 11 = : (2) We have two isomorhism classes: E 0 1 = E 0 g 2 = E 0 g 4 = E 0 g = E 0 g 3 = E 0 g 5 = : Corollary 3.4. Let =2q +1. (1) If q 1 (mod 4), then there are two isomorhism classes of ellitic curves over F : [E 0 1] [E 0 2 ]: (2) If q =3, then there are two isomorhism classes of ellitic curves over F : [E 0 1] [E 0 3 ]:

9 A CRITERION ON PRIMITIVE ROOTS MODULO 37 (3) If q (mod 20), then there are two isomorhism classes of ellitic curves over F : [E 0 1] [E 0 10]: (4) If q 79, 139, 279, 359, 419, 499, 519, 639, 799, 939, 1079, 1399 (mod 1540), then there are twoisomorhism classes of ellitic curves over F : [E 0 1] [E 0 11]: Proof. (1) By Theorem 1.2, if q 1 (mod 4), then 2 is the rimitive root modulo. Since q is odd, 3 (mod 4) for all. By Corollary 3.3, we have two isomorhism classes. (2), (3), (4) follow by Theorem 1.2 and Corollary 3.3. Corollary 3.5. Let =4q +1. There are four isomorhism classes of ellitic curves over F : [E 0 1] [E 0 2 ] [E 0 4 ] [E 0 8 ]: Corollary 3.6. Let = 8q +1 with > 41. There are four isomorhism classes of ellitic curves over F : [E 0 1] [E 0 3 ] [E 0 9 ] [E 0 27]: Examle 3.7. Let E 0 2 : y 2 = x 3 +2x over F 13. Then E 0 2 is isomorhic to E 0 6 : y 2 = x 3 +6x and E 0 5 : y 2 = x 3 +5x. In fact, E 0 2(F 13 )=fo (0 0) (1 4) (1 9) (2 5) (2 8) (11 1) (11 12) (12 6) (12 7)g: Using by Proosition 3.1, and E 0 6(F 13 )=fo (0 0) (10 7) (10 6) (7 12) (7 1) (6 5) (6 8) (3 4) (3 9)g E 0 5(F 13 )=fo (0 0) (9 9) (9 4) (5 8) (5 5) (8 12) (8 1) (4 7) (4 6)g: References [1] R. K. Guy, Unsolved roblems in number theory, Sringer-Verlag, [2] D. R. Heath-Brown, Artin's conjecture for rimitive roots, Quart. J. Math., Oxford Ser. (2) 37 (1986), [3] D. H. Lehmer and Emma Lehmer, On runs of residues, Proc. Amer. Math. Soc. 13 (1962), [4] D. H. Lehmer, Emma Lehmer and W. H. Mills, Pairs of consecutive owerresidues, Canad. J. Math. 15 (1963), [5] D. H. Lehmer, Emma Lehmer, W. H. Mills and J. L. Selfridge, Machine roof of a theorem on cubic residues, Math. Comut. 16 (1962),

10 38 HWASIN PARK, JOONGSOO PARK AND DAEYEOUL KIM [6] A. Menezes, Ellitic curve ublic key crytosystems, Kluwer Academic Publ., [7] J. Silverman, The arithmetic of ellitic curves, Sringer-Verlag, New York, [8] E. Waterhouse, Abelian varieties over nite elds, Ann. Sci. Ecole Norm. Su. 2 (1969), 521{560. Deartment of Mathematics, Chonbuk National University, Chonju, Chonbuk, , Korea Deartment of Mathematics, Woosuk University, Samlae, Wanju, Chonbuk, , Korea jsark@core.woosuk.ac.kr Deartment of Mathematics, Chonbuk National University, Chonju, Chonbuk, , Korea dykim@math.chonbuk.ac.kr