On the theory of cubic residues and nonresidues

Transcription

1 ACTA ARITHMETICA LXXXIV On the theory of cubic residues and nonresidues by Zhi-Hong Sun Huaiyin 1. Introduction. Let Z be the set of integers, ω 1 + / and Z[ω] {a + bω a, b Z}. For π a + bω Z[ω] the nor of π is given by Nπ ππ a ab + b. When π od we say that π is riary. If π Z[ω], Nπ > 1 and π ± od we ay write π ±π 1... π r, where π 1,..., π r are riary ries. For α Z[ω] the cubic Jacobi sybol α π is defined by α α α..., π π 1 π r where α π t is the cubic residue character of α odulo π t which is given by { α 0 if πt α, ω i if α Nπt 1/ ω i od π t. π t According to [IR,. 15, 1] the cubic Jacobi sybol has the following roerties: 1.1 If a, b Z and a + bω od then ω a+bω ωa+b+1/. 1. If a, b Z and a + bω od then 1 ω a+bω ωa+1/. 1. If π, λ Z[ω] and π, λ ± od then λ π π λ. The assertion 1. is now called the general cubic recirocity law; it was first roved by G. Eisenstein. Let be a rie of the for n+1. It is well known that there are unique integers L and M such that 4 L +7M with L 1 od. It follows that L M od and therefore 1/ 1, 1 L M / or 1 + L M / od for any integer 0 od. In 187 Jacobi [J] established the following rational cubic recirocity law Matheatics Subject Classification: 11A15, 11E5. [91]

2 9 Z. H. Sun Theore 1.1 Jacobi. Let q be a rie of the for n + 1, q and 4q L + 7M. Then q is a cubic residue odulo if and only if LM L M/LM + L M is a cubic residue odulo q. In 1958, using the eriod equation of degree, E. Leher [L1] gave the following criterion for cubic residuacity. Theore 1. E. Leher. If q is an odd rie different fro then q is a cubic residue of if and only if either LM 0 od q or L µm od q, where µ satisfies the congruence µ u u u + 1 od q with u 0, 1, 1, 1 Legendre sybol. od q and u+1u q 1. Here q is the In 1975 K. S. Willias [W1] showed how to choose the sign of M so that 1/ 1 L M / od when is a cubic nonresidue odulo. Let ε d be the fundaental unit in the quadratic field Q d. In 1970 s E. Leher [L], [L4] began to study criteria for ε d to be a cubic residue odulo, where is a rie of the for n + 1 satisfying d 1. Since the work of Euler, Gauss, Jacobi and Eisenstein see [IR,. 1] it is known that cubic congruences are connected with binary quadratic fors. In 199 B. K. Searan and K. S. Willias [SW] showed that is a cubic residue odulo if and only if can be reresented by one of the third coosition owers of riitive integral binary quadratic fors of discriinant 7, where is a rie greater than for which 0 od. Let be a ositive integer, and Z the set of those rational nubers whose denoinator is rie to. Insired by the above work of Jacobi, Leher and Willias we introduce the subsets C 0, C 1 and C of Z for 0 od, where { } C i k k ω ω i, k Z for i 0, 1,. In Sections and we concentrate on the structure and roerties of C 0, C 1 and C. Here are soe tyical results: 1.4 Let be a rie of the for n + 1 and hence 4 L + 7M for soe L, M Z and L 1 od. If q is a rie such that M 0 od q and i {0, 1, } then q 1/ 1 L i M / od if and only if L/M C i q.

3 Cubic residues and nonresidues Let be a rie for which 1 od, t od t Z, k Z, k + 0 od and i {0, 1, }. Then k C i if and only if 1/ i k t 1 t od. k + t 1.6 Let be a rie greater than, k Z and k + 0 od. Then k C 0 if and only if k x 9x x od for soe integer x. If q is also a rie of the for n + 1 and 4q L + 7M L, M Z with L 1 od, in view of 1.4 and 1.5 we see that i 1 L/M q 1/ od if and only if LM L q 1/ M LM + L M 1 L /M i od q. This generalizes Jacobi s result. Cobining 1.4 with 1.6 gives a sile criterion for cubic residuacity which iroves Leher s result. Section 4 is devoted to cubic congruences. Here are two ain results: 1.7 If > is a rie, a, b Z, ab and s b 4a od for soe s Z, then the congruence x ax ab 0 od is solvable if and only if s/b C Suose that is a rie greater than and that N is the nuber of values of x +Ax +Bx+C odulo, where A, B, C Z and x runs over all integers. If A B od then N /. If A B od then N + / or according as 1 od or od. In Section 5 the criteria for sd C i i 0, 1, are given in ters of binary quadratic fors, where > is a rie, d Z, d + and sd d od. In articular, sufficient and necessary conditions for sd C 0 are described in the cases d 1,, 5, 6, 7 and 15. As a consequence we obtain criteria for ε 6, ε 15, ε 1 to be cubic residues odulo. In Section 6 we ainly deterine u /a, b odulo, where > is a rie and {u n a, b} is the Lucas sequence given by u 0 a, b 0, u 1 a, b 1 and u n+1 a, b bu n a, b au n 1 a, b n 1. In articular, we obtain F / od and P / od, where {F n} and {P n } denote the Fibonacci sequence and Pell sequence resectively.

4 94 Z. H. Sun To illustrate the connections in the above work I state the following result: 1.9 Let be a rie for which 5 1, and q a rie of the for n + 1 satisfying L + 15M 0 od, where L and M are deterined by 4q L + 7M L, M Z. Then the following stateents are equivalent: a is a cubic residue odulo q. b s 15 C 0. c ε / is a cubic residue odulo. d The congruence x + x od is solvable. e F 1/. f x + 15y for soe integers x and y. For later convenience we list the following notations: ω 1 + /, Z the set of integers, Z + the set of natural nubers, Z[ω] the set {a + bω a, b Z}, Nπ the nor of π, Q the set of rational nubers, Z the set of those rational nubers whose denoinator is rie to, [x] the greatest integer not exceeding x, [x] the set {k k x od, k Z }, a, b the greatest coon divisor of a and b, [a, b] the least coon ultile of a and b, n divides n, n does not divide n, a the Legendre sybol, α π the cubic Jacobi sybol.. Basic roerties of C i. Let Z + and 0 od. For a, b Z it is clear that there are unique integers a 0, b 0 {0, 1,..., 1} satisfying a a 0 od and b b 0 od. Fro this we ay define a + bω a0 + b 0 ω a, a 0, and for > 1. When 1 define a, 1 and a + bω One can easily verify the following facts: 1..1 If a, b, c, d Z then a + bω c + dω a + bωc + dω. If n Z and, n 1 then n 1.. If a, b Z 1 then a + bω a + bω a + bω 1 1..

5 Cubic residues and nonresidues 95 Definition.1. Suose Z + and 0 od. For i 0, 1, define { C i k k ω ω i, k Z }. Fro the above definition it is easy to rove the following results:.4 C 0 C 1 C {k k +, 1, k Z }..5 k C 0 if and only if k C 0..6 k C 1 if and only if k C. Exale.1. Set C i C i {k / < k /, k Z} for i 0, 1,. Then C 0 5 {0}, C 1 5 {1, }; C 0 7 {0}, C 1 7 { 1, }; C 0 11 {0, 5, 5}, C 0 1 {0, 4, 4}, C 0 17 {0, 1, 1,, }, C 0 19 {0, 1, 1,, }, C 1 11 { 1,,, 4}; C 1 1 {1,,, 5}; C 1 17 {, 4, 5, 6, 7, 8}; C 1 19 {, 5, 6, 7, 8, 9}. Proosition.1. Suose Z + and 0 od. Then 0 C 0. P r o o f. Since 1 + ω we see that 0 C ω ω Lea.1. Suose that Z +, 0 od, k 1, k Z, k1 + k +, 1, and is the greatest divisor of for which, k 1 + k 1. Then k ω k ω ω k1 k k 1 +k P r o o f. Since k ωk ω k 1 k + k 1 + k 1 + ω it is seen that k ω k ω. k1 k + k 1 + k 1 + ω k1 k k 1 +k ω k1 k + k 1 + k 1 + ω /.

