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THE UNIVERSITY OF CALGARY
A Cubic Extension of the Lucas Functions
by
Eric L. F. Roettger
A THESIS
SUBMITTED TO THE FACULTY OF GRADUATE STUDIES
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
DEPARTMENT OF MATHEMATICS AND STATISTICS
CALGARY, ALBERTA
January, 2009
c© Eric L. F. Roettger 2009
THE UNIVERSITY OF CALGARY
FACULTY OF GRADUATE STUDIES
The undersigned certify that they have read, and recommend to the Faculty of
Graduate Studies for acceptance, a thesis entitled “A Cubic Extension of the Lucas
Functions” submitted by Eric L. F. Roettger in partial fulfillment of the requirements
for the degree of DOCTOR OF PHILOSOPHY.
Supervisor, Dr. H. C. WilliamsDept. of Mathematics and Statistics
Co–Supervisor, Dr. S. MullerDept. of Mathematics and StatisticsUniversity of Wyoming
Dr. R. ScheidlerDept. of Mathematics and Statistics
Dr. M. BauerDept. of Mathematics and Statistics
Dr. W. TittelDept. of Physics and Astronomy
Dr. C. BallotLaboratoire Nicolas OresmeUniversite de Caen
Date
ii
Abstract
From 1876 to 1878 Lucas developed his theory of the functions Vn and Un, which
now bear his name. He was particularly interested in how these functions could
be employed in proving the primality of certain large integers, and as part of his
investigations succeeded in demonstrating that the Mersenne number 2127 − 1 is
a prime. Vn and Un can be expressed in terms of the nth powers of the zeros of
a quadratic polynomial, and throughout his writings Lucas speculated about the
possible extension of these functions to those which could be expressed in terms of
the nth powers of the zeros of a cubic polynomial. Indeed, at the end of his life he
stated that “by searching for the addition formulas of the numerical functions which
originate from recurrence sequences of the third or fourth degree, and by studying in
a general way the laws of residues of these functions for prime moduli. . . we would
arrive at important new properties of prime numbers.”
In this thesis we discuss a pair of functions that are easily expressed as certain
symmetric polynomials of the zeros of a cubic polynomial and were undoubtedly
known to Lucas. We show how their properties seem to underlie the theory that
Lucas was seeking. We do this by deriving a number of results which show how
the combinatorial and arithmetic aspects of these functions provide an extension
of Lucas’ theory. Furthermore, we develop many new results, which illustrate the
striking analogy between our functions and those of Lucas. We also argue that, while
Lucas very likely never developed this theory, it was certainly within his abilities to
do so.
iii
Acknowledgments
I am endlessly indebted to my supervisor Dr. H. C. Williams. Without him I have
no doubt this thesis would never have been completed. Beyond his valuable criticism
throughout the writing of this thesis (and the endless rounds of corrections), Hugh
provided me with much-needed guidance and counsel in all parts of my life. He also
provided me with a considerable amount of money through the years, for which I
am also properly grateful. I can sincerely say that no student has ever had a better
supervisor than I.
Thanks to my co-supervisor Dr. Siguna Muller, for her encouragement and for
providing me with such a remarkable thesis topic. Many thanks to my committee
members, Dr. Christian Ballot, Dr. Mark Bauer, Dr. Renate Scheidler, and Dr. Wolf-
gang Tittel for the time they invested in reading my thesis and suggesting changes.
I would like to thank and acknowledge the Natural Sciences and Engineering
Research Council of Canada for funding received. Thanks also to the Faculty of
Graduate Studies and the entire Department of Mathematics at the University of
Calgary.
Many thanks to my colleagues at the University of Calgary Mathematics Depart-
ment: Aaron Christie, for his proofreading; to Pieter Rozenhart, my office mate who
always helped me solve elementary number theory problems; Alan Silvester, who is a
LATEXwizard; and finally Kjell Wooding, who I must thank for all the conversations
we had over beer or, to a lesser extent, over coffee.
Finally, I owe thanks to my family. My father Joe, mother Shirley, sister Jennelle,
and brother David. I thank them for their support throughout this entire venture.
iv
Table of Contents
Approval Page ii
Abstract iii
Acknowledgments iv
Table of Contents v
1 The Problem 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Commentary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4 Previous Extensions of the Lucas Functions . . . . . . . . . . . . . . 181.5 Our Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2 Lucas Sequences 252.1 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.2 Computation of Un, Vn . . . . . . . . . . . . . . . . . . . . . . . . . . 272.3 Arithmetic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.4 Primality Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3 A New Attempt to Generalize the Lucas Sequences 463.1 De Longchamps’ Method . . . . . . . . . . . . . . . . . . . . . . . . . 463.2 Another Cubic Generalization . . . . . . . . . . . . . . . . . . . . . . 473.3 Our Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.4 Addition Formulas for Wn and Cn . . . . . . . . . . . . . . . . . . . . 563.5 Multiplication Formulas for Wn and Cn . . . . . . . . . . . . . . . . . 653.6 Calculating Generalized Lucas Sequences . . . . . . . . . . . . . . . . 70
4 Arithmetic Properties of {Cn} and {Wn} 794.1 Introductory Arithmetic Results . . . . . . . . . . . . . . . . . . . . . 794.2 Preliminary Results for the Law of Repetition for {Cn} . . . . . . . . 944.3 The Polynomial Km(x) . . . . . . . . . . . . . . . . . . . . . . . . . . 964.4 The Law of Repetition for {Cn} . . . . . . . . . . . . . . . . . . . . . 1044.5 The Law of Apparition for {Cn} . . . . . . . . . . . . . . . . . . . . . 1134.6 Solutions of the Cubic . . . . . . . . . . . . . . . . . . . . . . . . . . 115
v
5 Arithmetic Properties of {Dn} 1275.1 Preliminary Results for the Law of Repetition for {Dn} . . . . . . . . 1275.2 The Law of Repetition for {Dn} . . . . . . . . . . . . . . . . . . . . . 1305.3 The Law of Apparition for {Dn} . . . . . . . . . . . . . . . . . . . . . 142
6 Arithmetic Properties of {En} 1546.1 Preliminary Results for {En} . . . . . . . . . . . . . . . . . . . . . . 1546.2 A Law of Apparition for {En} . . . . . . . . . . . . . . . . . . . . . . 1646.3 Further Observations on the Law of Apparition for {En} . . . . . . . 167
7 Primality Testing 1827.1 An Analogue of Lucas’ Fundamental Theorem . . . . . . . . . . . . . 1827.2 The Case of T (N) = N2 +N + 1 . . . . . . . . . . . . . . . . . . . . 1887.3 The Primality of L . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1927.4 The Case of T (N) = N2 − 1 . . . . . . . . . . . . . . . . . . . . . . . 1977.5 Primality Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
8 Conclusion 2048.1 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.2 Improvements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2058.3 Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
Bibliography 207
A 215
vi
Chapter 1
The Problem
1.1 Introduction
Let P and Q be coprime integers and α, β be the zeros of the polynomial f(x) =
x2 − Px+Q where α 6= P . The Lucas functions Un and Vn are defined by:
Un = (αn − βn)/(α− β), Vn = αn + βn.
Since both Un and Vn are symmetric functions of the zeros of a polynomial with
integer coefficients they must be integers for all non-negative integral values of n.
Furthermore, they must both satisfy the simple linear recurrence:
Xn+1 = PXn −QXn−1.
Since U0 = 0, U1 = 1, V0 = 2, V1 = P , this recurrence can be used to compute Un
and Vn for any integral value of n.
From 1876 until about 1880, Edouard Lucas discovered many properties of these
functions. Indeed, it was during this period that he used these properties to develop
tests for the primality of large integers, including what is now called the Lucas-
Lehmer test for the primality of Mersenne numbers. (See section 5.4 of [Wil98].)
These tests were usually sufficiency tests, which could be used to prove whether a
number N of a certain special form is a prime. As Lucas well realized these test
were quite novel for their time, because instead of having to trial divide N by a large
1
2
number of integers, for example all the primes less than√N , it was only necessary
to compute some integer S and test whether N | S.
It is important to recognize, however, that Lucas found many other applications
of his functions. He was particularly struck by the similarity of his functions with
the sine and cosine functions; in fact, he noticed that if i is used to denote a zero of
x2 + 1, then
Un = (2Qn/2/√−∆) sin[(ni/2) log(α/β)] and (1.1)
Vn = 2Qn/2 cos[(ni/2) log(α/β)], (1.2)
where ∆ = (α−β)2 = P 2−4Q. As both sine and cosine are singly periodic functions
with period 2π, Lucas (see Section 26 of [Luc78]) regarded Un and Vn as simply
periodic numerical functions, where for any particular modulus m, the (numerical)
period in this case is the least positive integer p such that both
Un+p ≡ Un (mod m) and Vn+p ≡ Vn (mod m)
hold.
Throughout his several papers on Un and Vn, Lucas alluded to the problem of
extending or generalizing these functions and offered various suggestions by which
this might be done. However, in spite of these ideas, he seems never to have produced
any consistent theory that was analogous to his work on the Lucas functions. The
purpose of this thesis is to provide an extension of the Lucas functions which makes
use of the zeros of a cubic polynomial and to develop a theory which is very much
analogous to that of the Lucas functions. The functions that we will discuss were
almost certainly known to Lucas and the techniques that we will employ would have
3
been available to him; thus, it is conceivable that he might have developed some or
much of this theory himself. However, we must emphasize here that the evidence
that Lucas was thinking exactly along these lines is at best circumstantial.
1.2 Sources
In much of his published work on the Un and Vn functions, Lucas mentioned the
problem of extending or generalizing them. In what follows, we present the most im-
portant quotes, which are relevant to this theme. It should be pointed out that Lucas
frequently repeated himself, so we will give only one of any repeated statements.
The first of these comes from [Luc76].
The objective that we intend in this note is to show the identity of for-
mulas concerning certain numerical functions of the roots of an equation
of the second degree with rational coefficients with those which connect
to the circular functions, and to indicate, more generally, the identity
of formulas concerning the numerical functions of the roots of an alge-
braic equation of the fourth or any degree with those which connect to the
elliptic or abelian transcendentals.
In his memoir [Luc78], the most extensive and important work which Lucas
devoted to the Lucas functions, we have several interesting quotes.
This memoir has as its objective the study of symmetric functions of the
roots of an equation of the second degree, and its application to the theory
of prime numbers. We will first show the complete analogy of these sym-
metric functions with the circular and hyperbolic functions. We then show
4
the connection that exists between the symmetric functions and the theory
of determinants, combinations, continued fractions, divisibility, divisors
of quadratic forms, continued radicands, division of the circumference of
the circle, indeterminate analysis of the second degree, quadratic residues,
decomposition of large numbers into their prime factors, etc. This method
is the point of departure of a more complete study, of the properties of
the symmetric functions of the roots of an algebraic equation of any de-
gree with rational coefficients, in their relation to the theories of elliptic
and Abelian functions, of power residues, and indeterminate analysis of
higher degrees. (Section 1)
We will complete this section with the proof of formulas of extreme im-
portance, because these will serve later as a basis for the theory of the
numerical functions of double period, derived from the consideration of
the symmetric functions of the roots of third and fourth degree equations
with rational coefficients. (Section 9)
We see that the coefficients of the binomial raised to the power p are
integers and divisible by p, whenever p denotes a prime number, except
for the coefficients of the pth powers. On denoting by α, β, γ, . . . λ, any
n integers one has therefore
[α + β + γ + · · ·+ λ]p − [αp + βp + γp + · · ·+ λp] ≡ 0 (mod p),
and, for α = β = γ = · · · = λ = 1, one obtains
np − n ≡ 0 (mod p).
5
It is this congruence which contains Fermat’s theorem that one can gen-
eralize in the following manner, which is different from Euler’s approach.
If α, β, γ, . . . λ, denote the qth powers of the roots of an equation with
integral coefficients, and Sq their sum, the first part of the preceding con-
gruence represents the product by p of a symmetric function, integral and
with integral coefficients, of the roots, and, as a consequence of the coef-
ficients of the proposed equation. One has therefore
Spq ≡ Spq (mod p),
and by applying the Theorem of Fermat,
Spq ≡ Sq (mod p).
The study of the prime divisors of the numerical function Sn and of some
others which are analogous is very important; one has in particular, for
n = 1 and S1 = 0, as in the equation
x3 = x+ 1,
the congruence
Sp ≡ 0 (mod p);
and thence deduces conversely that if, in the case of S1 = 0, one has
Sn divisible by p for n = p and not previously, the number p is prime.
Indeed, suppose p is equal, for example, to the product of two primes g
and h. One has
Sgh ≡ Sh (mod g)
Sgh ≡ Sg (mod h);
6
as a consequence, if one finds that
Sgh ≡ 0 (mod gh),
one will also have
Sg ≡ 0 (mod h),
Sh ≡ 0 (mod g),
and, by the demonstrated theorem,
Sg ≡ Sh ≡ 0 (mod gh).
Thus Sgh would not be the first of the numbers Sn divisible by gh. One
can obtain, in this fashion, a great many theorems serving, like that of
Wilson, to verify prime numbers. We will leave aside, for the moment,
the curious and new developments that we have thus found, in order
to consider only those which are derived from simply periodic numerical
functions. (Section 21)
We have further indicated (Sections 9 and 21) a first generalization of the
principal idea of this memoir in the study of recurrence sequences which
arise from the symmetric functions of the roots of algebraic equations of
the third and fourth degree and, more generally, of the roots of equations
of any degree with rational coefficients. One finds, in particular, in the
study of the function
Un = ∆(an, bn, cn, . . . )/∆(a, b, c, . . . )
7
in which a, b, c designate the roots of the equation, and ∆(a, b, c, . . . ), the
alternating function of the roots, or the square root of the discriminant
of the equation, the generalization of the principal formulas contained in
the first part of this work. (Section 29)
In later writing concerning this memoir, Lucas [Luc80] remarked,
Since the publication of this work, the author has added to it twenty
other sections, as yet unpublished, which altogether form the arithmetical
theory of the symmetric functions of the roots of equations of the second
degree. The author hopes to find the time to write up in a similar manner
the theory of doubly periodic functions, in their connection to symmetric
functions of the roots of equations of the third and fourth degree, and with
elliptic functions.
Finally, in the year of his death, Lucas [Luc91a] wrote at much greater length
concerning the problem of generalizing his functions.
But we think that we should stress in particular research concerning
linear recurring sequences of various orders and its connection to the
theory of elliptic and abelian functions. In several papers published in
the Comptes rendues de l’association, from the meetings in Clermont,
Nancy, Paris and le Havre, in the Actes de l’Academie royale des Sci-
ences de Turin and of Saint Petersburg, in Nouvelle Correspondance
mathematique, in the Journal de Sylvester in Baltimore [American
Journal of Mathematics], etc., we have demonstrated the analogy and,
8
as it were, the identity of the circular and hyperbolic functions with the
numerical functions of the second order, whose characteristic polynomi-
als are of degree two. (See Chapters 17 and 18 in our book.) To every
trigonometric formula corresponds a formula for these functions, and
conversely.
We had hoped to find in this study, through the prime decomposition of
the expressions (an ± bn), a demonstration of Fermat’s last theorem con-
cerning the impossibility of solving in integers the indeterminate equation
xp + yp + zp = 0
in which it suffices to assume that p denotes a prime. Although Kum-
mer treated this equation masterfully some time ago, it has still not been
completely solved, since many of the exponents p cannot be dealt with by
his admirable analysis.
But if this research plan has not up to now provided the solution of this
celebrated problem, it does allow us to obtain a number of Wilsonian the-
orems, that is to say the necessary and sufficient conditions that a given
p of twenty or thirty digits must satisfy to be prime when one knows the
decomposition in prime factors of one of the numbers p ± 1. Further-
more, this method guides us to the notion of periodicity of the residues
for prime or composite moduli. Therefore, there is every reason to search
for formulas analogous to the addition and multiplication formulas for
the numerical functions which are derived from recurrences whose char-
acteristic polynomials are of degree three and four. These formulas find
9
their origin in the theory of elliptic functions, and we encounter some of
them in a beautiful memoir of Moutard.
One rediscovers these recurrence sequences by generalizing the theory of
linear substitutions, described by Serret in his Cours d’Algebre superieure
(4th edition, vol II, pp. 356-412) in a very particular form. If one con-
siders n linear homogeneous forms in n variables provided by the linear
substitutions in which the coefficients λ, µ, ν, . . . , are constants, the
forms xp+1, yp+1, zp+1, . . . , are expressed as functions of x, y, z, . . . ,
and the coefficients of the variables x, y, z, . . . , this produces a sequence
of linear recurrences having for their characteristic equation
U =
∣∣∣∣∣∣∣∣∣∣∣∣∣
λ1 − u µ1 ν1 . . .
λ2 µ2 − u ν2 . . .
λ3 µ3 ν3 − u . . .
......
.... . .
∣∣∣∣∣∣∣∣∣∣∣∣∣= 0
where u denotes the variable. The ratios of consecutive functions, or
of coefficients corresponding to two consecutive functions, have for their
limits under certain conditions of convergence, the root of largest modulus
of the equation U = 0. One can therefore generalize in an infinitude
of ways Bernoulli’s approximation technique for calculating the roots of
equations. This method is developed in the first volume of Legendre’s
Theorie des nombres, but only for a very particular case.
In Addition X Sur l’extraction des racines pour les moyennes (p. 506)
of our book, we have pointed out a new process for obtaining roots of any
10
index.
Furthermore, this process relates to the preceding theories and to lin-
ear substitutions. We think that, by developing these new methods, by
searching for the addition and multiplication formulas of the numerical
functions which originate from recurrence sequences of the third and of
the fourth degree, and by studying in a general way the laws of the residues
of these functions for prime moduli, according to their aspect, their char-
acter (cubic or biquadratic) for the discriminant of the equation U = 0,
that we would arrive at important new properties of prime numbers. And
perhaps the complete proof of Fermat’s last theorem is just a consequence
of the famous theorem of Jacobi concerning the impossibility of more than
two periods for holomorphic functions of a single complex variable.
This is pretty much all that has survived of Lucas’ writings on this problem. He
may have been contemplating doing more in later volumes of his book, Theorie des
nombres [Luc91b], only the first volume of which he completed (see Chapter 6 of
[Dec99]), but there is little evidence to suggest this. However, in the introduction to
this book, he wrote,
The theory of recurrent sequences is an inexhaustible mine which con-
tains all the properties of numbers; by calculating the successive terms of
such sequences, decomposing them into their prime factors and seeking
out by experimentation the laws of appearance and reproduction of the
prime numbers, one can advance in a systematic manner the study of the
11
properties of numbers and their application to all branches of mathemat-
ics.
After Lucas’ untimely death, there seemed to be little interest in the problem of
generalizing his functions. His old friend, C.-A. Laisant tried to kindle some interest
through a question in L’Intermediaire des mathematiciens [Lai96].
We know how much the famous theorem of Fermat concerning the im-
possibility of the identity xn + yn = zn, in integers, has so preoccupied
mathematicians. We can no longer ignore that the greater part of the
work of Ed. Lucas on the theory of numbers had for its object, direct
or indirect, the quest for a demonstration of this theorem. In particu-
lar, he published a very interesting memoir Sur la theorie des fonctions
numeriques simplement periodiques. These functions Un and Vn arise
from the equation of the second degree and present striking analogies with
the sine and cosine functions.
In seeking to generalize these ideas, Lucas later sent a Communication
to the Societe mathematique de France, which unfortunately was not
inserted, no Note having been put in by the author, on three numerical
functions, arising from the equation of the third degree, offering very great
analogies with the elliptic functions, and exhibiting the property of being
doubly periodic.
Shortly after, during a conversation that I had with him, Lucas said
to me: “if one could establish that my doubly periodic functions only
have two distinct periods, the theorem of Fermat would be proved.” And
12
during a meeting, assuming this hypothesis, he justified his statement by
a demonstration which was easy for me to follow, but of which I have
totally lost any memory, being far from suspecting his approaching death.
I recall only that the case of the exponent 2 was isolated, just as it should
be, in a very precise manner.
It is possible that Lucas had made similar Communications to other
colleagues, more favoured from the point of view of memory and more
attentive than I had been. In this case, I make an appeal to them through
the present question to consult their memories. It may also be possible
that the members of the Societe mathematique might be able to recover
the three numerical functions of which I have spoken, and concerning
which I have not found any indication in the papers left by the author.
This would assuredly be a very interesting gap to fill.
Later in 1913, Laisant [Bel24] updated his 1896 question through a response to
a letter written by D. E. Smith at the behest of E. T. Bell. Bell wanted to know
whether there was anything beyond what is asserted in the above-mentioned memoir
concerning the connection between recurring series and elliptic functions.
The mathematical papers of Lucas, after his death, were entrusted to a
commission of three members: M. Delannoy, Lemoine and myself. We
found in them the necessary elements for the publication of the last two
volumes (iii and iv) of the Recreations Mathematiques, and of the
volume Arithemique Amusant.
The remainder consisted of scattered notes which, in our estimation,
13
were not available for publication . . . . I found no trace of the subject
about which you particularly enquire, and I regret it keenly. I had studied
with great interest the memoir Sur les fonctions numeriques simplement
periodiques, and I often chatted over it with the author. These functions
U , V , derived from the equation of the second degree present curious
analogies with the sine and cosine. Lucas has also considered three func-
tions derived from the equation of the third degree, on which he once made
a communication to the Mathematical Society of France. I can find no
trace of this communication, and I have lost all memory of it. I recall
only the definition of one of these functions; it was an + bn + cn, a, b,
c, being the roots of the equation. From the point of view of periodicity
these functions exhibited the closest analogies with sn, cn, dn of elliptic
functions. They presented certain characters of double periodicity.
Lucas, in a conversation at his house, said to me: ‘if we could prove that
these functions admit only a single system of periods, Fermat’s Theorem
would be demonstrated.’ And, making this assumption, he developed this
proof for me in less than a quarter of an hour. Now I have completely
lost all recollection of it. That was some months before his death, which
I did not in the least anticipate. Since then I proposed a question on this
subject in this Intermediare. It has remained unanswered . . . . I am
more and more confirmed in my conviction that Lucas’ premature death
was an irreparable loss to the science of numbers. . . .
In response to a 1930 letter written by Duncan Harkin, Bell [Bel30] responded,
14
. . . you ask about a feasible generalization through elliptic functions of
the Functions of Lucas. If I knew how to do this, I [s]hould be far from
telling you, as I have tried my darndest to make some real progress on
it myself for the past twentyeight [sic] years. It is a tough nut of the
first order. Whoever cracks it will make a contribution to the theory of
numbers on a par with anything Fermat did. Go to it!
1.3 Commentary
On examining the material in the previous section, we note several properties of
Lucas’ investigation into his functions and those that he might have considered as
proper generalizations. We certainly see that he was interested in functions satisfying
linear recurring sequences; these functions should be symmetric functions of the zeros
of a defining polynomial with rational (in practice, usually integral) coefficients, and
there is more than one function to be considered. He seems to have been particularly
interested in defining polynomials of degree three or four. He indicated the need to
find addition and multiplication formulas involving these functions; this is certainly
what he did in order to prove the many properties of his own functions. His method
of approach was to use empirical methods to attempt to elucidate what the laws
of apparition and repetition for these functions would be, and from this material
he should be able, as he did in the case of Un and Vn, to derive primality testing
algorithms.
However, there was another aspect of this study: periodicity. As mentioned in
Section 1.1, Lucas was very impressed by the close analogy between his functions
15
and the circular functions—functions that are singly periodic—and derived a kind
of numerical periodicity for his functions. It is clear from what little he did write
on this matter that he considered an attempt at generalizing his functions should
begin by looking at doubly periodic functions, such as elliptic functions. It seems
that Lucas believed (probably by analogy) that the numerical functions that would
be derived through this analysis should exhibit the property of being doubly (nu-
merically) periodic. However, it is not clear what this property would have been.
Suppose we try the following definition of such a function.
Definition 1.1. Let H(x) be a function of an integer variable x such that H(x) is
also an integer. We will say that H(x) is doubly numerically periodic modulo m, if
there exists a pair of positive integers p1 and p2 such that p1 < p2, p1 does not divide
p2 and p1 is the least positive integer such that
H(n+ p1) ≡ H(n) (mod m) and H(n+ p2) ≡ H(n) (mod m)
for all sufficiently large values of n.
This seems to be the direct doubly periodic analog of a singly periodic numerical
function like Un or Vn, but no such function can exist. For suppose that for some m
such a function H(x) does exist. By the definition, we must have positive integers
p1 and p2 such that
H(n+ rp1 + sp2) ≡ H(n) (mod m)
for any fixed pair of integers r, s and all n sufficiently large. Suppose we specify
r, s to be integers such that rp1 + sp2 = d, where d = (p1, p2). Since p1 does not
16
divide p2, we must have d < p1. However, for all sufficiently large n we must have
H(n+ d) ≡ H(n) (mod m) with 0 < d < p1, a contradiction to the definition of p1.
Thus, there are no doubly period numerical functions that would be similar to
the Lucas functions. Nevertheless, Lucas’ intuition that elliptic functions would be
helpful was moving him in a very productive direction. Unfortunately, he did not
possess the mathematical knowledge, nor did such knowledge exist until the 20th
century, to take advantage of his rather vague ideas. This is explained in some detail
in Chapter 17 of [Wil98] and needs no further elaboration here. What is important
to note is that his belief that linear recurring sequences would play a role in this
approach led him to a dead end.
If, instead, we focus our attention on certain symmetric functions of the zeros α,
β, γ of a cubic polynomial which also satisfy linear recurrences, we would certainly
examine Sn = αn + βn + γn, which is what Lucas did for the particular polynomial
x3−x−1. This particular sequence S ′n is now called the Perrin sequence, and was the
focus of much attention by Adams and Shanks [AS82]. It seems first to have been
considered by Catalan in 1861 (see Chapter 8 of [Dec99]). Catalan, who denoted the
sequence by An, believed that An is or is not divisible by n according to whether
or not n is a prime. This could easily be converted into a primality test that would
execute in polynomial time in log n, but unfortunately, Catalan’s assertion is untrue.
For example, we see in [AS82] that 271441(= 5212) divides A271441; many other
examples of this phenomenon are also provided. We should remark here that some
of the work in [AS82] was later extended by Szekeres [Sze96]. Decaillot [Dec99] has
raised the interesting possibility that Lucas was aware of Catalan’s work before he
(Lucas) embarked on his work in primality testing and that it might have inspired him
17
in his investigations. However, Lucas, who knew Catalan and who is usually most
punctilious about assigning priority, nowhere mentions Catalan’s work. Also, his
result concerning this matter is more carefully stated than Catalan’s, even though the
proof is incomplete. Perhaps Lucas did not want to embarrass Catalan by mentioning
his less circumspect work.
Laisant raised the intriguing possibility that Lucas was considering three func-
tions that were symmetric functions of α, β, γ, one of which was Sn; what were
the other two functions? In his attempt to interpret Lucas’ writings, Bell [Bel24]
considered three functions, which he denoted as xn, yn, zn. These can be most easily
described by the equation
αn = xn + ynα + znα2,
with similar expressions involving β and γ. Clearly, these functions are symmetric
functions of α, β, γ. However, none of these functions is Sn; furthermore, these
functions were known to Lucas (see pp. 305–306 of [Luc91b]), who mentioned them
in a more general context without further comment. If these were the functions he
was thinking about, it seems peculiar that he would not have mentioned something
about them. Further properties of Bell’s xn, yn, zn were discussed by Ward [War31a]
and Mendelsohn [Men62].
It is possible that Lucas had intended to publish his findings concerning the
extension of his functions in one of the later volumes of Theorie des nombres. We
know that he was considering the publication of additional books in this series (see
the latter part of Chapter 6 of [Dec99]), and Harkin [Har57] has pointed out a short
table of contents for Volume II: Divisibility and Algebraic Irreducibility, Binomial
Congruences and Primitive roots. However, in response to a question raised by G.
18
de Rocquigny concerning the possible appearance of the second and third volumes
of Theorie des nombres, Delannoy, Laisant and Lemoine [DLL95] replied,
A careful examination of the papers left by Ed. Lucas has led us to this
conclusion, that contrary to our first hopes, it would be very difficult to
publish a continuation to the Theorie des nombres, of which only the
first volume has appeared. Nevertheless, the author’s notes, certain pas-
sages of his correspondence, and the reprinting of some of his little known
memoirs, would constitute an interesting volume for those interested in
the higher arithmetic. This is a project which has not been completely
abandoned, but whose realization will not be soon, whenever it does hap-
pen.
In spite of the lack of information concerning it, the problem of extending or
generalizing the Lucas functions has inspired a great deal of work. Some early
attempts at this are mentioned in Chapter XVII of the first volume of [Dic19]. In
the next section, we will briefly describe some of these and some of the more modern
investigations into this problem.
1.4 Previous Extensions of the Lucas Functions
One of the earliest attempts to extend the Lucas functions was done in 1880 by de
Longchamps [dL80]. If we put R = αβγ, where α, β, γ are the zeros of a cubic
19
polynomial f(x), de Longchamps considered Dn, En and Sn, where
RnDn = (αn + βn)(βn + γn)(γn + αn),
RnEn =(αn − βn)(βn − γn)(γn − αn)
(α− β)(β − γ)(γ − α),
Sn = αn + βn + γn,
to be the degree three recurring function analogs of the Lucas functions. Are these
the three functions that Laisant mentioned? They would certainly have been known
to Lucas because he was the session chair for the talk in which de Longchamps
presented his results. In fact he (de Longchamps) showed how to express Dn and
En in terms of the coefficients of f(x). However, Lucas would likely not have been
comfortable with the fact that the first two of these functions are not necessarily
integral. Also, as we have seen in Section 1.2, Lucas had certainly mentioned the
function ∆(αn, βn, γn)/∆(α, β, γ) that de Longchamps denoted by RnEn. This seems
to be all that de Longchamps wrote concerning this topic because the list of his papers
in [Laz07] does not contain any other paper devoted to this subject.
The next work done on this problem was that of Pierce [Pie16] in 1916. He
defined the two functions
Sm =∏
(1 + αmi ) and ∆m =∏
(1− αmi ),
where the product is taken over all the zeros of a given polynomial f(x) with integral
coefficients. Pierce obtained several number theoretic results concerning these func-
tions, particularly when the degree of f(x) is three. Later, Lehmer [Leh33] extended
some of Pierce’s results, showing, among other things, that each satisfies a linear
recurrence relation. Indeed, Lehmer [Leh71], [Leh68] made use of these functions in
20
a test, which makes use of the factors of N2 + N + 1 to demonstrate the primality
of N . Unfortunately, Pierce’s functions are difficult to compute, which means that
using them is not very practical. While Pierce’s work represents a kind of extension
of the Lucas functions, it is very unlikely that Lucas was thinking in this direction,
because nowhere in his work does he allude to anything like these functions. Also,
Pierce’s functions do not become Un and Vn when f(x) is of degree two.
In 1929, Carmichael advocated to study the functions which he denoted by Gn
and Hn. Although it appears that Carmichael was unaware of this, Lucas had
mentioned both in his published work. The function Gn occurs as Un on page 306
of [Luc91b] (also, in the cubic case Gn is the same as Bell’s zn) and Hn is the same
as ∆(an, bn, cn, . . . )/∆(a, b, c, . . . ). Carmichael stated that an investigation of the
properties of these two functions would lead to two generalizations of Lucas’ Un
function, but he did not follow up on this remark.
Lehmer [Leh30] extended the Lucas functions by replacing the parameter P by√R, where R is an integer coprime to Q; however, the resulting sequences are no
longer integers for all n. Lehmer’s functions were later generalized by Williams
[Wil76], but in spite of the successes of the theory of Lehmer’s extension and its
generalization, there is no reason to believe that this was the direction in which
Lucas was looking to extend his functions.
Carmichael [Car20], Engstrom [Eng31] and Ward [War31b], [War31c], [War33],
[War36], [War37], [War55] investigated the arithmetical theory of linear recurring
sequences, but they did not produce a set of functions which were analogous to
Lucas’ Un and Vn. One of the most important properties of Lucas’ function Un is
that it satisfies the condition of being a divisibility sequence; that is, the sequence
21
of integers {Un} (n > 0) is such that if m | n, then Um | Un. Lucas was very
aware of this property of Un and made heavy use of it in developing his theory.
Both Hall [Hal36] and Ward were interested in the problem of whether any function
satisfying a linear recurrence could also be a divisibility sequence. While they did
not succeed in answering this question completely (this was done later by Bezivin,
Petho and van der Poorten [BPvdP90]), they did show that this would be a very
unlikely property for a sequence satisfying a third order recurrence unless it was a
very uninteresting sequence, such as a special sequence satisfying Un+3 = RUn. In-
deed, one of the simplest, non-trivial, linear divisibility sequence after Lucas’ Un is
∆(αn, βn, γn)/∆(α, β, γ), where α, β, γ are the zeros of a cubic polynomial with inte-
gral coefficients. Ward, who was Bell’s PhD student, seems to have contracted Bell’s
enthusiasm for extending Lucas’ functions. In fact, he coined the term “lucasian”
for any function satisfying a linear recurrence which was also a divisibility sequence.
In [War38] he discussed two candidates for lucasian sequences, one of which we will
discuss in great detail in this work. Of all the individuals who worked on the problem
of extending the Lucas functions, he seems to have made the most progress. While
we have mentioned only a few of his publications here, there are many more that are
also of some relevance to this discussion and we urge the interested reader to consult
the list of his published papers in [Leh93].
Williams [Wil69], [Wil72a], [Wil77] generalized the Lucas functions, but while his
functions satisfy a linear recurrence, they are not symmetric functions of the zeros of
a polynomial f(x). Furthermore, they are not always integers unless the coefficients
of f(x) obey certain properties. Again, these functions do not seem to be those for
which Lucas was searching. Although in the case where f(x) is of degree three, it is
22
possible to use certain of these functions to extend (1.1) and (1.2), by making use of
the tresine and cotresine functions of Graves [Gra47].
1.5 Our Objective
While many researchers have looked directly or peripherally at the problem of ex-
tending Lucas’ functions, none of them seems to have produced the kind of results
that Lucas was seeking. In what follows, we will offer a new suggestion as to how
Lucas might have wanted to extend his functions. This is based on a very simple
variant of Longchamps’ original suggestion.
We begin with a cubic polynomial f(x) = x3 − Px2 + Qx − R, where P , Q, R
are integers and we put
δ = (α− β)(β − γ)(γ − α), ∆ = δ2 = P 2Q2 − 4Q3 − 4RP 3 + 18PQR− 27R2,
where α, β, γ are the zeros of f(x). We will assume that δ 6= 0. We next define Cn
and Wn by
δCn = (αn − βn)(βn − γn)(γn − αn)
= (αnβ2n + βnγ2n + γnα2n)− (α2nβn + β2nγn + γ2nαn)
Wn = (αnβ2n + βnγ2n + γnα2n) + (α2nβn + β2nγn + γ2nαn).
Note that Cn is the same as Lucas’ ∆(αn, βn, γn)/∆(α, β, γ)(= RnEn) and Wn =
Ln − 2Rn, where
Ln = RnDn = (αn + βn)(βn + γn)(γn + αn).
23
Both Cn and Wn are symmetric functions of α, β, γ and are therefore integers for
all non-negative values of n. It is these functions that we will use as our extensions
of the Lucas functions Un and Vn. Observe that {Cn} is a divisibility sequence.
In Chapter 2 we will list the most important properties of the Lucas functions Un
and Vn; most of these were known to Lucas, and can be found in his memoir [Luc78].
It would be reasonable to expect that he would want to extend these results, and
this seems to be the tenor of his remarks in Section 1.2 above. In the succeeding
chapters we will develop analogous results involving Cn and Wn. These will include,
among several other items, the addition formulas, the multiplication formulas, the
laws of apparition and repetition and some primality testing results. What is most
remarkable in this entire investigation is the need for only two functions, not three.
The main tools that we will employ would have been known to Lucas. For
example, he would have needed the fundamental theorem of symmetric polynomials,
but he indicated in several places (see, for example, Section 21 of [Luc78] above),
that he was well aware of this result. We will make a great deal of use of Waring’s
theorem, but this was described in great detail by Lucas in Chapter XV of [Luc91b].
We will also use the theory of finite fields, but this would have been known (at least
the amount that he would need) to Lucas through the second volume of Serret’s
Cours d’Algebre superieure [Ser79], with which Lucas was quite familiar (see p.
vii of [Luc91b]). To develop our law of repetition, we require a small amount of
algebraic number theory to prove Theorem 4.18. Lucas might have been aware of
some of this material because he claims in part CLIX of [Luc80] that he was working,
together with a M. Tastavin, on producing a French translation of the third edition of
Dirichlet–Dedekind’s Vorlesungen uber die Zahlentheorie. Unfortunately, this volume
24
never appeared, but the result that we need could easily have been deduced by Lucas,
even though the proof might not have been completely rigorous. In the Appendix, we
provide an alternate, more elementary proof of Theorem 4.18, which Lucas should
have been able to deduce. We also make use of derivatives to establish a certain
identity that will be useful in our investigation into the law of repetition, but Lucas
often did this himself. See, for example, Section XVII of [Luc78].
Chapter 2
Lucas Sequences
Given the polynomial x2 − Px + Q, where P , Q are coprime integers, the Lucas
functions Un and Vn are defined by
Un = Un(P,Q) = (αn − βn)/(α− β), (2.1)
Vn = Vn(P,Q) = αn + βn, (2.2)
where α and β are the zeros of the given polynomial. Further let ∆ = δ2 = (α−β)2 =
P 2 − 4Q.
