CONGRUENCES CONCERNING LUCAS SEQUENCES ZHI-HONG SUN

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1 Int. J. Number Theory 004, no., CONGRUENCES CONCERNING LUCAS SEQUENCES ZHI-HONG SUN School of Mathematical Sciences Huaiyin Normal University Huaian, Jiangsu 00, P.R. China htt:// Received 8 May 0 Acceted Setember 0 Published 9 November 0 Let be a rime greater than. In this aer, by using exansions and congruences for Lucas sequences and the theory of cubic residues and cubic congruences, we establish some congruences for [/4] /m and [/] /m modulo, where [x] is the greatest integer not exceeding x, and m is a rational -adic integer with m 0 mod. Keywords: Congruence; binomial coefficient; Lucas sequence; binary quadratic form. Mathematics Subject Classification 00: A07, B9, A5, E5, 05A0. Introduction Congruences involving binomial coefficients are interesting, and they are connected with Fermat quotients, Lucas sequences, Legendre olynomials and binary quadratic forms. In 006 Adamchu [] conjectured that for any rime mod, 0 mod. So far this conjecture is still oen. In 00 Z.W. Sun and Tauraso [6] roved that for any rime >, mod,

2 where a is the Jacobi symbol. In 0 the author [] roved the following conjecture of Z.W. Sun: / { 0 mod if, 5, 6 mod 7, 4x mod if x + 7y,, 4 mod 7, where is an odd rime and x and y are integers. Let Z and N be the set of integers and the set of ositive integers, resectively. For a rime let Z denote the set of those rational numbers whose denominator is not divisible by. For a, b, c Z and a rime, if there are integers x and y such that ax + bxy + cy, we briefly write that ax + bxy + cy. Let {P n x} be the Legendre olynomials given by P 0 x, P x x and n + P n+ x n + xp n x np n x n. For any rime > and t Z, in [] and [] the author showed that P [ ] t P [ 4 ] t [/] [/4] t 54 mod, t mod, 8 where [x] is the greatest integer not exceeding x. Recently the author [-] also established many congruences for / /m and / /m mod, where m Z and m 0 mod. Such congruences are concerned with binary quadratic forms. Let > 5 be a rime. In [8] Zhao, Pan and Sun obtained the congruence 6 /m 5 / mod. In [5] Z.W. Sun investigated mod for m 0 mod. He gave exlicit congruences in the cases m 6, 7, 8, 9,,, 7, Suose that > is a rime and {0,,..., }. It is easy to see that 0 mod 4 0 mod 0 mod Thus, for any m Z with m 0 mod, [/4] m / mod and m for for for [/] < <, 4 < <, < <. m / m mod. Let be a rime greater than and m Z with m 0 mod. Insired by the above wor, in this aer we study congruences for [/4] /m and [/] /m

3 modulo. Such congruences are concerned with Lucas sequences, binary quadratic forms and the theory of cubic residues and cubic congruences. As examles, we have the following tyical results:. Let be a rime such that ±, ±4 mod 7. Then [/4] 4 7 [/4] mod.. Let be a rime of the form 4 + and so c + d with c, d Z and d. Let a Z with 6a +. Then /4 4 c 4ad mod if 6a + a 6a, + 0 mod if 6a +.. Let > be a rime. If, then x [/] mod is the unique solution of the congruence x + x + 0 mod. If, then [/] mod if x + xy + 5y, 8x + 7xy + 8y, 9x 0y/y mod if x + xy + 4y, 87x + 9y/y mod if 9x + 5xy + y 9.. Congruences for [/4] /m mod For any numbers P and Q, let {U n P, Q} and {V n P, Q} be the Lucas sequences given by U 0 P, Q 0, U P, Q, U n+ P, Q P U n P, Q QU n P, Q n, V 0 P, Q, V P, Q P, V n+ P, Q P V n P, Q QV n P, Q n. It is well nown that see [7] { P + P n P P } n if P 0, U n P, Q P n P n if P 0, P + P n P P n. V n P, Q + In articular, we have. U n, n and U n a + b, ab an b n a b for a b.