6 96 Z. H. Sun When, we have k1 k + k 1 + k 1 + ω / 1. Now assue that > and that is a rie divisor of /. It is clear that k 1 + k 0 od and therefore that k1 k + k 1 + k 1 + ω k1 k k 1 1. Thus, k1 k + k 1 + k 1 + ω / This coletes the roof. / k1 k + k 1 + k 1 + ω 1. Proosition.. Let be a ositive integer not divisible by, and i {0, 1, }. i If k, k Z and kk od then k C i if and only if k C i. ii If k 1, k C i and k 1 + k, 1 then k 1 k /k 1 + k C i. P r o o f. Since k, 1, by Proosition.1 we have k ω k k ω + k + kω 1 + ω k ω k ω So i follows. To rove ii, we note that k1 k k 1 +k ω k1 k k 1 +k 1 ω k1 k k 1 +k ω k1 k k 1 +k ω k ω k ω by Lea.1 ω i ω i ω i..

7 Cubic residues and nonresidues 97 Proosition.. Let Z + with 0 od, and k Z with k 1k +, 1.Then k 9k/k C 0. P r o o f. Clearly, k+1+ω k+1+ωk +k1+ω k 9k+k 1+ω. Thus, k 9k k ω The roof is now colete. k 9k + k 1 + ω k ω 1. Proosition.4. Let 1, Z + be such that 1 0 od, k Z and i {0, 1, }. If 1 od [9, k + ] then k C i 1 if and only if k C i. P r o o f. Write k ω 1 j ω s 1 ω t π 1... π r, where π 1,..., π r are riary ries in Z[ω]. Since k ωk ω k + it is seen that k + 0 od π i i 1,..., r. Using Proosition.1 and 1.1 we find ω1 ω ω1 ω ω ω 1 and 1 Hence, k ω 1 This roves the result. 1 j ω s t ω1 ω ω 1 1 s t r i1 k ω 1. π i 1 t 1 ω r πi i1 1 s t r i1 Now we oint out the connections between C i i {0, 1, } and cubic residues. Theore.1. Let 1 od be a rie, 4 L + 7M L, M Z with L 1 od, and α β Z + with ax{d d, d, 6M 1} and 6, 1. Then, for i 0, 1,, i 1 L/M 1/ od π i.

8 98 Z. H. Sun if and only if L/M C i, where i {0, 1, } is deterined by i + βm od if α or M, i i + βm + 1 r+s od if α 1 r od and L 1 s 1 M od 4. P r o o f. Set π L + M/ + Mω. Then π Z[ω]. Clearly π od and Nπ. Thus, i 1 L/M 1/ od i 1 L/M 1/ ω i od π α β ω1 ω ω i π π π π α α ω i βm ω i+βm π by 1.1, 1. and 1.. Now let us calculate π. Obviously π 1 for 1. Assue that > 1 and that q is a rie divisor of. It is clear that q M and so that q L. Thus, π L + M/ + Mω L/ 1. q q On the other hand, q L + M/ + Mω So π α α π L + M/ Mω 1 + ω ω q if M, ω if M and L M od 4, ω ω 1 if M and L M od 4. 1 if α or M, ω 1s 1 α ω 1s+r if α 1 r and 4 L 1 s 1 M.

9 Cubic residues and nonresidues 99 Putting the above together we see that i 1 L/M π 1/ od This concludes the roof. 1 π ω i+βm α ω i. Corollary.1. Let and q be distinct ries greater than, 1 od and 4 L + 7M L, M Z with L 1 od. If q M then q 1/ 1 od. If q M and i {0, 1, } then i 1 L/M q 1/ od if and only if L/M C i q. Reark.1. According to Theore.1 the value of 1/ od can be coletely deterined. The secial cases, were treated by E. Leher [L] and K. S. Willias [W1] resectively. When is a rie for which,,, it follows fro Corollary.1 that 1/ od deends only on L/M od. This iortant fact was first observed by Jacobi [J], and roved by E. Leher [L1] and K. S. Willias [W1]. Lea.. Let be a rie and k Z. i If 1 od and so λλ with λ Z[ω] and λ od then k ω k + k 1 ω. λ ii If od then k ω k + / k ω +1/ od. P r o o f. Suose λλ 1 od with λ Z[ω] and λ od. Fro the roerties of the cubic residue character it is seen that k ω k ω k ω λ λ k ω k 1 ω λ λ k ω k 1 ω λ λ k + k 1 ω λ For ii, we note that k ω k ω k ω k 1 ω od.

10 00 Z. H. Sun and so k ω k ω 1/ k ω k 1 ω / k ω / k ω +1/ k + / k ω +1/ od. Now we are ready to give Theore.. Let be a rie, i {0, 1, } and k Z with k + 0 od. i If 1 od and so t od for soe t Z then k C i if and only if 1/ i k t 1 t od. k + t ii If od then k C i if and only if +1/ k 1 ω ω i od. k ω P r o o f. Suose 1 od, 4 L + 7M L, M Z and L 1 od. Since L/M t od we ay choose M so that L Mt od. Set λ L + M/ + Mω. Then λ Z[ω] and λ od. Clearly Nλ and 1 L/M ω od λ. Thus, by Lea. we have k ω k + k 1 ω λ k + k 1 ω 1/ k + k + L M 1/ k + t k t 1/ od λ. It then follows that k ω k C i ω i i 1 t k + t k t 1/ od λ

11 Cubic residues and nonresidues 01 i 1 t k + t k t 1/ od 1/ i k t 1 t od. k + t This roves i. Now consider ii. Note that k ω k 1 ω od. Using Lea. we see that k ω k 1 ω / k ω / k ω +1/ This coletes the roof. / k ω + +1 k 1 ω k ω +1/ k 1 ω k ω k ω +1/ k 1 ω od. k ω k ω 1 k 1 ω Fro Theore. we have the following rational cubic recirocity law. Corollary.. Let and q be distinct ries, 1 od, 4 L + 7M L, M Z, L 1 od, q > and i {0, 1, }. i If q 1 od and hence 4q L + 7M L, M Z then i 1 L/M q 1/ od if and only if LM L q 1/ M LM + L M 1 L /M i od q. ii If q od then i 1 L/M q 1/ od if and only if q+1/ L M 6Mω ω i od q. L + M + 6Mω P r o o f. If q M, it follows fro Corollary.1 that q 1/ 1 od. If q M, using Corollary.1 and Theore. we see that

12 0 Z. H. Sun i 1 L/M q 1/ od L M C iq L/M L /M q 1/ 1 L /M i L/M + L /M od q if q 1 od, q+1/ L/M 1 ω ω i od q if q od. L/M ω This coletes the roof. Reark.. In the case i 0 Corollary.i was first observed by Jacobi [J], and Corollary.ii can be deduced fro [W1]. Insired by K. Burde s rational biquadratic recirocity, H. von Lienen see [Li], [Bu] established the first rational cubic recirocity law. Theore.. Let > be a rie, k Z and k + 0 od. i If 1 od and so t od for soe t Z then k C 0 if and only if k + k + t is a cubic residue od. ii If od then k C 0 if and only if r + 1/ k + 1 r +1/ od 1 r k + +1/ + 1 k + / od. P r o o f. If 1 od, it follows fro Theore. that 1/ k t k C 0 1 od k + t k + k + t 1/ 1 od k + k + t is a cubic residue od. This roves i. Now consider ii. For i 0, 1, set +1/ r + 1/ k + 1 A i. r r i od Then A 0 + A 1 + A 1 + k + 1/ +1/ and hence +1/ k ω A 0 + A 1 ω + A ω A 0 A + A 1 A ω