2.1 Identities
Lucas sequences satisfy many well-known identities, several of which will be men-
tioned herein. For further information the reader is referred to standard works such
as [Wil98] and [Rib89].
For a fixed m both Un, Vn satisfy the following equality
Xn+2m = VmXn+m −QmXn, (2.3)
where U0 = 0, U1 = 1, V0 = 2 and V1 = P .
Substituting n−m for n in (2.3) gives both
Un+m = VmUn −QmUn−m and Vn+m = VmVn −QmVn−m. (2.4)
25
26
If the mth and nth terms are known, the (m + n)th term of a Lucas function
may be found using the addition formulas presented below:
2Um+n = VmUn + UmVn, (2.5)
2Vm+n = VmVn + ∆UmUn. (2.6)
Now, if the facts QnU−n = −Un and QnV−n = Vn are used in (2.5) and (2.6), the
following subtraction formulas can be derived:
2QmUn−m = UnVm − VnUm, (2.7)
2QmVn−m = VnVm −∆UnUm. (2.8)
Subtracting (2.5) and (2.7) yields
Un+m = VnUm +QmUn−m. (2.9)
Doing the same with (2.6) and (2.8) gives
Vm+n = ∆UnUm +QmVn−m. (2.10)
Furthermore, by writing Un, Vn, ∆ in terms of α, β, it is clear that
V 2n −∆U2
n = 4Qn (2.11)
A doubling formula for Un follows from (2.5) and a doubling formula for Vn follows
from (2.6) and (2.11) to yield
U2n = VnUn, (2.12)
V2n = V 2n − 2Qn = ∆U2
n + 2Qn. (2.13)
27
Of key importance to later sections, the following multiplication formulas for Un
and Vn can be obtained by use of the fact that Vn+δUn = 2αn. From this we see that
2m−1[Vmn + δUmn] = [Vn + δUn]m and then expanding using the binomial theorem,
we obtain
2m−1Umn =
b(m−1)/2c∑i=0
(m
2i+ 1
)∆iU2i+1
n V m−2i−1n , (2.14)
2m−1Vmn =
bm/2c∑i=0
(m
2i
)∆iU2i
n Vm−2in . (2.15)
2.2 Computation of Un, Vn
Often we are interested in calculating Un or Vn, for some particular value of n.
Although this can clearly be done via the formulas (2.1), (2.2) or equation (2.3),
both methods are too slow for practical purposes. A faster method is presented
here.
Let (b0b1 . . . bk)2 be the binary representation of m ∈ Z+ such that b0 = 1,
bi ∈ {0, 1} for (1 ≤ i ≤ k) and k = blog2mc. The following formulas for Lucas
sequences depend on identities (4.2.22) and (4.2.24) from [Wil98]:
U2n = 2Un+1Un − PU2n,
U2n+1 = U2n+1 −QU2
n,
U2n+2 = PU2n+1 − 2QUnUn+1.
Now, if P0 = {1, P} and
Pi+1 =
{2AB − PA2, B2 −QA2} if bi+1 = 0,
{B2 −QA2, PB2 − 2QAB} if bi+1 = 1,
28
where Pi = {A,B}, then Pk = {Um, Um+1}. Moreover, one may use Pk to compute
Vm, as
Vm = 2Um+1 − PUm.
Hence Um (Vm) can be computed in O(logm) multiplications and additions. Note
that this result can be employed to compute Um (Vm) (mod N) quickly by defining
Pi+1 =
{2AB − PA2, B2 −QA2} (mod N) if bi+1 = 0,
{B2 −QA2, PB2 − 2QAB} (mod N) if bi+1 = 1.
This is a more useful result as the growth of Um is exponential.
Now let
Wn ≡ Q−nV2n (mod N).
Then clearly
W1 ≡ P 2Q−1 − 2 (mod N)
and by (2.13)
W2n ≡ W 2n − 2 (mod N).
Further, by (2.4) replacing n by 2n+ 2 and m by 2n, we have
V4n+2 = V2nV2n+2 −Q2nV2;
so then
W2n+1 ≡ WnWn+1 −W1 (mod N).
In this case define P0 = {W1,W2} and
Pi+1 =
{A2 − 2, AB −W1} (mod N) if bi+1 = 0,
{AB −W1, B2 − 2} (mod N) if bi+1 = 1,
29
where Pi = {A,B}, then
Pk = {Wm,Wm+1}.
This method for finding {Wn,Wn+1} (mod N) is faster to compute than the
previous method for Un or Vn (mod N). Moreover, this method may also be used
to find a particular value of Un or Vn (mod N) as follows. First,
V2h ≡ QhWh (mod N),
and by (2.3) with m = 1 and n = 2h+ 1 we have
PV2h+1 ≡ Qh+1(Wh+1 +Wh) (mod N).
Also, by (2.10) with m = 1 and n = 2h+ 1,
∆U2h+1 ≡ Qh+1(Wh+1 −Wh) (mod N).
To complete this a formula for U2h (mod N) is needed in terms of Wh and Wh+1.
This is achieved by the use of (2.6) with m = 2h and n = 2 to see
2V2+2h = (P 2 − 2Q)V2h + ∆PU2h,
and hence
∆PU2h ≡ Qh(2QWh+1 − (P 2 − 2Q)Wh) (mod N).
2.3 Arithmetic Results
The identities from the previous section may be employed to construct arithmetic
results for Lucas sequences. The global arithmetic results presented here are stan-
dard.
30
Definition 2.1. If a and b are not both zero, then the greatest common divisor (gcd)
of a and b is defined to be the largest integer that divides both a and b, denoted by
(a, b).
Definition 2.2. If a and b are nonzero integers, then the least common multiple
(lcm) of a and b is defined to be the least positive integer l such that a | l and b | l,
denoted by [a, b].
To begin, by the use of equation (2.11) it may be shown that
(Un, Vn) | 2Qn. (2.16)
If (P,Q) = 1, then by setting m = 1 in equation (2.3) and by induction it is clear
that for n > 0, we have
(Un, Q) = (Vn, Q) = 1, (2.17)
and hence for any n ≥ 0
(Un, Vn) | 2. (2.18)
Furthermore, it is not difficult to show that {Un} is a divisibility sequence; i.e.
Um | Un, when m | n. (2.19)
Note that if n = ms, then
Un(P,Q) =αn − βn
α− β=αms − βms
α− β
=αm − βm
α− β· α
ms − βms
αm − βm= Um(P,Q) · Us(Vm, Qm).
31
Definition 2.3. Given m ∈ Z, let ω be the least positive integer, if it exists, such
that m | Uω. This value is called the rank of apparition of m, denoted by ω(m).
The next theorem is the first local theorem to be seen here. It does however have
a global result as a corollary.
Theorem 2.4. Let (Q,m) = 1 and ω = ω(m). If m | Un for some n > 0, then
ω | n.
Proof. Put
n = qω + r where 0 ≤ r < ω.
If r = 0, then we are done. Thus we assume r > 0. Also,
n = (q + 1)ω − (ω − r).
Since either q or q + 1 is even, without loss of generality, let
n = qω + s where 2 | q and |s| < ω.
Setting m = qω2
+ s and n = qω2
in (2.4) produces
Un = U( qω2
+s)+ qω2
= U qω2V qω
2+s −Q
qω2
+sU−s.
Now since m | Un and m | U qω2⇒ m | Q qω
2+sU−s. Note that s may be positive or
negative. If s > 0, then U−s = Q−sUs ⇒ m | Q qω2 Us. Hence m | U|s|. But |s| < ω,
so by the minimality of ω, it must be that s = 0⇒ ω | n.
Corollary 2.4.1. If m, n > 0 and d = (m,n), then
(Um, Un) = |Ud|.
32
Proof. Let G = (Um, Un), then Ud | G, since Ud | Um and Ud | Un. Let ω = ω(G)
be the rank of apparition of G. Then by Theorem 2.4 ω | m and ω | n⇒ ω | d⇒
G | Ud. Thus, G = |Ud|.
The following theorem is a global result of Carmichael, and may be found as a
corollary to Theorem 17 in [Car13].
Theorem 2.5. If m, n ≥ 1, then
(Umn/Un, Un) | m.
Proof. Let r = bm/2c, then it may be shown that
(Umn/Un, Un) | mQnr.
From the identity (4.2.41) of [Wil98]
U(2r+1)n = Un
r∑j=0
2r + 1
r − j
(r + j
r − j − 1
)Qn(r−j)∆jU2j
n
with m = 2r + 1, we get
U(2r+1)n/Un ≡ (2r + 1)Qnr (mod Un).
Thus, if 2 - m, then
(Umn/Un, Un) | mQnr.
Also, from the identity (4.2.43) of [Wil98] we can write
U2rn = Vn
r−1∑j=0
(r + j
r − j − 1
)Qn(r−j−1)∆jU2j+1
n .
So, if m = 2r, then
U2rn/Un ≡ rVnQn(r−1) (mod Un);
33
thus
(U2rn/Un, Un) = (rVnQn(r−1), Un).
Now
(rVnQn(r−1), Un) | rQn(r−1)(Vn, Un);
hence, it follows from (2.16) that
(Umn/Un, Un) | mQnr
when 2 | m. Lastly, since (P,Q) = 1, we have
(Umn/Un, Un) | m
from (2.17).
We are often interested in values of n for which a prime p divides Un. It will be
assumed that p - Q. Notice that if p | Q, then p - P and
Un ≡ P n−1 (mod p).
Thus, p | U0 and p - Un for n ≥ 1. The following theorem provides us with what is
called the law of repetition for a prime p.
Theorem 2.6. If p is a prime and for λ > 0, we have pλ 6= 2 and pλ || Un, then
pλ+1 || Upn. If pλ = 2, then pλ+1 | Upn.
Proof. Let pλ || Un for some λ ≥ 1. If p = 2, then by (2.11) 2 | Vn, and since
U2n = UnVn, we get 2λ+1 | U2n.
Now if λ > 1, and p = 2, then Q is odd, hence by (2.11) 2 || Vn; thus in this case
2λ+1 || U2n.
34
If p is an odd prime, then by (2.14) with m = p, one has
2p−1Upn ≡ pUnVp−1n (mod pλ+2).
Since p - Q, then p - Vn by (2.11); thus
pλ+1 || Upn.
Definition 2.7. Let ε(n) be the Jacobi symbol of (∆/n).
The following theorem is called the law of apparition for a prime p. Let ε = ε(p)
for the remainder of the chapter.
Theorem 2.8. If p is a prime such that p - 2Q, then p | Up−ε.
Proof. First, note that p |(pi
)for i 6= 0, p. So by (2.14) and (2.15), with n = 1 and
m = p, the following congruences exist
2p−1Up ≡ ∆p−12 (mod p), 2p−1Vp ≡ P p (mod p).
So by Fermat’s little theorem and Euler’s criterion for quadratic residuacity
Up ≡ ε (mod p), Vp ≡ P (mod p). (2.20)
Thus if ε = 0, then p | Up−ε. If ε 6= 0, then we can use (2.5), (2.6), (2.7)and (2.8) to
deduce
2Q1+ε2 Up−ε ≡ PUp − εVp (mod p), (2.21)
2Q1+ε2 Vp−ε ≡ PVp − ε∆Up (mod p). (2.22)
Thus by (2.20) p | Up−ε when p is odd.
35
In the sequel we will need the following result.
Vp−ε ≡ 2Q1−ε2 (mod p). (2.23)
This follows easily from (2.22).
We have similar arithmetic results for {Vn}; many of these were possibly not
known to Lucas, but might have appeared in the subsequent literature (see, for
example, [Mul01]). In any event, we make no claims of originality of these results.
Observe the following short lemma that will be called upon in the next two theorems.
Lemma 2.9. If 2 | P , then 2 | Vn for all n ≥ 0. If 2 - P and 2 | Q, then 2 | Vn
only for n = 0. If 2 - P and 2 - Q, then 2 | Vn ⇔ 3 | n.
Proof. Certainly, if 2 | P , then 2 | Vk for all k ≥ 0. If 2 - P , then since V1 = P and
Vk+1 ≡ PVk (mod Q), we see that if 2 | Q, then 2 - Vn for n > 0. If 2 - P and 2 - Q,
it follows by using induction on (2.3) that 2 | Vk if and only if 3 | k.
Note that from the above lemma we can easily see that if 2 | Vn, then 2 | Vtn for
all t ∈ N. It is known that {Un} is a divisibility sequence, but this is not necessarily
true for {Vn}; however, we have the following weaker results provided by the next
two theorems.
Theorem 2.10. If m | n and 2 - nm
, then Vm | Vn.
Proof. Since m | n and 2 - nm
, then n = km where k odd, i.e. k = 2r + 1 for some
r ∈ Z. From (2.6)
2Vn = 2Vkm = 2V(2r+1)m = 2V2rm+m = VmV2rm + ∆UmU2rm.
36
Since V 2m −∆U2
m = 4Qm, if 2 | Vm, then 2 | ∆Um. Also, Vm | U2m and U2m | U2rm
implies Vm | U2rm, hence, Vm | 2Vkm. If 2 | Vm, then by Lemma 2.9, 2Vm | 2Vkm ⇒
Vm | Vkm. On the other hand, if 2 - Vm, then Vm | Vkm.
Note that if r | Vn and n > 0, then since (Vn, Q) = 1, it must be that (r,Q) = 1.
Also, since V−n = Vn/Qn, we may write r | V−n. This simply means that r divides
the integral numerator of the fraction V−n.
Theorem 2.11. If m | n and 2 | nm
, then (Vm, Vn) | 2.
Proof. We first employ (2.13) and (2.17) to observe that
V2m = V 2m − 2Qm ⇒ (Vm, V2m) = (2Qm, Vm) = (2, Vm).
Hence (Vm, V2m) | 2. Now assume (Vm, V2km) | 2, this is certainly true for k = 1,
then since
V(2k+2)m = VmV(2k+1)m −QmV2km,
we find that (V(2k+2)m, Vm) = (QmV2km, Vm) = (V2km, Vm). Thus the result follows
by induction.
Corollary 2.11.1. If 2 | m, then (Vn, Vmn) = (2, Vn).
The rank of apparition has been introduced for {Un}, and we might expect to
have something similar for {Vn}. But the situation may exist where r - Vn for every
n ∈ Z, hence the following modified definition for the {Vn} case is needed.
Definition 2.12. Suppose r | Vn (n > 0). Denote by ρ(r) the least positive integer
ρ such that r | Vρ.
37
In order to say something about ρ(r) for r | Vn, the result below is needed first.
Lemma 2.13. If r | Vn and r | Vm, then r | V2km+n (k ≥ 1).
Proof. By (2.6) we have
2V2km+n = V2kmVn + ∆U2kmUn.
Now r | Vm ⇒ r | U2m by (2.12). Consequently, r | U2km for any integral k ≥ 1.
Thus, since r | Vn and r | U2km we have r|2V2km+n. If 2 - r, the desired result is
obtained. On the other hand, if 2 | r, then 2 | Vm, and by Lemma 2.9, 2 | V2km.
Also, 2 | Vn and 2 | ∆Un, as V 2n + ∆U2
n = 4Qn. Hence 2r | 2V2km+n ⇒ r | V2km+n.
The theorem below is a local arithmetic result for {Vn} and is very similar to
Theorem 2.4 as seen for {Un}, though the method of proof is different.
Theorem 2.14. If r | Vn (n > 0), then ρ(r) | n.
Proof. Let 2µ || (n, ρ). Then 2µ || n or 2µ || ρ. Suppose, 2µ || n, then 2µ || d,
where d = (2ρ, n). There exist x, y ∈ Z such that
d = 2ρx+ ny ⇒ d
2µ=
2ρx
2µ+ny
2µ.
Now, 2 | 2ρ2µ
and 2 - d2µ⇒ 2 - y so by Theorem 2.10 r | Vyn. Let 2k || 2x, then
Vd = V2k 2x
2kρ+yn, and since r | Vρ again, by Theorem 2.10 r | V 2x
2kρ because 2 - 2x/2k.
Thus, r | Vd by the previous lemma. Hence, d ≥ ρ. But d | 2ρ⇒ d2µ| 2ρ
2µand since
d2µ
is odd, we get d2µ| ρ
2µwhich means that d | ρ ⇒ d = ρ. Since d | n, we have
completed the proof for this case.
38
Next suppose that 2µ || ρ and 2µ+1 | n. Put d = (ρ, 2n)⇒ 2µ || d. There exist
x, y ∈ Z such that
d = ρx+ 2ny ⇒ d
2µ=ρx
2µ+
2ny
2µ.
Here, 2 - d2µ
and 2 | 2n2µ⇒ 2 - x⇒ r | Vxρ. Let 2k || 2y (y ≥ 1). So, Vd = V2k 2ny
2k+xρ,
and since r | V 2y
2kn we get r | Vd by the previous lemma. This implies d ≥ ρ. But
d | ρ ⇒ d = ρ. Also, as d | 2n we must have ρ | 2n, which means that ρ2µ| 2n
2µ.
Since 2 - ρ2µ
, then ρ2µ| n
2µ⇒ ρ | n.
A consequence of Corollary 2.11.1 is the following helpful result which will be
called upon in three of the next four theorems.
Lemma 2.15. If r > 2 and r | Vn, then 2 - nρ(r)
.
Proof. Suppose 2 | nρ(r)
, then n = mρ, where m is even. Thus, by Corollary 2.11.1
(Vρ, Vn) = (Vρ, Vmρ) = (2, Vρ) ≤ 2.
This is a contradiction since r | (Vρ, Vn) and r > 2.
The next two theorems cover what can be said about r1r2 | Vs for some s, when
we know that r1 | Vn and r2 | Vm. The results here really depend on how many
factors of 2 the quantities m and n have and hence there are two cases: the first
case, 2µ || n and 2µ || m, is addressed in Theorem 2.16 and the second case, 2µ || n
and 2ν || n (µ 6= ν), in Theorem 2.17.
Theorem 2.16. If r1 | Vm, r2 | Vn, (r1, r2) = 1, 2µ || m and 2µ || n, then
r1r2 | V[m,n].
39
Proof. Note, [m,n]m
and [m,n]n
are both odd. It follows from Theorem 2.10 that
r1 | Vm [m,n]m
and r2 | Vn [m,n]n
.
Since (r1, r2) = 1, r1r2 | V[m,n].
Theorem 2.17. If r1 | Vm, r2 | Vn, (r1, r2) = 1, r1, r2 > 2, 2µ || m and 2ν || n
(µ 6= ν), then r1r2 - Vk for every k ∈ Z.
Proof. Let ρ1 = ρ(r1) and ρ2 = ρ(r2). Since, r1, r2 > 2, by Theorem 2.14 and Lemma
2.15,
ρ1 | m, ρ2 | n and 2 -m
ρ1
, 2 -n
ρ2
.
Thus, 2µ || ρ1, 2ν || ρ2. If r1r2 | Vs, then 2 - sρ1
and 2 - sρ2⇒ 2µ || s and 2ν || s.
This is obviously a contradiction, so the wanted result is obtained.
It has been established that for m, n > 0, (Um, Un) = |U(m,n)|. A similar, but
more complicated result exists for (Vm, Vn) and again it is dependent on the number
of factors of 2 for m and n. The two cases are covered in the next two theorems.
Theorem 2.18. If 2µ || m and 2µ || n, then
(Vm, Vn) = |V(m,n)|.
Proof. Since 2µ || m and 2µ || n, then 2 - m(m,n)
, 2 - n(m,n)
⇒ V(m,n) | Vm, V(m,n) | Vn
by Theorem 2.10. Put d = (Vm, Vn), then V(m,n) | d.
Now note that ρ(d) | m, ρ(d) | n ⇒ ρ(d) | (m,n). By Lemma 2.15, for d > 2,
it must be that 2 - mρ(d)
, 2 - nρ(d)⇒ 2 - (m,n)
ρ(d). So then, by Theorem 2.10, d | V(m,n) ⇒
(Vm, Vn) = |V(m,n)|.
40
If d = 1, then we are done. If d = 2, then 2 | Vn and 2 | Vm so that, by Lemma
2.9, either 2 | P and 2 | V(m,n), or 2 - P and 3 | (m,n). Thus 2 | V(m,n).
Theorem 2.19. If 2µ || m and 2ν || n (µ 6= ν), then (Vm, Vn) | 2.
Proof. Let d = (Vm, Vn), then for d > 2, by Lemma 2.15 one must have 2 - mρ(d)
,
2 - nρ(d)⇒ 2λ || m, 2λ || n, when 2λ || ρ(d). This is clearly a contradiction, thus
(Vm, Vn) | 2, if µ 6= ν.
The short theorem below, which was known to Lucas, is of interest because it
provides some insight into the characteristics of an odd prime p, when p | Vn.
Theorem 2.20. If p is an odd prime and p | Vn, then p ≡ ±1 (mod 2ν+1) where
2ν || n.
Proof. If p | Vn, then p - Q by (2.17). Also, by (2.11), we see that p - ∆ and p - Un.
However, we do have p | U2n by (2.12). Thus, if ω is the rank of apparition of p, then
ω | 2n by Theorem 2.4 and ω - n. Also, ω | p ± 1 by Theorem 2.8. So if 2ν || n,
then 2ν+1 | ω ⇒ p ≡ ±1 (mod 2ν+1).
We have shown, in Theorem 2.8, that for a prime p where p - 2∆Q, we have
p | Up−ε. Thus, Up−ε = U p−ε2V p−ε
2and so p | U p−ε
2or p | V p−ε
2, but not both. The
question of which one is divisible by p is answered by the following theorem, called
Euler’s criterion for the Lucas functions. This result was not known to Lucas and
was first proved in a more general setting by Lehmer [Leh30].
41
Theorem 2.21. If p is a prime such that p - 2∆Q, then
p | U p−ε2⇔ (Q/p) = 1,
p | V p−ε2⇔ (Q/p) = −1.
Proof. Setting n = p− ε in (2.13) yields
Vp−ε = V 2p−ε2
− 2Qp−ε2 .
So then by (2.23),
V 2p−ε2
≡ 2Q1−ε2 + 2Q
p−ε2 ≡ 2Q
1−ε2 + 2Q
1−ε2 Q
p−12 ≡ 2Q
1−ε2 (1 + (Q/p)) (mod p).
Thus p | V p−ε2
if and only if (Q/p) = −1.
2.4 Primality Testing
Lucas’ main purpose for his investigation into the sequences now named for him was
to find new methods for the discovery of primes. This can be seen in the following
result, which Lucas called his fundamental theorem.
Theorem 2.22. Suppose N is an odd integer. Let T = T (N) = N + 1 or N − 1. If
N | UT but N - UT/d for all d such that d < T and d | T , then N is a prime.
It was Lehmer [Leh27], who realized that this theorem could be rewritten as
follows.
Theorem 2.23. Suppose N is an odd integer. Let T = T (N) = N + 1 or N − 1. If
N | UT but N - UT/q for each prime divisor q of T , then N is a prime.
42
We also have the following corollary.
Corollary 2.23.1. Suppose N is an odd integer and T = T (N) = N + 1 or N − 1.
If N | UT and N | UT/UT/q for each prime divisor q of T , then N is a prime.
Proof. By Theorem 2.5, we have
(UT/UT/q, UT/q) | q.
Thus if N | UT/UT/q, then N - UT/q. The result follows by the theorem.
The following theorem is called the Lucas-Lehmer theorem. Lucas used a result
similar to this one to implement a primality test for Mersenne numbers.
Theorem 2.24. If N = A2n − 1, n ≥ 3, 0 < A < 2n, 2 - A, and the Jacobi symbols
(∆/N) = (Q/N) = −1, then N is a prime if and only if
N | VN+12
(P,Q).
Proof. Suppose N | VN+12
(P,Q). Let p be some prime such that p | N , then p ≡ ±1
(mod 2n) by Theorem 2.20. So p = k2n±1 for some k ∈ Z. Assume N is composite,
then without loss of generality N = pq, where p = k2n + 1, q = l2n− 1 and l, k > 0.
Thus
A2n − 1 = N = pq = (k2n + 1)(l2n − 1) = (kl2n + l − k)2n − 1;
in particular
A = kl2n − k + l.
43
Now if l ≥ k, then (kl2n−k+ l) ≥ 2n ⇒ A ≥ 2n, which is a contradiction as A < 2n.
On the other hand, if l < k, then l + 1 ≤ k and since l2n − 1 > 0 we have
kl2n − k + l = k(l2n − 1) + l ≥ (l + 1)(l2n − 1) + 1
= (l2 + l)2n − l ≥ 2n+1 − 1 ≥ 2n.
Again, this is a contradiction as A < 2n. So N is a prime.
Now suppose N is a prime. Since (∆/N) = −1, then
UN+1 ≡ 0 (mod N)
by the law of apparition Theorem 2.8. Further, since (Q/N) = −1 we have
N | VN+12
(P,Q)
by Euler’s criterion in Theorem 2.21.
Corollary 2.24.1. Suppose A = 1 and 2 - n, n ≥ 3. Put Q = −2, P ≡ 2 (mod N).
Then N is a prime if and only if
N | VN+12
(2,−2).
Proof. Since ∆ ≡ 12 (mod N), we have (∆/N) = (Q/N) = −1.
Put S0 = 4, Sj+1 = S2j − 2. Then
N | VN+12
(2,−2)⇔ N |Sn−2,
as
V2j(2,−2) = 22j−1
Sj−1.
44
Thus, if N is a Mersenne number, we have that N is a prime if and only if N | Sn−2.
It is this corollary that provides an efficient test for Mersenne primes; for further
information see [Leh35]. In fact, the largest prime ever found by hand calculation1,
M127 = 2127 − 1, was found by Lucas in 1876 using a result similar to this corollary.
This was most remarkable as M127 is a 39 digit number. Strangely Lucas seems to
have lacked confidence in this result, despite the robustness of a positive outcome. It
is believed that Lucas only performed this test once as it has been estimated that he
spent between 170 and 300 hours performing the necessary calculations. The world’s
largest known primes are still of the Mersenne form and continue to be found by use
of the Lucas-Lehmer test via the GIMPS project (the great internet Mersenne prime
search) [gim]. There are only 46 Mersenne primes known to date, the largest being
M43112609 which is an impressive 12978189 decimal digit number. For large primes of
other forms we direct the reader to the website the prime pages maintained by Boris
Iskra [Isk].
We conclude this chapter by characterizing all the values of P and Q modulo a
prime p ≡ −1 (mod 4) for which (∆p
) = (Qp
) = −1. We use the notation α to denote
the conjugate of α ∈ Q(√
∆) and we use N(α) = αα to denote the norm of α and
Tr(α) = α + α, to denote the trace of α.
Theorem 2.25. Let p be a prime such that p ≡ −1 (mod 4). There exist P , Q such
that (∆p
) = (Qp
) = −1 if and only if Q ≡ N(λ), P ≡ Tr(λ) (mod p), where λ ∈ Z[i]
and (N(λ)p
) = −1.
Proof. Suppose Q ≡ N(λ), P ≡ Tr(λ) (mod p) and (N(λ)p
) = −1. Then (Qp
) =
1Lucas did not actually write out the calculations but made a game of it, for details see [WS94]
45
(N(λ)p
) = −1 and ∆ = P 2 − 4Q = Tr(λ)2 − 4 N(λ). If λ = a + bi, then Tr(λ) = 2a
and N(λ) = a2 + b2, hence Tr(λ)2 − 4 N(λ) = −4b2 and (∆p
) = (−4b2
p) = (−1
p) = −1.
Now, suppose (∆p
) = (Qp
) = −1. We have (P2−4Qp
) = −1, thus (4Q−P 2
p) = 1.
Hence, there exists some c (mod p) such that 4Q − P 2 ≡ c2 (mod p) and hence
Q ≡ (2−1P )2 + (2−1c)2 (mod p). Putting λ = a + bi, where a ≡ 2−1P , b ≡ 2−1c
(mod p), we get λ ∈ Z[i], P ≡ Tr(λ), Q ≡ N(λ) (mod p) and (N(λ)p
) = −1.
Chapter 3
A New Attempt to Generalize the Lucas
Sequences
3.1 De Longchamps’ Method
Perhaps the oldest cubic generalization of Lucas sequences was provided by Gastone
Gohierre de Longchamps. If we let α, β, γ be the zeros of X3 − PX2 + QX − R,
where P,Q,R are integers, then we can define the following sequences suggested by
de Longchamps (1880)[dL80],
RnDn = (αn + βn)(βn + γn)(γn + αn),
RnEn = (αn − βn)(βn − γn)(γn − αn)/[(α− β)(β − γ)(γ − α)],
Sn = αn + βn + γn,
Notice that if we let δ = (α − β)(β − γ)(γ − α), then δ2 = ∆ = Q2P 2 − 4Q3 −
4RP 3 + 18PQR− 27R2, where ∆ is the discriminant of the above mentioned cubic.
Also note that α + β + γ = P , αβ + βγ + γα = Q and αβγ = R.
For the sake of clarity let us denote RnDn = Ln, RnEn = Cn and Sn = An, so
this notation will match the other generalizations. De Longchamps’ work yielded a
few interesting results, including the multiplicative formula
C2n = LnCn.
46
47
He also developed the following identities
Ln = Rn(σn + τn + 2) and δCn = Rn(σn − τn),
where
σn =αn
βn+βn
γn+γn
αnand τn =
βn
αn+αn
γn+γn
βn.
However, it should be stated that neither σn nor τn are integer sequences.
If we let Sn = αnβ2n + βnγ2n + γnα2n, Tn = α2nβn + β2nγn + γ2nαn, then
δCn = Sn − Tn and Ln = Sn + Tn + 2Rn.
Also,
SnTn = RnA3n +B3
n − 6RnAnBn + 9R2n
= RnA3n +B3n + 3R2n (see Theorem 3.5),
where Bn is defined in the next section.
3.2 Another Cubic Generalization
In an attempt to develop a theory analogous to that of Lucas functions the following
method was proposed by Williams (1998) [Wil98]. Again, there are three sequences
defined in this generalization. As in the last method, let α, β, γ be the zeros of
X3 − PX2 +QX −R, where P,Q,R are integers. Now define
An = αn + βn + γn, (3.1)
Bn = αnβn + βnγn + γnαn, (3.2)
Cn =
(αn − βn
α− β
)(βn − γn
β − γ
)(γn − αn
γ − α
). (3.3)
48
Rather than a second order linear recurrence as in Theorem 2.3 for the Lucas
case, there is the following result for An and Bn.
Theorem 3.1. The sequences An and Bn respectively satisfy the following third order
recurrence formulas,
tn+3 = Ptn+2 −Qtn+1 +Rtn,
tn+3 = Qtn+2 −RPtn+1 +R2tn.
Proof. First let
tn = c1αn + c2β
n + c3γn,
where ci are constants. Then
tn+3 = c1αn+3 + c2β
n+3 + c3γn+3
= c1αnα3 + c2β
nβ3 + c3γnγ3.
Using the fact that α, β and γ are the roots of the polynomial X3−PX2 +QX−R,
we can observe
α3 = Pα2 −Qα +R
β3 = Pβ2 −Qβ +R
γ3 = Pγ2 −Qγ +R.
49
Substituting these equalities into our original equation for tn+3 gives
tn+3 = c1αn(Pα2 −Qα +R) + c2β
n(Pβ2 −Qβ +R)
+ c3γn(Pγ2 −Qγ +R)
= P (c1αn+2 + c2β
n+2 + c3γn+2)−Q(c1α
n+1 + c2βn+1 + c3γ
n+1)
+ R(c1αn + c2β
n + c3γn)
= Ptn+2 −Qtn+1 +Rtn.
The recurrence relation for Bn follows from substituting αβ for α, βγ for β and
γα for γ in the above argument.
The recurrence relation for Cn is not as simple as that for An and Bn and will be
covered in a later section.
Some easily verified identities for generalized Lucas functions follow in the theo-
rems below. The next result is a nice generalization of the facts Un = −QnU−n and
Vn = QnV−n.
Theorem 3.2.
An = RnB−n,
Bn = RnA−n and
Cn = −R2nC−n.
Proof.
RnB−n = αnβnγn(α−nβ−n + β−nγ−n + γ−nα−n
)= γn + αn + βn = An.
50
The proof for Bn and Cn follow by the same method, that is, writing both sides of
the equation in terms of α, β and γ.
If we write (2.11) as ∆U2n = V 2
n − 4Qn, then the following theorem is a useful
generalization for this cubic case.
Theorem 3.3.
∆C2n = A2
nB2n + 18AnBnR
n − 4B3n − 4A3
nRn − 27R2n,
27∆C2n = 4(A2
n − 3Bn)3 − (27Rn + 2A3n − 9AnBn)2.
There are also doubling formulas analogous to (2.12) and (2.13) and tripling
formulas.
Theorem 3.4.
A2n = A2n − 2Bn,
B2n = B2n − 2RnAn,
C2n = Cn(AnBn −Rn).
Theorem 3.5.
A3n = A3n − 3AnBn + 3Rn,
B3n = B3n − 3RnAnBn + 3R2n,
C3n = Cn(A2nB
2n −B3
n −RnA3n).
More general than the doubling or tripling formulas, there is the following theo-
rem that provides some addition formulas for An and Bn.
51
Theorem 3.6.
An+m = AnAm − (BnAm−n −RnAm−2n)
Bn+m = BnBm −Rn(AnBm−n −RnBm−2n).
Proof.
AnAm − (BnAm−n −RnAm−2n) = (αn + βn + γn)(αm + βm + γm)
− [(αnβn + βnγn + γnαn)(αm−n + βm−n + γm−n)
− αnβnγn(αm−2n + βm−2n + γm−2n)]
= [αn+m + βn+m + γn+m + αmβn + γnαm + αnβm + γnβm + αnγm + βnγm]
− [(αmβn + αm−nβnγn + γnαm + αnβm + βmγn + γnαnβm−n
+ γm−nαnβn + βnγm + γmαn)− (αm−nβnγn + γnαnβm−n + γm−nαnβn)]
= αn+m + βn+m + γn+m = An+m.
Similar methods are used to show Bn+m = BnBm −Rn(AnBm−n −RnBm−2n).
Corollary 3.6.1.
σn+m = σnσm − τnσm−n + σm−2n
τn+m = τnτm − σnτm−n + τm−2n.
Proof. These identities follow from the previous theorem by setting R = 1 and
replacing α by α/β, β by β/γ and γ by γ/α.
52
Note that historically Corollary 3.6.1 was discovered by de Longchamps in his
original paper [dL80].
The addition identities together with the doubling and tripling formulas may
then be used to derive the following results.
Theorem 3.7.
A5n = A5n − 5A3
nBn + 5A2nR
n + 5AnB2n − 5BnR
n
B5n = B5n − 5B3
nAnRn + 5B2
nR2n + 5BnA
2nR
2n − 5AnR3n.
Proof. Replace the identities A2n, B2n and A3n from Theorem 3.5 into the addition
formula for An+m from Theorem 3.6 while setting m = 4n. Similarly use identi-
ties B2n, A2n, and B3n from Theorem 3.5 into the addition formula for Bn+m from
Theorem 3.6 setting m = 4n.
Corollary 3.7.1.
σ5n = σ5n − 5σ3
nτn + 5σnτ2n + 5σ2
n − 5τn
τ5n = τ 5n − 5τ 3
nσn + 5τnσ2n + 5τ 2
n − 5σn.
Proof. These identities follow from the previous theorem by setting R = 1 and
replacing α by α/β, β by β/γ and γ by γ/α.
3.3 Our Generalization
Let α1, α2, . . . , αm be the roots of the degree m polynomial Xm − Pm−1Xm−1 +
Pm−2Xm−2 − · · · + (−1)mP0, where Pm−1, . . . , P0 are integers. Further if we let
53
δ =∏
1≤i<j≤m(αj − αi) then ∆ = δ2 is the discriminant of the above polynomial. It
will be assumed that ∆ 6= 0. Lastly, let
V =
1 αn1 α2n1 . . . α
(m−1)n1
1 αn2 α2n2 . . . α
(m−1)n2
1 αn3 α2n3 . . . α
(m−1)n3
......
.... . .
...
1 αnm α2nm . . . α
(m−1)nm
where V is a Vandermonde matrix. Then we can define generalized Lucas sequences
of degree m as follows:
δCn = detV
=∏
1≤i<j≤m
(αnj − αni ).
Or, using the Leibniz formula,
δCn =∑σ∈Sm
sgn(σ)αn(σ(1)−1)1 . . . αn(σ(m)−1)
m
and we define Wn by
Wn =∑σ∈Sm
αn(σ(1)−1)1 . . . αn(σ(m)−1)
m ,
where Sm denotes the set of permutations of {1, 2, . . . ,m}, and sgn(σ) denotes the
sign of the permutation σ.
It can be readily verified for the case m = 2 that this generalization is, in fact,
just the historic Lucas sequence, that is, Cn = Un and Wn = Vn. Note that this
54
generalization only relies on the use of two sequences as in the original case, not the
expected three as for the cubic case.
In an effort to achieve simplicity and clarity with the new generalization, we will
restrict ourselves to the case where m = 3. As usual, let α, β, γ be the same as
in the previous generalizations where P , Q and R are their elementary symmetric
functions. We can put
δCn =(αnβ2n + βnγ2n + γnα2n
)−(α2nβn + β2nγn + γ2nαn
)and
Wn =(αnβ2n + βnγ2n + γnα2n
)+(α2nβn + β2nγn + γ2nαn
).
Theorem 3.8. For a fixed m, the sequences Cn and Wn satisfy the recurrence for-
mula
Xn+6m = a1Xn+5m − a2Xn+4m + a3Xn+3m − a4Xn+2m + a5Xn+m − a6Xn,
where
a1 = Wm, a2 = (W 2m −∆C2
m)/4 +RmWm,
a3 = Rm(W2m + 2RmWm + 2R2m), a4 = R2ma2,
a5 = R4ma1, a6 = R6m.