4 As usual, the sequences F n U n, and L n V n, are called the Fibonacci sequence and the Lucas sequence, resectively. It is easily seen that see [, Lemma.7] or [7, ]. U n+ P, Q P U n P, Q+V n P, Q, QU n P, Q P U n P, Q V n P, Q. Lemma. [7, 4..9]. For n N we have U n+ P, Q n n + Q n P. n Lemma. [, Lemma.]. Let be an odd rime and {0,,..., [ ]}. 4 Then [ ] [ ] mod Lemma.. Let be an odd rime and {0,,..., }. Then / 4 mod. Proof. It is clear that 4 mod. Theorem.. Let be an odd rime and P, Q Z with P Q 0 mod. Then [/4] [/4] 4 P Q [/4] U 64Q + 4 Q P 4P U + Proof. Using Lemmas. and. we see that Note that [ 4 U [ 4 ]+ P, Q [/4] [ 4 ] + [ 4 ] P, Q mod. Q [ 4 ] P P, Q mod, [/4] 4 Q [ 4 ] P mod. 64 Q ] + + /. We deduce the first result. 4

5 Using Lemma. we see that V P, P /4 P + Q P P Q + P Q + Q [/4] [/4] [/4] P Q 4 4 P By aealing to [7, Lemma.ii] we have U + P, Q V 4 Q mod. 4P P [/4] P, P /4 This comletes the roof. Corollary.. Let be an odd rime. Then [/4] 4 6 mod. 4 Q mod. 4P Proof. Taing P and Q in Theorem. and then alying. we deduce the result. Theorem.. Let be an odd rime and x Z with x 0, mod. i If mod 4, then [/4] 4 x x 4 6 ii If mod 4, then [/4] 4 6x 4 x [/4] [/4] 4 6x 4 Proof. If mod 4, by Theorem. we have [/4] 4 x 6 U +, x 4 x 4 5 mod. mod. 6 x [/4] 4 mod. 6x

6 If mod 4, from [7, Lemma.i] we now that U, x U, x mod. Now alying Theorem. and the fact /4 we deduce that [/4] 4 6x x [/4] U [/4] x [ 4 ] x [ 4 ], x x [/4] U, x 4 mod. 6 x So the theorem is roved. Theorem.. Let be an odd rime and P, Q Z with P QP 4Q 0 mod and Q. i If 4Q P, then [/4] P 64Q 0 mod. Q 0 mod. 4P ii If P 4Q, then [/4] Proof. Since Q, it is well nown [] that U P 4Q / This together with Theorem. yields the result. P, Q 0 mod. As an examle, taing P and Q in Theorem.i we see that. / mod for any rime 7 mod 8. Theorem.4. Let be an odd rime. Then [/4] mod if mod 8, 8 4 mod if mod 8, 0 mod if 5 mod 8, mod if 7 mod 8. Proof. Taing P and Q in Theorem. we obtain [/4] 4 6 U + /, mod. Now alying [4, Theorem.] or [7,.7-.8] we deduce the result. Theorem.5. Let > 5 be a rime. Then [/4] 4 [ +5 0 ] 5 [ 4 ] mod if,, 7, 9 mod 0, 64 [ +5 0 ] 5 4 mod if, 9 mod 0, 0 mod if, 7 mod 0 6

7 and [/4] 4 [ +5 0 ] 5 [ 4 ] mod if, 9,, 9 mod 0, [ +5 0 ] 5 4 mod if, 7 mod 0, 0 mod if, 7 mod 0. Proof. Taing P and Q in Theorem. we obtain [/4] 64 F + mod and [/4] 4 F + mod. Now alying [4, Corollaries - and Theorem ] we deduce the result. Theorem.6. Let be an odd rime with 7. i If mod 4, then.4 [/4] 4 0 mod if ±, ±5, ±6, ±7 mod 7, 7 /4 mod if ±, ±4 mod 7, 7 /4 mod if ±, ±8 mod 7. ii If mod 4, then 7 /4 mod if ±, ±4 mod 7, [/4] 4 7 /4 mod if ±, ±8 mod 7, 4 7 /4 mod if ±, ±5 mod 7, 4 7 /4 mod if ±6, ±7 mod 7. Proof. Taing P 8 and Q in Theorem. we see that By., U + [/4] 4 8, 4U U + 8, mod. 8, + V 8,. From the above and [0, Corol- lary 4.5] we deduce the result. Lemma.4 [5, Lemma.4]. Let be an odd rime and P, Q Z with QP 0 mod. If Q and c Q mod for c Z, then and U + P, Q P c mod if P 4Q, 0 mod if P 4Q 0 mod if P 4Q, U P, Q P c mod if P 4Q. c 7

