CONGRUENCES CONCERNING LUCAS SEQUENCES ZHI-HONG SUN

Similar documents
The Fibonacci Quarterly 44(2006), no.2, PRIMALITY TESTS FOR NUMBERS OF THE FORM k 2 m ± 1. Zhi-Hong Sun

A CRITERION FOR POLYNOMIALS TO BE CONGRUENT TO THE PRODUCT OF LINEAR POLYNOMIALS (mod p) ZHI-HONG SUN

Legendre polynomials and Jacobsthal sums

On the number of representations of n by ax 2 + by(y 1)/2, ax 2 + by(3y 1)/2 and ax(x 1)/2 + by(3y 1)/2

Some sophisticated congruences involving Fibonacci numbers

arxiv: v5 [math.nt] 22 Aug 2013

Zhi-Wei Sun Department of Mathematics, Nanjing University Nanjing , People s Republic of China

Congruences involving Bernoulli and Euler numbers Zhi-Hong Sun

#A45 INTEGERS 12 (2012) SUPERCONGRUENCES FOR A TRUNCATED HYPERGEOMETRIC SERIES

Department of Mathematics, Nanjing University Nanjing , People s Republic of China

ON THE LEAST SIGNIFICANT p ADIC DIGITS OF CERTAIN LUCAS NUMBERS

Advances in Applied Mathematics 48(2012), Constructing x 2 for primes p = ax 2 + by 2

On Arithmetic Properties of Bell Numbers, Delannoy Numbers and Schröder Numbers

p-regular functions and congruences for Bernoulli and Euler numbers

arxiv: v2 [math.nt] 9 Oct 2018

Diophantine Equations and Congruences

x 2 a mod m. has a solution. Theorem 13.2 (Euler s Criterion). Let p be an odd prime. The congruence x 2 1 mod p,

The Hasse Minkowski Theorem Lee Dicker University of Minnesota, REU Summer 2001

Nanjing Univ. J. Math. Biquarterly 32(2015), no. 2, NEW SERIES FOR SOME SPECIAL VALUES OF L-FUNCTIONS

DIVISIBILITY CRITERIA FOR CLASS NUMBERS OF IMAGINARY QUADRATIC FIELDS

CONSECUTIVE NUMBERS WITHTHESAMELEGENDRESYMBOL

CERIAS Tech Report The period of the Bell numbers modulo a prime by Peter Montgomery, Sangil Nahm, Samuel Wagstaff Jr Center for Education

SUPERCONGRUENCES INVOLVING PRODUCTS OF TWO BINOMIAL COEFFICIENTS

Math 4400/6400 Homework #8 solutions. 1. Let P be an odd integer (not necessarily prime). Show that modulo 2,

Super congruences involving binomial coefficients and new series for famous constants

BOUNDS FOR THE SIZE OF SETS WITH THE PROPERTY D(n) Andrej Dujella University of Zagreb, Croatia

Primes - Problem Sheet 5 - Solutions

f(r) = a d n) d + + a0 = 0

192 VOLUME 55, NUMBER 5

When do Fibonacci invertible classes modulo M form a subgroup?

A LLT-like test for proving the primality of Fermat numbers.

6 Binary Quadratic forms

Introduction to Arithmetic Geometry Fall 2013 Lecture #10 10/8/2013

On the Diophantine Equation x 2 = 4q n 4q m + 9

Class Numbers and Iwasawa Invariants of Certain Totally Real Number Fields

We collect some results that might be covered in a first course in algebraic number theory.

Binary quadratic forms and sums of triangular numbers

1. Introduction. = (2n)! (n N)

Apéry Numbers, Franel Numbers and Binary Quadratic Forms

Small Zeros of Quadratic Forms Mod P m

Some generalizations of a supercongruence of van Hamme

QUADRATIC RECIPROCITY

The inverse Goldbach problem

Almost All Palindromes Are Composite

Solvability and Number of Roots of Bi-Quadratic Equations over p adic Fields

A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS. 1. Abstract

Verifying Two Conjectures on Generalized Elite Primes

Zhi-Wei Sun Department of Mathematics, Nanjing University Nanjing , People s Republic of China

MAT 311 Solutions to Final Exam Practice

Representing Integers as the Sum of Two Squares in the Ring Z n

An Overview of Witt Vectors

Practice Final Solutions

#A47 INTEGERS 15 (2015) QUADRATIC DIOPHANTINE EQUATIONS WITH INFINITELY MANY SOLUTIONS IN POSITIVE INTEGERS

Applicable Analysis and Discrete Mathematics available online at HENSEL CODES OF SQUARE ROOTS OF P-ADIC NUMBERS

Quadratic Residues, Quadratic Reciprocity. 2 4 So we may as well start with x 2 a mod p. p 1 1 mod p a 2 ±1 mod p

p-adic Properties of Lengyel s Numbers

Arithmetic Consequences of Jacobi s Two-Squares Theorem

MATH 3240Q Introduction to Number Theory Homework 7

Colloq. Math. 145(2016), no. 1, ON SOME UNIVERSAL SUMS OF GENERALIZED POLYGONAL NUMBERS. 1. Introduction. x(x 1) (1.1) p m (x) = (m 2) + x.

A p-adic analogue of a formula of Ramanujan

CONGRUENCES CONCERNING LEGENDRE POLYNOMIALS III

QUADRATIC RECIPROCITY

When do the Fibonacci invertible classes modulo M form a subgroup?