The proof of the above theorem follows on noting that both Cn and Wn are
linear combinations of αmβ2m, βmγ2m, γmα2m, α2mβm, β2mγm, γ2mαm and these 6
quantities are the zeros of
x6 − a1x5 + a2x
4 − a3x3 + a4x
2 − a5x+ a6.
55
Returning to de Longchamps’ work, we can make the following observations.
Letting Sn = αnβ2n+βnγ2n+γnα2n, Tn = α2nβn+β2nγn+γ2nαn as before, we have
δCn = Sn − Tn, Wn = Sn + Tn,
Ln = Sn + Tn + 2Rn and Wn = Ln − 2Rn = AnBn − 3Rn.
It is also true that
Sn = Rnσn and Tn = Rnτn.
The relations above combine to yield the important formulas
Sn = Rnσn =Wn + δCn
2and Tn = Rnτn =
Wn − δCn2
. (3.4)
One can easily verify that
W 2n −∆C2
n
4∈ Z
by noting
W 2n −∆C2
n
4= SnTn = 3R2n +RnA3n +B3n (3.5)
= RnA3n +B3
n − 6RnAnBn + 9R2n ∈ Z. (3.6)
Theorem 3.9.
R2nC−n = −Cn and R2nW−n = Wn.
Note that in the above theorem R2n is the logical analogue to Qn in the identities
QnU−n = −Un and QnV−n = Vn
for the quadratic case.
56
3.4 Addition Formulas for Wn and Cn
As in the historic generalizations for Lucas sequences, there exist addition formulas
for Cn and Wn. These formulas build on de Longchamps’ work, and are analogues
of (2.5) and (2.6).
Theorem 3.10.
2W2n+m = WnWn+m + ∆CnCn+m −Rn(WnWm −∆CnCm − 2R2mWn−m)
2C2n+m = Cn+mWn + CnWn+m −Rn(CmWn − CnWm + 2R2mCn−m).
Proof. First, it is clear that
(Wn + δCn)(Wn+m + δCn+m) = WnWn+m + δCnWn+m + δCn+mWn + ∆CnCn+m.
Using the fact that Rnσn = Wn+δCn2
we have
(Wn + δCn)(Wn+m + δCn+m) = (2Rnσn)(2Rn+mσn+m)
= 4R2n+mσnσn+m.
Corollary 3.6.1 and the fact σ−n = τn yield
σnσn+m = σ2n+m + τnσm − τn−m.
Hence
(Wn + δCn)(Wn+m + δCn+m) = 4R2n+m(σ2n+m + τnσm − τn−m)
= 4R2n+m
(W2n+m + δC2n+m
2R2n+m+Wn − δCn
2Rn
Wm + δCm2Rm
− Wn−m − δCn−m2Rn−m
)= 2W2n+m + 2δC2n+m +Rn(WnWm − δCnWm + δCmWn −∆CnCm
− 2R2mWn−m + 2δR2mCn−m).
57
Thus we may conclude
WnWn+m + δCnWn+m + δCn+mWn + ∆CnCn+m
= 2W2n+m + 2δC2n+m + δRn(−CnWm + CmWn + 2R2mCn−m)
+ Rn(WnWm −∆CnCm − 2R2mWn−m).
We next use the identity Rnτn = Wn−δCn2
and manipulate (Wn−δCn)(Wn+m−δCn+m)
with the additive identity for τ in Corollary 3.6.1. By adding and subtracting the
resulting formula from that given above, we get
2W2n+m = WnWn+m + ∆CnCn+m −Rn(WnWm −∆CnCm − 2R2mWn−m)
and
2C2n+m = Cn+mWn + CnWn+m −Rn(CmWn − CnWm + 2R2mCn−m).
There are the following special cases of the previous theorem.
Corollary 3.10.1.
2W2n = ∆C2n +W 2
n − 4RnWn,
C2n = Cn(Wn + 2Rn) = CnLn,
4W3n = 3∆C2n(Wn + 2Rn) +W 2
n(Wn − 6Rn) + 24R3n,
4C3n = Cn(∆C2n + 3W 2
n).
The next corollary is only a slight modification of the previous theorem, but it
does put the identities in a nicer form by removing the subtractions in the subscripts.
58
Corollary 3.10.2.
2Wn+3m = ∆CmCn+2m +WmWn+2m −RmWmWn+m +Rm∆CmCn+m + 2R3mWn
2Cn+3m = WmCn+2m + CmWn+2m −RmWmCn+m +RmCmWn+m − 2R3mCn.
Proof. Use Theorem 3.10 and replace n by m and m by n+m.
Theorem 3.11.
4R2n−1PQ = W 2n −∆C2
n + 2(Wn+1Cn − Cn+1Wn)− 2R(Wn+1Cn−1
− Wn−1Cn+1) + 2R2(WnCn−1 −Wn−1Cn).
Proof. Replacing n with n+ r in the equations from Theorem 3.10 returns
2W2n+2r+m = Wn+rWn+r+m + ∆Cn+rCn+r+m
− Rn+r(Wn+rWm −∆Cn+rCm − 2R2mWn+r−m) (3.7)
2C2n+2r+m = Cn+r+mWn+r + Cn+rWn+r+m
− Rn+r(CmWn+r − Cn+rWm + 2R2mCn+r−m). (3.8)
Put m = −r in (3.7) and (3.8) to obtain
2W2n+r = Wn+rWn + ∆Cn+rCn
− Rn+r(Wn+rW−r −∆Cn+rC−r − 2R−2rWn+2r)
2C2n+r = CnWn+r + Cn+rWn
− Rn+r(C−rWn+r − Cn+rW−r + 2R−2rCn+2r).
59
Then use W−r = Wr/R2r, C−r = −Cr/R2r to get
2W2n+r = Wn+rWn + ∆Cn+rCn
− Rn−rWn+rWr −Rn−r∆Cn+rCr + 2Rn−rWn+2r (3.9)
2C2n+r = CnWn+r + Cn+rWn
− Rn−rCrWn+r +Rn−rCn+rWr − 2Rn−rCn+2r. (3.10)
Setting m = 0 in (3.7) and (3.8) we get
2W2n+2r = W 2n+r + ∆C2
n+r − (W0 − 2)Rn+rWn+r (3.11)
2C2n+2r = 2Wn+rCn+r + (W0 − 2)Rn+rCn+r. (3.12)
If we put m = n in (3.8) and m = n+ 2r in the second identity in Theorem 3.10 we
get
2C3n+2r = Wn+rC2n+r + Cn+rWn+2r −Rn+rWn+rCn +Rn+rCn+rWn − 2R3n+rCr
2C3n+2r = WnC2n+2r + CnW2n+2r −RnWnCn+2r +RnCnWn+2r + 2R3nC2r.
It follows by equating the right hand sides of the previous two equations and doubling
that
2Wn+rC2n+r +2Cn+rWn+2r − 2Rn+rWn+rCn + 2Rn+rCn+rWn − 4R3n+rCr
= 2WnC2n+2r + 2CnW2n+2r − 2RnWnCn+2r + 2RnCnWn+2r + 4R3nC2r.
Rearrange this to obtain
4R3n(C2r +RrCr)
= 2Wn+rC2n+r + 2Cn+rWn+2r − 2WnC2n+2r − 2CnW2n+2r
− 2Rn+r(Wn+rCn − Cn+rWn) + 2Rn(WnCn+2r − CnWn+2r). (3.13)
60
Now, using (3.9) and (3.10), notice that
2Wn+rC2n+r + 2Cn+rWn+2r
= Wn+r(Wn+rCn + Cn+rWn +Rn−rWn+rCr +Rn−rCn+rWr − 2Rn−rCn+2r)
+ Cn+r(Wn+rWn + ∆Cn+rCn −Rn−rWn+rWr −Rn−r∆Cn+rCr + 2Rn−rWn+2r)
= Cn(W 2n+r + ∆C2
n+r) + 2WnWn+rCn+r +Rn−r(W 2n+r −∆C2
n+r)Cr
+ 2Rn−r(Wn+2rCn+r − Cn+2rWr).
Similarly by (3.11) and (3.12)
2WnC2n+2r + 2CnW2n+2r
= Cn(W 2n+r + ∆C2
n+r) + 2WnWn+rCn+r +Rn+r(W0 − 2)(WnCn+r − CnWn+r).
Using the last two identities we see
2Wn+rC2n+r + 2Cn+rWn+2r − 2WnC2n+2r − 2CnW2n+2r
= Rn−r(W 2n+r −∆C2
n+r)Cr + 2Rn−r(Wn+2rCn+r − Cn+2rWr)
+ Rn+r(W0 − 2)(CnWn+r −WnCn+r).
Now use the above to modify (3.13) as follows
4R3n(C2r +RrCr)
= Rn−r(W 2n+r −∆C2
n+r)Cr + 2Rn−r(Wn+2rCn+r − Cn+2rWr)
+ Rn+r(W0 − 2)(CnWn+r −WnCn+r)− 2Rn+r(Wn+rCn − Cn+rWn)
+ 2Rn(WnCn+2r − CnWn+2r)
= Rn−r(W 2n+r −∆C2
n+r)Cr + 2Rn−r(Wn+2rCn+r − Cn+2rWr)
+ Rn+r(W0 − 4)(CnWn+r −WnCn+r) + 2Rn(WnCn+2r − CnWn+2r).
61
Dividing both sides of this equation by Rn−r gives
4R2n+r(C2r +RrCr)
= (W 2n+r −∆C2
n+r)Cr + 2(Wn+2rCn+r − Cn+2rWr)
+ R2r(W0 − 4)(CnWn+r −WnCn+r)− 2Rr(CnWn+2r −WnCn+2r). (3.14)
Putting r = 1 and replacing n by n− 1 in (3.14) yields
4R2n−1(C2 +RC1) = W 2n −∆C2
n + 2(Wn+1Cn − Cn+1Wn)− 2R(Wn+1Cn−1
− Wn−1Cn+1) +R2(W0 − 4)(WnCn−1 −Wn−1Cn).
Noting C2 = W1 + 2R, W1 = PQ − 3R, W0 = 6 and C1 = 1 ⇒ C2 + RC1 = PQ
completes the proof.
This formula is an extension the Lucas identity (2.11)
V 2n −∆U2
n = 4Q′n
where Vn = Vn(P ′, Q′) and Un = Un(P ′, Q′). This can be justified as follows. Since
V−n = Vn/Q′n and U−n = −Un/Q′n, we see that R2 corresponds to Q′. Using the
identity
2Q′mUn−m = VmUn − UmVn
we can see
−2Q′n = Vn+1Un − Un+1Vn when m = n+ 1 and n = n,
−2Q′n−1 = VnUn−1 − UnVn−1 when m = n and n = n− 1,
−2Q′n−1P ′ = Vn+1Un−1 − Un+1Vn−1 when m = n+ 1 and n = n− 1.
62
Replacing Q′ by R2 in the above returns
Vn+1Un − Un+1Vn = −2R2n,
VnUn−1 − UnVn−1 = −2R2n−2,
Vn+1Un−1 − Un+1Vn−1 = −2R2n−2P ′.
Also note that U2 + RU1 = P ′ + R. Using the above and replacing Wm by Vm and
Cm by Um into the identity in Theorem 3.11 we see
V 2n −∆U2
n = 4R2n−1(P ′ +R)− 2(−2R2n) + 2R(−2R2n−2P ′) + 2R2(−2R2n−2)
= 4R2n.
It is not surprising that Theorem 3.11 involves 6 objects: Wn−1, Wn, Wn+1, Cn−1,
Cn, Cn+1, as one may recall that both {Wn} and {Cn} satisfy a degree 6 recurrence.
By similar methods we can develop and justify another generalization of the same
Lucas identity V 2n −∆U2
n = 4Q′n in the following theorem.
Theorem 3.12.
4R2n−1(P 2Q2 − 2Q3 − 2RP 3 + 5PQR− 6R2) =
−(W 2n −∆C2
n)W1 + 2(Wn+1Wn −∆Cn+1Cn) + 2R(Wn+1Wn−1 −∆Cn−1Cn+1)
+2R2(WnWn−1 −∆Cn−1Cn).
Proof. If we put m = n in (3.7) and m = n+ 2r in the first identity of Theorem 3.10
we get
2W3n+r = Wn+rW2n+r + ∆Cn+rC2n+r
− Rn+rWn+rWn +Rn+r∆Cn+rCn + 2R3n+rWr
63
2W3n+r = WnW2n+2r + ∆CnC2n+2r
− RnWn+2rWn +Rn∆Cn+2rCn + 2R3nW2r.
Equate the right hand sides, double, then rearrange to obtain
4R3n(W2r −RrWr) = 2Wn+rW2n+r + 2∆Cn+rC2n+r − 2WnW2n+2r − 2∆CnC2n+2r
− 2Rn+r(Wn+rWn −∆Cn+rCn)
+ 2Rn(Wn+2rWn −∆Cn+2rCn). (3.15)
Similarly, use (3.9) and (3.10) to see
2Wn+rW2n+r + 2∆Cn+rC2n+r
= (W 2n+r + ∆C2
n+r)Wn + 2∆Wn+rCn+rCn −Rn−r(W 2n+r −∆C2
n+r)Wr
+ 2Rn−r(Wn+2rWn+r −∆Cn+rCn+2r).
By (3.11) and (3.12) we have
2WnW2n+2r + 2∆CnC2n+2r
= Wn(W 2n+r + ∆C2
n+r) + 2∆Wn+rCn+rCn
− (W0 − 2)Rn+r(WnWn+r −∆CnCn+r).
Using (3.15) and the above, we have
4R2n+r(W2r +RrWr) = −(W 2n+r −∆C2
n+r)Wr + 2(Wn+2rWn+r −∆Cn+2rCn+r)
+ 2R(Wn+2rWn −∆Cn+2rCn) +R2r(W0 − 4)
(Wn+rWn −∆Cn+rCn).
64
Using the above and replacing r by 1 and n by n− 1 we have
4R2n−1(W2 +RW1) = −(W 2n −∆C2
n)W1 + 2(Wn+1Wn −∆Cn+1Cn) + 2R(Wn+1Wn−1
− ∆Cn−1Cn+1) +R2(W0 − 4)(WnWn−1 −∆Cn−1Cn).
Using the identities W2 = 12∆+ 1
2W 2
1−2RW1 and W1 = PQ−3R to show W2+RW1 =
P 2Q2 − 2Q3 − 2RP 3 + 5PQR− 6R2 completes the proof.
The formula from the above theorem is another logical extension of the Lucas
identity
V 2n −∆U2
n = 4Q′n
where Vn = Vn(P ′, Q′) and Un = Un(P ′, Q′). Again, this can be justified as follows.
Since V−n = Vn/Q′n and U−n = −Un/Q′n we see that R2 corresponds to Q′. Using
the identity
2Q′mVn−m = VnVm −∆UnUm
we can see
2Q′nV1 = Vn+1Vn −∆Un+1Un when m = n and n = n+ 1,
2Q′n−1V2 = Vn+1Vn−1 −∆Un+1Un−1 when m = n− 1 and n = n+ 1,
2Q′n−1V1 = VnVn−1 −∆UnUn−1 when m = n− 1 and n = n.
Again, replace Q′ by R2 in the above to obtain
Vn+1Vn −∆Un+1Un = 2R2nV1,
Vn+1Vn−1 −∆Un+1Un−1 = 2R2n−2V2,
VnVn−1 −∆UnUn−1 = 2R2n−2V1.
65
It is easily verified that V2−RV1 = P ′2− 2R2−RP ′. The above facts and replacing
Wm by Vm and Cm by Um into the equation in Theorem 3.12 yield
P ′(V 2n −∆U2
n) = −4R2n−1(P ′2 − 2R2 −RP ′) + 2(2R2nP ′
+ 2R(2R2n−2(P ′2 − 2R2))− 2R2(2R2n−2P ′)
= 4R2nP ′.
Replacing R2 with Q′ and dividing both sides by P ′ completes the analogy, giving
V 2n −∆U2
n = 4Q′n.
In view of the importance that the quantity Wn − 6Rn will assume in later
chapters, we also point out that from Theorem 3.3 it is easy to deduce that
(Wn − 6Rn)2 + 3∆C2n = 4(A2
n − 3Bn)(B2n − 3RnAn).
3.5 Multiplication Formulas for Wn and Cn
Theorem 3.13.
16C5n/Cn = ∆2C4n + 20R2n∆C2
n + 20Rn∆C2nWn + 10∆C2
nW2n + 80R3nWn
−20R2nW 2n − 20RnW 3
n + 5W 4n + 80R4n,
16W5n = 5∆2C4n(Wn + 2Rn) + 10∆C2
n(W 3n − 2R2nWn + 4R3n) +Wn(W 4
n
−10RnW 3n + 20R2nW 2
n + 40R3nWn − 80R4n).
Proof. Replace σn and τn with the equations in (3.4) and place them in the first
66
identity from Corollary 3.7.1 to see
W5n + δC5n
2R5n=
(Wn + δCn
2Rn
)5
− 5
(Wn + δCn
2Rn
)3(Wn − δCn
2Rn
)+ 5
(Wn + δCn
2Rn
)(Wn − δCn
2Rn
)2
+ 5
(Wn + δCn
2Rn
)2
− 5
(Wn − δCn
2Rn
).
Multiply both sides by 32R5n to see
16(W5n + δC5n) = (Wn + δCn)5 − 10R(Wn + δCn)3(Wn − δCn) + 20R2n(Wn + δCn)
(Wn − δCn)2 + 40R3n(Wn + δCn)2 − 80R4n(Wn − δCn)
= W 5n + 5δW 4
nCn + 10∆W 3nC
2n + 10δ∆W 2
nC3n + 5∆2WnC
4n + δ∆2C5
n
−10R(W 4n −∆2C4
n) + 20R2n(W 3n −∆WnC
2n − δW 2
nCn + δ∆C3n)
+40R3n(W 3n + ∆C2
n + 2δWnCn)− 80R4n(Wn − δCn).
Equating the irrational parts and rearranging completes the proof. If δ ∈ Z, then
we may use W5n−δC5n
2R5n and the second identity from Corollary 3.7.1 to complete the
proof.
A more general multiplicative result is shown in the following theorem and this
result is our analogue to (2.14) and (2.15). It is at this point where our generalization
begins to outperform the others. This is because other generalizations are missing
the necessary multiplication formulas needed in order to develop arithmetic results.
Theorem 3.14. For any integers m ≥ 0 we have
Wmn =∑ (−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!Rn(λ0+λ3)Qλ2
n Vλ1−λ2(Pn, Qn) (3.16)
CmnCn
=∑ (−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!Rn(λ0+λ3)Qλ2
n Uλ1−λ2(Pn, Qn). (3.17)
67
Here the sum is extended over the values λi ∈ Z such that
λ0, λ1, λ2, λ3 ≥ 0, λ0 + λ1 + λ2 + λ3 = m, λ1 + 2λ2 + 3λ3 = m,
Uk is the Lucas function Uk(Pn, Qn) and Pn = Wn, Qn = (W 2n −∆C2
n)/4.
Proof. First note σ1 = α/β + β/γ + γ/α =∑ri, where the sum is over the three
quantities r1 = α/β, r2 = β/γ, and r3 = γ/α. Thus σ1 is the first elementary
function of degree three involving these three terms. Also τ1 = β/α + γ/β + α/γ =∑i 6=j rirj. Thus τ1 is the second elementary function of degree three. Finally note∑i 6=j 6=k rirjrk = r1r2r3 = 1. Hence we can use Waring’s theorem (see, for example,
[Mac15]) to see that
σn = (α/β)n + (β/γ)n + (γ/α)n =∑
λ1,λ2,λ3
(−1)n+kn(k − 1)!
λ1!λ2!λ3!σλ1
1 τλ21 ,
where λ1, λ2, λ3 ≥ 0, λ1 + λ2 + λ3 = k and λ1 + 2λ2 + 3λ3 = n.
Setting λ0 = n − k so (−1)n+k = (−1)n−k = (−1)λ0 we can write the previous
identity as
σn =∑
λ0,λ1,λ2,λ3
(−1)λ0n(n− λ0 − 1)!
λ1!λ2!λ3!σλ1
1 τλ21 ,
where λ0, λ1, λ2, λ3 ≥ 0, λ0 + λ1 + λ2 + λ3 = n and λ1 + 2λ2 + 3λ3 = n.
Similarly, we can use Waring’s theorem to derive
σmn =∑
λ0,λ1,λ2,λ3
(−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!σλ1n τ
λ2n , (3.18)
τmn =∑
λ0,λ1,λ2,λ3
(−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!τλ1n σλ2
n , (3.19)
where λ0, λ1, λ2, λ3 ≥ 0, λ0 + λ1 + λ2 + λ3 = m and λ1 + 2λ2 + 3λ3 = m. This is the
sum as stated in the theorem. Now, since Smn = Rmnσmn and Tmn = Rmnτmn, we
68
obtain
Wmn = Smn + Tmn =∑
λ0,λ1,λ2,λ3
(−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!Rmn(σλ1
n τλ2n + σλ2
n τλ1n ).
Or considering the term following the coefficient we obtain
Rmn(σλ1n τ
λ2n + σλ2
n τλ1n ) = R(m−λ1−λ2)n(Rnλ1σλ1
n Rnλ2τλ2
n +Rnλ2σλ2n R
nλ1τλ1n )
= R(λ0+λ3)n(Sλ1n T
λ2n + Sλ2
n Tλ1n ).
Now we will employ some well-known results for Lucas sequences; that is,
Sλn =
(Wn + δCn
2
)λ=Vλ + δnUλ
2,
T λn =
(Wn − δCn
2
)λ=Vλ − δnUλ
2,
where U = U(Pn, Qn), V = V (Pn, Qn), ∆n = ∆C2n, δn = δCn and Pn, Qn are as
stated in the theorem.
So
Sλ1n T
λ2n + Sλ2
n Tλ1n =
Vλ1 + δnUλ1
2
Vλ2 − δnUλ2
2+Vλ2 + δnUλ2
2
Vλ1 − δnUλ1
2
=Vλ1Vλ2 − ∆nUλ1Uλ2
2.
To complete the proof of the first identity, use the following identity known for Lucas
sequences:
2QmVn−m = VnVm −∆UnUm,
replacing n = λ1, m = λ2 and ∆ = ∆n. Hence
Sλ1n T
λ2n + Sλ2
n Tλ1n = Qλ2
n Vλ1−λ2 .
69
The second identity is proven similarly by expanding δCn = Sn−Tn via Waring’s
theorem as follows,
δCmn = Smn − Tmn =∑
λ0,λ1,λ2,λ3
(−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!Rmn(σλ1
n τλ2n − σλ2
n τλ1n ).
Now, using substitutions for the U , V as before,
Rmn(σλ1n τ
λ2n − σλ2
n τλ1n ) = R(m−λ1−λ2)n(Rnλ1σλ1
n Rnλ2τλ2
n −Rnλ2σλ2n R
nλ1τλ1n )
= R(λ0+λ3)n(Sλ1n T
λ2n − Sλ2
n Tλ1n )
=Vλ1 + δnUλ1
2
Vλ2 − δnUλ2
2− Vλ2 + δnUλ2
2
Vλ1 − δnUλ1
2
=δn2
(Uλ1Vλ2 − Uλ2Vλ1).
Again, we will use an identity for Lucas sequences to complete the proof,
Uλ1Vλ2 − Uλ2Vλ1 = 2Qλ2n Uλ1−λ2
and we replace δn by δCn.
The corollary below states some special cases.
Corollary 3.14.1.
C5n/Cn = 5R4n + 5R3nPn − 5R2nQn − 5RnPnQn + P 4n − 3P 2
nQn + Q2n,
C6n/Cn = 8R3n(P 2n − Qn)− 6RnQn(P 2
n − Qn) + P 5n − 4P 3
nQn + 3PnQ2n,
C7n/Cn = 7R6n − 7R5nPn − 21R4nQn + 7R3nPn(P 2n − Qn) + 14R2nQ2
n
+7RnPnQn(2Qn − P 2n) + P 6
n − 5P 4nQn + 6P 2
nQ2n − Q3
n.
It is of interest that we can use the multiplicative identity for Cmn to show we
may calculate Cm as a sum of Lucas sequences.
70
Corollary 3.14.2.
Cm = m∑
λ0,λ1,λ2,λ3
(−1)λ0(m− λ0 − 1)!
λ1!λ2!λ3!Rλ0+λ3Qλ2
1 Uλ1−λ2(P1, Q1),
where P1 = PQ− 3R and Q1 = (P 21 −∆)/4 = RP 3 +Q3 − 6PQR + 9R2.
It is the general multiplication formulas that allow us to proceed with this cubic
generalization. With them we are able to develop arithmetic properties for Cn and
Wn in Chapter 4. Once we have arithmetic properties some primality testing can be
done.
3.6 Calculating Generalized Lucas Sequences
By Theorem 3.8 we know that both the {Wn} and {Cn} sequences satisfy the recur-
rence formula
Znm+6n = a1Znm+5n − a2Znm+4n + a3Znm+3n − a4Znm+2n + a5Znm+n − a6Znm,
where
a1 = Wn, a2 = (W 2n −∆C2
n)/4 +RnWn,
a3 = Rn(W2n + 2RnWn + 2R2n), a4 = R2na2,
a5 = R4na1, a6 = R6n.
Also, we have in Corollary 3.10.1 the result that 2W2n = ∆C2n +W 2
n − 4RnWn. Now
put
Xk =Wk
2Rkand Dk =
∆k
4R2k=
∆C2k
4R2k,
71
then
a1
Rn= 2Xn,
a2
R2n=
(W 2n −∆C2
n)
4R2n+RnWn
R2n= X2
n + 2Xn − Dn,
a3
R3n=
Rn(W2n + 2RnWn + 2R2n)
R3n=W2n
R2n+
2Wn
Rn+ 2
=∆C2
n +W 2n − 4RnWn
2R2n+
2Wn
Rn+ 2 =
∆C2n +W 2
n
2R2n− 2Wn
Rn+
2Wn
Rn+ 2
= 2∆C2
n
4R2n+ 2
W 2n
4R2n+ 2 = 2(X2
n + Dn + 1),
a4
R4n=
R2na2
R4n=
a2
R2n= X2
n + 2Xn − Dn,a5
R5n=R4na1
R5n=
a1
Rn= 2Xn,
a6
R6n= 1.
Hence
X(m+6)n = 2XnX(m+5)n − (X2n − Dn + 2Xn)X(m+4)n + 2(X2
n + Dn + 1)X(m+3)n
− (X2n − Dn + 2Xn)X(m+2)n + 2XnX(m+1)n −Xmn,
where
X0 = 3, Xn =Wn
2Rn, X2n =
W2n
2R2n=
∆C2n +W 2
n − 4RnWn
4R2n= X2
n + Dn − 2Xn,
X3n =W3n
2R3n=
W 3n
8R3n+ 3
∆C2nWn
8R3n+ 3
Rn∆C2n
4R3n− 3
RnW 2n
4R3n+
6R3n
2R3n
=W 3n
8R3n+ 3
∆C2n
4R2n
Wn
2Rn+ 3
∆C2n
4R2n− 3
W 2n
4R2n+ 3
= X3n + 3DnXn + 3Dn − 3X2
n + 3.
Also,
X−mn =W−mn2R−mn
=Wmn
R2(mn)/2R−mn =
Wmn
2Rmn= Xmn.
It follows that Xmn = Fm(Xn, Dn), where
Fm+6 = 2XnFm+5 − (X2n − Dn + 2Xn)Fm+4 + 2(X2
n + Dn + 1)Fm+3
− (X2n − Dn + 2Xn)Fm+2 + 2XnFm+1 − Fm,
72
and
F0 = 3, F1 = Xn, F2 = X2n + Dn − 2Xn, F3 = X3
n + 3DnXn + 3Dn − 3X2n + 3,
Fm(Xn, Dn) = F−m(Xn, Dn).
Furthermore, if we put Ym,n = CmnCnRmn−n
, then Ym,n = Gm(Xn, Dn) where
Gm+6 = 2XnGm+5 − (X2n − Dn + 2Xn)Gm+4 + 2(X2
n + Dn + 1)Gm+3
− (X2n − Dn + 2Xn)Gm+2 + 2XnGm+1 −Gm,
and
G0 = 0, G1 = Y1,n =Cn
CnR1−1= 1,
G2 =C2n
CnRn=CnWn
CnRn+
2RnCnCnRn
= 2Xn + 2,
G3 =C3n
CnR2n=
3CnW2n
4CnR2n+
∆C3n
4CnR2n=
3W 2n
4R2n+
∆C2n
4R2n= 3X2
n + Dn.
Also,
Y−m,n =C−mn
CnR−mn−n=−CmnR2mn
/CnR−mn−n = −CmnR
mn+n
CnR2mn= − Cmn
CnRmn−n = −Ym,n.
So
Gm(Xn, Dn) = −G−m(Xn, Dn).
Note that
Dmn =∆C2
mn
4R2mn= ∆
C2n
C2n
C2mn
4R2mn=
∆C2n
4R2n
C2mn
C2n(Rmn−n)2
= DnG2m.
Thus we have found that if we put
Xn =Wn
2Rn, Dn =
∆C2n
4R2nand Ym,n =
CmnCnRmn−n ,
73
then
Xmn = Fm(Xn, Dn), Ym,n = Gm(Xn, Dn),
where Fm, Gm ∈ Z[x, y] satisfy
Zm+6 = 2xZm+5 − (x2 + 2x− y)Zm+4 + 2(x2 + y + 1)Zm+3
− (x2 + 2x− y)Zm+2 + 2xZm+1 − Zm, (3.20)
and
F0 = 3, F1 = x, F2 = x2 − 2x+ y, F3 = x3 + 3yx+ 3y − 3x2 + 3,
G0 = 0, G1 = 1, G2 = 2x+ 2, G3 = 3x2 + y,
F−m(x, y) = Fm(x, y) and G−m(x, y) = −Gm(x, y).
Consider the equation
z3 − (x+√y)z2 + (x−√y)z − 1 = 0. (3.21)
Let λ, µ, ν be the zeros of (3.21). Then clearly λ−1, µ−1, ν−1 are the zeros of
z3 − (x−√y)z2 + (x+√y)z − 1 = 0
and therefore λ, µ, ν, λ−1, µ−1, ν−1 are the zeros of
(z3 − (x+√y)z2 + (x−√y)z − 1)(z3 − (x−√y)z2 + (x+
√y)z − 1)
= z6 − 2xz5 + (x2 + 2x− y)z4 − (2x2 + 2y + 2)z3 + (x2 + 2x− y)z2 − 2xz + 1.
Note that λµν = 1. From this, the boundary conditions on Fm and Gm and the
fact that they satisfy the recurrence (3.20), we find that
Fm = Fm(x, y) =1
2(λm + µm + νm + λ−m + µ−m + ν−m),
74
Gm = Gm(x, y) =λm + µm + νm − λ−m − µ−m − ν−m
2√y
for y 6= 0.
Since λ, µ, ν satisfy
zn+3 = (x+√y)zn+2 − (x−√y)zn+1 + zn,
z−(n+3) = (x−√y)z−(n+2) − (x+√y)z−(n+1) + z−n,
we can see that
Fn+3 =1
2(λn+3 + µn+3 + νn+3 + λ−(n+3) + µ−(n+3) + ν−(n+3))
=1
2((x+
√y)λn+2 − (x−√y)λn+1 + λn + (x+
√y)µn+2
−(x−√y)µn+1 + µn + (x+√y)νn+2 − (x−√y)νn+1
+νn + (x−√y)λ−(n+2) − (x+√y)λ−(n+1) + λ−n
+(x−√y)µ−(n+2) − (x+√y)µ−(n+1) + µ−n + (x−√y)ν−(n+2)
−(x+√y)ν−(n+1) + ν−n)
=1
2(xλn+2 +
√yλn+2 − xλn+1 +
√yλn+1 + λn
+xµn+2 +√yµn+2 − xµn+1 +
√yµn+1 + µn
+xνn+2 +√yνn+2 − xνn+1 +
√yνn+1 + νn
+xλ−(n+2) −√yλ−(n+2) − xλ−(n+1) −√yλ−(n+1) + λ−n
+xµ−(n+2) −√yµ−(n+2) − xµ−(n+1) −√yµ−(n+1) + µ−n
+xν−(n+2) −√yν−(n+2) − xν−(n+1) −√yν−(n+1) + ν−n)
= xFn+2 − xFn+1 + yGn+2 + yGn+1 + Fn.
By the same method one can easily verify that
Gn+3 = xGn+2 − xGn+1 + Fn+2 + Fn+1 +Gn.
75
If we put Hn = λn + µn + νn then by Theorem 3.6 we have that
Hn+m = HnHm −H−nHm−n +Hm−2n. (3.22)
Putting m = n+ 1 in equation (3.22), we get
H2n+1 = HnHn+1 −H−nH1 +H−n+1. (3.23)
Now using the facts
Fn =1
2(Hn +H−n) and Gn =
1
2√y
(Hn −H−n),
yields
Hn = Fn +√yGn and H−n = Fn −
√yGn.
Substituting this into (3.23) we get
F2n+1 = FnFn+1 + yGnGn+1 − xFn + yGn + Fn−1,
G2n+1 = GnFn+1 +Gn+1Fn + xGn − Fn −Gn−1.
Or more generally, using (3.22) and replacing m by m+ n, we can derive
F2n+m = yGn(Gm+n +Gm) + Fn(Fm+n − Fm) + Fm−n,
G2n+m = Gn(Fm+n + Fm) + Fn(Gm+n −Gm) +Gm−n.
The cost of computing F2n+m, G2n+m from
{Fm+n, Gm+n, Fm, Gm, Fn, Gn, Fm−n, Gm−n}
is 5 multiplications.
76
Now since
Fn+2 = xFn+1 − xFn + yGn+1 + yGn + Fn−1,
Gn+2 = xGn+1 − xGn + Fn+1 + Fn +Gn−1,
we get
Fn−1 = Fn+2 − xFn+1 + xFn − yGn+1 − yGn,
Gn−1 = Gn+2 − xGn+1 + xGn − Fn+1 − Fn.
Hence
F2n+1 = FnFn+1 + yGnGn+1 − xFn+1 − yGn+1 + Fn+2
= Fn+1(Fn − x) + yGn+1(Gn − 1) + Fn+2, (3.24)
G2n+1 = GnFn+1 +Gn+1Fn −Gn+2 + xGn+1 + Fn+1
= Fn+1(Gn + 1) +Gn+1(Fn + x)−Gn+2. (3.25)
Also, if we replace n by n+ 1 in the above, then
F2n+3 = Fn+1(Fn+2 − x)− yGn+1(Gn+2 − 1) + Fn, (3.26)
= G2n+3 = Fn+1(Gn+2 + 1) +Gn+1(Fn+2 + x)−Gn. (3.27)
We can also set m = n in (3.22) to get
H2n = H2n − 3H−n +H−n = H2
n − 2H−n and H−2n = H2−n − 2Hn.
77
Using this, we can easily obtain
F2n = F 2n + yG2
n − 2Fn = Fn(Fn − 2) + yG2n, (3.28)
G2n = 2Gn(Fn + 1), (3.29)
F2n+2 = Fn+1(Fn+1 − 2) + yG2n+1, (3.30)
G2n+2 = 2Gn+1(Fn+1 + 1). (3.31)
Thus given the sextet
Sn = {Fn, Fn+1, Fn+2, Gn, Gn+1, Gn+2}
we can compute
S2n+1 = {F2n+1, F2n+2, F2n+3, G2n+1, G2n+2, G2n+3}
using (3.24), (3.25), (3.26), (3.27), (3.30), (3.31) with 12 multiplications. If one is
not careful it may appear as though we need to do 14 multiplications, but yGn+1
occurs 3 times and we need only calculate it once. We are also able to compute
S2n = {F2n, F2n+1, F2n+2, G2n, G2n+1, G2n+2}
using (3.24), (3.25), (3.28), (3.29), (3.30), (3.31) with 12 multiplications.
These observations can now be used to compute (by F ) Xmn, Ym,n (mod r) for a
given modulus r, given Xn, Dn in O(logm) modular multiplications. We begin with
S1 = {F1, F2, F3, G1, G2, G3} (mod r),
which can be computed using Xn, Dn only. We then compute
Sm = {Fm, Fm+1, Fm+2, Gm, Gm+1, Gm+2} (mod r)
78
as follows. Let (b0b1 . . . bk)2 = m be the binary representation of m such that b0 6= 0.
Set P0 = S1 and for i = 0 to i = k − 1
Pi+1 =
S2n (mod r) if bi+1 = 0
S2n+1 (mod r) if bi+1 = 1.
Then Pk = Sm. This gives us Xmn ≡ Fm (mod r), Ym,n ≡ Gm (mod r) and Dmn =
DnG2m (mod r).
Thus, if k = dlogme we need to perform 12k modular multiplications to com-
pute Sm (mod r). To compute tm ≡ am (mod r) requires on average 32k modular
multiplications. Thus, for a given m computing Sm is 8 times more expensive than
computing tm.
Chapter 4
Arithmetic Properties of {Cn} and {Wn}
4.1 Introductory Arithmetic Results
To continue our generalization we need to develop arithmetic results, both global
and local, that are logical analogues of the arithmetic results seen in Chapter 2 for
Lucas sequences.
Lemma 4.1. If (Q,R) = 1, then (Bn, R) = 1, for n > 0.
Proof. First note B0 = 3, B1 = Q and B2 = Q2 −RP . Also for n ≥ 0
Bn+3 = QBn+2 −RPBn+1 +R2Bn.