8 Theorem.7. Let be an odd rime and a Z with 6a mod. Then [/4] 4 0 mod if 6a, a 4a mod if 6a. Proof. Putting P 8a and Q in Theorem. we deduce that [/4] 4 By Lemma.4, [ 4 ] U + 8a, a [/4] U + [ 4 ] 8a 4a 8a, mod. mod if 6a, 0 mod if 6a. Now combining all the above we obtain the result. Theorem.8. Let be a rime of the form 4 + and c + d with c, d Z and d. Let b, m Z with gcdb, m and mb + 4m. Then m /4 4 b b 64m /4 4 m 4b bc + md b + 4m mod if b and b +4m, b c+md + d 8 b c + md mod if 4 b and b +4m, b + m / mc b d b 4 + m mod if 4 b and b +4m, 0 mod if b +4m. In articular, for b 8a and m we have /4 4 a a /4 c 4ad 6a a mod if 6a + 0 mod if 6a + Proof. Putting P b and Q m in Theorem. we see that U + b, m m,. 4 4 b b 4 4 m mod. 64m 4b 8

9 Now alying [0, Theorem.] we deduce the result. Theorem.9. Let be a rime of the form 4+ and a Z with +6a 6a. Let c + d with c, d Z and d. Then [/8] 8 a 4 4 Proof. Since [/8] 4a + c 4ad mod 6a + 6a if +6a, 4a mod if 6a +6a, c 4ad 6a mod if +6a, 6a + 0 mod if 6a +6a. 8 a 4 4 /4 4 a + from Theorems.7 and.8 we deduce the result.. Congruences for [/] /4 4 a, /m mod.. Lemma. [,.90]. Let > be a rime and {,,..., [ ]}. Then [ ] + [ ] mod. 7 Theorem.. Let > be a rime and a, b Z with ab 0 mod. Then [/] b a a [ ] U [ ]+ 9b, a mod. Proof. Using Lemmas. and. we see that for P, Q Z with P Q 0 mod, U [ ]+ P, Q [/] [ ] + [ ] Q [ ] P [/] Q [ P ] mod. 7Q Now taing P 9b and Q a in. we deduce the result. Lemma.. For n N we have U n, n n. Proof. Set ω + /. By., U n, ωn ω n ω ω Theorem.. Let > be a rime. Then n n. 9

10 [/] mod if ± mod 9, 7 mod if ± mod 9, 0 mod if ±4 mod 9. Proof. Taing a and b in Theorem. and then alying Lemma. we 9 deduce that [/] 7 [ ] U [ [ ]+, [ ] ] + mod. This yields the result. Remar.. Let > be a rime. By. and. we have and [/] 4 [ ] U [ 7 ]+, [ ] [ ] + [/] [ ] U [ 7 ]+, mod. mod Lemma.. Let > be a rime and P, Q Z with P Q 0 mod. Then Q U P, Q mod if P, U [ ]+ P, Q Q U P, Q mod if P. + Proof. Since [ ] + and P ± P 4Q Q, we see that U [ ]+ P, Q Since P ± P P { P + P [ P P 4Q/ ]+ P P { P + P P P / P Q P P P + P / }. Q P ± P P ± P ± P 4Q P 0 [ ]+ } P P mod,

11 from the above we have U [ ]+ P, Q Q / P {P + P 4Q P P P 4Q P If P 4Q, from the above we deduce that P P P + P } mod. If P 4Q U [ ]+ P, Q Q U +, from the above and the fact P ± P 4Q P, Q mod. Q P P 4Q/ we see that U [ ]+ P, Q Q U P, Q mod. So the lemma is roved. Theorem.. Let > be a rime and a, b Z with ab 0 mod. Then [/] b a [ ]+ U 9b, a mod if a a 8b, a [ ] U 9b, a mod if a + 8b. Proof. From Theorem. and Lemma. we see that [/] b a a [ ] a a [ ] U [ ]+ 9b, a 8b U 9b, a mod if a, a [ ] a 8b U 9b, a mod if a. + To see the result we note that [ ] and so a [ ] a [ ] a [ ] mod. Corollary.. Let > 5 be a rime. Then [/] 7 F mod if 5, F + mod if 5. Proof. Taing a and b 9 in Theorem. we obtain the result.