13 Cubic residues and nonresidues 0 +1/ k + A 0 + A 1 +1/ k + + A 0 + A 1 ω. In view of Lea.i we obtain k ω +1/ k + / +1/ +1/ k + k + A 0 + A 1 + A 0 + A 1 ω od. If k+1+ω 1, it is clear that +1/ k + A 0 + A / k + / od, +1/ k + A 0 + A 1 od and therefore that +1/ k + A 0 + k + / od. If k+1+ω ω, then we have +1/ k + A 0 + A 1 od, +1/ k + A 0 + A / k + / od and hence +1/ k +.7 A 0 1 k + / od. If k+1+ω ω, one can siilarly rove that.7 holds. Now, by the above, ii follows and the roof is colete. Corollary.. Let be the roduct of ries of the for n+1, and hence t od for soe t Z. If x Z and xx 1, 1 then x +1 x 1 t C 0.

14 04 Z. H. Sun P r o o f. Write 1... r, where 1,..., r are ries of the for n + 1. For i 1,..., r it is clear that t od i. Thus, x + 1 x x 1 t x 1x 1 x 1 0 od i and so x + 1 x x 1 t x 1 t + t 1x x 1 x t x 1 x t x od i. 1 Alying Theore.i we find x +1 x 1 t C 0 i and hence x +1 x 1 t ω r x +1 x 1 t ω 1. i This is the result. i1. The structure of C i. In this section we introduce the sets C 0, C 1 and C, and study their grou structure. As an alication we confir a conjecture due to K. S. Willias [W1]. Definition.1. Let be a rie, k Z, [k] {x x k od, x Z } and [ ] {n/, n Z,, n}. Define C 0 {[k] k C 0 } {[ ] }, C 1 {[k] k C 1 } and C {[k] k C }. As an exale, taking 5 we have C 05 {[0] 5, [ ] 5 }, C 15 {[1] 5, [] 5 } and C 5 {[ 1] 5, [ ] 5 }. Let be a rie greater than, { Z/Z if 1 od, D Z[ω]/Z[ω] if od be the residue class ring odulo, and U the ultilicative grou of D. It is well known that U is a cyclic grou of order / 1. Denote the unique subgrou of order of U by G. Then G is also a cyclic grou. So {g g 1 1 od, g n 1 od n 1,...,, g Z} if 1 od, S {g g +1 1 od, g n 1 od n 1,...,, g Z[ω]} if od.

15 We are now ready to give Cubic residues and nonresidues 05 Theore.1. Let be a rie greater than and g S. For i 0, 1, we have i C i. ii C i {[ ] 1 + g / gr+i + 1 g r+i 1 r 0, 1,..., } / 1. P r o o f. Suose k Z with k + 0 od. If 1 od, it is clear that 1 g 1/ od. For i {0, 1, } it follows fro Theore. that k g 1/ 1/ k C i g i 1/ od k 1 g 1/ k g 1/ k 1 g 1/ gr+i od for soe r {0, 1,..., 1/ 1} k 1 + g 1/ gr+i + 1 g r+i od 1 for soe r {0, 1,..., 4/}. If od, it is clear that g +1/ ω or ω od. For i {0, 1, } it follows fro Theore. that k 1 g +1/ +1/ k C i g i+1/ od k g +1/ k 1 g+1/ k g +1/ gr+i od for soe r {0, 1,..., + 1/ 1} k 1 + g +1/ gr+i + 1 g r+i od 1 for soe r {0, 1,..., /}. To conclude the roof, we note that [ 1 + g / g 0+0 ] + 1 g [ ] and that g r1+i + 1 g r 1+i g r 1+i g r +i 1 gr+i + 1 g r +i 1 rovided r 1 r od /. od

16 06 Z. H. Sun Corollary.1. Let > be a rie, and R a colete residue syste odulo. Then k 1 od. k C 1 R P r o o f. Let g S and /. It follows fro Theore.1 that 1 k 1 + g g r g r+1 1 k C 1 R r g + g r+1 od. 1 Since 1 r0 we find 1 1 g r+1 1 r0 1 r0 1 g g r g 1 1 s0 1 + k k C 1 R s1 s0 r0 g r+1 s g s g sr 1 g 1 g s 1 gs 1 g s 1 + g + g g g g 1 1 s0 g 1 g 1 Note that 1 + g + g g 1/g 1 0 od. g 1 od. We are done. 1 g s r0 g sr od, Reark.1. Corollary.1 is equivalent to a result conjectured by K. S. Willias [W1]. Corollary.. Let > be a rie, and R a colete residue syste odulo. Then { k k k C + 1 R, 1} { k k } k C + 1 R, 1. 6

17 Cubic residues and nonresidues 07 P r o o f. Let g S. In view of Theore.1 we can write C 1 R { k r r 0, 1,..., } / 1, where k r 1 + g / gr g r+1 od. 1 Fro this it follows that kr 1 + 4g / + 4g / 1 + g r g r g r+1 4 gr+1 1 g r+1 od 1 and so k r + k r + 1/ 4 1/ g 1 r+1 g r+1 1 g r+1 1 r+1 1 g 1 r+1 od. Thus, k n + k n+1 + This roves the corollary. g r+1 1 g r+1 1 r+1 g r+1 1 g r+1 1 g r+1 1 g r for n 0, 1,..., /6 1. Theore.. Let be a rie greater than. For [k], [k ] C 0 C 1 C define [ kk [k] [k ] ] k + k [k] [ ] [ ] [k] [k]. Then C 0 C 1 C fors a cyclic grou of order, and C 0 is a subgrou of order /. Moreover, C 0, C 1 and C are the three distinct cosets of C 0. P r o o f. Suose g S. Fro Theore.1 we know that { } C 0 C 1 C [k r ] r 0, 1,..., 1, where [k r ] [ 1 + g / gr + 1 g r 1 ].

18 08 Z. H. Sun Since [ ] ki k j k i + k j 1 + g [ [ we see that / 1 + g / 1 + g 1 + g [ ] ki k j [k i ] [k j ] k i + k j gi + 1 g i 1 gj + 1 g j 1 g i + 1 g i 1 + gj + 1 g j 1 / gi + 1g j g i 1g j ] 1 g i + 1g j 1 + g i 1g j + 1 ] / gi+j + 1 g i+j 1, [k i+j ], where x denotes the least nonnegative residue of x odulo By the above, C 0 C 1 C is a cyclic grou generated by [k 1 ]. Alying Theore.1 we see that C 0 is a cyclic grou generated by [k ], and that C 0, C 1 and C are the three cosets of C 0. The roof is now colete. Corollary.. Let be a rie greater than. Then {[ x C 0 ] } 9x x x {0, 1,..., 1}, x od. P r o o f. Clearly [ 1 ] [ ] C 0. Suose k Z. It follows fro Theore. that [k] C 0 [k] [x] [x] [x] for soe [x] C 0 C 1 C [ x ] [ x ] 9x [k] [x] x x for soe integer x So the result follows. satisfying x + 0 od and x {0, 1,..., 1}. Corollary.4. Let > be a rie, i {0, 1, } and [k i ] C i. For [k] C 0 define [ ] kki ϕ[k] ϕ[ ] [k i ]. k + k i Then ϕ is a one-to-one corresondence fro C 0 to C i..