By induction Bn ≡ Qn (mod R) for n > 0. The result follows immediately.
We also can produce a somewhat similar result which involves An instead of Bn.
Theorem 4.2. If (Q,R) = 1, then (An, R,∆) | 4.
Proof. Let p be any odd prime such that p | (∆, R). From the formula for ∆, we see
that p | Q2P 2 − 4Q3, and since (Q,R) = 1, we must have that p | P 2 − 4Q. Now
A0 = 3, A1 = P , A2 = P 2 − 2Q and
Ak+2 ≡ PAk+1 −QAk (mod R).
Since Q ≡ P 2/4 (mod p), we get
An ≡ P n/2n−1 (mod p)
79
80
by induction on n. Since p - P , we have p - An and p - (An, R,∆).
Next, suppose that 2ν || (An, R,∆). When ν > 2, we see that 2 | P and since Q
is odd, we must have P/2 odd. Since
Ak+2 ≡ PAk+1 −QAk (mod 4)
and Q ≡ P/2 ≡ 1 (mod 2), 2 || A1 and 2 || A2, we find by induction on n that
2 || An. This is a contradiction to the fact ν > 2.
Corollary 4.2.1. If (Q,R) = 1, then (Wn, R,∆) | 4.
Proof. Since Wn = AnBn − 3Rn, we get (Wn, R) = (AnBn, R) = (An, R) by Lemma
4.1. Hence, (Wn, R,∆) | 4 by the previous theorem.
To prove the next lemma we first note that since
27∆ = 4(P 2 − 2Q)3 − (2P 3 − 9PQ+ 27R)2,
we must have ∆ ≡ 0, 1 (mod 4).
Lemma 4.3. If 2 - R and 2α || (Wn, Cn), then α ∈ {0, 1}, and if 2 | Wn, then
Qn = W 2n−∆C2
n
4is odd.
Proof. If 2 - Wn we are done. Suppose 2 | Wn. Since 2 - R we must have 2 - AnBn,
as AnBn = Wn + 3Rn. So both An and Bn are odd and then 2 | A2n − 3Bn.
By Theorem 3.3 we have
27∆C2n = 4(A2
n − 3Bn)3 − (2A3n − 9Wn)2.
Hence
27∆C2n ≡ −(2A3
n − 9Wn)2 (mod 8).
81
Now, since 2A3n − 9Wn ≡ 2−Wn (mod 4), we get
27∆C2n ≡ −(2−Wn)2 (mod 8).
If 2 || Wn, then 8 | ∆C2n and Qn is odd. If 4 | Wn, then 27∆C2
n ≡ −4 (mod 8) and
Qn is odd, thus, if 4 | ∆ then Cn ≡ 1 (mod 2). If ∆ ≡ 1 (mod 4), then Cn ≡ 2
(mod 4).
Lemma 4.4. If 2 | R, 2 - Q and 2α || (Wn, Cn), then α ∈ {0, 1}, and if 2 | Wn,
then Qn is odd.
Proof. If 2 - Wn we are done. If 2 | Wn, then 2 | AnBn. Since 2 - Q we know 2 - Bn
and 2 | An. We may then observe that A2n − 3Bn is odd and thus
27∆C2n ≡ 4− (2A3
n − 9Wn)2 (mod 8)
≡ 4− (−9Wn)2 (mod 8)
≡ 4−W 2n (mod 8).
If 2 || Wn, then 8 | ∆C2n and Qn is odd. If 4 | Wn, then 27∆C2
n ≡ 4 (mod 8) or
−∆C2n ≡ 4 (mod 8) which implies Cn is odd or 2 || Cn; in either case Qn is odd.
From the above results we have the following theorem.
Theorem 4.5. If (Q,R) = 1 and 2α || (Wn, Cn), then α ∈ {0, 1}. If 2 | Wn, then
Qn is odd.
The following result is a clear analogue of (2.18).
82
Lemma 4.6. If (Q,R) = 1, then
(Wn, Cn, R) | 2.
Proof. If (Wn, Cn, R) = 1, we are done. Let p be any prime such that p | (Wn, Cn, R).
Since p | Wn and p | Cn we must have p | W 2n − ∆C2
n. Observe by equation (3.5)
W 2n − ∆C2
n = 4SnTn = 4(3R2n + RnA3n + B3n) ⇒ p | 4B3n. Also, B3n = B3n −
3RnAnBn + 3R2n, so p | 4B3n. Since (Q,R) = 1, we have (Bn, R) = 1 by Lemma
4.1 but this implies p - Bn ⇒ p = 2. Indeed, by Lemma 4.4, we must have
(Wn, Cn, R) | 2.
Furthermore, it is not difficult to show that, like {Un}, {Cn} is a divisibility
sequence; i.e.
Cm | Cn, when m | n. (4.1)
Note that if n = ms, then
Cn(P,Q,R) =(αn − βn)(βn − γn)(γn − αn)
(α− β)(β − γ)(γ − α)=
(αms − βms)(βms − γms)(γms − αms)(α− β)(β − γ)(γ − α)
=(αm − βm)(βm − γm)(γm − αm)
(α− β)(β − γ)(γ − α)· (αms − βms)(βms − γms)(γms − αms)
(αm − βm)(βm − γm)(γm − αm)
= Cm(P,Q,R) · Cs(Am, Bm, Rm).
Definition 4.7. Given m ∈ Z, let r be the least positive integer, if it exists, such
that m | Cr. This value is called the rank of apparition of m for the sequence {Cn}
and will be denoted by r(m).
In Theorem 2.4 for the classic Lucas case, we had that if m | Uk, then r(m) | k.
However, this is not necessarily true for {Cn}. It may be that m | Ck, yet r(m) - k.
83
Definition 4.8. Let r1 be the least positive integer for which p | Cr1. For i =
1, 2, . . . , k define ri+1, if it exists, to be the least positive integer such that p | Cri+1,
ri+1 > ri and rj - ri+1 for any j ≤ i+ 1. We define r1, r2, . . . , rk to be the ranks of
apparition for {Cn}.
It will become clear that the number of ranks of apparition is finite.
For example, if we let P = 1, Q = 2, R = 3 and p = 7, then {Cn} has two ranks
of apparition for the prime 7. In fact, C3 ≡ 0 (mod p) and C7 ≡ 0 (mod p). Also,
if we let P = 3, Q = 9, R = 7 and p = 31, then {Cn} has three ranks of apparition.
Here, C6 ≡ 0 (mod p), C10 ≡ 0 (mod p) and C15 ≡ 0 (mod p).
Our sequence {Cn} also fails to satisfy the generalization of Corollary 2.4.1 where
if d = (m,n) then
(Um, Un) = |Ud|.
It can be that
(Cm, Cn) 6= |Cd|,
and d = (m,n). For example, if P = 3, Q = 9, R = 7, then (C6, C10) = 22 · 5 · 31
and C2 = 22 · 5.
We can, however, reach a relatively close analogue to Carmichael’s result seen in
Theorem 2.5. To do so we must first derive several preliminary arithmetic results in
the next lemmas and theorems.
Lemma 4.9. If
Im =∑ m(−1)λ0(m− λ0 − 1)!
λ1!λ2!λ3!(4.2)
is summed over all λ0, λ1, λ2, λ3 ∈ Z≥0 such that λ0+λ1+λ2+λ3 = m, λ1+2λ2+3λ3 =
m and λ1 6≡ λ2 (mod 2), then Im ≡ m (mod 2).
84
Proof. Put I ′m equal to the right side of (4.2), where we insist that λ1 ≡ λ2 (mod 2).
By Waring’s theorem
αm + βm + γm =∑ m(−1)λ0(m− λ0 − 1)!
λ1!λ2!λ3!P λ1Qλ2Rλ3 ,
where the sum is over all λi satisfying the constraints in the statement of the lemma
except that of λ1 6≡ λ2 (mod 2). This is true as P = α + β + γ, Q = αβ + βγ + γα
and R = αβγ.
Putting P = Q = R = 1 we get
αm + βm + γm = I ′m + Im
so I ′m + Im = 1m + im + (−i)m ≡ 1 (mod 2), where i2 = −1. Putting P = Q = −1,
R = 1 we have α = −1, β = −1,γ = 1, yielding I ′m − Im = 2(−1)m + 1. It then
follows that
Im =im + (−i)m
2− (−1)m ≡ m (mod 2).
Let Pn = Wn and Qn = (W 2n −∆C2
n)/4 for the remainder of this section. We will
now give a series of results that will be useful in the next chapter.
Theorem 4.10. If 2 | Pn and 2 - Qn, then CmnCn≡ m (mod 2).
Proof. First note that, by equation (2.7), we can derive
−Qλ2n Uλ1−λ2 = Qλ1
n Uλ2−λ1 (4.3)
and if k ≥ 0, Uk(Pn, Qn) ≡ Uk(2, 1) ≡ k (mod 2). It follows that
Qλ1n Uλ2−λ1 ≡ λ2 − λ1 (mod 2)
85
for λ2 ≥ λ1 and λ2 < λ1. Hence by the multiplication formula for Cmn, we get
CmnCn≡∑ m(−1)λ0(m− λ0 − 1)!
λ1!λ2!λ3!R(λ0+λ3)n (mod 2)
where the sum is as in (3.17) with the extra condition λ1 6≡ λ2 (mod 2).
If 2 | R, then each of the terms in the expression for Cmn/Cn modulo 2 is even
unless λ0 + λ3 = 0, λ0 + λ1 + λ2 + λ3 = m, λ1 + 2λ2 + 3λ3 = m and 2 - (λ1 + λ2).
These conditions imply λ0 = λ3 = 0⇒ λ2 = 0⇒ λ1 = m⇒ 2 - m. In this case
CmnCn
≡ m(−1)λ0(m− λ0 − 1)!
λ1!λ2!λ3!R(λ0+λ3)n
≡ 1 (mod 2).
If 2 - R, then by Lemma 4.9
CmnCn≡ Im ≡ m (mod 2).
Lemma 4.11. If CrnCn≡ 0 (mod k) for all n > 0, then Cmn
Cn≡ 0 (mod k) if r | m.
Proof. Let m = rs. Then it is easy to see that
CmnCn
=CrsnCn
=CrsnCrn
.CrnCn≡ 0 (mod k).
Theorem 4.12. If 2 - R, 2 | Pn, 2 | Qn, then
CmnCn
≡ m (mod 2) if 3 - m
CmnCn
≡ 0 (mod 2) if 3 | m.
86
Proof. Note that Qλ2n Uλ1−λ2 ≡ 0 (mod 2) unless λ2 = 0, λ1 = 1 or λ1 = 0, λ2 = 1.
We now consider these cases where Qλ2n Uλ1−λ2 6≡ 0 (mod 2).
It follows that if m ≡ 1 (mod 3) it must be that λ2 = 0, λ1 = 1, λ3 = m−13
and
λ0 = 2m−13
. So m− λ0 − 1 = m−13
. This implies
CnmCn≡ m
(m−1
3
)!(
m−13
)!≡ m (mod 2).
Similarly, if m ≡ −1 (mod 3), we can only have λ2 = 1, λ1 = 0, λ3 = m−23
and
λ0 = 2m−13
. Hence m− λ0 − 1 = m−23
, which gives
CmnCn≡ m (mod 2).
If 3 | m we can never have λ2 = 0, λ1 = 1, or λ1 = 0, λ2 = 1. Thus,
CnmCn≡ 0 (mod 2).
Theorem 4.13. If 2 - R, 2 - Pn and 2 - Qn, then
CmnCn≡ 1 (mod 2) if 3 - m
CmnCn≡ 0 (mod 2) if 3 | m.
Proof. First note that
Ut(Pn, Qn) ≡ Ut(1, 1) ≡
0 if 3 | t
1 if 3 - t(mod 2).
Hence
Qλ2n Uλ1−λ2 ≡
0 if 3 | λ1 − λ2
1 if 3 - λ1 − λ2
(mod 2).
87
Since λ1 + 2λ2 + 3λ3 = m, we see that if λ1 ≡ λ2 (mod 3), then 3 | m. It follows
that if 3 - m, we know λ1 6≡ λ2 (mod 3) and we get
CmnCn≡∑ (−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!= Im + I ′m ≡ 1 (mod 2).
We note that C3n
Cn= (∆C2
n + 3W 2n)/4 = P 2
n − Qn ≡ 0 (mod 2), thus in the case
where 3 | m, it follows from Lemma 4.11 that CmnCn≡ 0 (mod 2).
Theorem 4.14. If 2 - R, 2 - Pn and 2 | Qn, then
CmnCn≡ 1 (mod 2) if 7 - m
CmnCn≡ 0 (mod 2) if 7 | m.
Proof. First, we use the fact that
Ut(Pn, Qn) ≡ Ut(1, 0) (mod 2).
This implies
Ut(Pn, Qn) ≡ 1 (mod 2) if t > 0.
Now, Qλ2n Uλ1−λ2 ≡ 0 (mod 2) if λ1 ≥ λ2 > 0 or λ2 ≥ λ1 > 0. Thus, if λ1 = 0, λ2 ≥ 1
or λ2 = 0, λ1 ≥ 1, then
Qλ2n Uλ1−λ2 ≡ 1 (mod 2)
as QtnU−t = −Ut.
88
It follows that
CmnCn
≡∑
λ0 + λ1 + λ3 = m
λ1 + 3λ3 = m
(−1)λ0m(m− λ0 − 1)!
λ1!λ3!
+∑
λ0 + λ2 + λ3 = m
2λ2 + 3λ3 = m
(−1)λ0m(m− λ0 − 1)!
λ2!λ3!(mod 2).
Now let us restrict ourselves to the case where P = 1, Q = 0, R = 1 and α, β, γ
are the zeros of X3 − PX2 +QX −R. Then
Am = αm + βm + γm =∑
λ0 + λ1 + λ3 = m
λ1 + 3λ3 = m
(−1)λ0m(m− λ0 − 1)!
λ1!λ3!.
Also let P ′ = 0, Q′ = 1, R′ = 1 and α′, β′, γ′ be the zeros of X3−P ′X2 +Q′X −R′,
then
A′m = α′m + β′m + γ′m =∑
λ0 + λ2 + λ3 = m
2λ2 + 3λ3 = m
(−1)λ0m(m− λ0 − 1)!
λ2!λ3!.
Clearly, the above two identities follow from Waring’s theorem.
Now it is clear that A′0 = 3, A′1 = P ′ = 0, A′2 = P ′2 − 2Q′ = −2 and
A′n+3 = P ′A′n+2 −Q′A′n+1 +R′A′n ≡ A′n+1 + A′n (mod 2).
89
Thus,
A′m ≡ 1 (mod 2) if m ≡ 0, 3, 5, 6 (mod 7)
A′m ≡ 0 (mod 2) if m ≡ 1, 2, 4 (mod 7).
Similarly, one can show that
Am ≡ 1 (mod 2) if m ≡ 0, 1, 2, 4 (mod 7)
Am ≡ 0 (mod 2) if m ≡ 3, 5, 6 (mod 7).
Hence
Am + A′m ≡ 1 (mod 2) if 7 - m
Am + A′m ≡ 0 (mod 2) if 7 | m.
Since CmnCn≡ Am + A′m (mod 2), we are done.
We now assume that (Q,R) = 1, and we recall from Theorem 4.5 that if 2 | Pn,
then 2 - Qn. Putting n = 1, from these results, we now know that
Cm ≡ m (mod 2)
when 2 | P1, 2 - Q1. Put r = 2 in this case.
If 2 - R, 2 - P1, 2 - Q1, put r = 3 and note that
Cm ≡ 1 (mod 2) if 3 - m,
Cm ≡ 0 (mod 2) if 3 | m.
90
If 2 - R, 2 - P1, 2 | Q1, put r = 7 and note that
Cm ≡ 1 (mod 2) if 7 - m,
Cm ≡ 0 (mod 2) if 7 | m.
There remains the case of 2 - P1 and 2 | R. We have 2 - Q; thus, by Corollary
3.14.2, we get 2 - Q1 and
Cm ≡ Um(1, 1) (mod 2).
In this case, we put r = 3 and we have
Cm ≡ 1 (mod 2) if 3 - m,
Cm ≡ 0 (mod 2) if 3 | m.
Hence, we have proved the following theorem.
Theorem 4.15. If (Q,R) = 1, there always exists a minimal r > 1 such that 2 | Cr.
Furthermore, if 2 | Cn, then r | n.
Lemma 4.16. If m, n ≥ 1, then
CmnCn≡ 0 (mod (Pn, Qn)) if 3 | m
CmnCn≡ mR(m−1)n (mod (Pn, Qn)) if 3 - m.
Proof. Recall for Uk(Pn, Qn), U−k = −Uk/Qkn. By (4.3) we see that if λ1 ≥ λ2, then
Qλ2n Uλ1−λ2 ≡ 0 (mod Qn) if λ2 ≥ 1,
Qλ2n Uλ1−λ2 ≡ Uλ1 ≡ P λ1−1
n (mod Qn) if λ2 = 0 6= λ1,
Qλ2n Uλ1−λ2 ≡ 0 (mod Qn) if λ1 = λ2.
91
For λ2 ≥ λ1, we have the following results:
−Qλ1n Uλ2−λ1 ≡ 0 (mod Qn) if λ1 ≥ 1,
−Qλ1n Uλ2−λ1 ≡ −Uλ2 ≡ −P λ2−1
n (mod Qn) if λ1 = 0 6= λ2,
−Qλ1n Uλ2−λ1 ≡ 0 (mod Qn) if λ1 = λ2.
Thus Qλ2n Uλ1−λ2 ≡ 0 (mod Qn) unless λ1 ≥ 1, λ2 = 0 or λ1 = 0, λ2 ≥ 1. But since
(Pn, Qn) will divide Pn, we get Qλ2n Uλ1−λ2 ≡ 0 (mod (Pn, Qn)) unless λ1 = 1, λ2 = 0
or λ1 = 0, λ2 = 1.
If 3 | m, then 3 | λ1 + 2λ2 and neither λ1 = 1, λ2 = 0 nor λ1 = 0, λ2 = 1 can
occur. Thus, from Theorem 3.14 we can conclude that
CmnCn≡ 0 (mod (Pn, Qn)) if 3 | m.
If 3 - m, then either m ≡ 1 (mod 3) or m ≡ 2 (mod 3). If m ≡ 1 (mod 3),
then we must have λ1 = 1, λ2 = 0, in which case Qλ2n Uλ1−λ2 = 1, λ3 = m−1
3,
λ3 = m− λ0 − 1, and λ0 = 2m−13
. Applying the above to Theorem 3.14 yields
CmnCn≡ mR(m−1)n (mod (Pn, Qn)) if m ≡ 1 (mod 3).
Similarly, if m ≡ 2 (mod 3), then it must be that λ1 = 0, λ2 = 1. Then we can
see −Qλ1n Uλ2−λ1 = Qλ2
n Uλ1−λ2 = QnU−1 = −U1 = −1, λ3 = m−23
, λ0 = 2m−13
which is
odd. Again use Theorem 3.14 to get
CmnCn≡ mR(m−1)n (mod (Pn, Qn)) if m ≡ 2 (mod 3).
Which completes the proof.
Corollary 4.16.1. If 3 - m, then
(Cmn/Cn, Pn, Qn) | mRn(m−1).
92
Proof. This follows immediately from Lemma 4.16.
Theorem 4.17. If 2 - R or if 2 | R and 2 - Q, then
(Cmn/Cn,Wn, Cn) | mRn(m−1) when 3 - m.
Proof. We have by Corollary 4.16.1 that if 3 - m, then
(Cmn/Cn, Pn, Qn) | mRn(m−1) ⇒ (Cmn/Cn,Wn, (W2n −∆C2
n)/4) | mRn(m−1).
We divide our proof into 2 cases.
Case 1: 2 - Wn. In this case
(4Cmn/Cn, 4Wn,W2n −∆C2
n) | 4mRn(m−1) ⇒ (Cmn/Cn,Wn, Cn) | 4mRn(m−1).
But since Wn is odd (Cmn/Cn,Wn, Cn) | mRn(m−1).
Case 2: 2 | Wn. In this case we have 2 | Pn and Qn is odd, so by Theorem 4.10
CmnCn≡ m (mod 2).
Also, we know by Theorem 4.5 that if 2s || (Wn, Cn) then s ∈ {0, 1}.
Since (Cmn/Cn,Wn, Cn) | 4mRn(m−1) we see that (Cmn/Cn,Wn, Cn) | mRn(m−1)
when m is odd. Otherwise, if m is even and 2s || (Cmn/Cn,Wn, Cn), then s ∈ {0, 1}.
If s = 0, then (Cmn/Cn,Wn, Cn)|mRn(m−1). If s = 1, then 2 || (Cmn/Cn,Wn, Cn),
which implies (Cmn/Cn,Wn, Cn) | mRn(m−1).
We have been working towards the following corollary which is somewhat anal-
ogous to Carmichael’s result seen in Theorem 2.5. We will derive a closer analogue
in Chapter 5.
93
Corollary 4.17.1. If (Q,R) = 1, then
(Cmn/Cn,Wn, Cn) | m when 3 - m.
Proof. If (Wn, Cn, R) = 1, we are done. Let p be any prime such that p | (Wn, Cn, R).
By Lemma 4.6 we can only have p = 2. Since 2 | Wn we have Qn odd and then by
Theorem 4.10 we have
CmnCn≡ m (mod 2).
Now if 2 - m, then
(Cmn/Cn,Wn, Cn, R) = 1⇒ (Cmn/Cn,Wn, Cn) | m.
If 2 | m, then
(Cmn/Cn,Wn, Cn, Rn(m−1)) = 2⇒ (Cmn/2Cn,Wn/2, Cn/2, R
n(m−1)/2) = 1.
We have then
(Cmn/2Cn,Wn/2, Cn/2) | mRn(m−1)/2⇒ (Cmn/2Cn,Wn/2, Cn/2) | m
⇒ (Cmn/Cn,Wn, Cn) | 2m.
But 2 || (Cmn/Cn,Wn, Cn) and 2 | m ⇒ (Cmn/Cn,Wn, Cn) | m.
We have seen that many of Lucas’ results have analogues when we assume that
(Q,R) = 1. This is similar to Lucas’ condition that (P,Q) = 1, and we will assume
for the remainder of this work that (Q,R) = 1.
94
4.2 Preliminary Results for the Law of Repetition for {Cn}
We will now require some elementary results in algebraic number theory to develop
the proof of the following theorem. This result will be of some importance in estab-
lishing a law of repetition for {Cn}.
Theorem 4.18. If p is a prime and p - 6R∆, p | Cn and p | Wn−6Rn, then p3 | Cn
and p2 | Wn − 6Rn.
Proof. Let α, β, γ be the distinct (p - ∆) zeros of x3 − Px2 + Qx − R and put
L = Q(α). If we put K = Q(α, β), then K is the normal closure of L and is, of
course, Galois. Put δ = (α − β)(β − γ)(γ − α), λ1 = αn − βn, λ2 = βn − γn and
λ3 = γn − αn. Note λ1 + λ2 + λ3 = 0.
Since δCn = λ1λ2λ3, we note that if p is prime ideal divisor of (p) in K, then
p | (λ1λ2λ3). We also note that the discriminant of L must divide ∆. It follows
(see for example, Theorem 86 of [Hil98]), that since p - ∆, then p cannot divide the
discriminant of K. Thus, in K we must have
(p) =k∏i=1
pi,
where the prime ideals pi (i = 1, 2, . . . , k) are all distinct, that is (pi, pj) = O, the
maximal order of K, for i 6= j. Since p | (λ1λ2λ3), we must have p | (λ1) or p | (λ2)
or p | (λ3). Without loss of generality, suppose p | (λ1).
Since
Wn − 6Rn = 2βn(αn − γn)2 − (αn − βn)(βn − γn)(αn + γn)
and p | Wn − 6Rn, we have
2βn(αn − γn)2 ≡ (αn − βn)(βn − γn)(αn + γn) ≡ 0 (mod p).
95
Since (p, 2R) = 1 and R = αβγ, we must have p - β and hence αn ≡ γn
(mod p) ⇒ p | (λ3). Also since λ2 = −λ1 − λ3 and (λ1) ≡ (λ3) ≡ 0 (mod p) we
must have (λ2) ≡ 0 (mod p). Hence p3 | (λ1λ2λ3). Since ((p), (δ)) = O, we get
p3 | ((λ1λ2λ3)/δ).
Since p3i | ((λ1λ2λ3)/δ) for i = 1, 2, . . . , k and the pi for i = 1, 2, . . . , k are distinct
prime ideals, we must have
k∏i=1
p3i | ((λ1λ2λ3)/δ).
Thus (p3) | (Cn)⇒ p3 | Cn. Note that we also have p2 | Wn − 6Rn.
Suppose again that the prime p - 6∆R and that pµ || Cn, pν || Wn− 6Rn, where
µ, ν ≥ 1. Let p be any of the distinct ideals which lie over (p) in K. Since
pµ || (αn − βn)(βn − γn)(γn − αn),
we may assume without loss of generality that
pµ1 || αn − βn, pµ2 || βn − γn, pµ3 || γn − αn,
where µ1 + µ2 + µ3 = µ and µ1 ≥ µ2 ≥ µ3. Since
γn − αn = −(βn − γn)− (αn − βn),
we see that pµ2 | γn − αn and µ2 ≤ µ3. Hence µ2 = µ3. If µ1 > µ2, then since
Wn − 6Rn = 2βn(αn − γn)2 − (αn − βn)(βn − γn)(αn + γn)
and µ1 +µ2 > 2µ2, we must have p2µ2 || Wn−6Rn and 2µ2 < µ. Thus, ν = 2µ2 < µ.
If µ1 = µ2, then 3 | µ and µ1 = µ2 = µ3 = µ/3. This seems to suggest that the case
of ν > µ would occur less frequently than the case of ν ≤ µ.
96
Put Dn = (Wn − 6Rn, Cn). Since (Wn, Cn, R) | 2 by Lemma 4.6, we see that
if p 6= 2 and p | Dn, then p - R. Further results on Dn will be developed in the
following chapter. We also have the following theorem for the case where p = 2.
Theorem 4.19. If 2 - R∆ and 16 | Cn, then 8 | Wn − 6Rn.
Proof. We note that if 2 - R and 4 | Cn, then 2 || Wn by Theorem 4.5. Thus Wn −
6Rn ≡ 0 (mod 4). Now since 2 - ∆, we have (2) is the product of distinct prime ideals
in K. Let p be any one of these prime ideals. Since 16 | (αn−βn)(βn−γn)(γn−αn),
we must have p4 | (αn − βn)(βn − γn)(γn − αn). Without loss of generality, let
pµ1 || (αn−βn), pµ2 || (βn−γn) and pµ3 || (γn−αn), where µ1 ≥ µ2 ≥ µ3. We must
have µ1 + µ2 + µ3 = 4⇒ µ1 ≥ 2 and µ2 ≥ 1.
Since p2 | Wn − 6Rn, we get
p2 | 2βn(αn − γn)2 − (αn − βn)(βn − γn)(αn + γn).
Since p3 | (αn − βn)(βn − γn), we get p2 | 2βn(αn − γn)2 ⇒ p | (αn − γn)2 ⇒
p | αn − γn ⇒ p3 | Wn − 6Rn. It follows that 8 | Wn − 6Rn.
4.3 The Polynomial Km(x)
We now introduce the polynomials Hm(X, Y ) and Km(X). We will develop some
properties of these polynomials which will help us to produce the law of repetition
for {Cn}. Put
Hm(X, Y ) =∑ (−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!Xλ1−λ2Y 2λ2
and
Km(X) =∑ (−1)λ0m(m− λ0 − 1)!(λ1 − λ2)
λ1!λ2!λ3!Xλ1+λ2−1. (4.4)
97
As before, the sums are extended over the values λi ∈ Z such that
λ0, λ1, λ2, λ3 ≥ 0, λ0 + λ1 + λ2 + λ3 = m, λ1 + 2λ2 + 3λ3 = m.
Note that
Km(X) =∂Hm(X, Y )
∂X
∣∣∣Y=X
.
Consider
FX,Y (Z) = Z3 −XZ2 + (Y 2/X)Z − 1
and let α1(X, Y ), α2(X, Y ), α3(X, Y ) be the three (not necessarily distinct) zeros of
FX,Y (Z). By Waring’s theorem
Hm(X, Y ) = α1(X, Y )m + α2(X, Y )m + α3(X, Y )m.
Hence
∂Hm(X, Y )
∂X= m
(α1(X, Y )m−1∂α1(X, Y )
∂X+ α2(X, Y )m−1∂α2(X, Y )
∂X
+α3(X, Y )m−1∂α3(X, Y )
∂X
).
For convenience we will now write α1, α2 and α3, to denote α1(X, Y ), α2(X, Y ) and
α3(X, Y ), respectively. One can easily see then,
α1 + α2 + α3 = X, α1α2 + α2α3 + α3α1 = Y 2/X and α1α2α3 = 1.
Following Lagrange, we put
θ1(X, Y ) = α1 + ζα2 + ζ2α3,
θ2(X, Y ) = α1 + ζ2α2 + ζα3,
98
where ζ is a primitive cube root of unity (ζ2 + ζ + 1 = 0). Then
α1 =1
3(X + θ1 + θ2) ,
α2 =1
3
(X + ζ2θ1 + ζθ2
),
α3 =1
3
(X + ζθ1 + ζ2θ2
).
Further, one can easily show
θ1θ2 = X2 − 3Y 2/X and θ31 + θ3
2 = 2X3 − 9Y 2 + 27.
With these two identities we can show
θ31(X, Y ) =
2X3 − 9Y 2 + 27 +√
∆(X, Y )
2and
θ32(X, Y ) =
2X3 − 9Y 2 + 27−√
∆(X, Y )
2,
where
∆(X, Y ) = (2X3 − 9Y 2 + 27)2 − 4(X2 − 3Y 2/X)3
= −27(Y 4 − 4Y 6/X3 − 4X3 + 18Y 2 − 27).
Note that
∆(X, Y )∣∣∣Y=X
= −27(X4 − 8X3 + 18X2 − 27
)= −27 (X − 3)2 (X2 − 2X − 3
).
Put
D(X) = X2 − 2X − 3 = (X − 3)(X + 1).
Now, observe
∂∆(X, Y )
∂X= −27(−12X2 + 12Y 6/X4)⇒ ∂∆(X, Y )
∂X
∣∣∣Y=X
= 0.
99
It is easy to see that
∂θ1(X, Y )
∂X=
1
3θ−2
1
(6X2 +
1
2∆−1/2(X, Y )
∂∆(X, Y )
∂X
)/2.
Thus
∂θ1(X, Y )
∂X
∣∣∣Y=X
= θ−21 (X,X)X2.
Similarly
∂θ2(X, Y )
∂X
∣∣∣Y=X
= θ−22 (X,X)X2.
Since
θ1(X,X)θ2(X,X) = X2 − 3X,
we get
θ−11 (X,X) =
θ2(X,X)
X2 − 3Xand θ−1
2 (X,X) =θ1(X,X)
X2 − 3X.
So we then have
∂θ1(X, Y )
∂X
∣∣∣Y=X
=θ2
2(X,X)
(X − 3)2and
∂θ2(X, Y )
∂X
∣∣∣Y=X
=θ2
1(X,X)
(X − 3)2.
Now
θ1(X,X) = α1(X,X) + ζα2(X,X) + ζ2α3(X,X) and
FX,X(Z) = Z3 −XZ2 +XZ − 1.
Hence, we may put
α1(X,X) = 1, α2(X,X) =X − 1 +
√D(X)
2, α3(X,X) =
X − 1−√D(X)
2.
We then have
θ1(X,X) =3−X + (ζ − ζ2)
√D(X)
2,
θ2(X,X) =3−X − (ζ − ζ2)
√D(X)
2.
100
Upon squaring and using the fact (ζ − ζ2)2 = −3, we get
θ21(X,X) = (X − 3)
[−X − 3− (ζ − ζ2)
√D(X)
2
],
θ22(X,X) = (X − 3)
[−X − 3 + (ζ − ζ2)
√D(X)
2
].
Hence
∂θ1(X, Y )
∂X
∣∣∣Y=X
=−X − 3 + (ζ − ζ2)
√D(X)
2(X − 3),
∂θ2(X, Y )
∂X
∣∣∣Y=X
=−X − 3− (ζ − ζ2)
√D(X)
2(X − 3).
Now
∂α1(X, Y )
∂X
∣∣∣Y=X
=1
3
(1 +
∂θ1(X, Y )
∂X+∂θ2(X, Y )
∂X
) ∣∣∣Y=X
=1
3
(1 +−X − 3 + (ζ − ζ2)
√D(X)
2(X − 3)
+−X − 3− (ζ − ζ2)
√D(X)
2(X − 3)
)
=1
3
(1− X + 3
X − 3
)=−2
X − 3.
Also,
∂α2(X, Y )
∂X
∣∣∣Y=X
=1
3
(1 + ζ2∂θ1(X, Y )
∂X+ ζ
∂θ2(X, Y )
∂X
) ∣∣∣Y=X
=1
3
(1 +
X + 3 + (2− ζ − ζ2)√D(X)
2(X − 3)
)
=X − 1 +
√D(X)
2(X − 3)=α2(X,X)
X − 3,
∂α3(X, Y )
∂X
∣∣∣Y=X
=1
3
(1 +
X + 3− (2− ζ − ζ2)√D(X)
2(X − 3)
)=α3(X,X)
X − 3.
101
It follows that
Km(X) =∂Hm(X, Y )
∂X
∣∣∣Y=X
= mαm−11 (X,X)
(−2
X − 3
)+mαm−1
2 (X,X)
(α2(X,X)
X − 3
)+mαm−1
3 (X,X)
(α3(X,X)
X − 3
)=
m
X − 3[−2 + α2(X,X)m + α3(X,X)m] .
We then get the identity
Km(X) =m
X − 3
[(X − 1 +
√D(X)
2
)m
+
(X − 1−
√D(X)
2
)m
− 2
].
If m is odd, then using identity (4.2.44) from [Wil98] gives
Vm(X − 1, 1) = α2(X,X)m + α3(X,X)m (4.5)
= (X − 1)
(m−1)/2∑j=0
((m− 1)/2 + j
(m− 1)/2− j
)Dj(X). (4.6)
In this case
Km(X) = m
1 +
(m−1)/2∑j=1
((m− 1)/2 + j
(m− 1)/2− j
)(X − 1)(X + 1)j(X − 3)j−1
.If m is even, then identity (4.2.42) from [Wil98] yields
Vm(X − 1, 1) = αm2 (X,X) + αm3 (X,X) =
m/2∑j=0
m
m/2− j
(m/2 + j
m/2− j
)Dj(X) (4.7)
= 2+
m/2∑j=1
m
m/2− j
(m/2 + j
m/2− j
)Dj(X).(4.8)
In this case
Km(X) = m
m/2∑j=1
m
m/2− j
(m/2 + j − 1
m/2− j − 1
)(X + 1)j(X − 3)j−1.
102
Theorem 4.20. Let X ∈ Z and p be a prime such that p > 3. If p - X − 3, then
Kp(X) ≡ p (mod p2).
If p | X − 3, then
Kp(X) ≡ p3 (mod p4).
Proof. From the top of page 100 we have
(X − 3)Kp(X) = p(−2 + αp2(X,X) + αp3(X,X)).
Since α2α3 = 1 and α2 + α3 = X − 1, we have
αp2(X,X) + αp3(X,X) = Vp(X − 1, 1) ≡ X − 1 (mod p).
Thus pVp(X − 1, 1) ≡ p(X − 1) (mod p2) and so we have
(X − 3)Kp(X) ≡ p(X − 3) (mod p2)⇒ Kp(X) ≡ p (mod p2)
when p - X − 3.
Now suppose that p | X − 3. We know that
Kp(X) = p
1 +
(p−1)/2∑j=1
((p− 1)/2 + j
(p− 1)/2− j
)(X − 1)(X + 1)j(X − 3)j−1
.Hence, if p ≥ 7,
Kp(X) ≡ p
[1 +
((p+ 1)/2
2
)(X2 − 1) +
((p+ 3)/2
4
)(X − 3)(X + 1)2(X − 1)
+
((p+ 5)/2
6
)(X − 3)2(X + 1)3(X − 1)
](mod p4).
Next note that
6
((p+ 1)/2
2
)+ 32
((p+ 3)/2
4
)= 6
(p2 − 1)
8+
(p2 − 9)(p2 − 1)
12
= p2
(p2 − 1
8
)≡ 0 (mod p2).
103
Also, ((p+ 1)/2
2
)+ 32
((p+ 3)/2
4
)+ 128
((p+ 5)/2
6
)
=p2 − 1
8+
(p2 − 9)(p2 − 1)
12+
(p2 − 25)(p2 − 9)(p2 − 1)
360
=
(p2 − 1
8
)(1 +
2(p2 − 9)
3+
(p2 − 25)(p2 − 9)
45
)=
(p2 − 1
8
)(p4 − 4p2
45
)=p2(p2 − 1)(p2 − 4)
360≡ 0 (mod p2).