12 Theorem.4. Let > 5 be a rime, and let ε, or 0 according as ± mod 9, ± mod 9 or ±4 mod 9. i If, 4 mod 5 and so x + 5y with x, y Z, then [/6] 6 [/] 7 ε { mod if y, 7 x 5y/0y mod if y x. ii If, 8 mod 5 and so 5x + y with x, y Z, then [/6] 6 [/] 7 ε { mod if y, 7 x + y/y mod if y x. Proof. By Theorem., [/6] 6 7 [/] 7 + [/] ε 7 + If x + 5y, 4 mod 5, by [6, Theorem 6.] we have mod. 7. F { 0 mod if y, x/5y mod if y x and L { mod if y, mod if y. If 5x + y, 8 mod 5, by [6, Theorem 6.] we have. F + { 0 mod if y, x/y mod if y x and L + { mod if y, mod if y. Note that F n± L n ± F n. From Corollary. and the above we deduce the result. Theorem.5. Let be an odd rime with,, 4, 8 mod 5. i If, 4 mod 5 and so x + 5y with x, y Z, then [/] { mod if y, x + 5y/0y mod if y x. ii If, 8 mod 5 and so 5x + y with x, y Z, then [/] { mod if y, x y/y mod if y x. Proof. It is nown that U n, F n F n L n. Thus, utting a b in Theorem. we see that [/] { U, F L mod if mod, U + +, F + + +L + mod if mod.

13 It is easily seen that F n± L n ± F n and L n± 5F n ± L n. Thus, if x + 5y, 4 mod 5, using. we see that F L 4 L F 5F If 5x + y, 8 mod 5, using. we see that F + + +L + 4 L + +F + 5F + { mod if y, L x + 5y/0y mod if y x. { mod if y, +L + x y/y mod if y x. Now combining all the above we obtain the result. Theorem.6. Let be an odd rime with. Then [/] mod if x + xy + 88y, 0x + 7xy + 0y, or x + xy + 8y, 5x + 0y/y mod if 5x + 7xy + 4y, 4x + y/y mod if 4x + 7xy + 0y 4, 5x + 8y/y mod if 5x + xy + 8y 5, 47x + 9y/y mod if 47x + 5xy + y 47. Proof. Taing b and a in Theorem. and alying. we see that [/] U, U, V, mod if, U + +, U +, + V +, mod if. Now alying [9, Corollary 6.7] we deduce the result. Theorem.7. Let be an odd rime with. Then 5 7 [/] mod if x + xy + 64y, x + xy + y, 8x + xy + 8y or 5x + 5xy + 4y, 7x + 74y/85y mod if 9x + 7xy + 4y 9, 6x + 65y/85y mod if 7x + 5xy + 0y 7, 6x + y/7y mod if 5x + 5xy + y, 99x 9y/85y mod if x + xy + 6y.

14 Proof. Taing b and a in Theorem. and alying. we see that [/] U 9, V U + +9, 9, 9U 9, mod if 85, V + Now alying [9, Corollary 6.9] we deduce the result. 9, + 9U + 9, mod if 85. Let u, v be the greatest common divisor of integers u and v. For a, b, c Z we use [a, b, c] to denote the equivalence class containing the form ax + bxy + cy. It is well nown that.4 [a, b, c] [c, b, a] [a, a + b, a + b + c] for Z. We also use Hd to denote the form class grou of discriminant d. Let ω + /. Following [6] and [9] we use a+bω m m to denote the cubic Jacobi symbol. For a rime > and Z with + 0 mod, using [6, Corollary 6.] we can easily determine ++ω. In articular, by [6, Proosition.] we have +ω. For later convenience, following [9] we introduce the following notation. Definition.. Suose u, v, d Z, dvu dv 0 and u, v. Let u dv α r W W, W and let w be the roduct of all distinct rime divisors of W. Define if d, mod 4, u, v, d if d mod 8, α > 0 and α 0, mod, otherwise, ord v+ if r and u, 9 if r and u, u, v, d if r, u and 9 u, otherwise and u, v, d u, v, d u, v, dw/u, w. Lemma.4 [9, Theorem 6. and Remar 6.]. Let > be a rime, and P, Q Z with Q and P 4Q. Assume P df d, f Z and ax + bxy + cy with a, b, c, x, y Z, a, 6 4Q/P, f and b 4ac d, where P/P, f, f/p, f, d. Then 0 mod if bf P,f P P,f +ω, a ax + by Q U / P, Q Q 6 mod if bf P,f P P,f +ω a dfy ω, ax + by Q Q 6 mod if bf P,f P P,f +ω a dfy ω 4