19 P r o o f. In view of Theore., So the result follows. Cubic residues and nonresidues 09 C i [k i ] C 0 {ϕ[k] [k] C 0}. Reark.. Corollaries. and.4 rovide a sile ethod of calculating C 0, C 1 and C for any rie >. 4. Cubic congruences. Let be a rie greater than. In this section we consider the general cubic congruence x + Ax + Bx + C 0 od, where A, B, C Z. In [St] Stickelberger showed that the nuber of solutions of x + Ax + Bx + C 0 od is given by D 0 or if 1, N D 1 if 1, where D A B 4B 4A C 7C + 18ABC. Since x +Ax +Bx+C x+ A A B x+ A + A 9AB + 7C, 9 7 it is enough to discuss the congruence x ax ab 0 od a, b Z. Lea 4.1. Assue that is a rie greater than, k Z and k + 0 od. Then k C 0 if and only if the congruence x 9k + x 18k + 0 od is solvable. Moreover, if k C 0 then the solutions of the above congruence are given by { + ktu1 u od if 1 od, x k + kωu1 u od if od, where t and u satisfy t od t Z and k t od u Z if 1 od, u k + t k 1 ω od u Z[ω] if od. k ω P r o o f. If k 0 od, then k C 0. Clearly, the congruence x 9k + x 18k + 0 od has the solutions x 6,, od. So the result is true when k. Now assue k 0 od. It follows fro Corollary. that k C 0 k s 9s s od for soe s {0, 1,..., 1} s ks 9s + k 0 od is solvable.

20 10 Z. H. Sun Set x k s s od. Then 9k s s ks 9s + k 9k 9k k s 9 k + s k s 9k 9k x + 9x + + x + x 9k + x 18k + od. So k C 0 if and only if x 9k + x 18k + 0 od is solvable. Let k C 0 and r t or 1 + ω according as 1 or od. Fro Theore. we know that So the congruence k r k + r / 1 od. u k r od k + r is solvable. Suose u k r k+r od and x + kru1 u od. Then and hence u 1 u u 1 u + u u k r r u + u k + r k + r od x 9k + x + kr u 1 u 9k + + kru u r k + r k r r u + u k + r k + r 9rk + r k ru u 18k + od. When od it is easily seen that u u 1 od and so that u u 1 od. Hence, k + kωu1 u 1 + ω k ω u1 u 1 ωk 1 ω u u u1 u 1 + ωk 1 ω u k + kωu1 u od. This shows that k + kωu1 u is congruent to an integer odulo. By the above, the lea is roved.

21 Cubic residues and nonresidues 11 Theore 4.1. Let > be a rie, a, b, s Z, ab 0 od and s b 4a od. Then the congruence x ax ab 0 od is solvable if and only if s/b C 0. Moreover, if s/b C 0 then the solutions of the above congruence are given by { 1 6 st bu1 u od if 1 od, x 1 6 s b + sωu1 u od if od, where t and u satisfy t od t Z and s bt od u Z if 1 od, u s + bt s b1 + ω od u Z[ω] if od. s + b1 + ω P r o o f. Set k s/b. Then k + 1a/b 0 od. It is clear that x 9k + x 18k + x 108a x 16a So the result follows fro Lea 4.1. b b 6 b b 6 x a b 6 x ab od. Corollary 4.1. Let > be a rie and a, b Z. Then the congruence x ax ab 0 od is unsolvable if and only if b 4a k b od for soe k C 1. P r o o f. If ab 0 od then 0 a 0 ab 0 od. If b 4a 0 od then b ab ab 0 od. So x ax ab 0 od is solvable when abb 4a 0 od. Now assue abb 4a 0 od. Since 4 a 7 ab b 4a 9a, using Stickelberger s result we see that x ax ab 0 od has one solution if b 4a 1. 1, there is an integer k such that k b 4a/b If b 4a od. Since k + 1a/b 0 od we have k C 0 C 1 C. Alying Theore 4.1 we see that x ax ab 0 od is solvable if and only if k C 0. So x ax ab 0 od is unsolvable if and only if b 4a k b od for soe k C 1 C. Since k C if and only if k C 1, by the above the corollary is roved. Reark 4.1. If is a rie greater than, a, b Z and b 4a 1, one can easily check that the unique solution of x ax ab 0 od is given by

22 1 Z. H. Sun x a / v + / od, where {v n } is defined by v 0, v 1 b and v n+1 bv n av n 1 n 1. As alications of Theore 4.1 we have Theore 4.. Let > be a rie,, n Z and n 0 od. Then {[ x ] x + n x n } od, x Z. P r o o f. Set b n/. Then b Z and b 0 od. Fro Corollary 4.1 and Theore.1i we see that {[ x ] x + n x Z, x n } od {[ ] x x + b x Z, x b } od We are done. {[a] x ax ab 0 od is solvable} {[a] x ax ab 0 od is unsolvable} {[a] b 4a k b od for soe k C 1 } {[ k + b ] } 1 k C 1. Theore 4.. Let > be a rie and A, B, C Z. Then {[x + Ax + Bx + C] x Z } P r o o f. Since x + Ax + Bx + C we see that {[x + Ax + Bx + C] x {0, 1,..., 1}} + if 1 od and A B od, if od and A B od, if A B od. x + A A B x + A + A 9AB + 7C 7

23 Cubic residues and nonresidues 1 {[x + Ax + Bx + C] x Z } {[ ] x A B x + A 9AB + 7C 7 } x Z {[ ] x A B x } x Z { } [t] x A B x t od is solvable, t Z if 1 od and A B od, { if od and A B od, [b] x A B 9 if A B od. x A B b 0 od is solvable} 9 Now suose A B od and a A B/9. By Corollaries 4.1 and. we get {[b] x ax ab 0 od is solvable} {[b] x ax ab 0 od is unsolvable} {[b] b 4a k b od for soe k C 1 } { [b] b 1a 1} k + od for soe k C {[ ] 1a k + 1a k +, k C 1 }. 6 Putting the above together yields the result. 5. Connections with binary quadratic fors. Let d be a squarefree integer, and a rie greater than satisfying d 1. In this section we obtain a criterion for sd C i i {0, 1, } in ters of the binary quadratic fors of discriinant 4d, where sd satisfies sd d od. Theore 5.1. Let be a rie greater than and ax +bxy +cy with a, b, c, x, y Z. If d b ac, a α r a 1 a 1, a 1, d+ β s d 1 d 1, d 1, a, d + 1 and ad + 0 od, then

24 14 Z. H. Sun ax + b + 1y + yω a 1 d 1 1 s ω ax y if y 0 od, ω f 1u if a 0 od and x uy od 9, ω f u if aax + by 0 od and x uy od 9, 1 if ax + by 0 od 9, ± ω d if ax + by ±y od 9, where f 1 u 1 bu + c b bu + c + r 1 1 b b f u 1 s au + b 1 d + 1 d 1 d and 1 + ac + α a 1 α a 1 β d 1 β d 1 au + b { 1 if 1 od, 1 if 1 od. au + b 4, P r o o f. For later use we first oint out the following facts: 1 i y 0 od. Indeed, if y then ax and so x. Thus, ax + bxy + cy 0 od. This is a contradiction. ii If π ax + by + y + yω then Nπ, a Nπ, d + 1. Indeed, clearly Nπ ππ ax + by + y a + d + y. Thus, Nπ, a d + y, a 1 and Nπ, d + a, d + 1. iii If A + Bω Z[ω] and A + Bω od then A+Bω ω B/. Indeed, since ω 1 ω, it follows fro 1.1 and 1. that A + Bω ω A + Bω 1 ω A + Bω ω A+B+1/ ω 4A+1/ ω B/. Now let π ax + b + 1y + yω. Since N1 ω and Nπ ax+by +y 0 od 9, there are integers i, k {0, 1} and j {0, 1, } such that π 1 i ω j 1 ω k π, where π Z[ω] and π od.