Since p 6= 3, by using
X2 − 1 = (X − 3)2 + 6(X − 3) + 8,
(X + 1)2(X − 1) = (X − 3)3 + 10(X − 3)2 + 32(X − 3) + 32,
(X + 1)3(X − 1) = (X − 3)4 + 14(X − 3)3 + 72(X − 3)2 + 160(X − 3) + 128,
we can rewrite Kp(X) (mod p4) as follows
Kp(X) ≡ p
[1 +
((p+ 1)/2
2
)((X − 3)2 + 6(X − 3) + 8
)+
((p+ 3)/2
4
)(X − 3)
((X − 3)3 + 10(X − 3)2 + 32(X − 3) + 32
)+
((p+ 5)/2
6
)(X − 3)2
((X − 3)4 + 14(X − 3)3 + 72(X − 3)2 + 160(X − 3) + 128
)]≡ p
[1 +
((p+ 1)/2
2
)8 + (X − 3)
(6
((p+ 1)/2
2
)+ 32
((p+ 3)/2
4
))+(X − 3)2
(((p+ 1)/2
2
)+ 32
((p+ 3)/2
4
)+ 128
((p+ 5)/2
6
))]≡ p
[1 + (p2 − 1) + (X − 3)
(p2(p2 − 1)
12
)+ (X − 3)2
(p2(p2 − 1)(p2 − 4)
360
)]≡ p3 (mod p4).
In the case of p = 5 we get
Kp(X) ≡ 5
[1 +
(3
2
)(X2 − 1) +
(4
4
)(X − 3)(X + 1)2(X − 1)
](mod p4).
104
So
Kp(X) ≡ 5[1 + 3
((X − 3)2 + 6(X − 3) + 8
)+(X − 3)
((X − 3)3 + 10(X − 3)2 + 32(X − 3) + 32
)]≡ 5
[25 + 50(X − 3) + 35(X − 3)2
]≡ 125 (mod p4).
4.4 The Law of Repetition for {Cn}
Note that
P 2n − 4Qn = ∆n = ∆C2
n and 4Qn ≡ W 2n (mod C2
n).
If m > 0, we can easily derive from the multiplicative properties of Lucas sequences
mentioned in Chapter 2 that
2m−1Um(P,Q) ≡ mPm−1 (mod ∆) and 2m−1Vm(P,Q) ≡ Pm (mod ∆).
Thus,
2m−1Um(Pn, Qn) ≡ mPm−1n (mod ∆C2
n) and
2m−1Vm(Pn, Qn) ≡ Pmn (mod ∆C2
n).
We may use the identity
2Qλ2n Uλ1−λ2(Pn, Qn) = Uλ1(Pn, Qn)Vλ2(Pn, Qn)− Vλ1(Pn, Qn)Uλ2(Pn, Qn)
105
as follows:
2λ1+λ2−1Qλ2n Uλ1−λ2(Pn, Qn) = 2λ1−1Uλ1(Pn, Qn)2λ2−1Vλ2(Pn, Qn)
−2λ1−1Vλ1(Pn, Qn)2λ2−1Uλ2(Pn, Qn)
≡ λ1Pλ1−1n P λ2
n − P λ1n λ2P
λ2−1n (mod ∆C2
n)
≡ (λ1 − λ2)P λ1+λ2−1n (mod ∆C2
n).
Similarly, we have
2λ1+λ2−1Qλ2n Vλ1−λ2(Pn, Qn) ≡ P λ1+λ2
n (mod ∆C2n).
If 2 - ∆Cn, then
Qλ2n Uλ1−λ2(Pn, Qn) ≡ (λ1 − λ2)(Pn/2)λ1+λ2−1 (mod ∆C2
n),
Qλ2n Vλ1−λ2(Pn, Qn) ≡ 2(Pn/2)λ1+λ2 (mod ∆C2
n).
When 2 | ∆Cn, we have 2 | Pn and Qn ≡ (Pn/2)2 (mod ∆C2n/4). In this case
we can show by induction that
Um(Pn, Qn) ≡ m(Pn/2)m−1 (mod ∆C2n/4),
Vm(Pn, Qn) ≡ 2(Pn/2)m (mod ∆C2n/4)
and therefore
Qλ2n Uλ1−λ2(Pn, Qn) ≡ (λ1 − λ2)(Pn/2)λ1+λ2−1 (mod ∆C2
n/4), (4.9)
Qλ2n Vλ1−λ2(Pn, Qn) ≡ 2(Pn/2)λ1+λ2 (mod ∆C2
n/4). (4.10)
It follows by (3.17) that
CmnCn
≡∑
λ0,λ1,λ2,λ3
((−1)λ0m(m− λ0 − 1)!Rn(m−λ1−λ2)
λ1!λ2!λ3!
)(
(λ1 − λ2)P λ1+λ2−1n
2λ1+λ2−1
)(mod Fn),
106
where
Fn =
∆C2n if 2 - Cn
∆C2n/4 if 2 | Cn.
(4.11)
This symbol Fn and the symbol Gn introduced near the beginning of Chapter 5
should not be confused with the Fn and Gn defined in Section 3.6. So we can use
the above and equation (4.4) to see
CmnCn≡ Rn(m−1)Km(Wn/2R
n) (mod Fn).
Thus, by our earlier results, if m is odd,
CmnCn
≡ mRn(m−1) +m
(m−1)/2∑j=1
((m− 1)/2 + j
(m− 1)/2− j
)2−2jRn(m−2j−1)(Wn − 2Rn)
(Wn + 2Rn)j(Wn − 6Rn)j−1 (mod Fn)
and if m is even,
CmnCn
≡ m
m/2∑j=1
m
m/2− j
(m/2 + j − 1
m/2− j − 1
)2−2j+1Rn(m−2j)(Wn + 2Rn)j
(Wn − 6Rn)j−1 (mod Fn).
If p 6= 2, p | Cn and p | R, then since p | p(p−λ0−1)!λ1!λ2!λ3!
and λ0 + λ3 6= 0 whenever
λ1 6= p, we get by equation (3.17) that
CpnCn≡ Up(Pn, Qn) (mod p2).
Then we may use (2.14) with m = p and n = 1 to see
Up(Pn, Qn) ≡ p(Wn/2)p−1 ≡ p (mod p2).
Thus if pλ || Cn, then pλ+µ || Cpµn.
107
Now suppose that p - 6R and p | Cn. It is easy to show by use of (3.16) that
Wpn ≡ Vp(Pn, Qn) ≡ Pn ≡ Wn (mod p).
Clearly, then, Wpn − 6Rpn ≡ Wn − 6Rn (mod p). If p - Wn − 6Rn, then p2 | Fn and
by Theorem 4.20
CpnCn≡ Rn(p−1)Kp(Wn/2R
n) ≡ Rn(p−1)p ≡ p (mod p2).
In this case if pλ || Cn, then pλ+µ || Cpµn.
On the other hand, if p | Wn − 6Rn and p - ∆, then by Theorem 4.18 we have
p3 | Cn ⇒ p3 | Fn. Also, if p | Wn − 6Rn and p | ∆, then p3 | Fn. We then have
by (4.9)
Qλ2n Uλ1−λ2(Pn, Qn) ≡ (λ1 − λ2)(Pn/2)λ1+λ2−1 (mod p3).
Further, since p | p(p−λ0−1)!λ1!λ2!λ3!
when λ1 6= p, we can say
CpnCn
= Up(Pn, Qn) +∑λ1 6=p
(−1)λ0p(p− λ0 − 1)!
λ1!λ2!λ3!Rn(p−λ1−λ2)Qλ2
n Uλ1−λ2(Pn, Qn)
≡ Up(Pn, Qn) +∑λ1 6=p
(−1)λ0p(p− λ0 − 1)!
λ1!λ2!λ3!Rn(p−λ1−λ2)(λ1 − λ2)(Pn/2)λ1+λ2−1
≡ Up(Pn, Qn) +Rn(p−1)[Kp(Wn/2R
n)− p(Pn/2Rn)p−1]
(mod p4).
Now,
2p−1Up(Pn, Qn) ≡ pP p−1n + ∆n
(p
3
)P p−3n (mod p4).
Since p3 | Cn we have p4 | ∆n, yielding
Up(Pn, Qn) ≡ p(Pn/2)p−1 (mod p4),
108
and so
CpnCn
≡ p(Pn/2)p−1 +Rn(p−1)Kp(Wn/2Rn)− p(Pn/2)p−1 (mod p4)
≡ Rn(p−1)Kp(Wn/2Rn) (mod p4)
≡ (Rn)p−1p3 ≡ p3 (mod p4).
Thus, in this case, if pλ || Cn, then pλ+3µ || Cpµn.
We can now state the law of repetition for {Cn}. Let pλ || Cn (λ ≥ 1).
• If p = 2, 3 then pλ+µ | Cpµn.
• If p 6= 2 and p | R, then pλ+µ || Cpµn.
• If p - R and p - Wn − 6Rn, then pλ+µ || Cpµn.
• If p - R and p | Wn − 6Rn, then pλ+3µ || Cpµn.
We next provide a closer examination of the case of p = 3: If 3 | Cn, 3 - R and
3 - Wn − 6Rn, then 3 - Wn. By Corollary 3.10.1 we have
4W3n = 3∆C2n(Wn + 2Rn) +W 2
n(Wn − 6Rn) + 24R3n,
4C3n = Cn(∆C2n + 3W 2
n).
We then can see
4C3n
Cn≡ ∆C2
n + 3W 2n (mod 9)⇒ 4
C3n
Cn≡ 3W 2
n (mod 9).
Hence 3 || C3n
Cnand 3 - W3n. Thus, if 3λ || Cn, then 3λ+µ || C3µn.
If 3 | Cn, 3 - R, 3 | Wn − 6Rn, then 3 | Wn ⇒ 3 | AnBn − 3Rn ⇒ 3 | AnBn.
Since
∆C2n = A2
nB2n + 18AnBnR
n − 4B3n − 4A3
nRn − 27Rn,
109
we see that if 3 | An, then 3 | Bn and if 3 | Bn, then 3 | An. So we have 9 | AnBn
and this implies 3 || Wn. Further 27 || 3W 2n and 3 | W3n. Thus, if 3λ || Cn and
λ ≥ 2, then
4C3n
Cn≡ 3W 2
n (mod 81).
It follows that 33 || C3n/Cn ⇒ 3λ+3µ || C3µn. If 3λ = 3, then 3λ+3µ | C3µn.
We next provide a similar examination of the p = 2 case.
Case 1. 2 | R.
If 4 | Cn, then since (W 2n −∆C2
n)/4 ∈ Z we have 2 | Wn; furthermore, 2 | AnBn as
AnBn = Wn + 3Rn. Also
27∆C2n = 4(A2
n − 3B2n)3 − (27Rn + 2A3
n − 9AnBn)2
= 4(A2n − 3B2
n)3 − (2A3n − 9Wn)2.
We have in this case that A2n−3B2
n is odd since 2 - Q⇒ 2 - Bn and therefore 2 | An.
We can use this to see that
27∆C2n ≡ 4−W 2
n (mod 8).
Thus 2 || Wn ⇒ 2 || Wn + 2Rn. But this means 2 || C2n/Cn since C2n = Cn(Wn +
2Rn). We may conclude 2µ+λ || C2µn, if 2λ || Cn and λ ≥ 2. If 2 | R and λ = 1, we
can only show that 2µ+λ | C2µn.
Case 2. 2 - R.
If 4 | Cn, then 2 | Wn ⇒ 2 | AnBn − 3Rn ⇒ 2 - AnBn ⇒ 2 | A2n − 3Bn. So
27∆C2n ≡ −(2A3
n − 9Wn)2 (mod 8).
110
Now
2A3n − 9Wn ≡ 2−Wn (mod 4)
⇒ (2A3n − 9Wn)2 ≡ (2−Wn)2 (mod 8)
⇒ 27∆C2n ≡ −(2−Wn)2 (mod 8).
If 4 | Cn, then 8 | (2 −Wn)2 ⇒ 4 | 2 −Wn ⇒ 2 || Wn. Hence 4 | Wn + 2Rn ⇒
4 | C2n/Cn ⇒ 2λ+2µ | C2µn if 2λ || Cn and λ ≥ 2. If 2 || Cn, then 2 | C2n/Cn and
4 | C2n. Hence 22µ | C2µn when µ ≥ 1.
Since (Wn, Cn, R) | 2, we can now revise the law of repetition for {Cn} in the
theorem below.
Theorem 4.21. Let pλ || Cn and (λ ≥ 1)
• If p = 2 we have two cases:
– If 2 | R, then
2λ+µ || C2µn for λ > 1,
2λ+µ | C2µn for λ = 1.
– If 2 - R, then
2λ+2µ | C2µn for λ > 1,
22µ | C2µn for λ = 1.
• If p 6= 2, then
pλ+µ || Cpµn when p - Wn − 6Rn.
111
• If p 6= 2 and pλ 6= 3, then
pλ+3µ || Cpµn when p | Wn − 6Rn.
• If pλ = 3, then
3λ+3µ | C3µn when p | Wn.
If p = 2 and 2 - ∆R, we have some additional special cases. When λ ≥ 4, we can
only have 8 || Wn−6Rn or 16 | Wn−6Rn by Theorem 4.19. If 16 | Wn−6Rn, then
since Wn + 2Rn = Wn− 6Rn + 8Rn, we see that 23 || Wn + 2Rn ⇒ 23 || C2n/Cn, as
C2n = Cn(Wn + 2Rn). Also, 16 | W2n − 6R2n. Thus 2λ+3µ || C2µn by induction.
On the other hand, if 8 || Wn − 6Rn, then since
2(W2n − 6R2n) = ∆C2n + (Wn − 6Rn)(Wn + 2Rn),
we observe that 32 | W2n − 6R2n. Further, since W2n − 6R2n + 8R2n = W2n + 2R2n
and 32 | W2n − 6R2n, we have 8 || W2n + 2R2n. Now, both (Wn − 6Rn)/8 and Rn
are odd. Thus
2 | Wn − 6Rn
8+Rn ⇒ 8 · 2 | 8
(Wn − 6Rn
8+Rn
)⇒ 16 | Wn + 2Rn.
Finally, if 2ν || Wn + 2Rn, we get 2ν || C2n/Cn ⇒ 2ν+λ || C2n, then
23(µ−1)+ν+λ || C2µn
by our earlier observation and induction on µ.
In the case of the law of repetition for the Lucas functions Un, we know that
pλ+µ || Unmpµ , if p - m and pλ || Un. Unfortunately, this result does not generalize
112
to Cn. For example, if p - Wn − 6Rn and p - 2R, it is possible for pλ || Cn and
pλ+1 | Cmn, where p - m. We note that(Wn
2Rn− 3
)CmnCn≡ m (−2 + Vm(Wn/2R
n − 1, 1)) (mod p).
If Vm(Wn/2Rn − 1, 1) ≡ 2 (mod p) and p - m, then p | Cmn/Cn. We also have the
equality of the two Legendre symbols((Wn/2R
n − 1)2 − 4
p
)=
((Wn − 6Rn)(Wn + 2Rn)
p
).
So if ((Wn − 6Rn)(Wn + 2Rn)
p
)= 1,
then
Vp−1(Wn/2Rn − 1, 1) ≡ 2 (mod p)⇒ p | C(p−1)n/Cn and p - p− 1.
Also, if ((Wn − 6Rn)(Wn + 2Rn)
p
)= −1,
then
Vp+1(Wn/2Rn − 1, 1) ≡ 2 (mod p)⇒ p | C(p+1)n/Cn and p - p+ 1.
Lastly, if ((Wn − 6Rn)(Wn + 2Rn)
p
)= 0,
then p | (Wn/2Rn−1)2−4⇒ Wn/2R
n−1 ≡ ±2 (mod p). In this case, V2(Wn/2Rn−
1, 1) ≡ (Wn/2Rn−1)2−2 ≡ (Wn/2R
n−1)2−4+2 ≡ 2 (mod p). So then p | C2n/Cn
and 2 - p.
113
4.5 The Law of Apparition for {Cn}
If a prime p divides R, it is easy to see that
Cn ≡ Qn−1Un(P,Q) (mod p),
in which case the theory reduces to that of the Lucas function Un(P,Q). We will
therefore assume p - R in what follows.
We recall that
27∆ = 4(P 2 − 3Q)3 − (27R + 2P 3 − 9QP )2.
When p | ∆ and p 6= 2, the splitting field of f(x) = x3 − Px2 + Qx − R ∈ Fp[x] is
Fp, and we have two possible cases.
Case one occurs when p | P 2−3Q. Here f(x) ≡ (x−a)3 (mod p) where a ≡ P/3
(mod p) (if p = 3, then 3 | P ). In this case we can put α = β = γ = a in Fp. Now
in Fp,
αn − βn
α− β= αn−1 + βαn−2 + β2αn−3 + · · ·+ βn−1
= nan−1,
it follows that
Cn ≡ n3a3(n−1) (mod p) and Wn ≡ 6a3n (mod p).
We may then conclude that p | Cn ⇔ p | n. Also, if p | Cn, then p | Wn − 6Rn.
Case two occurs when p - P 2 − 3Q. In this case f(x) ≡ (x− a)2(x− b) (mod p),
where
a ≡ PQ− 9R
2(P 2 − 3Q)(mod p) and b ≡ P 3 − 4PQ+ 9R
P 2 − 3Q(mod p). (4.12)
114
Hence we can put α = β = a 6= 0 and γ = b 6= 0 in Fp. Put P ′ ≡ P − a (mod p)
and Q′ ≡ a2 − Pa + Q (mod p). One can see that since a2b ≡ R (mod p), we get
ab ≡ R/a ≡ a2 − Pa + Q (mod p). Also, 2a + b ≡ P (mod p) ⇒ a + b ≡ P − a
(mod p). We use these results to obtain
Cn =
(αn − βn
α− β
)(βn − γn
β − γ
)(γn − αn
γ − α
)= nan−1
(an − bn
a− b
)2
in Fp. Thus,
Cn ≡ nan−1U2n(P ′, Q′) (mod p).
It is also true that (∆′
p) = 1, as
∆′ = P ′2 − 4Q′ ≡ (a− b)2 ≡ (27R + 2P 3 − 9PQ)2
4(P 2 − 3Q)2≡ P 2 − 3Q (mod p).
Thus p | Cn ⇔ p | n or p | Un(P ′, Q′) since p - a. If p | a or p | Q′, then p | R,
which is a contradiction. Since the rank of apparition of p in Un(P ′, Q′) is a divisor
r of p− 1, we can say p | Cn ⇔ either p | n or r | n. Since (r, p) = 1 we have two
ranks of apparition in this case. We also note that since
Wn − 6Rn ≡ 2an∆′U2n(P ′, Q′) (mod p)
we see that p | Wn − 6Rn if and only if n is a multiple of r.
We have already shown that r(2) always exists and is unique. The case for p = 3
can be handled explicitly by calculation. The results are given in Table 4.1, where
we assume 3 - R and P , Q, R are given modulo 3.
115
Table 4.1: Ranks of apparition for p = 3P Q R ∆ (mod 3) Cm ≡ 0 (mod 3) iff Corresponding Wm
2 2 2 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)2 2 1 0 m ≡ 0 (mod 2) or W2 ≡ 0 (mod 3) and
m ≡ 0 (mod 3) W3 ≡ 1 (mod 3)2 1 2 2 m ≡ 0 (mod 2) W2 ≡ 1 (mod 3)2 1 1 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)2 0 2 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)2 0 1 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)1 2 2 0 m ≡ 0 (mod 2) or W2 ≡ 0 (mod 3) and
m ≡ 0 (mod 3) W3 ≡ 2 (mod 3)1 2 1 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)1 1 2 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)1 1 1 2 m ≡ 0 (mod 2) W2 ≡ 1 (mod 3)1 0 2 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)1 0 1 2 m ≡ 0 (mod 4) W4 ≡ 1 (mod 3)0 2 2 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)0 2 1 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)0 1 2 2 m ≡ 0 (mod 4) W4 ≡ 1 (mod 3)0 1 1 2 m ≡ 0 (mod 4) W4 ≡ 1 (mod 3)0 0 2 0 m ≡ 0 (mod 3) W3 ≡ 0 (mod 3)0 0 1 0 m ≡ 0 (mod 3) W3 ≡ 0 (mod 3)
From this we see that there must always exist at least one rank of apparition for
3 in {Cn} as long as (Q,R) = 1. Also, r(3) ≤ 13 = 32 + 3 + 1. Note also that if 3 - ∆
and 3 | Cn, then r(3) | n.
4.6 Solutions of the Cubic
We now deal with those primes p such that p - 6∆R. The law of apparition for {Cn}
is more difficult than that for {Un}. This is largely due to the fact that {Cn} can have
multiple ranks of apparition, as has been seen. Just how many ranks of apparition
{Cn} actually has modulo a prime p, is dependent on the splitting behaviour of f(x)
116
modulo p. Following Adams and Shanks [AS82] we will characterize the primes that
do not divide 6∆R as follows.
Let f(x) = x3−Px2 +Qx−R and p - 6R∆. There are three possibilities for the
splitting field K of f(x) ∈ Fp[x] :
1. if K = Fp, we say that p is an S prime.
2. if K = Fp2 , we say that p is a Q prime.
3. if K = Fp3 , we say that p is an I prime.
Suppose G is the galois group of the polynomial f(x). There are four possibilities
for G: G1 = {1}, G2 a group of order 2, G3 a group of order 3 or G6 the dihedral
group of order 6. In G1 there is only one conjugacy class, in G2 there are two
conjugacy classes, in G3 there are two conjugacy classes and in G6 there are three
conjugacy classes. Let π(x) denote the number of primes up to x and for a fixed f(x)
denote by πS(x), πQ(x), πI(x) the number of S, Q, I primes up to x, respectively.
By the Chebotarev density theorem [SHWL96] we know that if G ' G1, then all the
primes are S primes. If G ' G2, then
limx→∞
πS(x)
π(x)=
1
2, and lim
x→∞
πQ(x)
π(x)=
1
2.
If G ' G3, then
limx→∞
πS(x)
π(x)=
1
3, and lim
x→∞
πI(x)
π(x)=
2
3.
Finally, if G ' G6, then
limx→∞
πS(x)
π(x)=
1
6, lim
x→∞
πQ(x)
π(x)=
1
2and lim
x→∞
πI(x)
π(x)=
1
3.
117
Determining which of these types of prime p is important, since its type dictates
where Cn = 0 in K. This is also an old problem and several references to how it
can be solved are mentioned in Chapter VIII of the first volume of [Dic19] (see also
[WZ74] and [Mul04]). As the results concerning this problem are widely scattered, we
will present a self-contained version here. Remember that δ = (α−β)(β−γ)(γ−α),
∆ = (α− β)2(β − γ)2(γ − α)2, and −27∆ = (2P 3 − 9QR + 27R)2 − 4(P 2 − 3Q)3.
If p is a Q prime, we may assume that α ∈ Fp, β, γ ∈ Fp2\Fp. Hence αp = α,
βp = γ and γp = β. Then, in K,
δp = (α− β)p(β − γ)p(γ − α)p
= (αp − βp)(βp − γp)(γp − αp)
= (α− γ)(γ − β)(β − α) = −δ.
So δp−1 = −1⇔ ∆(p−1)/2 = −1⇔ (∆p
) = −1.
In the other cases, we get either αp = α, βp = β, γp = γ or αp = β, βp = γ,
γp = α. In either case δp = δ ⇒ (∆p
) = 1. Thus p is a Q prime if and only if
(∆p
) = −1.
Now assume (∆p
) = 1, and let A = 2P 3 − 9QR + 27R and B = P 2 − 3Q. Then
−27∆ = A2 − 4B3. Let K be the splitting field of f(x) ∈ Fp[x]. Then K = Fp or
Fp3 . In either case K is a subfield of L = Fp6 . Now let F∗p6 = 〈λ〉 for some primitive
element λ of F∗p6 . Put ζ = λ(p6−1)/3, then ζ 6= 1 and ζ3 = 1⇒ (ζ − 1)(ζ2 + ζ + 1) =
0 ⇒ ζ2 + ζ + 1 = 0. This implies that αβ(ζ2 + ζ + 1) = 0 ⇒ ζ2αβ + ζαβ = −αβ.
Put
L1 = α + ζβ + ζ2γ ∈ L
L2 = α + ζ2β + ζγ ∈ L.
118
Then, L1L2 = B ∈ Fp and L31 + L3
2 = A ∈ Fp. Thus L31, L3
2 are the zeros of
x2 − Ax+B3 ∈ Fp[x].
Now since, (∆p
) = 1 and δ2 = ∆ we have δ ∈ Fp. If we put ρ = 3(ζ − ζ2)δ, then
ρ2 = −27∆ and ρ ∈ Fp2 . So ρ2 = −27∆ = A2 − 4B3 = (L31 + L3
2)2 − 4L31L
32. Notice
(L31−L3
2)2 = (L31)2− 2L3
1L32 + (L3
2)2 = (L31)2 + 2L3
1L32 + (L3
2)2− 4L31L
32 = A2− 4B3 =
−27∆⇒ L31 − L3
2 = ±ρ. This implies
2L31 = A± ρ
2L32 = A∓ ρ.
Now since L31, L3
2 are the zeros of x2 − Ax+B3 we have
Vn(A,B3) = (L31)n + (L3
2)n and Un(A,B3) =(L3
1)n − (L32)n
L31 − L3
2
.
From this we see that
2(L31)n = Vn(A,B3)± ρUn(A,B3) (4.13)
and
2(L32)n = Vn(A,B3)∓ ρUn(A,B3). (4.14)
Suppose p ≡ 1 (mod 3). If K = Fp (p is an S prime) then
Lp1 = αp + ζpβp + ζ2pγp = α + ζ3nζβ + ζ6nζ2γ = L1.
Similarly Lp2 = L2. This gives us L3( p−1
3)
1 = 1 and L3( p−1
3)
2 = 1.
If B 6= 0, then L1 6= 0 and L2 6= 0. If B = 0, then L1 = 0 or L2 = 0, but not both
since −27∆ = (L31 − L3
2)2. Without loss of generality assume L1 = 0, then L32 = A.
119
In this case p is an S prime if and only if Ap−13 ≡ 1 (mod p). Thus, if (∆
p) = 1,
p | B, then p is an S prime if and only if Ap−13 ≡ 1 (mod p).
If p - B, then L3( p−1
3)
1 = L3( p−1
3)
2 . Use the above and equations (4.13) and (4.14)
to see that U p−13
(A,B3) = 0.
Now suppose U p−13
(A,B3) ≡ 0 (mod p) ⇒ L3( p−1
3)
1 = L3( p−1
3)
2 in L by (4.13) and
(4.14). Since p - B, we have (L1/L2)p−1 = 1 ⇒ (L1/L2)p = (L1/L2), and therefore
L1/L2 ∈ Fp. Since L1L2 ∈ Fp we get L21, L2
2 ∈ Fp. Also (L31 − L3
2)2 ≡ −27∆
(mod p) and (−27∆p
) = (−3p
)(32
p)(∆
p) = 1 gives us L3
1 − L32 ∈ Fp and L3
1 − L32 6= 0. So
L31−L3
2 = (L1−L2)(L21 +L2
2 +L1L2) and L21 +L2
2 +L1L1 ∈ Fp ⇒ L1−L2 ∈ Fp. But
L21 − L2
2 ∈ Fp ⇒ L1 + L2 ∈ Fp ⇒ L1, L2 ∈ Fp. Since ζp = ζ, we have ζ ∈ Fp, and it
then follows that α, β, γ ∈ Fp ⇒ p is a S prime.
Suppose p ≡ −1 (mod 3) and K = Fp. Then B 6= 0. For if B = 0 we get
−27∆ = A2 − 4B3 = A2 and (−27∆p
) = 1. This is a contradiction since p ≡ −1
(mod 3)⇒ (−27∆p
) = −1. Now,
Lp1 = (α + ζβ + ζ2γ)p = αp + ζpβp + ζ2pγp.
Thus if p is an S prime, then Lp1 = L2 and Lp2 = L1 ⇒ Lp2
1 = L1 and Lp2
2 =
L2 ⇒ (L3(p−1)1 )(p+1)/3 = 1, (L
3(p−1)2 )(p+1)/3 = 1⇒ (L3
2/L31)(p+1)/3 = 1⇒ (L3
2)(p+1)/3 =
(L31)(p+1)/3 ⇒ U p+1
3(A,B3) = 0.
Now if U p+13
(A,B3) ≡ 0 (mod p), then L3 p+1
32 = L
3 p+13
1 in L⇒ (L2/L1)p+1 = 1⇒
Lp+12 = Lp+1
1 ⇒ (L2/L1)p = (L1/L2). So (L2/L1)p2
= (L2/L1). Hence L2/L1 ∈ Fp2
and L1/L2 ∈ Fp2 . Since ζp2
= ζ, we can employ our previous reasoning to establish
that αp2
= α, βp2
= β, γp2
= γ. Since (∆p
) = 1, we must have that α, β, γ are in
Fp or Fp3 . If p is an I prime, then αp = β, βp = γ, γp = α ⇒ αp2
= γ, βp2
= α,
120
γp2
= β. But then αp2
= γ = α, and p | ∆, which is a contradiction. Hence p is not
an I prime ⇒ p is an S prime.
It follows that if p ≡ ε (mod 3) (ε ∈ {−1, 1}), then if (∆p
) = 1 and p - B, we have
that p is an S prime if and only if U p−ε3
(A,B3) ≡ 0 (mod p).
Thus, we have proved the following theorem.
Theorem 4.22. Suppose p is a prime and p - 6∆R. If (∆p
) = −1, then p is a Q
prime. If (∆p
) = 1, p ≡ ε (mod 3), A = 2P 3 − 9QR + 27R, B = P 3 − 3Q and
p | U p−ε3
(A,B3), then p is an S prime; otherwise, p is an I prime.
We will now develop the law of apparition for a prime p in {Cn}. We first prove
the following simple lemma.
Lemma 4.23. Let p be a prime such that p - 2R∆, K be the splitting field for
f(x) ∈ Fp[x] and α, β, γ be the zeros of f(x) in K. If αn = βn and αm = βm, then
αs = βs, where n ≡ s (mod m).
Proof. We have n = qm + s and αqm+s = βqm+s. Since αqm = βqm and R 6= 0, we
get αs = βs.
We next determine the number of ranks of apparition of a Q prime.
Theorem 4.24. Let p be a Q prime and α, β, γ be the zeros of f(x) in Fp2, where
β 6∈ Fp. Then p | Cm if and only if βm = βpm.
Proof. Since p is a Q prime, f(x) has the 3 zeros, namely, α in Fp and β, γ in Fp2
such that αp = α, βp = γ, γp = β.
121
Assume p | Cm. It follows that αm = βm, βm = γm or γm = αm. If αm = βm,
then βm ∈ Fp ⇒ βpm = βm. If βm = γm, then βpm = γpm = βm. If γm = αm, then
βpm = γm = αm ⇒ βpm ∈ Fp ⇒ βp2m = βpm ⇒ βm = βpm.
On the other hand if βm = βpm, then βm = γm ⇒ p | Cm.
Corollary 4.24.1. If p is a Q prime, then p | Cp+1.
Proof. This follows directly from the theorem by noting,
βp+1 = ββp = βp2
βp = βp2+p = βp(p+1).
Corollary 4.24.2. Let p be a Q prime, then p can only have one rank, r, of appari-
tion in {Cn} and r | p+ 1.
Proof. Suppose r is the minimal rank of apparition of p and p | Cm. We must
have βm = βpm by Theorem 4.24. Put m = qr + s such that 0 < s < r. We
have βqr+s = βpqr+ps. Now βr = βrp ⇒ βqr = βqrp 6= 0. So βs = βps ⇒ p | Cs
contradicting the definition of r. It follows that s = 0 and r | m. Thus, there can
only be one rank of apparition r for p and r | p+ 1.
Corollary 4.24.3. If p is a Q prime and r is its rank of apparition in {Cn}, then
if p | Cn, we must have r | n.
Proof. The proof follows at once from Corollary 4.24.2; if there existed an n with
r - n then there would be another rank of apparition.
122
Note that if p is a Q prime then
Wp+1 ≡ 2α4βγ + 2β3γ3 + 2R3 (mod p).
This will not be useful to us here; however, we can see that
Wp2−1 ≡ 6 (mod p). (4.15)
Theorem 4.25. If p is an I prime, then p | Cp2+p+1.
Proof. Since p is an I prime,
αp = β, βp = γ and γp = α.
So
αp2+p+1 = βp
2+p+1 = γp2+p+1 = R.
Hence the result follows.
Under the same conditions we can see
Wp2+p+1 ≡ 6R3 (mod p).
Corollary 4.25.1. Let p be an I prime, then p can only have one rank, r, of appari-
tion in {Cn} and r | p2 + p+ 1.
Proof. Suppose r is the minimal rank of apparition of p for {Cn}. So, without loss
of generality αr = βr ⇒ αpr = βpr ⇒ βr = γr ⇒ αr = βr = γr. If p | Cm, then
αm = βm, βm = γm or γm = αm. Thus, if m = qr + s and 0 ≤ s < r, then p | Cs
by Lemma 4.23. By the definition of r, we must have s = 0 and r | m. Thus, there
can only be one rank of apparition of p in {Cn}. Furthermore, since p | Cp2+p+1, we
get r | p2 + p+ 1.
123
Corollary 4.25.2. If p is an I prime and r is its rank of apparition in {Cn}, then
if p | Cn, we must have r | n.
Thus, the situation with Q and I primes parallels that concerning primes that
divide Un. That is, we know that if a prime p divides Un, then the rank of apparition
ω = ω(p) of p in {Un} must divide n. However, the situation with S primes can be
different from this as we see below.
Theorem 4.26. Let p - 6∆R and p be an S prime, then p | Cp−1.
Proof. Since α, β, γ ∈ Fp ⇒ αp−1 = βp−1 = γp−1 = 1. Hence, p | Cp−1.
Once more we note that
Wp−1 ≡ 6 (mod p),
under these circumstances.
Corollary 4.26.1. Let p - 6∆R and p be an S prime, then p may have at most 3
ranks of apparition in {Cn} and each rank of apparition divides p− 1.
Proof. Let r1 be any rank of apparition of p in {Cn}. If p | Cr1 , then αr1 = βr1 or
βr1 = γr1 or γr1 = αr1 . Without loss of generality assume αr1 = βr1 . Since p is an S
prime, we know that αp−1 = βp−1. Suppose r1 is the least positive integer such that
αr1 = βr1 . Let p−1 = qr1+s such that 0 < s < r1. Then αqr1+s = βqr1+s ⇒ αs = βs.
This is a contradiction, thus r1 | p− 1.
Now suppose r1 is the minimal rank of apparition. Further, suppose p | Cr2 and
r1 - r2. We know αr2 6= βr2 , for if αr2 = βr2 , then r2 = r1q + s where 0 < s < r1 and
by Lemma 4.23 αs = βs, which is a contradiction. Thus without loss of generality
124
βr2 = γr2 and let us assume r2 is the least positive integer such that this is true.
We know by the above reasoning that r2 | p − 1. Continue in this fashion, letting
r3 | Cr3 and r1 - r3, r2 - r3. Again, by Lemma 4.23 γr3 = αr3 . Also, if we assume r3
to be the least positive integer such that γr3 = αr3 , then r3 | p − 1. Now if we try
to define r4 such that r4 | Cr4 and r1 - r4, r2 - r4, r3 - r4, then αr4 = βr4 , βr4 = γr4
or γr4 = αr4 . None of which is possible by Lemma 4.23. Thus there can be at most
3 ranks of apparition in this case.
Corollary 4.26.2. If p is an S prime and p | Cn, then at least one of the ranks of
apparition of p in {Cn} must divide n.
Proof. Without loss of generality we may suppose that αn = βn in K. We have
already seen that if r1 is the least positive integer for which αr1 = βr1 , then r1 is a
rank of apparition of p in {Cn}. Furthermore, by Lemma 4.23 we must have αs = βs,
where n = qr1 + s (0 ≤ s < r1). If s > 0, we get a contradiction to the definition of
r1; thus, s = 0 and r1 | n.
In the next theorem, we show that the case of 3 ranks of apparition can occur
infinitely often.
Theorem 4.27. Let p be a prime such that p = 2κk1k2 + 1, where k1, k2 are odd,
(k1, k2) = 1, and k1, k2 > 1. There exists a set of values for P , Q, R such that {Cn}
has 3 ranks of apparition.
Proof. We select any primitive root g of p and any integer r. In Fp, put
α = gr, β = gr+k1 , γ = gr+k1+k2 .
125
From these we can easily produce corresponding P , Q, R (mod p). Note that if we
put
r1 = 2κk2, r2 = 2κk1, r3 =2κ−1k1k2
(2κ−1k1k2, (k1 + k2)/2),
then
αr1 = βr1 , βr2 = γr2 , γr3 = αr3 .
Thus
Cr1 , Cr2 , Cr3 ≡ 0 (mod p).
Furthermore, none of r2, r2, r3 divides any of the others. Thus, there must be three
ranks of apparition for {Cn}.
We remark that there exists an infinitude of distinct primes satisfying the con-
ditions of the theorem. For if we put k2 = 2xk1 + 1 for a fixed odd k1 > 1, then by
Dirichlet’s theorem there must exist an infinitude of values of x for any fixed κ ≥ 1
such that
2κk1k2 + 1 = 2κ(2xk1 + 1)k1 + 1 = 2κ+1k21x+ 2κk1 + 1
is a prime.
Theorem 4.28. Let p be an S prime and p ≡ 1 (mod 3). Suppose that Rp−13 6≡ 1
(mod p). Then there is one and only one rank of apparition, r, of p such that r | p−13
.
Proof. Since p ≡ 1 (mod 3) we can let ζ2 +ζ+1 = 0 in Fp. Since Rp−13 6≡ 1 (mod p),
we know that
αp−13 = ζ i, β
p−13 = ζj, γ
p−13 = ζk,
where 3 - i+ j + k. Since i, j, k cannot all be the same or all different modulo 3, we
must have exactly two equal modulo 3. Without loss of generality suppose i = j 6= k.
126
Then αp−13 = β
p−13 ⇒ p | C p−1
3⇒ ∃ a rank of apparition r of p such that r | p−1
3.