15 and V / P, Q Q Q Q 6 mod if bf P,f P Q 6 mod if bf P,f P P,f +ω a P,f +ω a,. Moreover, the criteria for U / P, Q and V / P, Q mod are also true when a. Theorem.8. Let > be a rime with. Then [/] mod if x + xy + 5y, 8x + 7xy + 8y, 9x 0y/y mod if x + xy + 4y, 87x + 9y/y mod if 9x + 5xy + y 9. Proof. Putting a b in Theorem. and alying. we see that [/] + 9U 9, V 9, mod if, 6 9U + 9, + V + 9, mod if. Since 69 we have and 69. Thus is reresented by some class in H 07. From the theory of reduced forms we now that H 07 {[,, 5], [8, 7, 8], [4,, ], [4,, ], [,, 6], [,, 6]}. Using.4 one can easily see that [,, 6] [, 5, 9] [9, 5, ] and [8, 7, 8] [8,, ] [,, 8]. Note that 9 + ω 9 + ω + ω,, 9 + ω + + ω ω, ω ω ω. 9 9 Since 9,, 69 by Definition., utting P 9, Q, d 69, f and in Lemma.4 and alying the above we see that 0 mod if x + xy + 5y, 8x + 7xy + 8y, 6x + y 6 mod if x U 9, + xy + 4y, 69y 58x + 5y + 6 mod if 9x + 5xy + y 9 69y and V 9, /6 mod if x + xy + 5y +/6 mod if 8x + 7xy + 8y, /6 mod if x + xy + 4y, +/6 mod if 9x + 5xy + y. 5

16 Now combining all the above with the fact / mod we deduce the result. Theorem.9. Let > be a rime with. Then [/] mod if x + xy + 70y, 9x + 9xy + 0y or 8x + xy + 9y, 5x 4y/y mod if 5x + xy + 4y 5, x 4y/y mod if 7x + xy + 0y 7, 57x 8y/y mod if 9x + 5xy + 4y 9, 05x + 7y/y mod if 5x + xy + y. Proof. Putting a and b in Theorem. and alying. we see that [/] + 9U 9, V 9, mod if, 6 9U + 9, + V + 9, mod if. Since 9 we have and 9. Thus is reresented by some class in H 79. From the theory of reduced forms we now that H 79 {[,, 70], [9, 9, 0], [,, 5], [,, 5], [5,, 4], [5,, 4], [7,, 0], [7,, 0], [4,, 8], [4,, 8], [8,, 9], [8,, 9]}. Using.4 one can easily see that [,, 5] [5,, ], [4,, 8] [4, 5, 9] [9, 5, 4], [8,, 9] [8, 9, 5] [5, 9, 8] and [9, 9, 0] [0, 9, 9] [0,, ] [,, 0]. By [6, Examle.], 9 + ω 9 + ω + ω,, ω ω ω ω ω, ω + + ω 9 + ω ω, + + ω ω, ω ω ω, ω ω ω ω ω ω Since 9,, 9 by Definition., utting P 9, Q, d 9, f and 6

17 in Lemma.4 and alying the above we see that U 9, and V 9, 0 mod if x + xy + 70y, 9x + 9xy + 0y 0x + y 9y 4x + y 9y 8x + 5y 9y 70x + y 9y or 8x + xy + 9y, + 6 mod if 5x + xy + 4y 5, 6 mod if 7x + xy + 0y 7, 6 mod if 9x + 5xy + 4y 9, + 6 mod if 5x + xy + y /6 mod if x + xy + 70y, 9x + 9xy + 0y, +/6 mod if 8x + xy + 9y, +/6 mod if 5x + xy + 4y, 5x + xy + y, /6 mod if 7x + xy + 0y, 9x + 5xy + 4y. Now combining all the above we deduce the result. Theorem.0. Let > be a rime and a Z with a4 7a. Then x [/] a mod is the unique solution of the cubic congruence 7a 4x + x + 0 mod. Proof. As a4 7a we have 8a a a4 7a. Thus utting b a in Theorem. we obtain [/] { a a a U + U + From [8, Theorem.] or [9, Remar 6.] we now that and U 9a, a 7a 4 a a V 9a, a +9a, a mod if mod, 9a, a mod if mod. a 6 x 0 + x 0 + 8a mod a 6 x 0 6a mod, where x 0 is the unique solution of the congruence X 9aX 7a 0 mod. Hence 9aU 9a, a + V 9a, a 7