SOME VARIANTS OF LAGRANGE S FOUR SQUARES THEOREM

HASSE INVARIANTS FOR THE CLAUSEN ELLIPTIC CURVES

arxiv: v1 [math.nt] 23 Jan 2019

ON FREIMAN S 2.4-THEOREM

MATH342 Practice Exam

Ayşe Alaca, Şaban Alaca and Kenneth S. Williams School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada. Abstract.

Number Theory Naoki Sato

MA3H1 TOPICS IN NUMBER THEORY PART III

RECIPROCITY LAWS JEREMY BOOHER

RAMANUJAN-NAGELL CUBICS

Predictive criteria for the representation of primes by binary quadratic forms

arxiv: v1 [math.nt] 22 Jan 2019

THE LEAST PRIME QUADRATIC NONRESIDUE IN A PRESCRIBED RESIDUE CLASS MOD 4

Arithmetic and Metric Properties of p-adic Alternating Engel Series Expansions

A talk given at University of Wisconsin at Madison (April 6, 2006). Zhi-Wei Sun

Quadratic Reciprocity

Yi Wang Department of Applied Mathematics, Dalian University of Technology, Dalian , China (Submitted June 2002)

MODULAR FORMS, HYPERGEOMETRIC FUNCTIONS AND CONGRUENCES

Jonathan Sondow 209 West 97th Street, New York, New York

Genus theory and the factorization of class equations over F p

Infinitely Many Quadratic Diophantine Equations Solvable Everywhere Locally, But Not Solvable Globally

On Character Sums of Binary Quadratic Forms 1 2. Mei-Chu Chang 3. Abstract. We establish character sum bounds of the form.

t s (p). An Introduction

arxiv: v1 [math.nt] 4 Nov 2015

On the normality of p-ary bent functions

arxiv: v1 [math.nt] 9 Sep 2015

ON THE SUPERCONGRUENCE CONJECTURES OF VAN HAMME

Ohno-type relation for finite multiple zeta values

MATH 371 Class notes/outline September 24, 2013

SQUAREFREE VALUES OF QUADRATIC POLYNOMIALS COURSE NOTES, 2015

Jacobi symbols and application to primality

Mobius Functions, Legendre Symbols, and Discriminants

THE DIOPHANTINE EQUATION x 4 +1=Dy 2

THUE-VINOGRADOV AND INTEGERS OF THE FORM x 2 + Dy 2. Contents. Introduction Study of an Elementary Proof

On the smallest point on a diagonal quartic threefold

On the Solutions of the Equation x + Ax = B in with Coefficients from 3

MATH 361: NUMBER THEORY EIGHTH LECTURE

Transcription:

Int. J. Number Theory 004, no., 79-85. CONGRUENCES CONCERNING LUCAS SEQUENCES ZHI-HONG SUN School of Mathematical Sciences Huaiyin Normal University Huaian, Jiangsu 00, P.R. China zhihongsun@yahoo.com htt://www.hytc.edu.cn/xsjl/szh 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,

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, 8. 4 4 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

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.

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, 4..0-4..]. 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 [ ] + 4 4 [ ] mod. 64 4 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

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

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. /4 8 4 0 mod for any rime 7 mod 8. Theorem.4. Let be an odd rime. Then [/4] 6 8 4 mod if mod 8, 8 4 mod if mod 8, 0 mod if 5 mod 8, + 8 4 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

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

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 + 4 56a 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

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

[/] 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,

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.

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.

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.

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

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, 5 9 + ω 7 + + 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

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,, 9 9 + ω 4 + + ω 4 + + ω ω ω, 5 5 7 9 + ω + + ω 9 + ω ω, + + ω 5 5 7 7 ω, 5 9 + ω 9 + + ω ω, 9 9 9 + ω 4 + + ω 4 + + ω ω ω ω. 5 5 7 Since 9,, 9 by Definition., utting P 9, Q, d 9, f and 6

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

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

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 A066796 in 006, The On-Line Encycloedia of Integer Sequences, htt://oeis.org/a066796. [] D.H. Lehmer, An extended theory of Lucas functions, Ann. Math. 90,49-448. [] Z.H. Sun, The combinatorial sum n,r mod m n and its alications in number theory I, J. Nanjing Univ. Math. Biquarterly 999, 7-40. [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, 05-8. [5] Z.H. Sun, The combinatorial sum r mod m n and its alications in number theory III, J. Nanjing Univ. Math. Biquarterly 995, 90-0. [6] Z.H. Sun, On the theory of cubic residues and nonresidues, Acta Arith. 84998, 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, 4-89. [9] Z.H. Sun, Cubic residues and binary quadratic forms, J. Number Theory 4007, 6-04. [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, 95-99. [] Z.H. Sun, Congruences involving, J. Number Theory 0, 57-595. [] Z.H. Sun, Congruences concerning Legendre olynomials II, J. Number Theory 0, 950-976. 9

[4] Z.H. Sun and Z.W. Sun, Fibonacci numbers and Fermat s last theorem, Acta Arith. 6099, 7-88. [5] Z.W. Sun, Various congruences involving binomial coefficients and higher-order Catalan numbers, arxiv:0909.808v, htt://arxiv.org/abs/0909.808. [6] Z. W. Sun and R. Tauraso, New congruences for central binomial coefficients, Adv. in Al. Math. 4500, 5-48. [7] H.C. Williams, Édouard Lucas and Primality Testing, Canadian Mathematical Society Series of Monograhs and Advanced Texts, Vol., Wiley, New Yor, 998,. 74-9. [8] L.L. Zhao, H. Pan and Z.W. Sun, Some congruences for the second-order Catalan numbers, Proc. Amer. Math. Soc. 800, 7-46. 0