25 Cubic residues and nonresidues 15 Assue y t y 0 y 0 and π A + Bω. Then we have ax + b + 1y + yω 1 i ω j 1 ω k π j k k ω ω1 ω π j k k ω 1 + ω π j k ω a a j k ω d + y a π π π π j k ω r+s+t α+β a 1d 1 y 0 π π j k s r t ω π π α+β a 1 d 1y0 j k ω ω s r tb/ π 1 i ω j k 1 + ω k 1 α+β a 1 d 1y0 α+β a 1 d 1y0 j k k j α+β ω ω r+t sb/ ω π π π α+β a 1 d 1y0 a 1 d 1 y 0 by Proosition.1 j k j k ω ω r+t sb/ ω π α+β a 1 d 1y0 a 1 d 1 Note that π π y 0 1. j k 1 ω π α+β a 1 d 1 y ω r+t sb/ 0 a 1 d. 1 That is, 5.1 ax + b + 1y + yω a 1 d 1 Let us consider five cases. ω r+t sb/ ω α+β a 1 d 1 y 0 j k Case 1: y 0 od. Since ax + bxy + cy ax od we have ax 0 od and so r 0. Clearly π ax π. Hence j k 0 and B ax y. Fro 5.1 we see that ax + b + 1y + yω a 1 d 1 ω ax y t s 1 s ω ax y. Case : a 0 od. In this case, y 0 od. Since a, d + 1 we have b ac + d + 0 od and so ax + by 0 od..

26 16 Z. H. Sun If y 0 od, x uy od 9 and ax + by 0 od, then clearly ωπ y + ax by + yω y y π if au + b 1 od, ω π ax by + y ax + by + yω y y if au + b 1 od. Fro this and 5.1 it follows that ax + b + 1y + yω a 1 d 1 ω α+β a 1 d 1 y ω α+β a 1 d 1 y ω α+β a 1 d 1 y ω y ax by+yr s/ if au + b 1 od, ω au+b y ax+by+yr s/ if au + b 1 od au+b ω au+b s r/. Observing that x y a + b x y y + c y y + au + bu + c od 9, y we get ax + b + 1y + yω 5. a 1 d 1 au+b ω au+b s r/ y ω α+β a 1 d 1 au + bu + c au+b Since a 0 od we ust have d+ 0 od and so d b ac 1 od. Hence, ω and β d 1 ω α+β a 1 d 1 au + bu + c ω ω 1 d / ω d 1/+1 d + ω ω α a 1 β d 1 ω 1 1 α a α a 1 1 ω d 1 +1 ω 1 1 ω au + bu + c bu+c au +bu+c..

27 In view of 5., Cubic residues and nonresidues 17 ax + b + 1y + yω a 1 d 1 ω n 1, where n 1 r au + b au+b 1 1 α au + b α a 1 a 1 + d au + bu + c bu + c 1 r au + b + d + α b b α a 1 a au + bu + c. b bu + c Since rau au au r u b bu + c bc u 0 od 9 if a 0 od 9 or u 0 od, a bc 1 bc 0 od 9 bc if a ± od 9 and u bc od and d + b 1 ac b ac + 4 od 9, b we see that n 1 f 1 u od and so ax + b + 1y + yω a 1 d 1 ω f 1u. Case : aax + by 0 od. In this case, r 0. Suose x uy od 9. Then α a 1 au + bu + c a u + abu + ac au + b au+b + d au + b 1 d au + b 1 od 9. au + b Fro this and 5. it follows that

28 18 Z. H. Sun ax + b + 1y + yω a 1 d 1 ω β d 1 1 d au + b au+b where n 1 1 β d 1 au + b 1 d 1 d au + b au + b 1 au + b 1 au + b au + b s + β d 1 1 d 1 d 1 d au+b sau+b ω au+b / ω n, s au + b au+b 1. Since β d 1 1 d 1 1 d β d d d + β d 1 1 d 1 d 1 d β d 1 1 d β od 9, d 1 we obtain ax + b + 1y + yω a 1 d 1 ω n ω fu. Case 4: ax + by 0 od 9. Since ax + by 0 od we have a ax + by dy dy od 9. We clai that ady 0 od and so that r s t 0. If y 0 od then ax + bxy + cy ax + byx + bx + cyy 0 od. Thus, y 0 od. If a 0 od then dy a 0 od. Since y 0 od we have d 0 od and so d + 0 od. This contradicts the assution a, d + 1. Hence, a 0 od. By the above, ay 0 od and dy a od 9. So d 0 od. This roves the assertion. Now, it is easy to check that π π y ω1 ω ax + by ax + by y + + ω. y

29 So we have Cubic residues and nonresidues 19 r s t 0, j k 1 and B This together with 5.1 gives ax + b + 1y + yω a 1 d 1 ax + by. y ω 0 1. Case 5: ax + by ±y od 9. In this case, ax + by 0 od. By the above clai we have r s t 0. It is clear that π ax + by y ω 1 ω ax + by + y + y ω y π if ax + by y od 9, π ax + by ax + by y 1 ω + + y ω y if ax + by y od 9. Thus, by 5.1, ax + b + 1y + yω a 1 d 1 ω α+β a 1 d 1 y ω α+β a 1 d 1 y Since α a 1 β d 1 y ad + ax + bxy + cy y we obtain ω α+β a 1 d 1 y and hence if ax + by y od 9, if ax + by y od 9. d + ax + by dy y d d dy d + d y + od 9, d y This coletes the roof. ω d Fro Theore 5.1 we have ω 1 ax + b + 1y + yω a 1 d 1 1 d ± ω d. d ω d

30 0 Z. H. Sun Theore 5.. Let be a rie greater than, d { 1,, 5, 6, 7, 15}, d 1 and sd d od. Then sd C 0 if and only if can be reresented by one of the corresonding binary quadratic fors in Table 5.1. d Table 5.1 Binary quadratic fors 1 x + 81y, x + xy + 41y x + 16y, x + 81y 5 x + 405y, 5x + 81y, 10x + 10xy + 4y, x + xy + 0y 6 x + 54y, x + 7y 7 x + 567y, 7x + 81y, x + 0xy + 9y 15 x + 15y, 5x + 7y P r o o f. If d 1 then 1 1 and so x + y for soe x, y Z. Setting a 1, b 0 and c 1 in Theore 5.1 we get ω x + y + yω x y if y, u ω u / if x and x uy od 9, 1 if 9 x, ω 1 if x ±y od 9. Since y 0 od and s 1 ±x/y od we see that s 1 C 0 x/y C 0 x + y + yω x/y ω 9 x, 9 y or x y od 9. xy Clearly, x + y with x, y Z and 9 xy if and only if x y1 for soe x 1, y 1 Z. If x + y with x xy y od 9 then x 1 + x 1 y y1 for x x + 5 xy y and y1 1 9 x xy y. Conversely, if x 1 + x 1 y y1 with x 1, y 1 Z then x + y for x x 1 + 5y 1 and y x 1 4y 1. Also, x xy y od 9. By the above, s 1 C 0 if and only if x + 81y or x + xy + 41y for soe x, y Z. If d then 1 and so x + y for soe x, y Z. Using the fact that s ±x/y od and Theore 5.1 we see that x + y + yω s C x or 9 y 1 x y 1 or x y 1 x 1, y 1 Z.