Thus if αn = βn, then r | n. Suppose r1 is another rank of apparition of p such that
r1 6= r. If αr1 = βr1 , then r | r1 ⇒ r = r1, which is a contradiction.
Thus we must have βr1 = γr1 or αr1 = γr1 . If r1 | p−13
, then βp−13 = γ
p−13 or
αp−13 = γ
p−13 , neither of which is possible, as k is distinct modulo 3 from i and j.
Chapter 5
Arithmetic Properties of {Dn}
5.1 Preliminary Results for the Law of Repetition for {Dn}
While, as we have seen, Cn is analogous to the Lucas function Un in many respects,
there are a number of significant differences between the arithmetic behaviour of
Cn and Un. This is particularly the case in the law of repetition and the law of
apparition, where it is possible to have more than one rank of apparition for {Cn}.
In the law of repetition for a prime p such that p | Cn, it is important to know
whether or not p divides the quantity Wn − 6Rn. We also noted in Section 4.6 that
we often have the case of a prime p dividing both Cn and Wn− 6Rn. In view of this,
we define
Dn = gcd(Cn,Wn − 6Rn).
This is not as peculiar as it might seem at first. For if we look at the formula for
Wn − 6Rn in terms of α, β, γ, we see that the corresponding formula involving α, β
of the Lucas functions would be
α2n + β2n − 2αnβn = V2n − 2Qn.
This is because if we consider Wn − 6Rn to be a polynomial in αn, βn, γn, then it
is of degree three and the αnβnγn term is subtracted as many times as there are
terms in the expression for Wn. Hence, the degree two counter part to this would be
127
128
α2n + β2n − 2αnβn. However,
V2n − 2Qn = V 2n − 4Qn = ∆U2
n and (V2n − 2Qn, Un) = Un.
Notice that by Theorem 4.5 and Lemma 4.6 we have
(Dn, R) | 2. (5.1)
As we shall see below, it turns out that Dn has arithmetic properties which are much
more analogous to those of Un than does Cn. In order to derive the law of repetition
for {Dn} we will first develop some results for the sequence {Lm} in the same way
we derived a law of repetition for {Cn} by first using the sequence {Km}.
Let
Lm(X) =∑ (−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!Xλ1+λ2 ,
where the sum is extended over the values λi ∈ Z such that
λ0, λ1, λ2, λ3 ≥ 0, λ0 + λ1 + λ2 + λ3 = m, λ1 + 2λ2 + 3λ3 = m.
By Waring’s theorem,
Lm(X) = αm1 + αm2 + αm3 ,
where α1, α2, α3 are the zeros of Z3 −XZ2 +XZ − 1 such that
α1 = 1, α2 =X − 1 +
√D(X)
2, α3 =
X − 1−√D(X)
2,
and D(X) = (X − 3)(X + 1). We can then write
Lm(X) = 1 + αm2 + αm3 = 1 + Vm(X − 1, 1).
So if 2 - m, then by (4.6) we have
Vm(X − 1, 1) = V1
(m−1)/2∑j=0
((m− 1)/2 + j
(m− 1)/2− j
)Dj(X).
129
On the other hand, if 2 | m, then by (4.7) we have
Vm(X − 1, 1) =
m/2∑j=0
m
m/2− j
(m/2 + j − 1
m/2− j − 1
)Dj(X).
Now, by using results similar to those in Section 4.4 and noting Pn = Wn, we have
Wmn ≡ 2∑
λ0,λ1,λ2,λ3
(−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!
(Pn2
)λ1+λ2
Rn(m−λ1−λ2) (mod Fn)
where
Fn =
∆C2n if 2 - Cn
∆C2n/4 if 2 | Cn.
Let 2γ || Dn. Then Dn/2γ | Fn, 2 - Dn/2
γ and (Dn/2γ, R) = 1. Put Gn =
Dn/2γ. Then
Wmn ≡ 2Rmn∑ (−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!
(Wn
2Rn
)λ1+λ2
(mod Fn) (5.2)
≡ 2RmnLm(Wn/2Rn) (mod Fn). (5.3)
If m is odd,
Lm(Wn/2Rn) = 1 + (Wn/2R
n − 1)
(m−1)/2∑j=0
((m− 1)/2 + j
(m− 1)/2− j
)(Wn/2R
n + 1)j
(Wn/2Rn − 3)j.
Since Wn/2Rn − 3 ≡ 0 (mod Gn),
Lm(Wn/2Rn) ≡ 1 +Wn/2R
n − 1 ≡ 3 (mod Gn).
If m is even,
Lm(Wn/2Rn) = 1 +
m/2∑j=0
m
m/2− j
(m/2 + j − 1
m/2− j − 1
)(Wn/2R
n + 1)j
(Wn/2Rn − 3)j
≡ 3 (mod Gn).
130
Thus, Wmn ≡ 6Rmn (mod Gn)⇒ Gn | Wmn − 6Rmn.
It follows that if γ = 0, then Dn | Dnm.
If γ = 1, then since 2 | (Wn, Cn), we have 2 | Cmn, and since Qmn is an integer,
2 | Wmn; thus, Dn | Dmn.
If γ > 1, then 4 | Cn and 4 | Wn−6Rn. Recall that if 2α || (Wn, Cn), then α = 0
or 1 by Theorem 4.5. In this case α = 1 and 2 || Wn. It follows from 4 | Wn − 6Rn
that R must be odd. Thus, since 2γ ≥ γ + 2, we have 2γ | Fn and
Wmn ≡ 2Rmn∑ (−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!
(Wn
2Rn
)λ1+λ2
(mod 2γ)
≡ 6Rmn (mod 2γ),
so that 2γ | Wmn − 6Rmn ⇒ Dn | Wmn − 6Rmn ⇒ Dn | Dmn. Thus, if n | m, we
get Dn | Dm. Therefore, like {Un} and {Cn}, {Dn} is a divisibility sequence.
5.2 The Law of Repetition for {Dn}
The goal in this section is to develop a law of repetition for {Dn}.
Suppose pµ || Dn, µ ≥ 1 and p - 6∆. There are three cases to consider.
Case 1. pµ || Wn − 6Rn and pν || Cn such that ν > µ ≥ 1.
Note that µ ≥ 2 by Theorem 4.18, and hence 3µ ≥ 2µ + 2. We then have p2ν | Fn.
Thus, by (5.3)
Wpn ≡ 2RpnLp(Wn/2Rn) (mod p2ν) and
Lp(Wn/2Rn) ≡ 1 + (Wn/2R
n − 1)
[1 +
(p2 − 1
8
)(Wn/2R
n + 1)(Wn/2Rn − 3)+
(p2 − 1)(p2 − 9)
16 · 24(Wn/2R
n + 1)2(Wn/2Rn − 3)2
](mod p3µ).
131
Now, since
Wn/2Rn + 1 ≡ 4 (mod pµ) and p2µ | (Wn/2R
n − 3)2,
we have
(Wn/2Rn + 1)2(Wn/2R
n − 3)2 ≡ 16(Wn/2Rn − 3)2 (mod p3µ).
Thus
Lp(Wn/2Rn) ≡ 1 + (Wn/2R
n − 1)
[1 +
(p2 − 1
8
)(Wn/2R
n + 1)(Wn/2Rn − 3)+
(p2 − 1)(p2 − 9)
24(Wn/2R
n − 3)2
]≡ Wn/2R
n +
(p2 − 1
8
)((Wn/2R
n)2 − 1)(Wn/2Rn − 3) +
(p2 − 1)(p2 − 9)
24(Wn/2R
n − 1)(Wn/2Rn − 3)2 (mod p3µ).
Now
Wn/2Rn − 1 ≡ 2 (mod pµ) and p2µ | (Wn/2R
n − 3)2
so
(Wn/2Rn − 1)(Wn/2R
n − 3)2 ≡ 2(Wn/2Rn − 3)2 (mod p3µ).
Thus,
Lp(Wn/2Rn) ≡ Wn/2R
n +
(p2 − 1
8
)((Wn/2R
n)2 − 1)(Wn/2Rn − 3) +
(p2 − 1)(p2 − 9)
12(Wn/2R
n − 3)2 (mod p3µ).
Also, we have
(Wn/2Rn − 3)2 ≡ 0 (mod p2µ)⇒ (Wn/2R
n)2 − 6Wn/2Rn + 9 ≡ 0 (mod p2µ)
132
⇒ (Wn/2Rn)2 − 1 ≡ 6Wn/2R
n − 10 (mod p2µ).
Note, then, that
((Wn/2Rn)2 − 1)(Wn/2R
n − 3) ≡ (6Wn/2Rn − 10)(Wn/2R
n − 3) (mod p3µ).
This yields
Lp(Wn/2Rn) ≡ Wn/2R
n +
(p2 − 1
8
)(6Wn/2R
n − 10)(Wn/2Rn − 3) +
(p2 − 1)(p2 − 9)
12(Wn/2R
n − 3)2
≡[(Wn/2R
n − 3) +
(p2 − 1
8
)(6Wn/2R
n − 10)(Wn/2Rn − 3)+
(p2 − 1)(p2 − 9)
12(Wn/2R
n − 3)2
]+ 3 (mod p3µ).
This can be rewritten as
Lp(Wn/2Rn)− 3 ≡ (Wn/2R
n − 3)
[1 +
(p2 − 1)
4(3Wn/2R
n − 9 + 4)+
(p2 − 1)(p2 − 9)
12(Wn/2R
n − 3)
]≡ (Wn/2R
n − 3)
[1 + 3
(p2 − 1)
4(Wn/2R
n − 3) + 4(p2 − 1)
4+
(p2 − 1)
4
(p2 − 9)
3(Wn/2R
n − 3)
]≡ p2(Wn/2R
n−3) + (Wn/2Rn−3)2
[3
(p2 − 1)
4+
(p2 − 1)
4
(p2 − 9)
3
]≡ p2(Wn/2R
n − 3) +(p2 − 1)
4(Wn/2R
n − 3)2
[3 +
(p2 − 9)
3
]≡ p2(Wn/2R
n − 3) + p2 (p2 − 1)
12(Wn/2R
n − 3)2 (mod p3µ)
≡ p2(Wn/2Rn − 3) (mod p2µ+2).
Observe that
Wpn − 6Rpn ≡ 2RpnLp(Wn/2Rn)− 6Rpn (mod p2ν)
133
and that ν > µ⇒ ν ≥ µ+ 1⇒ 2ν ≥ 2µ+ 2. So
Wpn − 6Rpn ≡ 2Rpn
[p2(Wn/2R
n − 3) + p2 (p2 − 1)
12(Wn/2R
n − 3)2
](mod p2µ+2)
≡ 2Rpn[p2(Wn/2R
n − 3)]
(mod p2µ+2)
≡ (Rn)p−1p2(Wn − 6Rn) (mod p2µ+2)
≡ (1 + kp)p2(Wn − 6Rn) (mod p2µ+2),
for some k ∈ Z. Since pµ || Wn−6Rn and 2µ+2 > µ+2 we have pµ+2 || Wpn−6Rpn.
We also know that pµ+3 | Cpn; hence, pµ+2 || Dpn.
Case 2. pµ || Wn − 6Rn and pµ || Cn.
We then have p2µ | Fn which gives us
Wpn ≡ 2RpnLp(Wn/2Rn) (mod p2µ) and
Lp(Wn/2Rn) ≡ 1 + (Wn/2R
n − 1)
[1 +
((p+ 1)/2
2
)(Wn/2R
n + 1)(Wn/2Rn − 3)
]≡ Wn/2R
n +(p2 − 1)
8((Wn/2R
n)2 − 1)(Wn/2Rn − 3) (mod p2µ).
Now, note that since
Wn/2Rn + 1 ≡ 4 (mod pµ) and Wn/2R
n − 1 ≡ 2 (mod pµ),
we have
((Wn/2Rn)2 − 1) ≡ (Wn/2R
n − 1)(Wn/2Rn + 1) ≡ 8 (mod pµ).
Together with the fact that pµ | Wn/2Rn − 3, we see that
((Wn/2Rn)2 − 1)(Wn/2R
n − 3) ≡ 8(Wn/2Rn − 3) (mod p2µ).
134
So
Lp(Wn/2Rn) ≡ Wn/2R
n + (p2 − 1)(Wn/2Rn − 3) (mod p2µ).
Thus
Wpn − 6Rpn ≡ 2Rpn[Wn/2R
n + (p2 − 1)(Wn/2Rn − 3)
]− 6Rpn
≡ Rn(p−1)[Wn + (p2 − 1)(Wn − 6Rn)− 6Rn
]≡ p2(Wn − 6Rn) (mod p2µ).
Since by Theorem 4.18, µ ≥ 3, we get 2µ ≥ µ+ 3 and we can write
Wpn − 6Rpn ≡ p2(Wn − 6Rn) (mod pµ+3).
Using the fact pµ || Wn − 6Rn, we get pµ+2 || Wpn − 6Rpn. Again we know that
pµ+3 || Cpn ⇒ pµ+2 || Dpn.
Case 3. pν || Cn and pµ || Wn − 6Rn such that ν < µ.
In this case pν+3 || Cpn and
Wpn − 6Rpn ≡ p2(Wn − 6Rn) (mod p2ν)
≡ p2(Wn − 6Rn) ≡ 0 (mod pν+3).
Thus pν+3 | Wpn − 6Rpn ⇒ pν+3 || Dpn.
If p > 3, p | ∆, pµ || Wn−6Rn and pν || Cn such that ν ≥ µ, we have p2ν+1 | Fn.
Since
Qλ2n Vλ1−λ2(Pn, Qn) ≡ 2(Pn/2)λ1+λ2 (mod 22ν+1)
and p | p(p−λ0−1)!λ1!λ2!λ3!
except for λ1 = p we find, by using the reasoning in Section 4.4,
that
Wpn ≡ 2Rpn [Lp(Wn/2Rn)− (Wn/2R
n)p] + Vp(Pn, Qn) (mod p2ν+2).
135
We may use the fact that
Vp(Pn, Qn) ≡ Pn(Pn/2)p−1 + P p−2n ∆n
(p
2
)(mod p2ν+2)
≡ Pn(Pn/2)p−1 (mod p2ν+2)
to get
Wpn ≡ 2RpnLp(Wn/2Rn)− 2Rpn(Wn/2R
n)p +Wn(Wn/2)p−1
≡ 2RpnLp(Wn/2Rn) (mod p2ν+2).
Since
Wpn − 6Rpn ≡ p2R(p−1)n(Wn − 6Rn) (mod p2µ+2),
we get pµ+2 || Dpn when ν ≥ µ(≥ 1).
If ν < µ, then pν+3 | Wpn − 6Rpn and pν+3 || Dpn.
If p = 3, 3µ || Dn, 3µ || Wn − 6Rn and 3ν || Cn where ν ≥ µ, then we use
4(W3n − 6R3n) = 3∆C2n(Wn + 2Rn) +W 2
n(Wn − 6Rn).
We have 32ν+1 | 3∆C2n and
4(W3n − 6R3n) ≡ W 2n(Wn − 6Rn) (mod 32ν+1).
Suppose µ > 1. Since ν ≥ µ ≥ 2⇒ 2ν + 1 ≥ µ+ 3 we have
4(W3n − 6R3n) ≡ W 2n(Wn − 6Rn) (mod 3µ+3).
Since (Dn, R) | 2, and 9 | Wn − 6Rn, we must have 3 || Wn. Thus for µ > 1 we
have 3µ+2 || D3n. If µ = 1, all we can say is that 3µ+2 | D3n.
136
Now if p = 3, µ > 1, 3µ || Cn and 3ν || Wn− 6Rn such that ν > µ, we have that
3µ+3 || C3n and
4(W3n − 6R3n) ≡ W 2n(Wn − 6Rn) (mod 32µ+1)
≡ W 2n(Wn − 6Rn) (mod 3µ+3)
≡ 0 (mod 3µ+3),
hence 3µ+3 || D3n. If µ = 1, we can only say that 3µ+3 | D3n.
For the case where p = 2 and 2 - R we have the following. If 2 | Dn, then
2 | Wn and 2 | Cn, hence 2 | C2n/Cn as C2n = Cn(Wn + 2Rn). Also, since Qn =
(W 2n −∆C2
n)/4 is odd in this case, we get W 2n −∆C2
n ≡ 4 (mod 8) and
W 2n + ∆C2
n ≡ 4 + 2∆C2n ≡ 4 (mod 8).
Now
2W2n = ∆C2n +W 2
n − 4RnWn ≡ ∆C2n +W 2
n ≡ 4 (mod 8).
This means
W2n ≡ 2 (mod 4)⇒ 2 || Wn ⇒ 4 | W2n − 6R2n.
Now suppose further that 2m | Dn and m ≥ 2. So then
22m | (Wn − 6Rn)2 ⇒ W 2n ≡ 12RnWn − 36R2n (mod 22m).
This, together with the identity 2W2n = ∆C2n +W 2
n − 4RnWn yields,
2W2n ≡ −36R2n + 12RnWn − 4RnWn ≡ 8RnWn − 36R2n (mod 22m).
137
So then
2(W2n − 6R2n) ≡ 8RnWn − 36R2n − 12R2n (mod 22m)
≡ 8RnWn − 48R2n (mod 22m)
≡ 8Rn(Wn − 6Rn) (mod 22m).
Hence,
W2n − 6R2n ≡ 4Rn(Wn − 6Rn) (mod 22m−1),
and since m ≥ 2, we get 2m− 1 ≥ m+ 1, so
W2n − 6R2n ≡ 4Rn(Wn − 6Rn) (mod 2m+1).
Thus we can see 2m+1 | W2n − 6R2n. Thus, if 2m | Dn, then 2m+1 | D2n.
When m ≥ 4 we can say more. Here we have 16 | Wn−6Rn, so 2 - R, since other-
wise 4 | (Wn, Cn) and this is a contradiction because (Wn, Cn) | 2 when (Q,R) = 1
by Theorem 4.5. Now
Wn + 2Rn = Wn − 6Rn + 8Rn ⇒ 8 || Wn + 2Rn ⇒ 8 || C2n/Cn.
Also, if 2ν || Cn, then
2W2n ≡ W 2n − 4RnWn (mod 22ν),
or
2(W2n − 6R2n) ≡ (Wn − 6Rn)2 + 8Rn(Wn − 6Rn) (mod 22ν).
Now let 2µ || Wn − 6Rn.
Case 1. If ν ≥ µ, we get
2(W2n − 6R2n) ≡ 8Rn(Wn − 6Rn) (mod 22µ).
138
Hence
W2n − 6R2n ≡ 4Rn(Wn − 6Rn) (mod 22µ−1).
Now we use the fact µ ≥ 4 to see that
W2n − 6R2n ≡ 4Rn(Wn − 6Rn) (mod 2µ+3).
Hence 2µ+2 || W2n − 6R2n and 2ν+3 || C2n ⇒ 2m+2 || D2n ⇒ 4 || D2n/Dn when
m = µ. Thus by induction we get 22k || D2kn/Dn.
Case 2. If ν < µ, then we have 2ν+3 | W2n−6R2n and 2ν+3 || C2n ⇒ 2ν+3 || D2n ⇒
23 || D2n/Dn. However, since we do not know that 2ν+4 | W2n − 6R2n, the best we
can say in general is that, if 2γ || D2kn/Dn, then γ ≤ 3k.
Theorem 5.1. If p is a prime, pλ || Dn (λ ≥ 1) and p - m, then pλ || Dmn.
Proof. Suppose first that p 6= 2. If 2 - m, then since p | Dn, we get
CmnCn
≡ mRn(m−1) +m
(m2 − 1
8
)Rn(m−3)
4(Wn − 2Rn)(Wn + 2Rn) (mod p)
≡ mRn(m−1) +m
(m2 − 1
8
)Rn(m−3)
432R2n (mod p)
≡ m3Rn(m−1) (mod p).
If 2 | m, then
CmnCn
≡ m
(m
m/2− 1
(m/2
2
)Rn(m−2)
2(Wn + 2Rn)
)(mod p)
≡ m
((m
m/2− 1
)((m/2)(m/2− 1)
2
)Rn(m−2)
28Rn
)(mod p)
≡ m3Rn(m−1) (mod p).
Suppose pλ || Dn. There two cases: either pλ || Cn or pλ || Wn − 6Rn. If pλ || Cn
and p - m, then pλ+1 - Cmn, and we have pλ || Dmn.
139
On the other hand, if pλ || Wn − 6Rn, then pλ | Cn and p2λ | Fn. If 2 - m,
Lm(Wn/2Rn) ≡ 1 + (Wn/2R
n − 1) +
(m2 − 1
8
)(Wn/2R
n − 1)(Wn/2Rn + 1)
(Wn/2Rn − 3) (mod p2λ)
≡ Wn/2Rn +
(m2 − 1
8
)8(Wn/2R
n − 3) (mod p2λ)
≡ Wn/2rn +m2(Wn/2R
n − 3)− (Wn/2Rn − 3) (mod p2λ)
≡ m2(Wn/2Rn − 3) + 3 (mod pλ+1).
If 2 | m we get
Lm(Wn/2Rn) ≡ 1 +
m
m/2+
m
m/2− 1
(m/2
2
)(Wn/2R
n + 1)(Wn/2Rn − 3) (mod p2λ)
≡ 3 +
(m
m/2− 1
)(m/2(m/2− 1)
2
)4(Wn/2R
n − 3) (mod p2λ)
≡ m2(Wn/2Rn − 3) + 3 (mod pλ+1).
We can then see
Wmn − 6Rmn ≡ 2RmnLm(Wn/2Rn)− 6Rmn (mod pλ+1)
≡ 2Rmn (Lm(Wn/2Rn)− 3) (mod pλ+1)
≡ 2Rmn(m2(Wn/2R
n − 3))
(mod pλ+1).
But then pλ+1 | Wmn−6Rmn ⇒ p | m. Hence if p - m and pλ || Dn, then pλ || Dmn.
Now for the case of p = 2. Since 2 | Dn, we have 2 | Cn and 2 | Wn. This means
that 2 | Pn and 2 - Qn by Theorem 4.5. By Theorem 4.10, we have
Cmn/Cn ≡ m (mod 2).
Thus, if 2 - m, then 2 - Cmn/Cn. If 2λ || Cn, then 2λ+1 - Cmn and so 2λ+1 - Dmn
140
when 2 - m. If 2λ+1 | Cn, then 2λ || Wn− 6Rn. In this case 22λ | Fn. If 2 - m, then
Wmn ≡ 2RmnLm(Wn/2Rn) (mod 22λ)
≡ 2Rmn(m2(Wn/2Rn − 3) + 3) (mod 2λ+1).
Thus
Wmn − 6Rmn ≡ 2m2Rmn(Wn/2Rn − 3) (mod 2λ+1)
≡ m2Rm(n−1)(Wn − 6Rn) (mod 2λ+1).
Since 2λ || Wn − 6Rn, we get 2λ || Wmn − 6Rmn when m is odd. Hence 2λ+1 - Dmn
when 2 - m.
Thus, we have shown that if p is any prime and pλ || Dn, then pλ || Dmn when
p - m.
Our Law of Repetition for {Dn} is stated in the following theorem.
Theorem 5.2. If pλ || Dn (p 6= 2, pλ 6= 3), then
pλ+2 || Dpn when pλ || Wn − 6Rn and
pλ+3 || Dpn otherwise .
Also, pλ+2 | Dpn when pλ = 3 and pλ+1 | Dpn when p = 2. Furthermore, pλ+1 - Dmn
if p - m.
Notice that if p 6= 2, 3 and pλ || Wn − 6Rn, pλ || Dn, then pλ+2 || Wpn − 6Rpn
and therefore
pλ+2µ || Dpµn.
141
However, if pλ || Dn and pλ+1 | Wn − 6Rn, it is not necessarily the case that
pλ+4 || Wpn − 6Rpn.
The best we are able to show is that pλ+3 | Wpn−6Rpn. If pλ+3 || Wpn−6Rpn, then
we return to the previous condition and by induction we get
(pλ+1+2µ =)pλ+3+2(µ−1) || Dpµn.
Of course, this latter situation would never occur if the case of
pλ || Dn and pλ+1 | Wn − 6Rn
could not happen. We have given some reason in Chapter 4 to believe that this
might be an infrequent occurrence, but unfortunately it does happen. For example,
if P = 257, Q = 2004 and R = 5389, then 73 || C6 and 74 | W6 − 6R6.
Thus, we cannot provide as complete a law of repetition for {Dn} as we were
able to do for {Cn}. However, if pλ || Cn, pλ+κ || Wn − 6Rn and κ < λ − 2, it can
be shown that
pλ+3µ || Cpµn and pλ+κ+2µ || Wpµn − 6Rpµn.
Hence, we get
pλ+3µ || Dpµn for µ ≤ κ.
Note that if µ = κ, then λ+ κ+ 2µ = λ+ 3µ and we return to the previous case. It
follows, then, that
pλ+κ+2µ || Dpµn
when µ > κ. Unfortunately if κ ≥ λ − 2, it seems to be difficult to formulate a
comprehensive law of repetition.
142
5.3 The Law of Apparition for {Dn}
Definition 5.3. Let p be a prime and ω(p) be the least positive integer n, if it exists,
such that p | Dn. We call this the rank of apparition of p in {Dn}.
The next theorems build towards a result very comparable to Theorem 2.4. What
is remarkable is that this is a result that did not hold for {Cn}. Hence, with the help
of {Dn} we are able to establish a more convincing analogue.
Lemma 5.4. Suppose p is a prime, p - 2R∆ and K is the splitting field of f(x) in
Fp[x]. If α, β, γ are the zeros of f(x) in K, then p | Dn if and only if αn = βn = γn
in K.
Proof. (⇒) If p | Dn, then p | Cn and we may assume with no loss of generality
that αn = βn. Since
Wn − 6Rn = 2βn(αn − γn)2 − (αn − βn)(βn − γn)(αn + γn),
it follows that
2βn(αn − γn)2 ≡ (αn − βn)(βn − γn)(αn + γn) (mod p),
as p | Wn − 6Rn and hence αn = γn.
(⇐)On the other hand, if αn = βn = γn, then it is clear that p | Dn.
We are now able to present an important result concerning ω(p).
Theorem 5.5. Suppose p is a prime such that p - 2R∆ and suppose further that
ω = ω(p) exists for {Dn}. If p | Dn, then ω | n.
143
Proof. Since p | Dn and p | Dω we have αn = βn = γn and αω = βω = γω in K by
Lemma 5.4. If ω - n, then n = ωq + r, where 0 < r < ω. By Lemma 4.23, we get
αr = βr = γr and p | Dr by Lemma 5.4. Since this contradicts the definition of ω,
we must have r = 0 and ω | n.
We next note that if p | R and p 6= 2, then ω(n) does not exist. If 2 | R, then
2 - Q. By our results in Theorem 4.15, there exists a rank of apparition r for {Cn}
of 2. Since 2 | Cn ⇒ 2 | Wn, we see that ω = ω(2) = r and 2 | Dn if and only if
r | n.
Theorem 5.6. If p - R, then ω(p) must exist. Further, if p | Dn, then ω(p) | n.
Proof. We have already seen that this holds if p = 2. We now turn our attention to
the case of p = 3. If 3 | Cn, then by Corollary 3.10.1 we have
8W4n ≡ 4W 22n − 16R2nW2n (mod 3)
≡ 2W2n(2W2n − 8R2n) (mod 3)
≡ (W 2n − 4RnWn)((W 2
n − 4RnWn − 8R2n) (mod 3)
≡ Wn(Wn −Rn)((W 2n + 2RnWn +R2n) (mod 3)
≡ Wn(Wn −Rn)(Wn +Rn)2 (mod 3)
≡ 0 (mod 3),
since 3 - R, then 3 must divide one of Wn, Wn − Rn or Wn + Rn. Thus, since r(3)
exists and is unique when 3 - ∆, we see by Table 4.1 that ω(3) exists. By Theorem
5.5, we also see that ω(3) | n if p | Dn. If p - 6∆R, then p is either an S prime,
I prime or Q prime. If p is an I prime, then αp = β, βp = γ, γp = α in Fp3 , the
144
splitting field of f(x). Hence
αp2+p+1 = βp
2+p+1 = γp2+p+1 = R
and p | Dp2+p+1.
If p is an S prime, then αp−1 = βp−1 = γp−1 = 1 in the splitting field Fp of f(x).
Hence, p | Dp−1 by Lemma 5.4.
If p is a Q prime, then αp2−1 = 1, βp
2−1 = (βp+1)p−1 = (γβ)p−1 = (R/α)p−1 = 1,
γp2−1 = 1 and p | Dp2−1. Thus, if p - 6∆R, then ω(p) exists, and if p | Dn, then
ω(p) | n by Theorem 5.5.
When p | ∆ and p 6= 2, we have seen that there are two cases:
Case 1. p | P 2 − 3Q.
In this case, we have seen in Section 4.3 that there is a unique r = r(p) = p, and
p | Cn if and only if r | n. Also, since, in this case, p | Wn− 6Rn whenever p | Cn,
we have ω = r and ω | n if p | Dn.
Case 2. p - P 2 − 3Q.
In this case, we know that by our results in Section 4.3 that p | Dn if and only if r | n.
Here r is the least positive integer such that p | ar−br, where a, b are as in equation
(4.12). Thus ω(p) = r and r | p− 1. Note that in both cases p | Dn ⇒ ω(p) | n.
Thus if p - R, we have shown that ω(p) always exists and ω(p) | n whenever
p | Dn.
Corollary 5.6.1. If p is a prime and ω(p) exists, then ω(p) ≤ p2 + p+ 1.
If p is an I prime, we have a very simple result connecting divisibility of Cn and
Dn by p.
145
Theorem 5.7. If p is an I prime, then p | Cn ⇔ p | Dn.
Proof. Clearly, if p | Dn, then p | Cn. Let α, β, γ be the zeros of f(x) in K = Fp3 .
In K we must have αn = βn ⇒ αpn = βpn ⇒ βn = γn ⇒ αn = βn = γn. It follows
that Wn = 6αnβnγn = 6Rn in K. Thus, p | Wn − 6Rn ⇒ p | Dn.
Since for any Q prime p we know that p | Cp+1, it is of some interest to determine
under what conditions p | Dp+1. We require a simple lemma.
Lemma 5.8. Let α, β, γ be the zeros of f(x) in Fp2 where αp = α, βp = γ, γp = β.
If p | Q3 −RP 3, then α3 = R in Fp2.
Proof. Suppose p | Q3 − RP 3. If p | P , then p | Q and f(x) ≡ x3 − R (mod p);
hence, α3 = R in Fp ⊆ Fp2 . If p - P , then
f(x) ≡ x3 − Px2 +Qx− (Q/P )3 (mod p)
and
P 3f(x) ≡ (Px−Q)(P 2x2 + (PQ− P 3)x+Q2) (mod p).
Because β, γ 6∈ Fp it follows that Pα−Q = 0 in Fp; hence,
α3 = Q3/P 3 = R in Fp2 .
We are now able to show that for a fixed triple P , Q, R, there can only be a finite
number of Q primes such that p | Dp+1.
Theorem 5.9. If p is a Q prime, then p | Dp+1 if and only if p | Q3 −RP 3.
146
Proof. Since p is a Q prime by Corollary 4.24.1, we have p | Cp+1. In Fp2 , we have
Wp+1 = 2R2 + 2(βγ)3 + 2Rα3;
hence, since α 6= 0 (R 6= 0), we have
Wp+1 − 6Rp+1 = 2(βγ)3 + 2Rα3 − 4R2
= 2((R/α)3 +Rα3 − 2R2)
=2R
α3(α3 −R)2.
It follows that p | Wp+1−6Rp+1 if and only if α3 = R in Fp2 . By Lemma 5.8 we know
that if p | Q3 − RP 3, then α3 = R. If α3 = R, then since α3 − Pα2 +Qα− R = 0,
we get Pα2 −Qα = 0 and hence Pα−Q = 0. If P = 0 in Fp2 , then p | Q3 − RP 3.
If P 6= 0, then α = Q/P and R = (Q/P )3 ⇒ p | Q3 −RP 3.
Suppose p - R and pα | Dn. Let ω = ω(p) and let ω(pα) denote the least positive
integer k such that pα | Dk. If pα | Dω, put ν = 0; otherwise define ν ∈ Z≥0 by
pα | Dpνω(p), pα - Dpν−1ω(p).
By our previous results concerning the law of repetition for {Dn} such a ν must
exist.
Theorem 5.10. If p - R, pα | Dn, then ω(pα) = pνω(p) and ω(pα) | n.
Proof. Since p | Dn, we must have n = mω(p) for some m ∈ N. Suppose pγ || m
and put m = m′pγ (p - m′). Since pα - Dpν−1ω(p) we must have pα - Dm′pν−1ω(p) ⇒
γ > ν − 1. But pα | Dm′pνω(p) ⇒ γ = ν.
Furthermore, since pα | Dpνω(p) we must have ω(pα) = pνω(p) and ω(pα) | n.
147
Theorem 5.11. Suppose m | Dn. Denote by ω(m) the least positive integer such
that m | Dω(m). Let
m =k∏i=1
pαii ,
then ω(m) = lcm[ω(pαii ); i = 1, 2, . . . k].
Proof. Clearly ω(pαii ) | ω(m) for i = 1, 2, . . . k. Since Dn is a divisibility sequence
the result follows.
We may now prove the following theorem, which very much resembles Corollary
2.4.1. Again, as seen in Chapter 4, this is another result that did not hold for {Cn}.
Theorem 5.12.
(Dn, Dm) = D(m,n).
Proof. Since Dn is a divisibility sequence we have D(m,n) | Dn, D(m,n) | Dm ⇒
D(m,n) | (Dn, Dm). Let pα || (Dn, Dm), then ω(pα) exists and ω(pα) | n, ω(pα) | m
hence ω(pα) | (m,n)⇒ pα | D(m,n). Thus (Dn, Dm) = D(m,n).
In Chapter 4 we were able to develop a result somewhat akin to Carmichael’s
result in Theorem 2.5. Surprisingly, if we look at {Dn} rather than {Cn}, we can in
fact do better. We have that
CmnCn≡ m3Rn(m−1) (mod (Fn,Wn − 6Rn)),
where Fn is as in (4.11). If 2ν || Cn and ν > 1, then Cn | Fn and
CmnCn≡ m3Rn(m−1) (mod Dn). (5.4)
148
If 2 || Cn, then CmnCn≡ m (mod 2) (Theorem 4.10) and Cn/2 | Fn so
CmnCn≡ m3Rn(m−1) (mod (Cn/2,Wn − 6Rn)).
Now (Dn, R) | 2. If (Dn, R) = 1, then (CmnCn
, Dn) | m3.
If (Dn, R) = 2, then by Lemma 4.4, 2 || Dn. We then have (CmnCn
, Dn/2) | m3.
Since (Dn/2, R) = 1 and (CmnCn
, 2) | m we have (CmnCn
, Dn) | m3.
We next examine (DmnDn
, Dn). Let pα || (DmnDn
, Dn). We will show that pα | m3
when p is a prime and α ≥ 1. This means of course that
(Dmn
Dn
, Dn) | m3.
We first observe that if p - m, then p - Dmn/Dn, which is a contradiction to Theorem
5.1, so p | m. If α < 4, then pα | m3. If α ≥ 4, then by the law of repetition for
Dn, we know that pλ || Dmn with λ ≤ 3µ + ν where pν || Dn (ν ≥ 4) and pµ | m.
Thus if pγ || Dmn/Dn, then γ = λ − ν ≤ 3µ ⇒ pγ | m3. We now have the desired
analogue of Theorem 2.5.
Theorem 5.13. If m, n ≥ 1, then
(Dmn/Dn, Dn) | m3.
The next two theorems will be needed in Chapter 6 to produce an analogue to
Euler’s criterion for the Lucas functions.
Theorem 5.14. Let p be an I prime and p ≡ 1 (mod 3), then
C p2+p+13
≡ 0 (mod p) and W p2+p+13
≡ 6Rp2+p+1
3 (mod p)
if and only if Rp−13 ≡ 1 (mod p).
149
Proof. We have α, β, γ as the zeros of f(x) in Fp3 where β = αp, γ = βp. In Fp3 we
get R = αp2+p+1 and
Rp−13 = (αp
2+p+1)p−13 = α
p3−13 .
Hence
αp2+p+1
3 Rp−13 = α
p2+p+13 α
p3−13 = α
p2+p+13
+ p3−13 = α
p3+p2+p3 = αp(
p2+p+13
).
This yields
βp2+p+1
3 = αp2+p+1
3 Rp−13 .
Similarly, one may show
γp2+p+1
3 = βp2+p+1
3 Rp−13 or γ
p2+p+13 = α
p2+p+13 R
2(p−1)3 .
Now if Rp−13 ≡ 1 (mod p), then p | C p2+p+1
3
. If Rp−13 6≡ 1 (mod p), then p - C p2+p+1
3
.
Then Rp−1 = 1 and
αp2+p+1
3 − βp2+p+1
3 = αp2+p+1
3 [1−Rp−13 ],
βp2+p+1
3 − γp2+p+1
3 = αp2+p+1
3 [Rp−13 −R
2(p−1)3 ],
γp2+p+1
3 − αp2+p+1
3 = αp2+p+1
3 [R2(p−1)
3 − 1]
imply
∆C2p2+p+1
3
= −27R2 6= 0.