18 7a 4 7a 4 a a a 6 9a x 0 + x 0 + 8a + 7a 4x 0 6a a 6 x 0 9ax 0 a mod. Now utting b a in Theorem. and alying. and the above we deduce that [/] a a a U a + 9a, a + a / 7a 4 a 9aU a7a 4 x 0 9ax 0 a mod. As x 0 9ax 0 + 7a mod we see that 9a, a + V a 6 x 0 9ax 0 a x 0 9ax 0 ax 0 + 9a a4 7ax 0 mod. 9a, a Hence [/] a a7a 4 x 0 9ax 0 a x 0 x 0 + 9a mod. Set x x 0 x 0 +9a. Then x 0 9ax x+ and 7a 4x + x + x 0 4 7a x 0 + 9a x 0 x 0 + 9a + 7ax 0 9ax 0 7a. x 0 + 9a As X x 0 mod is the unique solution of X 9aX 7a 0 mod we see that X x [/] a mod is the unique solution of 7a 4X + X + 0 mod. This roves the theorem. Remar.. Let > be a rime. By Theorems.8-.0 we have: [/] 0 mod x + xy + 5y or 8x + 7xy + 8y, [/] 0 mod 8

19 x + xy + 70y, 8x + xy + 9y or 9x + 9xy + 0y. Acnowledgment The author is suorted by the Natural Sciences Foundation of China grant no. 76. References [] A. Adamchu, Comments on OEIS A in 006, The On-Line Encycloedia of Integer Sequences, htt://oeis.org/a [] D.H. Lehmer, An extended theory of Lucas functions, Ann. Math. 90, [] Z.H. Sun, The combinatorial sum n,r mod m n and its alications in number theory I, J. Nanjing Univ. Math. Biquarterly 999, [4] Z.H. Sun, The combinatorial sum n,r mod m n and its alications in number theory II, J. Nanjing Univ. Math. Biquarterly 099, [5] Z.H. Sun, The combinatorial sum r mod m n and its alications in number theory III, J. Nanjing Univ. Math. Biquarterly 995, [6] Z.H. Sun, On the theory of cubic residues and nonresidues, Acta Arith , 9-5. [7] Z.H. Sun, Values of Lucas sequences modulo rimes, Rocy Mountain J. Math. 00, -45. [8] Z.H. Sun, Cubic and quartic congruences modulo a rime, J. Number Theory 000, [9] Z.H. Sun, Cubic residues and binary quadratic forms, J. Number Theory 4007, [0] Z.H. Sun, On the quadratic character of quadratic units, J. Number Theory 8008,95-5. [] Z.H. Sun, Congruences concerning Legendre olynomials, Proc. Amer. Math. Soc. 90, [] Z.H. Sun, Congruences involving, J. Number Theory 0, [] Z.H. Sun, Congruences concerning Legendre olynomials II, J. Number Theory 0,

20 [4] Z.H. Sun and Z.W. Sun, Fibonacci numbers and Fermat s last theorem, Acta Arith. 6099, [5] Z.W. Sun, Various congruences involving binomial coefficients and higher-order Catalan numbers, arxiv: v, htt://arxiv.org/abs/ [6] Z. W. Sun and R. Tauraso, New congruences for central binomial coefficients, Adv. in Al. Math. 4500, [7] H.C. Williams, Édouard Lucas and Primality Testing, Canadian Mathematical Society Series of Monograhs and Advanced Texts, Vol., Wiley, New Yor, 998, [8] L.L. Zhao, H. Pan and Z.W. Sun, Some congruences for the second-order Catalan numbers, Proc. Amer. Math. Soc. 800,