31 Cubic residues and nonresidues 1 If d 5 then 5 1 and so x + 5y or x + xy + y for soe x, y Z. Using Theore 5.1 we see that x/y C 0 9 x or 9 y if x s 5 C 0 + 5y, 1 + x/y C 0 9 x or x y od 9 if x + xy + y. This yields the result. If d 6 then 6 1 and so x + 6y or x + y for soe x, y Z. Alying Theore 5.1 we get { x/y C0 y if x s 6 C 0 + 6y, x/y C 0 y if x + y. This gives the result. If d 7 then 7 1 and so x +7y for soe x, y Z. Alying Theore 5.1 we see that s 7 C 0 x/y C 0 9 x, 9 y or x 4 y od 9. xy This yields the desired result. If d 15 then 15 1 and hence x + 15y or 5x + y for soe x, y Z. In view of Theore 5.1 we get { x/y C0 y if x s 15 C y, 5x/y C 0 y if 5x + y. This deduces the result. Cobining the above we rove the theore. Corollary 5.1. Let be a rie greater than. i If 1 1 then x + 6x od is solvable if and only if x + 81y or x + xy + 41y for soe x, y Z. ii If 1 then x 9x 18 0 od is solvable if and only if x + 16y or x + 81y for soe x, y Z. iii If 6 1 then x + x + 0 od is solvable if and only if x + 54y or x + 7y for soe x, y Z. iv If 15 1 then x + x od is solvable if and only if x + 15y or 5x + 7y for soe x, y Z. P r o o f. If 1 1, then s 1 1 od for soe s 1 Z. Set a and b. Then 6s 1 b 4a od. Fro Theores 4.1 and 5. we see that

32 Z. H. Sun x + 6x od is solvable 6s 1/ C 0 s 1 C 0 by Proosition.i x + 81y or x + xy + 41y x, y Z. This roves i. Siilarly, by using Theores 4.1 and 5. one can rove ii iv. Reark 5.1. Kronecker [K] showed that x + x od is solvable for rie satisfying 1 1 if and only if x + 1y for soe integers x and y. In 197, E. Leher [L] roved Corollary 5.1iv in the case 1 od. For recent iortant aers along this line one ay consult [WH] and [SW]. Corollary 5.. Let be a rie of the for n + 1, and ε d denote the fundaental unit of the quadratic field Q d. i If d {,, 5} and d 1 then εd is a cubic residue od if and only if x + 7dy for soe integers x and y. ii If 6 1 then ε is a cubic residue od if and only if x + 16y for soe integers x and y. iii If 15 1 then ε is a cubic residue od if and only if x +405y or 10x +10xy +4y for soe integers x and y. iv If 1 1 then ε is a cubic residue od if and only if x + 567y or 7x + 81y for soe integers x and y. P r o o f. Suose t od, d 1 and s d d od. By Theore.i, s d C 0 if and only if s d t/s d + t is a cubic residue od. Observing that s d t s d + t s d s dt + t s d t and that s dt d od we find that 5. s d C 0 d + + d d Clearly, d + + s dt d od is a cubic residue od. ε , ε +, ε / , ε , ε , ε

33 Cubic residues and nonresidues Hence, cobining Theore 5. with 5. in the cases d 6, 1, 15,, 5, 7 gives the result. Reark 5.. Corollary 5.i was known by E. Leher [L], and the rest of Corollary 5. is new. For a general result on the cubic character of quadratic units one ay consult [We]. Theore 5.. Let be a rie of the for n + 1, 4 L + 7M L, M Z, and qd qd > a rie divisor of L 9dM or dl + 81M. i If k Z then q k is a cubic residue od if and only if k 1k 1 is a cubic residue od q k. ii If d { 1,, 5, 6, 7, 15} then qd is a cubic residue od if and only if qd can be reresented by one of the corresonding binary quadratic fors in Table 5.1. P r o o f. Suose sd d od qd. We first clai that 5.4 qd is a cubic residue od sd C 0 qd. If qd d then qd LM and qd sd. Fro Proosition.1 and Corollary.1 we see that sd C 0 qd and that qd is a cubic residue od. If qd d then qd LM. Otherwise, 4 L +7M 0 od qd. Since we have L d od qd or M 9M d od qd L sd ± L M or sd ±9M od qd. L Now, alying Corollary.1 and Proosition.i we see that qd is a cubic residue od L M C 0qd sd C 0 qd. This roves the assertion. Now let us consider i. Suose d k for soe k Z. If k ±1 od qd then d od qd and so 4 L + 7M 0 od qd. This ilies qd. So qd is a cubic residue od. If k ±1 od qd, by 5. and 5.4 we see that qd is a cubic residue od sd C 0 qd k + + k k k + 1 is a cubic residue od qd k 1 k 1k 1 is a cubic residue od qd. This roves i. ii follows fro 5.4 and Theore 5..

34 4 Z. H. Sun Reark 5.. When qd is a rie divisor of L + 9dM Federighi and Roll [FR] conjectured Theore 5.ii in the cases d 6, 15. Ph. Barkan [Ba] showed how to rove their conjecture about ries qd 1 od. In 199, using class field theory Searan and Willias [SW] roved the following iortant result: 5.5 Suose > is a rie and x + Ax + Bx + C A, B, C Z is irreducible over the rational field Q. If the discriinant D A B 4B 4A C 7C +18ABC is not a erfect square such that D 1, and HD is the for class grou of classes of riitive, integral binary quadratic fors of discriinant D, then the cubic congruence x + Ax + Bx + C 0 od is solvable if and only if can be reresented by one of the third coosition owers of fors in HD. Fro 5.5 we have Lea 5.1. Assue that > is a rie,, n Z, n, n, n, 4, 64, 108, 50, n/ 7 {k k Z} and n/ 7 1. Then the cubic congruence x n x n 0 od is solvable if and only if can be reresented by one of the third owers of riitive integral binary quadratic fors of discriinant n/ 7n. P r o o f. Clearly the discriinant of x n x n is given by D 4 n 7 n n 7 n. Since D is not a square, by 5.5 it is sufficient to rove that x n x n 0 for any integer x. If t Z and t n t n 0, then n st for soe s Z. Since n 0 od we have st 0 and so t st s 0. This ilies t s. Write s rt. Then t r 4 t rt 0. Naely, 4t rt r 0. It then follows that 4t kr for soe k Z. Observing that r s/t 0 we find k4 r 8 and hence k {±1, ±, ±4, ±8}. Since n r kr r 4 k k 4 8 k k k and k {±1, ±, ±4, ±8} we get n {, 54, 0, 64, 4, 108, 50}. This contradicts the assution. Thus, x n x n 0 for any x Z. This coletes the roof. Now we can give Theore 5.4. If > is a rie, d Z, d, d od, d {k k Z}, d 1, sd d od and 18d + n with, n Z, then sd C 0 if and only if can be reresented by one of