150
Now note
W p2+p+13
= α2(p2+p+1)
3 βp2+p+1
3 + β2(p2+p+1)
3 αp2+p+1
3 + β2(p2+p+1)
3 γp2+p+1
3 +
γ2(p2+p+1)
3 βp2+p+1
3 + γ2(p2+p+1)
3 αp2+p+1
3 + α2(p2+p+1)
3 γp2+p+1
3
= α2(p2+p+1)
3 αp2+p+1
3 Rp−13 + α
2(p2+p+1)3 α
p2+p+13 R
2(p−1)3 +
α2(p2+p+1)
3 R2(p−1)
3 αp2+p+1
3 R2(p−1)
3 + α2(p2+p+1)
3 Rp−13 α
p2+p+13 R
p−13 +
α2(p2+p+1)
3 Rp−13 α
p2+p+13 + α
2(p2+p+1)3 α
p2+p+13 R
2(p−1)3
= 3R[Rp−13 +R
2(p−1)3 ].
If Rp−13 ≡ 1 (mod p), then WP2+p+1
3
− 6R ≡ 0 (mod p). Also, since 3 | p + 2, we
have R(p+2)(p−1)
3 ≡ 1 (mod p) ⇒ Rp2+p+1
3 ≡ R (mod p). Thus if Rp−13 ≡ 1 (mod p),
we get p | W p2+p+13
− 6Rp2+p+1
3 . If Rp−13 6≡ 1 (mod p), then R
2(p−1)3 +R
p−13 ≡ −1 6≡ 2
(mod p) and p - W p2+p+13
− 6Rp2+p+1
3 as W p2+p+13
≡ −3Rp2+p+1
3 (mod p).
Theorem 5.15. Let p (≡ 1 (mod 3)) be a Q prime, then p | D p2−13
if and only if
Rp−13 ≡ 1 (mod p).
Proof. We have α, β, γ as the zeros of f(x) in Fp2 such that
αp = α, βp = γ and γp = β
and
αp−1 = 1, αp+1 = α2, βp+1 = βγ and γp+1 = βγ.
Now βγ = R/α and
βp2−1
3 = γp2−1
3 = (βγ)p−13 =
(R
α
) p−13
=R
p−13
αp−13
· α2(p−1)
3
α2(p−1)
3
= Rp−13 α
2(p−1)3
= Rp−13 (α2)
p−13 = R
p−13 (αp+1)
p−13 = R
p−13 α
p2−13 .
151
We know that p | C p2−13
and
W p2−13
≡ 2αp2−1(1 +R
p−13 +R
2(p−1)3 ) ≡ 2(1 +R
p−13 +R
2(p−1)3 ) (mod p).
Hence
W p2−13
− 6Rp2−1
3 ≡ W p2−13
− 6R2(p−1)
3 ≡ 2(
1−R2(p−1)
3
)2
(mod p).
Thus p | W p2−13
− 6Rp2−1
3 ⇔ Rp−13 ≡ 1 (mod p).
A companion result to Theorem 5.15 for p ≡ −1 (mod 3) is given below.
Theorem 5.16. If p is a Q prime and p ≡ −1 (mod 3), then
p | D p2−13
if and only if p | C p+13.
Proof. If p | C p+13
, then
αp+13 = β
p+13 , β
p+13 = γ
p+13 or α
p+13 = γ
p+13 .
In the first case, since αp+13 ∈ Fp, we get β
p2−13 = 1 = α
p2−13 . In the second case,
βp2−1
3 = (βp−1)p+13 = (γ/β)
p+13 = 1 = (β/γ)
p+13 = (γp−1)
p+13 = γ
p2−13 .
In the third case, since αp+13 ∈ Fp, we get γ
p2−13 = 1 = α
p2−13 . Since (αβγ)
p2−13 = 1,
we get αp2−1
3 = βp2−1
3 = γp2−1
3 in all of the three cases. It follows that p | C p2−13
and
p | W p2−13
− 6Rp2−1
3 .
Conversely, if p | D p2−13
, then
1 = αp2−1
3 = βp2−1
3 , βp2−1
3 = γp2−1
3 or 1 = αp2−1
3 = γp2−1
3 .
152
In the first and the last of these cases we get
(βp−1)p+13 = (γ/β)
p+13 = 1 or (γp−1)
p+13 = (β/γ)
p+13 = 1.
In either case p | C p+13
. In the remaining case we get
(γ/β)p+13 = (β/γ)
p+13 or β
2(p+1)3 = γ
2(p+1)3 .
If βp+13 = γ
p+13 , we have p | C p+1
3. If β
p+13 = −γ p+1
3 , then
βp+1 = −γp+1 ⇒ βγ = −γβ,
which is impossible.
There remains the problem of dealing with S primes. If p is an S prime and p ≡ 1
(mod 3) the determination of when p | D p−13
is provided in the following result.
Theorem 5.17. Let p be an S prime and p ≡ 1 (mod 3). Then p | D p−13
if and
only if p | C p−13
and Rp−13 ≡ 1 (mod p).
Proof. Since p is an S prime, we have zeros α, β, γ of f(x) in Fp such that
αp−13 = ζ i, β
p−13 = ζj, γ
p−13 = ζk,
where ζ2 + ζ + 1 = 0. If p | D p−13
, then p | C p−13
and two of i, j, k are the same
modulo 3. Suppose without loss of generality that i ≡ j (mod 3). If k 6≡ i (mod 3),
then
W p−13≡ 0 6≡ 6R
p−13 (mod p),
which is impossible. Thus, we must have i ≡ j ≡ k (mod 3) and Rp−13 ≡ 1 (mod p).
If Rp−13 ≡ 1 (mod p) and p | C p−1
3, then 3 | i + j + k and two of i, j, k are
the same modulo 3. Hence, all three of them must be the same modulo 3 and
p | W p−13− 6R
p−13 .
153
In the following corollary we will use the sequence An as defined in (3.1).
Corollary 5.17.1. If p is an S prime, p ≡ 1 (mod 3) and Rp−13 ≡ 1 (mod p), then
p | D p−13
if and only if p - A p−13
.
Proof. Since Rp−13 ≡ 1 (mod p), we know that i, j, k in the proof of the theorem are
either all the same or all distinct modulo 3. If they are all the same, then p - A p−13
and p | C p−13
, hence p | D p−13
by the above theorem. If they are all distinct, then
p | A p−13
and p - C p−13
, so p - D p−13
.
Chapter 6
Arithmetic Properties of {En}
6.1 Preliminary Results for {En}
While working on the sequences {Wn} and {Cn}, several results developed concern-
ing the sequence {En}, where En = gcd(Wn, Cn). This sequence has a number of
properties analogous to those of the Lucas sequence {Vn}. In the next several sections
we will develop these properties. We begin with a result analogous to (Un, Vn) | 2
for Lucas functions.
Theorem 6.1. If (Q,R) = 1, then (Dn, En) | 6.
Proof. Suppose p is any prime such that p | Dn and p | En. Since p | Wn−6Rn, we
must have p | 6Rn. Since (Dn, R) = (En, R) and (Dn, R) | 2 by (5.1), we can only
have p = 2, 3. If 32 | (Dn, En), then 3 | R, which is impossible. If 22 | (Dn, En), then
22 | En, which we have seen by Theorem 4.5 is also impossible. Hence (Dn, En) | 6.
It is readily apparent that equation (2.12) implies Vn | U2n. Similarly we have
En | D3n, (6.1)
which can be seen as follows. We can rework the identity
4W3n = 3∆C2n(Wn + 2Rn) +W 2
n(Wn − 6Rn) + 24R3n
154
155
to see that
W3n − 6R3n = (Wn − 6Rn)
(W 2n −∆C2
n
4
)+ ∆WnC
2n. (6.2)
Recall again from Theorem 4.5 that if (Q,R) = 1, then 2α || (Wn, Cn)⇒ α = 0 or 1
and if α = 1, then Qn = W 2n−∆C2
n
4is odd.
We are now ready to show En | D3n. Clearly we have Cn | C3n. If 2 - En,
then En | Qn ⇒ En | W3n − 6R3n by equation (6.2). If 2 | En, then En/2 is
odd and En/2 | Qn. Since 2 | Wn, then 2 | Wn − 6Rn ⇒ En | (Wn − 6Rn)Qn ⇒
En | W3n − 6R3n by equation (6.2). Since En | Cn and Cn | C3n, we get En | C3n
and En | W3n − 6R3n ⇒ En | D3n.
We next derive some useful results concerning the primes which can divide En.
Theorem 6.2. If (Q,R) = 1 and p > 3 is a prime dividing En, then p ≡ 1 (mod 3).
Proof. First note p | Wn ⇒ AnBn ≡ 3Rn (mod p). Also remember p - R as
(Wn, Cn, R) | 2. Since p | (Wn, Cn) and W 2n−∆C2
n
4∈ Z we have p | W
2n−∆C2
n
4. Replac-
ing W 2n by (AnBn− 3Rn)2 and ∆C2
n by A2nB
2n + 18AnBnR
n− 4B3n− 4A3
nRn− 27R2n
yields
p | 3B3n + A4
nBn − 3A2nB
2n ⇒ p | Bn(A4
n − 3A2nBn + 3B2
n).
Notice p - Bn; for p | Bn ⇒ p | R and this is not possible by Lemma 4.1. Thus,
p | A4n − 3A2
nBn + 3B2n ⇒ p | 4A4
n − 12A2nBn + 12B2
n ⇒ p | (2A2n − 3Bn)2 + 3B2
n.
But this implies (2A2n − 3Bn)2 ≡ −3B2
n (mod p)⇒ (−3B2n
p) = 1⇒ (−3
p) = 1.
156
Thus, if p is a prime such that p > 3, p ≡ −1 (mod 3) and p | D3n, we know
that p - En. However, as shown in the next theorem we can say that if p | D3n and
p | Cn, then p | Dn or p | En.
Theorem 6.3. Let p be a prime such that p > 3. If p | D3n and p | Cn, then p | Dn
or p | En.
Proof. From Corollary 3.10.1, we see that
4(W3n − 6R3n) = 3∆C2nWn + 6∆C2
nRn +W 3
n − 6W 2nR
n.
Thus, if p | Cn and p | D3n, then p | W 2n(Wn − 6Rn). If p - En, then p - Wn. It
follows, then, that p | Wn − 6Rn and, therefore, p | Dn.
Corollary 6.3.1. Let p be a prime such that p > 3 and p ≡ −1 (mod 3). If p | D3n,
then
p | Dn ⇔ p | Cn.
Proof. Since p ≡ −1 (mod 3), we cannot have p | En by Theorem 6.2. Thus, if
p | Cn, then p | Dn by Theorem 6.3.
Theorem 6.4. Let p be a prime such that p > 3 and p ≡ −1 (mod 3). If p | D3n,
then
p - Dn ⇔ p | C3n/Cn.
Proof. First we will show (⇐). Assume that p | C3n/Cn and p | Dn. Since p | Dn,
we have p | Cn. By Corollary 3.10.1, we have
4C3n/Cn = ∆C2n + 3W 2
n ;
157
thus, since p | Cn and p | C3n/Cn, we have p | Wn. Hence p | En and p ≡ −1
(mod 3), which is a contradiction to Theorem 6.2.
Now suppose that p | D3n and p - Dn. By Corollary 6.3.1, we cannot have p | Cn;
hence, if p | C3n, then p | C3n/Cn.
We also have a result which tells us when p | D3n and p - Cn.
Theorem 6.5. Suppose p is a prime such that p - 6∆. Then p | D3n and p - Cn if
and only if
Wn ≡ −3Rn, ∆C2n ≡ −27R2n (mod p).
Proof. Suppose Wn ≡ −3Rn, ∆C2n ≡ −27R2n (mod p). Since
4C3n/Cn = ∆C2n + 3W 2
n ≡ 0 (mod p),
we get p | C3n/Cn. If p | R, then p | (Cn,Wn), which is impossible because
(Wn, Cn, R) | 2
by Lemma 4.6 and p 6= 2. Thus, p - 3R and hence p - Cn. Also, since
4(W3n − 6R3n) = 3∆C2nWn + 6∆C2
nRn +W 2
n(Wn − 6Rn),
we find that p | W3n − 6R3n and hence p | D3n.
Now suppose that p | D3n and p - Cn. Since p | C3n, we get
p | Cn(∆C2n + 3W 2
n)
and it follows that
∆C2n ≡ −3W 2
n (mod p).
158
Since p - Cn and p - ∆, we cannot have p | Wn. Also, p | W3n − 6R3n implies
p | 3∆C2nWn + 6∆C2
nRn +W 2
n(Wn − 6Rn)
and therefore p | 8W 2n(Wn + 3Rn). Thus, we must have
Wn ≡ −3Rn, and ∆C2n ≡ −27R2n (mod p).
We next eliminate the possibility that an I prime could divide En.
Theorem 6.6. If p is an I prime, then p - En.
Proof. Consider the zeros, α, β, γ, of f(x) in Fp3 . We have αp = β, βp = γ, γp = α.
If p | Cn, then without loss of generality αn = βn in Fp3 . So then αpn = βpn ⇒ βn =
γn ⇒ αn = βn = γn. Hence Wn = 6α3n = 6Rn in Fp3 . It follows that Wn ≡ 6Rn
(mod p). Since (Wn, Cn, R) | 2 by Lemma 4.6, we must have p - En.
Theorem 6.2 can now be generalized.
Theorem 6.7. If p is a prime, p > 3 and p | En, then p ≡ 1 (mod 3ν+1), where
3ν || n.
Proof. Since (En, R) | 2, we must have p - R. Suppose p | En. Then we know that
p cannot be an I prime by Theorem 6.6 and p ≡ 1 (mod 3) by Theorem 6.2. We
also have p | D3n. If p | Dn, then p | Wn and p | Wn− 6Rn ⇒ p | 6Rn, which is a
contradiction. Hence p - Dn. Let ω be the rank of apparition of p in {Dn}; we have
ω(p) | 3n, ω(p) - n, as p | D3n and p - Dn. So if 3ν || n, then 3ν+1 | ω(p). Also,
159
since p is not an I prime and p - 6R, we have ω(p) | p or ω(p) | p2 − 1, by results
seen in Theorem 5.6 for the S and Q prime cases. Hence,
ω(p) | (p− 1)(p+ 1)⇒ 3ν+1 | (p− 1)(p+ 1).
Since 3 | ω(p), we know ω(p) - p. Also, since 3 | p− 1, we must have
3ν+1 | p− 1⇒ p ≡ 1 (mod 3ν+1).
Lucas showed (Theorem 2.20) that if p is an odd prime such that p | Vn, then p ≡
±1 (mod 2ν+1), where 2ν || n. We next produce an analogue of this result. Recall
that Vn = U2n/Un. We will consider those primes p 6= 2, 3 such that p | D3n/Dn.
Theorem 6.8. If p is a prime, p > 3 and p | D3n/Dn, then p ≡ ±1 (mod 3ν+1),
where 3ν || n.
Proof. Since p | D3n, we see that if p | R, then p = 2 by (5.1), which is not possible.
Thus, p - R. Also, since (D3n/Dn, Dn) | 27 by Theorem 5.13, we cannot have p | Dn.
It follows by the same reasoning used in the proof of Theorem 6.7, that
3ν+1 | (p− 1)(p+ 1).
Hence p ≡ ±1 (mod 3ν+1).
We next produce some conditions on those primes p such that p - En and p is not
an I prime.
Theorem 6.9. If p (≡ 1 (mod 3)) is a Q prime or an S prime, then p - En if 3 - p−13
and Rp−13 ≡ 1 (mod p).
160
Proof. Let K be the splitting field of f(x) ∈ Fp[x] and let α(∈ Fp), β, γ, be the
zeros of f(x) in K. If p | Cn, then either αn = βn or βn = γn or αn = γn in K.
Without loss of generality assume βn = γn; then Wn = 2α2nβn + 2αnβ2n + 2β3n =
2βn(β2n + αnβn + α2n) in K. Suppose p | En. Since p - R we have βn 6= 0; thus,
since p | Wn, there exists ζ ∈ K such that ζ2 + ζ + 1 = 0 and βn = ζαn. We can
then see that
Rn = αnβnγn = ζ2α3n ⇒ Rn( p−13
) = ζ2( p−13
).
Thus, if 3 - p−13
, we cannot have Rp−13 = 1 in Fp. Hence, if R
p−13 ≡ 1 (mod p) and
3 - p−13
, then p - En.
Notice also that if 3 | n and 3 - p−13
, then p - En. Further, if 3 - n, 3 | p−13
and
Rp−13 6≡ 1 (mod p), then p - En.
We will next derive a law of repetition for {En}.
Theorem 6.10. If r | En and (r, 2) = 1, then
Wmn ≡
mWnR
(m−1)n (mod r2) if m ≡ 1 (mod 3)
−mWnR(m−1)n (mod r2) if m ≡ −1 (mod 3)
6Rmn (mod r2) if m ≡ 0 (mod 3)
and
CmnCn≡
mR(m−1)n (mod r) if m ≡ 1,−1 (mod 3)
0 (mod r) if m ≡ 0 (mod 3).
Proof. Our assumptions directly imply r | Wn and r2 | W2n−∆C2
n
4. As before, let Pn =
Wn and Qn = W 2n−∆C2
n
4. Then it can be shown by induction that rk−1 | Uk(Pn, Qn)
and rk | Vk(Pn, Qn).
161
Now use equations (2.7) and (2.8) as written below
2Qλ2n Uλ1−λ2 = Uλ1Vλ2 − Vλ1Uλ2 and 2Qλ2
n Vλ1−λ2 = Vλ1Vλ2 − ∆nUλ1Uλ2
to see that
rλ1+λ2−1 | Qλ2n Uλ1−λ2(Pn, Qn) and rλ1+λ2 | Qλ2
n Vλ1−λ2(Pn, Qn).
We now consider Wmn and Cmn/Cn under the three conditions for m (mod 3).
If λ1 + λ2 ≥ 2, then certainly
r | Qλ2n Uλ1−λ2(Pn, Qn) and r2 | Qλ2
n Vλ1−λ2(Pn, Qn).
The remaining cases are:
1. if λ1 = 1, λ2 = 0, then m ≡ 1 (mod 3),
2. if λ1 = 0, λ2 = 1, then m ≡ −1 (mod 3),
3. if λ1 = 0, λ2 = 0, then m ≡ 0 (mod 3).
Case 1. m ≡ 1 (mod 3). By use of Theorem 3.14 we have
Wmn ≡ (−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!Qλ2n Vλ1−λ2(Pn, Qn) (mod r2).
This can be further simplified using the facts λ1 = 1, λ2 = 0 and λ0 +λ1 +λ2 +λ3 =
m⇒ λ0 +λ3 = m−1⇒ m−1−λ0 = λ3. Further, we can see λ1 +2λ2 +3λ3 = m⇒
m = 3λ3 − 1 ⇒ m ≡ λ3 − 1 (mod 2). Since λ0 = m − 1 − λ3 ⇒ λ0 ≡ m − 1 − λ3
(mod 2) and m ≡ λ3 − 1 (mod 2)⇒ 2 | λ0. So
Wmn ≡ mWnR(m−1)n (mod r2).
162
Similarly,
CmnCn≡ mR(m−1)n (mod r).
Case 2. m ≡ −1 (mod 3).
Now since λ1 = 0, λ2 = 1 and λ0+λ1+λ2+λ3 = m⇒ λ0+λ3 = m−1⇒ m−1−λ0 =
λ3. Also λ0 = m − 1 − λ3 ⇒ λ0 ≡ m− 1 − λ3 (mod 2) but λ1 + 2λ2 + 3λ3 = m ⇒
2 + 3λ3 = m⇒ λ3 ≡ m (mod 2)⇒ 2 - λ0. So
Wmn ≡ (−1)mR(m−1)nQnV−1(Pn, Qn) (mod r2)
≡ −mR(m−1)nV1(Pn, Qn) (mod r2)
≡ −mR(m−1)nWn (mod r2).
Similarly,
CmnCn≡ mR(m−1)n (mod r).
Case 3. m ≡ 0 (mod 3).
Here λ1 = λ2 = 0⇒ λ0 +λ3 = m and 3λ3 = m. So λ3 ≡ m (mod 2) and λ0 ≡ m−λ3
(mod 2) yields λ0 ≡ 0 (mod 2). Hence
Wmn ≡ (−1)λ0m(m− λ0 − 1)!
λ1!λ2!λ3!R(λ0+λ3)nQλ2
n Vλ1−λ2(Pn, Qn) (mod r2)
≡ (−1)0 3λ3(λ3 − 1)!
λ3!R(m)nQ0
nV0(Pn, Qn) (mod r2)
≡ 6Rmn (mod r2).
Similarly, since U0 = 0
CmnCn≡ 0 (mod r).
163
Corollary 6.10.1. If p is any prime such that p > 3 and p | En, then p - Emn when
3 | m.
Corollary 6.10.2. If p is any odd prime and pµ || En (µ ≥ 1), then pµ || Emn when
p - m and 3 - m.
The case of p = 3 is contained in the next two corollaries.
Corollary 6.10.3. If 3µ || En, then 3 || Emn when 3 | m.
Corollary 6.10.4. If 3µ || En, then 3µ || Emn when 3 - m.
Of course, when p = 2, we know that 2µ || En ⇒ µ = 0 or 1. From this,
Corollary 3.10.1 and Theorem 4.5, we see that if 2 || En, then 2 || E2n. Also, since
4 | W 2n −∆C2
n, we see that 2 | En ⇔ 2 | Cn.
Theorem 6.11. If p is a prime such that p > 3 and pµ || En, then pµ+1 || Epn for
µ ≥ 1.
Proof. We note that p | p(p−λ0−1)!λ1!λ2!λ3!
unless λ1 = p, λ0 = λ2 = λ3 = 0.
If λ1 = p, then
Qλ2n Uλ1−λ2(Pn, Qn) = Up(Pn, Qn) and Qλ2
n Vλ1−λ2(Pn, Qn) = Vp(Pn, Qn).
Also, by the previous theorem, (pµ)p | Vp(Pn, Qn) and (pµ)p−1 | Up(Pn, Qn). Now
since p > 3 we know that 2µ+ 1 < µ(p− 1) and thus
Up(Pn, Qn) ≡ Vp(Pn, Qn) ≡ 0 (mod p2µ−1).
Furthermore, p2µ | Qλ2n Vλ1−λ2 for λ1 + λ2 ≥ 2⇒ p2µ+1 | p(p−λ0−1)!
λ1!λ2!λ3!Qλ2n Vλ1−λ2 .
164
Similarly, pµ | Qλ2n Uλ1−λ2(Pn, Qn) for λ1 + λ2 ≥ 2⇒ pµ+1 | p(p−λ0−1)!
λ1!λ2!λ3!Qλ2n Uλ1−λ2 .
Thus we need only concern ourselves with the case λ1 + λ2 = 1, which yields
Wpn ≡ ±pR(p−1)nWn (mod p2µ+1) andCpnCn≡ pR(p−1)n (mod pµ+1).
We may then conclude that if pµ || En, then pµ+1 || Epn.
Also, notice that if p > 3, p - En and p - ∆, then p - Epn. For if p - Cn, then
∆n = ∆C2n ⇒ p - ∆n ⇒ Up(Pn, Qn) 6≡ 0 (mod p). But Cpn
Cn≡ Up(Pn, Qn) (mod p),
so p - Cpn. Also, since if p | Cn, then p | Wn ⇔ p | Wpn, we see that p - Epn when
p - Wn.
6.2 A Law of Apparition for {En}
It will be seen here that {En} behaves in much the same way as {Vn}. By employing
{En}, we will be able to extend more of the results for {Vn} from Chapter 2 that
were, until now, missing. We must first deal with the case of p = 2. Since 2 | En ⇔
2 | Cn, from our results in Section 4.1, there always exists some minimal ρ such
that 2 | Eρ and 2 | En if and only if ρ | n. We next consider the case of a general
modulus. We note by the first two identities in Corollary 3.10.1 that Cn | C2n and
W2n ≡ (∆C2n +W 2
n)/2 (mod Wn). Since En is either odd or 2 || En, it is easy to see
that En | E2n.
Lemma 6.12. If r | Em, r | En and (r, 2) = 1, then r | E3n+m.
165
Proof. We interchange m and n in Corollary 3.10.2 to obtain
2W3n+m = WnW2n+m + ∆CnC2n+m −RnWnWn+m +Rn∆CnCn+m + 2R3nWm
2C3n+m = WnC2n+m + CnW2n+m −RnWnCn+m +RnCnWn+m − 2R3nCm.
The proof follows immediately.
We will say r | En when n < 0 if r | E|n|.
Lemma 6.13. If r | Em and r | En, then r | E3n+m.
Proof. Note that if 2 | r, then 2 - r2
as 4 - (Wn, Cn) as seen in Theorem 4.5. By
the previous lemma we have r2| E3n+m. Now since W 2
n−∆C2n
4∈ Z we have 2 | En ⇔
2 | Cn. So we need only show 2 | Cm and 2 | Cn implies 2 | C3n+m. There exists
some minimal ρ such that 2 | Cρ. Moreover, if 2 | Cn, then ρ | n. Thus n = k1ρ
and m = k2ρ. Hence 2 | C(3k1+k2)ρ ⇒ 2 | C3n+m, which concludes the proof.
Theorem 6.14. If r | Em and r | En, then r | E3kn+m for k ≥ 1.
Proof. Proceed by induction using the previous lemma. Clearly it is true for k = 1.
Assume this is true for r | E3kn+m. We also have r | En. So by the previous lemma,
if we replace m by 3kn+m, we have r | E3n+3kn+m ⇒ r | E3(k+1)n+m.
Corollary 6.14.1. If 3 - m, then En | Emn.
Proof. Since 3 - m, we must have m = 3k + 1 or m = 3k + 2. Since En | E2n, we
have En | E3kn+n and En | E3kn+2n. Hence En | Emn, when 3 - m.
We are now able to provide an analogue to Theorem 2.14.
166
Theorem 6.15. Suppose r | En, (n > 0), then there must be a least positive ρ = ρ(r)
such that r | Eρ. Further, ρ | n.
Proof. Clearly, since r | En, such a value for ρ must exist. Let 3µ || (n, ρ).
Case 1: 3µ || n
In this case put d1 = (3ρ, n)⇒ 3µ || d1. Since
d1 = 3ρx+ ny where x, y ∈ Z
we get
d1
3µ=
3ρx
3µ+ny
3µ.
Thus 3 - y as 3 - d13µ
and 3 | 3ρ3µ
.
Let 3k || x. Since r | E xρ
3kand r | Eyn by Corollary 6.10.2 we see from Theorem
6.14 that
r | E3ρx+yn ⇒ r | Ed1 .
We know d1 ≥ ρ by minimality of ρ. But d1 | 3ρ ⇒ d13µ| 3ρ
3µ. Since 3 - d1
3µwe have
d1 | ρ⇒ ρ = d1. Since d1 | n, we have ρ | n.
Case 2: 3µ || ρ
In this case put d2 = (ρ, 3n)⇒ 3µ || d2. Hence
d2 = 3nz + ρw where z, w ∈ Z.
Thus
d2
3µ=
3z
3µ+ρw
3µ⇒ 3 - w.
Reasoning as before with 3k || z, we get
r | E3nz+ρw ⇒ r | Ed2 ⇒ d2 ≥ ρ.
167
Since d2 | ρ⇒ d2 = ρ. Also ρ | 3n⇒ ρ3µ| n
3µ⇒ ρ | n.
6.3 Further Observations on the Law of Apparition for {En}
We next examine the problem of determining some m such that p | Em when p is
a prime and (p, 6R∆) = 1. Although it may not seem obvious, the next theorem is
part of our extension of Euler’s criterion, this will become more apparent later in
the chapter as several theorems concerning {En} and {Dn} lend themselves to the
generalization of this result. We note that if p is a Q prime, then Wp2−1 ≡ 6 (mod p)
by (4.15). Hence p - Ep2−1. However, it is possible that p | E p2−13
. We also recall
from Theorem 6.2 that if p | En for any n, then p ≡ 1 (mod 3).
Theorem 6.16. If p (≡ 1 (mod 3)) is a Q prime, then
p | E p2−13
⇔ Rp−13 6≡ 1 (mod p).
Proof. We have α, β, γ as the zeros of f(x) in Fp2 such that
αp = α, βp = γ and γp = β.
Hence,
αp−1 = 1, αp+1 = α2, βp+1 = βγ and γp+1 = βγ.
Now βγ = R/α and
βp2−1
3 = γp2−1
3 = (βγ)p−13 =
(R
α
) p−13
=R
p−13
αp−13
· α2(p−1)
3
α2(p−1)
3
= Rp−13 α
2(p−1)3
= Rp−13 (α2)
p−13 = R
p−13 (αp+1)
p−13 = R
p−13 α
p2−13 .
168
It then follows that p | C p2−13
and
W p2−13
≡ 2αp2−1(1 +R
p−13 +R
2(p−1)3 ) ≡ 2(1 +R
p−13 +R
2(p−1)3 ) (mod p).
We can then see
Rp−13 6≡ 1 (mod p)⇔ 1 +R
p−13 +R2 p−1
3 ≡ 0 (mod p)⇔ p | W p2−13
.
We next consider the possibility that p | Ep+1 when p is a Q prime. In this
case p 6= 3 and p ≡ 1 (mod 3) by Theorem 6.2. We already know that p | Cp+1
by Corollary 4.24.1. Let α, β, γ be the zeros of f(x) in Fp2 , then αp = α, βp = γ,
γp = β. Then we have αp+1 = α2, βp+1 = γβ, γp+1 = γβ. So
Wp+1 = (αp+1β2p+2 + βp+1γ2p+2 + γp+1α2p+2) + (α2p+2βp+1 + β2p+2γp+1 + γ2p+2αp+1)
= (α2β2γ2 + β3γ3 + α4βγ) + (α4βγ + β3γ3 + β2γ2α2)
= 2(α4βγ + α2β2γ2 + β3γ3).
Hence, we can see that
Wp+1 ≡ 0 (mod p)⇔ α4βγ + (αβγ)2 + (βγ)3 = 0⇔ α2 = ζβγ or α2 = ζ2βγ
where ζ2 + ζ + 1 = 0. Note p ≡ 1 (mod 3)⇒ ζ ∈ Fp.
Now consider
M1 = (α2 − ζβγ)(β2 − ζαγ)(γ2 − ζαβ)
M2 = (α2 − ζ2βγ)(β2 − ζ2αγ)(γ2 − ζ2αβ).
Clearly then M1M2 = 0, when Wp+1 = 0, as Wp+1 = 0 if and only if α2 = ζβγ
or α2 = ζ2βγ. Also if M1M2 = 0, then Wp+1 = 0. This is obvious for the cases
169
α2−ζβγ = 0 and α2−ζ2βγ = 0, so suppose β2−ζαβ = 0⇒ β2(p+1) = ζp+1(αβ)p+1 ⇒
(βγ)2 = ζp+1α2βγ ⇒ βγ = ζp+1α2 ⇒ α2 = ζ−(p+1)βγ ⇒ α2 = ζβγ. Similarly for
any of the other 3 factors of M1M2.
Now
M1 = (α2 − ζβγ)(β2 − ζαγ)(γ2 − ζαβ)
= α2β2γ2 − ζα3γ3 − ζβ3γ3 − ζα3β3 + ζ2αβγ4 + ζ2α4βγ + ζ2αβ4γ − ζ3ζ2α2β2γ2
= ζ2(αβγ)(α3 + β3 + γ3)− ζ((αβ)3 + (βγ)3 + (γα)3)
= ζ2RA3 − ζB3.
Also M2 = ζRA3 − ζ2B3. So
M1M2 = (ζ2RA3 − ζB3)(ζRA3 − ζ2B3)
= ζ3R2A23 − ζ2RA3B3 − ζ4RA3B3 + ζ3B2
3
= R2A23 − ζ2RA3B3 − ζRA3B3 +B2
3
= (RA3)2 +RA3B3 +B23 .
Hence, p | Wp+1 ⇔ p | (RA3)2 + RA3B3 + B23 . Thus, if p is a Q prime, then
p | Ep+1 ⇔ p | M1M2.
Remember A3 = P 3−3PQ+3R and B3 = Q3−3PQR+3R2. Since only a finite
number of primes can divide M1M2, there can only be a finite number of Q primes
p such that p | Wp+1, which means that there can only be a finite number of primes
p such that p | Wr, where r is the rank of apparition of p for {Cn}. For if p | Wr,
then p | Ep+1 because r | p+ 1 and 3 - p+ 1, by Corollary 6.10.2.
We now produce a result similar to Theorem 6.16 for S primes. Let α, β, γ be
the zeros of f(x) in K, where p ≡ 1 (mod 3) is an S prime. Suppose that there exists
170
ρ = ρ(p) such that p | Eρ, ρ > 1 and ρ is minimal. Then ρ = kr1 where r1 is some
rank of apparition of p in {Cn}. Suppose further that r1 | 3n and
(α/β)n = ζ i, (β/γ)n = ζj, where ζ2 + ζ + 1 = 0 in K.
This is certainly the case when n = (p−1)/3. Note that αn = ζ iβn and βn = ζjγn ⇒
αn = ζ i+jγn. Also since (ζ − 1)(ζ2 + ζ + 1) = 0⇒ ζ3 = 1, we have α3n = β3n = γ3n.
Thus W3n = 6R3n (mod p).
We distinguish 3 cases:
Case 1: 3 - ij(i+ j)
Since αn = ζ iβn, βn = ζjγn, αn = ζ i+jγn and 3 - ij(i+ j) we see that αn, βn, γn
are pairwise distinct. Thus p - Cn ⇒ r1 - n ⇒ 3 - 3nr1
. Since p | Cρ and p | Wρ, we
must have p | W 3nr1ρ by Corollary 6.10.2 and hence p | W3nk. Since W3nk ≡ 6R3nk
(mod p), this is impossible.
Case 2: 3 divides only one of i, j, (i+ j)
In this case p | Cn, since without loss of generality αn = βn, so 3 | i and 3 - j.
Using αn = βn = ζjγn,
Wn = (α2nβn + β2nγn + γ2nαn) + (αnβ2n + βnγ2n + γnα2n)
= (ζ2jγ2nζjγn + ζ2jγ2nγn + γ2nζjγn) + (ζjγnζ2jγ2n + ζjγnγ2n + γnζ2jγ2n)
= 2γ3n(1 + ζj + ζ2j)
= 0 since 3 - j ⇒ 1 + ζj + ζ2j = 0.
So p | En. Thus ρ exists and ρ | n.
Case 3: 3 divides 2 of i, j, (i+ j)
171
Since 3 divides two of i, j, (i+ j), it divides all of them. Hence αn = βn = γn ⇒
p | Cn and Wn = 6Rn (mod p).
Subcase 1: r1 - n.
Since r1 | 3n, we know that 3 - 3nr1⇒ p | W3nk, which is impossible.
Subcase 2: r1 | n
If 3 - n, then 3 - nr1
and by Corollary 6.10.2 we have p | W nr1ρ ⇒ p | Wkn. This
is a contradiction, as Wkn ≡ 6Rkn (mod p). Thus, if ρ exists, then ρ = kr1, where
r1 | n and 3 | n and (α/β)n/3 = ζ i, (β/γ)n/3 = ζj. By repeating the above argument
we see that if ρ exists, then ρ | n/3.
The fact that whenever ρ exists, ρ | n implies in the case that n = (p − 1)/3,
that (α/β)(p−1)/3 = ζ i and (β/γ)(p−1)/3 = ζj and therefore ρ | p−13
. Also, Rp−13 =
(αβγ)p−13 = ζ i+2j, and 3 | ij(i2 − j2). So if 3 - ij(i + j), then 3 | (i − j). If 3
divides only one of i, j, i + j, then 3 - i − j. But if 3 divides all of i, j, i + j, then
3 | i− j. Thus ρ exists and ρ | p−13
if Rp−13 6≡ 1 (mod p). When R
p−13 ≡ 1 (mod p)
and 3 - p−13
, ρ does not exist. If Rp−13 ≡ 1 (mod p) and 3 | p−1
3, then, if ρ exists,
ρ | p−19
. We have proved the following theorem.
Theorem 6.17. If p is an S prime and p ≡ 1 (mod 3), then
p | E p−13⇔ R
p−13 6≡ 1 (mod p).
Also, if Rp−13 ≡ 1 (mod p) and p 6≡ 1 (mod 9), then ρ(p) cannot exist. If R
p−13 ≡ 1
(mod p) and p ≡ 1 (mod 9), then ρ | p−19
if ρ exists.
We also have the following result.
172
Theorem 6.18. If p is an S prime and p ≡ 1 (mod 3), then
p | E p2−13
⇔ p | E p−13.
Proof. Clearly, since 3 - p+1, p | E p2−13
when p | E p−13
. Next suppose that p | E p2−13
.
If α, β, γ are defined as above, we must have
αp2−1
3 = βp2−1
3 ⇒ (αp+1)p−13 = (βp+1)
p−13 ⇒ (α2)
p−13 = (β2)
p−13 .
So, we then have
(αp−13 )2 = (β
p−13 )2 ⇒ α
p−13 = ±β
p−13 .
If αp−13 = β
p−13 , then p | C p−1
3. If α
p−13 = −β p−1
3 , then
αp−1 = −βp−1 ⇒ 1 = −1,
which is impossible. Also, since p | W p2−13
, we get
2αp2−1
3 (α2(p2−1)
3 + γp2−1
3 αp2−1
3 + γ2(p2−1)
3 ) ≡ 0 (mod p).
Since, p - 2R, this means that
αp2−1
3 = ζγp2−1
3
for some ζ such that ζ2 + ζ + 1 = 0. It follows that
α2(p−1)
3 = ζγ2(p−1)
3 .
Since αp−1 = γp−1 = 1, we find by squaring that
αp−13 = ζ2γ
p−13 .
Since p | C p−13
, we get p | E p−13
.
173
By similar techniques we can also establish the next result.
Theorem 6.19. If p is an S prime and p ≡ 1 (mod 3), then
p | D p2−13
⇔ p | D p−13.