35 Cubic residues and nonresidues 5 the third owers of riitive integral binary quadratic fors of discriinant 9dn. P r o o f. It follows fro Lea 4.1 that sd C 0 x 9sd + x 18sd + 0 od is solvable x 9d + x 18d + 0 od is solvable y n y n 0 od is solvable y n y n 0 od is solvable. Since n/ 7n 9dn and n 18d +, 4, 64, 108, 50, alying Lea 5.1 we obtain the result. Corollary 5.. If is a rie, 1 od, k Z, k 0, ±1 od and k 1 n, n Z, then k+1 k 1 is a cubic residue od if and only if can be reresented by one of the third owers of riitive integral binary quadratic fors of discriinant 7k n. P r o o f. Suose d k and t od. Clearly, d 1, d od and kt d od. By Theore.i, kt C 0 if and only if k+1 kt t k 1 kt+t is a cubic residue od. Also, 18d + 18 k n and 9dn 7k n. So the result follows fro Theore 5.4. Corollary 5.4. Let be a rie of the for n + 1, 4 L + 7M L, M Z, d Z, d {k k Z}, d, d od and 18d + n, n Z. If qd is a rie divisor of L 9dM or dl + 81M satisfying qd, and qd d, then qd is a cubic residue od if and only if qd can be reresented by one of the third owers of riitive integral binary quadratic fors of discriinant 9dn. P r o o f. This is iediate fro 5.4 and Theore Alications to Lucas series. Let a and b be two real nubers. The Lucas sequences {u n a, b} and {v n a, b} are defined as follows: 6.1 u 0 a, b 0, u 1 a, b 1, u n+1 a, b bu n a, b au n 1 a, b n 1; 6. v 0 a, b, v 1 a, b b, v n+1 a, b bv n a, b av n 1 a, b n 1.

36 6 Z. H. Sun It is well known that n n 1 b + b 4a b b 4a 6. u n a, b b 4a b 4a 0 and that n n b + b 4a b b 4a 6.4 v n a, b +. Suose that is a rie greater than. It is the urose of this section to deterine u /a, b od and v /a, b od. Theore 6.1. Let > be a rie, a, b Z, ab, b 4a and s b 4a od. Then and u /a, b v 0 od if s/b C 0, ± s a [/] od if ±s/b C 1 { a /a, b [/] od if s/b C 0, a [/] od if s/b C 0. 1 P r o o f. Set k b/s. For n Z + it is clear that n n 1 b + b 4a b b 4a u n a, b b 4a [n 1/] n n b n r 1 b b 4a r + 1 4a r+1 r0 [n 1/] n n b n r 1 b 4a r r + 1 r0 [n 1/] r+1 n s1 + ω n b n r 1 r + 1 s1 + ω r0 n n b + s1 + ω/ b s1 + ω/ s1 + ω ω1 ω s n k ω s 6 n k 1 ω n od. Siilarly, v n a, b n n [n/] n b n r b 4a r r r0 [n/] n r r0 b n r s1 + ω r

37 Cubic residues and nonresidues 7 n b + s1 + ω/ b s1 + ω/ + 6 s n k ω n + k 1 ω n od. If 1 od, we ay write λλ with λ Z[ω] and λ od. By Lea.i, k + 1/ k 1 ω 1/ ± k ω 1/ k + k 1 ω k + k ω ± λ λ 1 k ω k ω ± od λ. Hence, by the above and Ferat s little theore we get ω1 ω u 1/ a, b s 1/ k + 1/ s 6 1 k ω k ω 0 od λ if k C 0, ± s a 1/ od λ if ±k C 1 and v 1/ a, b s 1/ k ω k + 1/ 6 { a 1/ od λ if k C 0, a 1/ od λ if k C 1 C. n k ω + 1 Since both sides of the above congruences are rational, the congruences are also true when λ is relaced by Nλ. If od, it follows fro Lea.ii that k ω +1/ ± k 1 ω +1/ k + / k ω k ω ± 1 Fro this and the above it follows that ω1 ω u +1/ a, b s +1/ k + / s 6 1 k ω k ω od.

38 8 Z. H. Sun 0 od if k C 0, ± s a / od if ±k C 1 and v +1/ a, b s +1/ k ω k ω k + / + 6 { a / od if k C 0, a / od if k C 1 C. To colete the roof, we note that k 6a + b 0 od 4a and k C i if and only if s/b C i by Proosition.i. 1 Corollary 6.1. Let > be a rie, k Z and kk + 0 od. Then 0 od if k C 0, 1 u /k + 9, 6 k k 9 [/] od if k C 1, 1 k k 9 [/] od if k C and v /k + 9, 6 { k + 9 [/] od if k C 0, k + 9 [/] od if k C 0. Corollary 6.. Let > be a rie, d Z, d od, d 1 and sd d od. Then sd C 0 if and only if u /d + 9, 6 0 od. P r o o f. Set k sd. Then u n k + 9, 6 u n d + 9, 6 od by 6.1. Hence the result follows fro Corollary 6.1. Corollary 6.. Let > be a rie, a, b Z, ab and b 4a 1. Then the congruence x ax ab 0 od is solvable if and only if u /a, b 0 od. P r o o f. Suose s b 4a od. It then follows fro Theore 6.1 that s/b C 0 if and only if u /a, b 0 od. This together with Theore 4.1 gives the result. Corollary 6.4. Let > be a rie, a, b Z, ab and b 4a 1. If b 4a is not a square and x ax ab is irreducible over

39 Cubic residues and nonresidues 9 Q, then u /a, b if and only if can be reresented by one of the third owers of riitive integral binary quadratic fors of discriinant 7a b 4a. P r o o f. Since the discriinant of x ax ab is 7a b 4a the result follows fro 5.5 and Corollary 6.. Let {F n } and {L n } be defined by and F 0 0, F 1 1, F n+1 F n + F n 1 n 1 L 0, L 1 1, L n+1 L n + L n 1 n 1. It is well known that {F n } is the Fibonacci sequence and that {L n } is the Lucas sequence. Fro Theores 5.1 and 6.1 we have Theore 6.. Let > 5 be a rie for which 15 1 and hence x +15y or 5x +y for soe x, y Z according as 1 od or od. Then 0 od if y 0 od, and F / L / x + y od if y x od od if y 0 od, od if y 0 od. P r o o f. Suose s x y. Then s 15 od. Since F n u n 1, 1 and L n v n 1, 1, it follows fro Theore 6.1 that F / 0 od if s C 0, ± s od if ±s C 1 and that 1 [/] od L / 1 [/] od if s C 0, if s C 0.

40 0 Z. H. Sun Fro Theore 5.1 we know that s ω sy + y + yω 1 if y 0 od, ω if x y od, ω if x y od. Hence, s C 0 if and only if y 0 od. If y 0 od then x ± y od and so ±s C1 by the above. Since s y x x x + y od + y we obtain F / x od 5y x y This coletes the roof. od if 1 od and x y od, if od and x y od. Lea 6.1. Let be a rie greater than, a, b Z, u n u n a, b, v n v n a, b and abb 4a 0 od. Then a u if and only if b 4a 1. b u n if and only if v n a n od. P r o o f. Fro [D] and [R] we know that b 6.5 u b 0 od, u 4a 4a od. Thus, u + b 4a bu au 1 b od 1 a bu u +1 b a 0 od. It then follows that u od if b 4a if 1, b 4a 1 b 4a b 4a 1. This roves art a. Now consider art b. According to [D] and [R] we have 6.6 u n u n v n, v n v n a n,

41 6.7 Thus, This concludes the roof. Cubic residues and nonresidues 1 v n b 4au n 4a n. u n v n 4a n od v n a n od. Using Lea 6.1 and Theore 6. we have Corollary 6.5. Let be a rie greater than 5. Then i F / if and only if can be reresented by x + 15y or 5x + 7y according as 1 od or od. ii F /6 if and only if can be reresented by x + 540y or 5x + 108y according as 1 od or od. P r o o f. It is well known that see [D], [SS], [R] F n F n for, n Z +. Thus F / F. If F / then F. Alying Lea 6.1 we find 15 1 and so A + 15B or 5A + B for soe A, B Z. It then follows fro Theore 6. that B. Hence x + 15y or 5x + 7y for soe x, y Z. Conversely, if is reresented by x + 15y or 5x + 7y, then Alying Theore 6. we find F /. This roves i. Let us consider ii. If F /6 then F and so 15 1 by Lea 6.1. If is reresented by x + 540y or 5x + 108y, we also have Hence, we ay assue 15 1 and so x + 15y or 5x + y for soe x, y Z. It then follows fro Lea 6.1 and Theore 6. that F /6 L 1 /6 / 1 and y { A + 15B 1 od 1 A, B Z if 1 od, 5A + 7B 5 od 1 A, B Z if od A + 15B or 5A + 7B with B 0 od x + 540y or 5x + 108y for soe x, y Z. This coletes the roof. Reark 6.1. In [L], [L4] E. Leher roved Corollary 6.5i in the case 1 od 1. For the criteria for F 1/4 if 1 od 4 is a rie one ay consult [L6], [SS].

42 Z. H. Sun Now we oint out siilar results for the Pell sequence. The Pell sequence {P n } and its coanion {Q n } are given by and P 0 0, P 1 1, P n+1 P n + P n 1 n 1 Q 0, Q 1, Q n+1 Q n + Q n 1 n 1. Clearly, P n u n 1, and Q n v n 1,. Using Theores 6.1 and 5.1 one can siilarly rove Theore 6.. Let > be a rie such that 6 1 and hence x + 6y or x + y for soe x, y Z according as 1 od or od. Then 0 od if y 0 od, and P / Q / x 1 + y od if y x od od if y 0 od, od if y 0 od. Reark 6.. For the values of P 1/ od and P +1/ od one ay consult [S1]. Corollary 6.6. Let be a rie greater than. Then i P / if and only if x + 54y or x + 7y for soe integers x and y according as 1 od or od. ii P /6 if and only if x + 16y or 8x + 8xy + 9y for soe integers x and y according as 1 od or od. The roof of Corollary 6.6 is siilar to the roof of Corollary 6.5. Reark 6.. Let 1 od 4 be a rie. Fro [L4], [S1] we know that P 1/4 if and only if x + y for soe integers x and y. Finally, we discuss the Lucas sequence {u n 1, 4}. Theore 6.4. Let 1 od 4 be a rie and hence x + y for soe integers x and y. Then

43 u /61, 4 and and Cubic residues and nonresidues /6 u /, 0 od if 9 xyx y, 1 x od if x y, y, 4y od 9 or y x od 9 y v /61, 4 /6 v /, 6 od if 9 xyx y, 6 od if 9 xyx y. P r o o f. By 6. and 6.4 we have n u n 1, n 4 n n 1 n u n, n v n 1, n + 4 n 1 + n + 1 n v n,. Since x y x y od it follows fro Proosition.i that x/y C i if and only if x/y C i. Thus, fro 6x/y 4 od and Theore 6.1 we get { 0 od if x/y C0, and u /, v ± y x [/] od if ±x/y C 1 { /, [/] od if x/y C 0, [/] od if x/y C 0. Fro the roof of Theore 5. we see that and that x/y C 0 9 xyx y x/y C 1 x y, y, 4y od 9 or y x od 9.

44 4 Z. H. Sun Now, cobining the above with the facts that 1 [/] and /6 [/] 1/ yields the desired result. od Corollary 6.7. Let > be a rie. Then u /61, 4 or u /, if and only if can be reresented by x + 81y or x + xy + 41y according as 1 od or od. Reark 6.4. Let > be a rie. Using the ethod in the roof of Corollary 6.5ii one can siilarly rove that u /6, if and only if can be reresented by 16x + 81y, x + 196y, 8x + 8xy + 41y or x 8xy + 41y. Acknowledgeents. I a grateful to the referee for suggesting any additional references. References [Ba] Ph. B a r k a n, Partitions quadratiques et cyclotoie, Sé. Delange Pisot Poitou 1975, Zbl4:1007. [Bu] K. B u r d e, Zur Herleitung von Rezirozitätsgesetzen unter Benutzung von endlichen Körern, J. Reine Angew. Math. 9/ , , MR57: [Ca] C. Cailler, Sur les congruences du troisièe degré, Enseign. Math , [C] A. Cauchy, Sur la résolution des équivalences dont les odules se réduisent à des nobres reiers, Exercises de Mathéatiques 4 189, 5 9. [CG] A. Cunningha and T. Gosset, 4-tic and -bic residuacity-tables, Messenger Math , 1 0. [D] L. E. Dickson, History of the Theory of Nubers, Vol. I, Chelsea, New York, 195, [FR] E. T. Federighi and R. G. Roll, Fibonacci Quart. 1966, [HW] R. H. Hudson and K. S. Willias, Resolution of abiguities in the evaluation of cubic and quartic Jacobsthal sus, Pacific J. Math , [IR] K. Ireland and M. Rosen, A Classical Introduction to Modern Nuber Theory, Sringer, New York, 198. [J] J. G. D. Jacobi, De residuis cubicis coentatio nuerosa, J. Reine Angew. Math. 187, [K] L. Kronecker, Werke, Vol. II, 9 and ; Vol. IV, 1 19, Chelsea, New York, [L1] E. Leher, Criteria for cubic and quartic residuacity, Matheatika , 0 9. [L], On Euler s criterion, J. Austral. Math. Soc /1961, art 1, 64 70, MR1:7191. [L], On the cubic character of quadratic units, J. Nuber Theory 5 197, 85 89, MR48:71.

45 Cubic residues and nonresidues 5 [L4] E. Leher, On the quartic character of quadratic units, J. Reine Angew. Math. 68/ , [L5], On the nuber of solutions of u k + D w od, Pacific J. Math , [L6], On the quadratic character of the Fibonacci root, Fibonacci Quart , 15 18, MR9:160. [Li] H. von Lienen, Reelle kubische und biquadratische Legendre-Sybole, J. Reine Angew. Math , [R] P. Ribenboi, The Book of Prie Nuber Records, nd ed., Sringer, Berlin, 1989, [Sh] D. Shanks, On Gauss and coosition I, in: Nuber Theory and Alications edited by R. A. Mollin, Dordrecht, Boston, 1989, , MR9e: [SW] B. K. Searan and K. S. Willias, The cubic congruence x +Ax +Bx+C 0 od and binary quadratic fors, J. London Math. Soc , , MR9j: [St] L. Stickelberger, Über eine neue Eigenschaft der Diskriinanten algebraischer Zahlkörer, Verhand. I. Internat. Math. Kongress 1897, Zürich, [S1] Z. H. S u n, Cobinatorial su n k0 n k and its alications in nuber k r od theory II, J. Nanjing Univ. Biquarterly , , MR94j:1101. [S], Suleents to the theory of biquadratic residues, subitted. [SS] Z. H. Sun and Z. W. Sun, Fibonacci nubers and Ferat s last theore, Acta Arith , [We] P. J. Weinberger, The cubic character of quadratic units, Proc. 197 Nuber Theory Conference, Univ. of Colorado, 197, 41 4, MR5:1067. [W1] K. S. Willias, On Euler s criterion for cubic non-residues, Proc. Aer. Math. Soc , [W] [WH], Cubic nonresidues od, Delta , 8, MR54:5095. K. S. Willias and R. H. Hudson, Reresentation of ries by the rincial for of discriinant D when the class nuber h D is, Acta Arith , Deartent of Matheatics Huaiyin Teachers College Huaiyin 001, Jiangsu, Peole s Reublic of China E-ail: zwsun@netra.nju.edu.cn Received on and in revised for on