Let νp(x) denote that value of ν such that pν || x, where x ∈ Z>0 and p is a
prime.
Theorem 6.20. Let p be an S prime and suppose that ρ = ρ(p) exists. In this case
ρ = kr1, where r1 is a rank of apparition of p in {Cn}. If r2 is any other rank of
apparition of p in {Cn}, then ν3(r2) > ν3(r1).
Proof. Since p | Eρ, we must have (α/β)ρ = ζ i, (β/γ)ρ = ζj in Fp, where ζ2 +ζ+1 =
0. Here 3 divides only one of i, j, i + j. Without loss of generality suppose 3 | i,
then βρ = ζjγρ, αρ = ζjγρ and 3 - j.
If r2 is another rank of apparition of p, then it is either the order of β/γ or α/γ
in Fp. So r2 | 3ρ and r2 - ρ. Thus if k = ν3(ρ) then ν3(r2) = k + 1. Since r1 | ρ, we
must have ν3(r1) ≤ ν3(ρ) = k < ν3(r2).
To proceed any further we will need the following results for Lucas sequences.
Because they seem to be difficult to locate in the literature, proofs are provided here.
Theorem 6.21. Let Vn, Un be the Lucas functions Vn(P,Q), Un(P,Q) and let p be
a prime such that p > 3 and (Qp
) = 1. Let ω be the rank of apparition of p in {Un}.
Let S2 ≡ Q (mod p). Then there exists a minimal λ such that
Vλ(P,Q) ≡ −Sλ (mod p)
if and only if 3 | ω. Furthermore, λ = ω/3 or 2ω/3.
174
Proof. (⇒) Suppose λ exists. Then V 2λ ≡ Qλ (mod p) and using U3n = Un(V 2
n −Qn)
we see p | U3λ. Also p - Uλ, for if p | Uλ, then V 2λ ≡ 4Qλ (mod p), as V 2
n −∆U2n =
4Qn, which is not possible. Thus ω | 3λ and ω - λ, so 3 | ω.
Put µ = ω/3. Then µ | λ. Again, use U3µ = Uµ(V 2µ − Qµ), knowing U3µ ≡ 0
(mod p) and Uµ 6≡ 0 (mod p), so V 2µ ≡ Qµ (mod p). Hence
Vµ(P,Q) ≡ ±Sµ (mod p).
If Vµ ≡ −Sµ (mod p), then λ = µ by minimality of λ.
If Vµ ≡ Sµ (mod p), then V2µ ≡ −S2µ (mod p), as
V2µ = V 2µ − 2Qµ
≡ Qµ − 2Qµ
≡ −S2µ (mod p).
Thus λ = 2µ by the minimality of λ.
(⇐) If 3 | ω, we must have
Vµ ≡ ±Sµ (mod p)
where µ = ω/3 as before. Thus Vµ ≡ −Sµ (mod p) or Vµ ≡ Sµ (mod p). Thus,
there exists a minimal λ such that Vλ(P,Q) ≡ −Sλ (mod p) holds, and we have
already seen that λ = µ or λ = 2µ.
One may also notice that if (∆p
) = 1, where ∆ = P 2 − 4Q and p ≡ 1 (mod 3),
then p | Up−1 ⇒ ω | p − 1. Let 3µ || p − 1. If 3 | ω, then ω - p−13µ
. Also if ω - p−13µ
,
then 3 | ω. Thus λ exists if and only if p - U p−13µ
.
175
Theorem 6.22. If ω is the rank of apparition of p in {Un} and 6 | ω, then Vω3≡ Q
ω3
(mod p).
Proof. Put ω = 6k. We have U6k ≡ 0 (mod p) and using U2n = UnVn we get
U2(3k) = U3kV3k ≡ 0 (mod p). So V3k ≡ 0 (mod p) as U3k 6≡ 0 (mod p). Moreover,
Vω2≡ 0 (mod p) because p - Uω
2. Now, since V2n = V 2
n − 2Qn if we set n = ω2⇒
Vω = V 2ω2− 2Q
ω2 . So Vω ≡ −2Q
ω2 (mod p).
We also know that V3n = V 3n−3QnVn; by setting n = ω
3, we get Vω = V 3
ω3−3Q
ω3 Vω
3.
So
−2Qω2 ≡ V 3
ω3− 3Q
ω3 Vω
3(mod p)
−2Qω2
Qω2
≡V 3ω3
Qω2
−3Q
ω3 Vω
3
Qω2
(mod p)
−2 ≡(Vω
3
Qω6
)3
− 3
(Vω
3
Qω6
)(mod p).
On putting T =Vω
3
Qω6
, we get T 3−3T+2 ≡ 0 (mod p) or (T−1)2(T+2) ≡ 0 (mod p).
If T + 2 ≡ 0 ⇒Vω
3
Qω6≡ −2 ⇒ Vω
3≡ −2Q
ω6 ⇒ V 2
ω3≡ 4Q2ω
6 (mod p). Using
V 2n −∆U2
n = 4Qn with n = ω3⇒ V 2
ω3−∆U2
ω3
= 4Qω3 ⇒ Uω
3≡ 0 (mod p), which is a
contradiction. So T ≡ 1 (mod p)⇒ Vω3≡ Q
ω6 (mod p).
We next establish some results concerning the divisibility of Wmn by a prime p
such that p | Cn.
Theorem 6.23. Suppose p is a prime such that p - 6R∆ and p | Cn. Then p | Wmn
if and only if Vm(Wn
2−Rn, R2n) ≡ −Rmn (mod p).
Proof. Since p | Cn, in K we must have βn = γn ⇒ Wn = 2(α2nβn + β2nαn + β3n)
176
and Rn = αnβ2n. We may then notice
Wn − 2Rn = 2(α2nβn + β3n)
and β3n(α2nβn) = α2nβ4n = R2n. Further, since βmn = γmn, we have
Wmn − 2Rmn = 2((α2nβn)m + (β3n)m).
Note that Vm(Wn
2− Rn, R2n) = α′m + β′m, where α′ = α2nβn, β′ = β3n. It follows
that
Wmn − 2Rmn = 2Vm(Wn
2−Rn, R2n).
Thus p | Wmn if and only if Vm(Wn
2−Rn, R2n) ≡ −Rmn (mod p).
Corollary 6.23.1. Suppose p is a prime such that p - 6R∆, Wn ≡ 6Rn (mod p)
and Cn ≡ 0 (mod p), then Wmn ≡ 6Rmn (mod p).
Proof. We know that
Wmn ≡ 2Rmn + 2Vm(Wn − 2Rn
2, R2n) (mod p).
Now Wn−2Rn
2≡ 6Rn−2Rn
2= 2Rn (mod p) and Vm(2S, S2) = 2Sm. By letting S =
2Rmn we get
Wmn ≡ 2Rmn + 2(2Rmn) = 6Rmn (mod p).
Corollary 6.23.2. Suppose p is a prime such that p - 6R∆, Wn ≡ −2Rn (mod p)
and Cn ≡ 0 (mod p), then
Wmn ≡
−2Rmn (mod p) if 2 | m
6Rmn (mod p) if 2 - m.
177
Proof. Here Wn−2Rn
2≡ −2Rn (mod p). Also Vm(−2S, S2) = 2(−1)mSm. If S = Rmn
and 2 - m, then
Wmn ≡ 2Rmn + 2(−2Rmn) ≡ −2Rmn (mod p).
Leaving S the same, but with 2 | m, we have
Wmn ≡ 2Rmn + 2(2Rmn) ≡ 6Rmn (mod p).
Theorem 6.24. Let p be an S prime or a Q prime. If p | Cn and
∆′ =
(Wn
2−Rn
)2
− 4R2n,
then (∆′
p) = 1 or 0.
Proof. If p is an S prime, we have βn = γn in K (= Fp) since p | Cn. So ∆′ =
(Wn
2−Rn)2−4R2n = (α2nβn+β3n)2−4(αnβ2n)2 = α4nβ2n+2α2nβ4n+β6n−4α2nβ4n =
α4nβ2n − 2α2nβ4n + β6n = (α2nβn − β3n)2. Thus (∆′
p) = 1, 0.
If p is a Q prime, we must have β, γ ∈ Fp2 , α ∈ Fp.
If p | Cn and we have αn = βn ⇒ αn = βn = γn ⇒ Wn = 6Rn ⇒ ∆′ =
(2Rn)2 − 4R2n = 0.
If p | Cn and we have βn = γn, then ∆′ = (α2nβn−β3n)2. Now (α2nβn−β3n)p =
α2pnβpn − β3pn = α2nγn − γ3n = α2nβn − β3n. Thus α2nβn − β3n ∈ Fp and (∆′
p) = 1.
Theorem 6.25. If p is a prime such that p | Cn, p - 6R∆ and (∆′
p) = 0, there is no
value of m such that p | Wmn.
178
Proof. If (∆′
p) = 0, then p | ∆′. Also, p | ∆′ ⇒ p | (Wn
2− Rn)2 − 4R2n ⇒ (Wn
2−
Rn)2 ≡ 4R2n (mod p) ⇒ Wn
2− Rn ≡ ±2Rn (mod p) ⇒ Wn ≡ 6Rn or Wn ≡ −2Rn.
Thus, by Corollaries 6.23.1 and 6.23.2 we have p - Wmn for all m.
If p is a Q prime, we know that p | Cp+1. Suppose p | Eρ, ρ > 1, ρ minimal,
then we must have p ≡ 1 (mod 3) by Theorem 6.2 and r | ρ ⇒ ρ = kr, where r is
the rank of apparition of p in {Cn}. Also r | p + 1 and if k > 1, by the minimality
of ρ we must have p - Wr.
Put
P ′ ≡ Wr/2−Rr and Q′ ≡ R2r (mod p).
Then p | Wρ ⇔ Vk(P′, Q′) ≡ −Rkr (mod p) by Theorem 6.21. Let ω′ be the rank
of apparition of p in Un(P ′, Q′). Now Vk(P′, Q′) ≡ −Rkr ⇒ V 2
k (P ′, Q′) ≡ Q′k
(mod p) ⇒ p | U3k(P′, Q′), p - Uk(P ′, Q′). The least possible value for k is ω′/3
and ω′|p−12
, as (Q′
p) = 1. But if Vω′/3(P ′, Q′) ≡ Rkr (mod p), where k = ω′/3, then
V2ω′/3(P ′, Q′) ≡ −R2kr (mod p). Hence k = ω′/3 or 2ω′/3. In either event, k | p−13
.
Theorem 6.26. If p is a Q prime and ρ exists, then ρ = kr, where r | p + 1 and
k | p−13
.
As mentioned earlier, if ρ(p) exists, then ρ = kr, where r is some rank of ap-
parition of p in {Cn}. If p is an S prime, then kr | p−13⇒ k | p−1
3r. If we put
P ′ ≡ Wr
2− Rr, Q′ ≡ R2r (mod p), then k exists if and only if (∆′
p) = 1 and 3 | ω′,
where ω′ is the rank of apparition of p in Un(P ′, Q′). Furthermore, k = ω′/3 or
2ω′/3. In particular by Theorem 6.22 we know that k = 2ω′/3 if 2 | ω′.
179
The theorem below provides an analogue to both Theorems 2.16 and 2.17. Rather
than consider highest powers of 2 as in the Lucas case, we consider powers of 3.
Theorem 6.27. Suppose (r1, r2) = 1 and r1 | Em, r2 | En. If 3µ || m and 3µ || n
(µ ≥ 0), then
r1r2 | E[m,n].
If 3µ || m and 3ν || n and µ 6= ν, then r1r2 - Es for any s ∈ Z.
Proof. When 3µ || m and 3µ || n, we have 3µ || [m,n]. It follows that 3 - [n,m]n
and
3 - [n,m]m
. Thus r1 | E[m,n] and r2 | E[m,n] by Corollary 6.14.1. Since (r1, r2) = 1 we
get r1r2 | E[m,n].
Now suppose that 3µ || m and 3ν || n and µ 6= ν. Put ρ1 = ρ(r1) and ρ2 = ρ(r2).
We know that ρ1 | m and ρ2 | n. Also, 3 - mρ1
, 3 - nρ2⇒ 3µ || ρ1, 3ν || ρ2.
Suppose for some s ∈ Z we get r1r2 | Es. Then ρ1 | s, ρ2 | s and 3 - sρ1
,
3 - sρ2
. But then 3µ || s and 3ν || s and hence µ = ν which is contrary to our
assumptions.
Put ρ = ρ(3), 3µ || Eρ. If ρ - n then 3 - En. If ρ | n and 3 | nρ, then 3 || En by
Corollary 6.10.3. If ρ | n and 3 - nρ, then 3µ || En by Corollary 6.10.4.
The next theorems parallel Theorems 2.18 and 2.19.
Theorem 6.28. Let 3µ || m, 3ν || n. If µ = ν, then
(Em, En) = E(m,n).
Proof. Let (Wm, Cm,Wn, Cn) = 2λ3κd such that (d, 6) = 1. Since 3µ || m and
3µ || n, we have 3 - m(m,n)
, 3 - n(m,n)
. Hence E(m,n) | Em and E(m,n) | En; conse-
quently, E(m,n) | 2λ3κd.
180
We know that λ = 0 or 1 by Theorem 4.5. If λ = 0, then ρ(2) - m or ρ(2) - n⇒
ρ(2) - (m,n) ⇒ 2 - E(m,n). If λ = 1, then ρ(2) | m, ρ(2) | n ⇒ ρ(2) | (m,n) ⇒
2 | E(m,n). Thus E(m,n) = 2λ3κ′d′ where (d′, 6) = 1, and κ′ ≤ κ, d′ | d.
If κ = 0, then ρ(3) does not exist ⇒ 3 - E(m,n), or ρ(3) exists and ρ(3) - m or
ρ(3) - n⇒ ρ(3) - (m,n)⇒ κ′ = 0.
If κ = 1, then ρ = ρ(3) exists and ρ | m, ρ | n. Without loss of generality
assume 3 || En. Let 3δ || Eρ. If δ = 1, then 3 || 2λ3κ′d′ ⇒ κ′ = κ. If δ > 1, then
3 | mρ
, 3 | nρ⇒ 3 || 2λ3κ
′d′ ⇒ κ′ = κ.
If κ > 1, then ρ(3κ) | m, ρ(3κ) | n and 3 - mρ(3κ)
, 3 - nρ(3κ)
. Now ρ(3κ) | (m,n)
and 3 - (m,n)ρ(3κ)
⇒ 3κ | 3κ′ ⇒ κ′ ≥ κ > 1. Since κ′ ≤ κ, we get κ = κ′.
Now since ρ(d) | m and ρ(d) | n, we must have ρ(d) | (m,n). Also, since 3 - mρ(d)
and 3 - nρ(d)
, we get 3 - (m,n)ρ(d)
. Hence, d | E(m,n) and so d | d′. Thus d = d′.
Theorem 6.29. If 3µ || m, 3ν || n and ν 6= µ, then
(Em, En) | 6.
Proof. If we are given p | (Em, En) and (p, 6) = 1, then ρ(p) | m and ρ(p) | n. Since
3 - mρ(p)
and 3 - nρ(p)
, we get 3µ || ρ(p), 3ν || ρ(p), which is impossible. If 3λ | (Em, En)
and λ > 1, then ρ(3λ) | m, ρ(3λ) | n. Also, 3 - mρ(3λ)
, 3 - nρ(3λ)
⇒ 3µ || ρ(3λ) and
3ν || ρ(3λ). Again, this is impossible. Further, by Theorem 4.5, if 2λ || (Em, En),
then λ ∈ {0, 1}.
Thus, we also have a result comparable to Theorem 2.19.
We restate Euler’s criterion for Un, Vn as follows:
181
Theorem 6.30. If p - 2∆Q, then
p | UT (p)2
⇔ Qp−12 ≡ 1 (mod p)
p | VT (p)2
⇔ Qp−12 ≡ −1 (mod p),
where T (p) = p− 1 if p splits in Q(α) and T (p) = p+ 1 otherwise.
By using Theorems 6.2, 6.16, 6.18, 5.14, 5.15, and 5.17, we can prove our analogue
of this result.
Theorem 6.31. Suppose p is a prime such that p - ∆R and p ≡ 1 (mod 3). Let
T (p) = p2 − 1 if p splits in Q(α) and T (p) = p2 + p + 1 otherwise. If p is a Q or I
prime, then
p | DT (p)3
⇔ Rp−13 ≡ 1 (mod p).
If p is an S prime, then
p | DT (p)3
⇔ Rp−13 ≡ 1 (mod p) and p | C p−1
3.
Also, if T (p) = p2 − 1, then
p | ET (p)3
⇔ Rp−13 6≡ 1 (mod p).
If T (p) = p2 + p+ 1, then
p - ET (p)3
.
Chapter 7
Primality Testing
7.1 An Analogue of Lucas’ Fundamental Theorem
As we have seen in Chapter 1, one of Lucas’ main purposes in attempting to extend
his functions was to find new primality tests. In this chapter we will explore how
the Wn and Cn functions can be used for producing such tests. We will first develop
some analogues of Theorem 2.22 of Chapter 2. We begin with a simple lemma.
Lemma 7.1. If k ≥ 2 and ri ≥ 5 for i = 1, 2, . . . , k, then(k∏i=1
r2i
)− 1 > 2
k∏i=1
(r2i + ri + 1
2
).
Proof. We note that
1 >1
54+ 2
(7
10
)2
;
hence
1 >1
52k+ 2
(7
10
)kfor k ≥ 2.
Now
1
52k≥
k∏i=1
1
r2i
and7
5= 1 +
2
5> 1 +
1
ri+
1
r2i
182
183
imply that
1 >k∏i=1
1
r2i
+ 2k∏i=1
(1 + 1
ri+ 1
r2i
2
)
⇒
(k∏i=1
r2i
)>
(k∏i=1
r2i
)(k∏i=1
1
r2i
)+
(k∏i=1
r2i
)(2
k∏i=1
(1 + 1
ri+ 1
r2i
2
))
⇒
(k∏i=1
r2i
)− 1 > 2
k∏i=1
(r2i + ri + 1
2
).
Theorem 7.2. Let N be an integer such that (N, 6) = 1. If N | DN2−1, N - DN2−1q
for all primes q such that q | N2 − 1 and (DN2−1q′, N) = 1 for some prime divisor q′
of N2 − 1, then N is a prime.
Proof. Clearly ω(N) exists and ω(N) | N2−1. Also, if ω(N) 6= N2−1, then N2−1 =
kω(N) where k > 1. If q is any divisor of k, then ω(N) | N2−1q⇒ N | DN2−1
q
, which
is a contradiction. Hence ω(N) = N2 − 1. Let
N =k∏i=1
pαii ,
where the primes pi are all distinct and exceed 4. Also, since N | DN2−1, then by
equation (5.1) and the fact 2 - N , we must have (N,R) = 1. We know by Theorem
5.11 that
ω(N) = lcm[ω(pαii ); i = 1, 2, . . . , k].
Since (pi, ω(N)) = 1 and by Theorem 5.10 ω(pαii ) = pνi ω(pi), we must have
ω(N) | lcm[ω(pi); i = 1, 2, . . . , k].
184
Let p be any prime divisor of N . We have p | DN2−1 and p - DN2−1q′
. Hence
ω(p) | N2 − 1 and ω(p) - N2−1q′⇒ q′ | ω(p). Hence
lcm[ω(pi); i = 1, 2, . . . , k] | q′k∏i=1
ω(pi)
q′.
Now by Corollary 5.6.1, for k ≥ 2 we have,
q′k∏i=1
ω(pi)
q′≤ q′
k∏i=1
p2i + pi + 1
q′≤ 2
k∏i=1
p2i + pi + 1
2
we get (k∏i=1
p2i
)− 1 = N2 − 1 ≤ 2
k∏i=1
p2i + pi + 1
2,
which is impossible by the previous lemma.
If k = 1, then N = pα and by Theorem 5.10 ω(N) = ω(pα) = pνω(p)⇒ N2−1 =
ω(p), since (p,N2−1) = 1. If α ≥ 2, then p4−1 ≤ p2+p+1, which is a contradiction.
Thus N = p, a prime.
By similar methods used to prove the previous theorem we also have the following
result.
Theorem 7.3. Let N be an integer such that (N, 6) = 1. If N | DN2+N+1 and
N - DN2+N+1q
for each prime divisor q of N2 +N + 1 and (DN2+N+1q′
, N) = 1 for some
prime divisor q′ of N2 +N + 1, then N is a prime.
We have proved our analogue of Theorem 2.22.
Theorem 7.4. Let N be an integer such that (N, 6) = 1 and let T = T (N) = N2−1
or N2+N+1. If N | DT and N - DTq
for each prime divisor q of T and (D Tq′, N) = 1
for some prime divisor q′ of T , then N is a prime.
185
The difficulty in providing this as a complete analogue to Lucas’ result is the need
to involve the prime q′, which is not needed in Theorem 2.22. This is because 2 | N±
1 and 2 | pi± 1, and any proof of Theorem 2.22 makes use of these observations. In
what follows next, we will modify Theorems 7.2 and 7.3 to eliminate the need for q′
in certain cases.
Suppose p is a prime, p - 6∆ and 3 | T (p). By Theorem 6.5, we know that if
m = T (p)/3, then p - Cm if and only if
Wm ≡ −3Rm, ∆C2m ≡ −27R2m (mod p).
We also have a result for an arbitrary modulus.
Lemma 7.5. Let (N, 6) = 1. If ∆C2n ≡ −27R2n (mod N) and Wn ≡ −3Rn
(mod N), then N | D3n and N - Cn.
Proof. We have ∆C2n + 3W 2
n ≡ 0 (mod N). Since 4C3n = Cn(∆C2n + 3W 2
n) we get
N | C3n. Also, 4W3n = 3∆C2n(Wn + 2Rn) +W 2
n(Wn − 6Rn) + 24R3n implies
4(W3n − 6R3n) = 3∆C2n(Wn + 2Rn) +W 2
n(Wn − 6Rn)
≡ −9W 2n(Wn + 2Rn) +W 2
n(Wn − 6Rn) (mod N)
≡ −9W 3n − 18RnW 2
n +W 3n − 6RnW 2
n (mod N)
≡ −8W 3n − 24RnW 2
n (mod N)
≡ −8W 2n(Wn + 3Rn) ≡ 0 (mod N).
Thus N | W3n − 6R3n ⇒ N | D3n. Now since (Wn, Cn, R) | 2 by Lemma 4.6 and
(N, 6) = 1 we have (N,R) = 1⇒ N - Cn.
We can use the last result to prove the following theorem.
186
Theorem 7.6. If N is odd, 3 | T (N), ∆C2T (N)
3
≡ −27R2T (N)
3 (mod N), WT (N)3
≡
−3RT (N)
3 (mod N), and N - CT (N)q
for each prime divisor q of T (N)3
, then N is a
prime.
Proof. By the previous lemma, we know that N | DT (N), N - DT (N)q
for all prime
divisors of T (N). By our earlier reasoning we have ω(N) = T (N). Also, since
(T (N), N) = 1,
ω(N) = lcm[ω(pi); i = 1, 2, . . . , k] if N =k∏i=1
pαii .
Let p be any prime divisor of N . If p | R, then by the conditions of the theorem
p | WT (N)3
and p | ∆CT (N)3
. Since p - CT (N)3
by Lemma 7.5, we must have p | ∆.
However, by Corollary 4.2.1 we can only have p = 2, which is not possible because
N is odd. Thus, (N,R) = 1. Also, if p | N and p | ∆, then p | R and p | WT (N)3
,
which is also impossible. It follows that (N, 6∆R) = 1. Now since p | CT (N) and
p - CT (N)3
, we get
p | ∆C2T (N)
3
+ 3W 2T (N)
3
and we know that p - ∆CT (N)3
WT (N)3
. Thus (−3∆p
) = 1. If p is an I prime, then
ω(p) | p2 + p + 1, (∆p
) = 1 and (−3p
) = 1, which means that p ≡ 1 (mod 3) and
3 | p2 +p+1. If p is a Q prime, then ω(p) | p2−1 and 3 | p2−1. If p is an S prime,
then ω(p) | p− 1 and p− 1 < (p2 − 1)/3 < (p2 + p+ 1)/3.
Thus,
lcm[ω(pi); i = 1, 2, . . . , k] ≤ 3k∏i+1
p2i + pi + 1
3.
That N is a prime now follows from our previous reasoning.
187
Notice that 3 | T (N) when T (N) = N2 − 1 and 3 | T (N) when T (N) = N2 +
N + 1 and N ≡ 1 (mod 3).
A more general result than Theorem 7.6 and one that is more in line with Lucas’
precept that the primality of N be established by showing that N divides certain
integers is provided in Theorem 7.8 below. In order to demonstrate this result we
need a simple lemma.
Lemma 7.7. Suppose N is odd and let m be any positive integer such that (m,N) =
1. If N | Cmn/Cn, then (N,Dn) = 1.
Proof. Suppose p is any prime divisor of Dn and N . Since p | Cn and p | Wn−6Rn,
we see by our results in Section 4.4, in particular equation (5.4), we have that
Cmn/Cn ≡ m3Rn(m−1) (mod p).
It follows that since p - m and p - R ((Dn, R) | 2), we must have p - Cmn/Cn,
contradicting N | Cmn/Cn.
We are now able to produce an analogue of Corollary 2.23.1.
Theorem 7.8. Let N be an integer such that (N, 6) = 1. If N | DT (N) and
N | CT (N)/CT (N)q
for each prime divisor q of T (N), then N is a prime.
Proof. Since (T (N), N) = 1, we have (q,N) = 1. By Lemma 7.7 we know that if p is
any prime divisor of N , then (N,DT (N)q
) = 1. Thus, N - DT (N)q
for all prime divisors
q of T (N) and (N,DT (N)
q′) = 1 for some (any) prime divisor q′ of T (N). The result
follows by Theorem 7.4.
188
We note here that Lucas himself ([Luc78], pp. 310-311) made use of the divisi-
bility of U3n/Un to produce a primality test for N = A3n−1. Also, the computation
of CT/CTq
can be done by using the methods of Section 3.6.
Unfortunately, results like Theorems 7.4 and 7.8 are of limited utility in primality
testing because we need to know the complete factorization of T (N), and this is often
not available to us. In the next section, we will consider some special cases when
T (N) = N2 +N + 1.
7.2 The Case of T (N) = N 2 +N + 1
We will deal here with the case of N2 +N + 1 = tL, where L is a prime.
Theorem 7.9. Let N2 + N + 1 = tL, where L is a prime. If t < N −√N + 1,
(N,Dt) = 1 and N | DN2+N+1, then N is a prime.
Proof. If N is composite, there must be a prime p such that p | N and p ≤√N .
Also, since ω(p) | N2 + N + 1, we must have ω(p) | tL. Certainly ω(p) 6= 1 and
since p - Dt, ω(p) - t. It follows that since ω(p) | tL and ω(p) - t, we must have
L | ω(p). Now, ω(p) ≤ p2 + p+ 1 by Corollary 5.6.1; hence
p2 + p+ 1 ≥ L =N2 +N + 1
t
or
t(p2 + p+ 1) ≥ N2 +N + 1 = (N +√N + 1)(N −
√N + 1).
Since p ≤√N and N+
√N+1 ≥ p2 +p+1, we get t ≥ N−
√N+1, a contradiction.
189
Corollary 7.9.1. If N is odd, N ≡ 1 (mod 3), L = (N2 + N + 1)/3 is a prime,
(N,D3) = 1, and N | DN2+N+1, then N is a prime.
Proof. If N is composite, then N ≥ 52 and 3 < 52 − 5 + 1.
We also have the following result.
Theorem 7.10. Let N2 + N + 1 = tL, where L is a prime. If L > t2 + t + 1,
(N,Dt) = 1 and N | DN2+N+1, then N is a prime.
Proof. From our proof of Theorem 7.9 we know that if p is any prime divisor of N ,
then L | ω(p). Thus, N2 + N + 1 | tω(p). Hence, if N is composite, then N = pr
(r > 1) and
p2r2 + pr + 1 ≤ tω(p) ≤ t(p2 + p+ 1);
hence, r < t. Now let q be any prime divisor of r. Then since q | N we have L | ω(q)
and
t2 + t+ 1 < L ≤ q2 + q + 1.
It follows that r ≥ q > t, a contradiction.
Now, suppose S is a fixed positive integer and N = tS + u, where t = u2 + u+ 1
and u ∈ Z. Then N2 +N + 1 = tL, where
L = tS2 + (2u+ 1)S + 1.
For such numbers, we have the following result.
Theorem 7.11. If N = tS + u, where S ≥ 2 and N > 4, we have t < N −√N + 1.
190
Proof. Since
N −√N + 1 =
1
4(2√N − 1)2 +
3
4,
we see that N −√N + 1 is an increasing function of N for N > 0. Thus, since
N ≥ 2t+ u > 0, we get
N −√N + 1 ≥ 2t+ u−
√2t+ u+ 1.
If u = 0, −1, −2, then 1 ≤ t ≤ 3 and N −√N + 1 > 4 − 2 + 1 > t. If u 6= 0, −1,
−2, then (u+ 1)2 ≥ 4 and
(u+ 1)4 ≥ 4(u+ 1)2 > 2u2 + 3u+ 2 = 2t+ u.
Thus,
(u+ 1)2 >√
2t+ u and hence 2t+ u−√
2t+ u > t.
From Theorems 7.9 and 7.11 we see that if tS2 + (2u+ 1)S + 1 is a prime, then
we can use the test of Theorem 7.9 to prove that tS + u is a prime. Of course, if
N = tS+u is a prime, it might not be an I prime and therefore T (N) 6= N2 +N +1;
consequently, this test would not be successful. Thus, we need to find values for P , Q,
R such that ifN = tS+u is a prime, thenN is an I prime for f(x) = x3−Px2+Qx−R.
It is well-known (see [Wil72b], [Leh58]) that if N is a prime and
Np−13 6≡ 1 (mod p),
where p is a prime (≡ 1 (mod 3)), 4p = r2 + 27s2 with r ≡ 1 (mod 3) and N - spr,
then the cubic congruence
x3 − 3px− pr ≡ 0 (mod N)
191
is irreducible; that is, N is an I prime for
P ≡ 0, Q ≡ −3p, R ≡ pr (mod N).
Notice that since (pr,N) = 1, there always exists some x such that (pr+xN,−3p) =
1; hence, the fact that (−3p, pr) = p 6= 1 does not have any affect on the validity of
our results. We have proved the following theorem.
Theorem 7.12. Let L = tS2 + (2u + 1)S + 1 be a prime and put N = tS + u.
Suppose (N, 6) = 1, p is a prime such that p ≡ 1 (mod 3), Np−13 6≡ 1 (mod p) and
4p = r2 + 27s2 with r ≡ 1 (mod 3) and (N, prs) = 1. If we put
P ≡ 0, Q ≡ −3p, R ≡ pr (mod N),
then N is a prime if and only if N | DN2+N+1 and (Dt, N) = 1.
Corollary 7.12.1. Let 2 - S, L = 3S2 − 3S + 1 be a prime and suppose that p is a
prime such that p ≡ 1 (mod 3) and
Np−13 6≡ 1 (mod p),
where N = 3S − 2. If we define r and s as above and (N, prs) = 1, then N is a
prime if and only if N | DN2+N+1, where
P ≡ 0, Q ≡ −3p, R ≡ pr (mod N).
Proof. This follows easily from the theorem by putting u = −2 and noting that
∆ ≡ 27(4p3 − p2r2) (mod N),
W1 = PQ− 3R ≡ −3pr (mod N),
4C3 = ∆ + 3W 21 ≡ −4(27p3) (mod N).
Hence, (C3, N) = 1.
192
Notice, that we need only perform O(logN(M(log2N))) operations to establish
the primality of N , once we know that L is a prime. This is much faster than several
other tests because it is not necessarily all that easy to find enough factors of N ± 1
to use the techniques of Brillhart, Lehmer and Selfridge [BLS75], which generalized
those of Lucas, to establish the primality of N .
7.3 The Primality of L
If we put
L = tS2 + (2u+ 1)S + 1, t = u2 + u+ 1 and N = tS + u,
the results of the last section allow us to establish the primality of N , when we have
already proved L is a prime. This is not a vacuous result because we certainly expect
by the Bateman–Horn conjecture [BH62] that there exists an infinitude of values of
S such that for a fixed u, L and N will both be prime. Also, for a fixed value of
S, we would expect that there exists an infinitude of values of u such that both L
and N will be prime. There remains, however, the difficulty of proving that L is a
prime. We notice, however, that S | L − 1. Suppose S = FG, where we know the
complete factorization of F . It is then possible, by using the methods of [BLS75] to
prove that L is either prime or that all the prime factors of L must have the form
kF + 1.
Theorem 7.13. If L = tS2+(2u+1)S+1 (t = u2+u+1), S = FG and all the prime
factors of L have the form kF + 1, then L is a prime when F > tG2 + |2u+ 1|G+ 2.
193
Proof. Suppose L is composite. We must have
L = (k1F + 1)(k2F + 1),
where k1, k2 ≥ 1. Hence,
tS2 + (2u+ 1)S = k1k2F2 + (k1 + k2)F,
tSG+ (2u+ 1)G = k1k2F + k1 + k2
and
(tG2 − k1k2)F = k1 + k2 − (2u+ 1)G.
If tG2 − k1k2 = 0, then
(k1 − k2)2 = (k1 + k2)2 − 4k1k2 = (2u+ 1)2G2 − 4tG2 = −3G2,
which is impossible; hence |tG2 − k1k2| ≥ 1. Also, since k1, k2 ≥ 1, we get (k1 −
1)(k2 − 1) ≥ 0 and hence k1k2 ≥ k1 + k2 − 1. Now
|tG2 − k1k2|F ≤ k1 + k2 + |2u+ 1|G ≤ k1k2 + 1 + |2u+ 1|G
= k1k2 − tG2 + tG2 + |2u+ 1|G+ 1;
hence,
F ≤ tG2 + |2u+ 1|G+ 2,
which is impossible.
Suppose we now consider the following simple example, where we put F = 2n,
G = 1. We get
Ln = (u2 + u+ 1)22n + (2u+ 1)2n + 1, Nn = (u2 + u+ 1)2n + u.
194
In this case if 2n > u2 + 3|u| + 4, we can easily establish (when it is the case) that
Ln is a prime. We can next use our earlier results to prove that Nn is a prime, when
that is the case. In the table below we provide all instances for various values of u
(−500 ≤ u ≤ 500) and n ≤ 1000 such that both Ln and Nn are prime.
195
Table 7.1: Values of u such that Ln and Nn are both prime (n ≤ 1000)n u n u
1 3, -3, -5, 13, 25, 31, -33, 37, -39, 55, -57, 41 -27, 111, -141-71, 79, -87 -159, 181, -183, 219, -221, -243, 43 -173
255, -255, 279, -281, 289, -291, 307, 325, 333, 46 -449-353, -369, 375, -395, -423, -435, -495 51 273
2 -11, 31, 55, 115, -191, -221, 271, 361 53 -53, 267, -2673 -3, 7, 19, 25, -33, 39, -51, -65, 79, 105, 117, 177, 55 25, 87, -327, 475,
-231, 259, -401, 483, 499 56 307, -4194 -5,31, 223, 277, -323, 367, 415 57 217, 361, 4835 3,-17, 19, 39, -39, -45, -65, 73, -95, -101, 58 -221
129, -137, -153, 165, 207, -233, 295, -297, 61 475-323, 339, -389, 417, 463, 481 65 -195
6 -65, 145, 259, -311 69 -21, 4157 9, -15, -95, 109, -243, 297, -297, 457, 459, -477 70 -58 -179, -209, -263, -395 72 3619 9,-65, 91, -227, 397, 471 73 44710 -5, 349 74 16911 13, 15, 25, 87, -111, 159, 199, 285, 309, -381 75 49912 -119, 205, 271 77 3913 25,-33, -285, 325, 349, -449 79 -22714 199,-281, -359,439 83 -25115 9, 25, 39, 105, -105, -107, 235, 313, 397, 415, -471 84 -3516 -89,277, -389, -395, -407 95 -6517 61, 73, -135, -141, 255, 321, 481 96 -23318 -101 121 2719 -123, -221, -255, -311, 487 123 -46720 31,-185,-269 137 -29721 -17, 81,-149, -413, 445 140 19322 -215, 319 143 40323 127, 129, 265, -275, -323, -335, -401, -437 207 12325 -267, -309, 499 211 -14127 39,-161 231 16328 181 360 -46730 235 399 11732 187, -209 407 23735 -381, 463 417 -25738 361 533 -40740 -365 819 289
196
If we put F = qn, where q is a prime and G = 1, we can once again easily establish
the primality of
Ln = (u2 + u+ 1)q2n + (2u+ 1)qn + 1
when Ln is a prime. However, if we specify q and u, it seems a very rare event to
have both Ln and Nn = (u2 + u+ 1)qn + u prime simultaneously. We illustrate this
rarity in the next table, we provide those few values of n for which Ln and Nn are
both prime for all n < 1000.
Table 7.2: Prime pairs for specific choices of u and qu q Ln Nn n < 1000
2 3 7 · 32n + 5 · 3n + 1 7 · 3n + 2 1-2 3 3 · 32n − 3 · 3n + 1 3 · 3n − 2 1, 4, 5-2 5 3 · 52n − 3 · 5n + 1 3 · 5n − 2 1, 2-2 7 3 · 72n − 3 · 7n + 1 3 · 7n − 2 1-2 11 3 · 112n − 3 · 11n + 1 3 · 11n − 2 1, 73 2 13 · 22n + 7 · 2n + 1 13 · 2n + 3 1, 5-3 2 7 · 22n − 5 · 2n + 1 7 · 2n − 3 1, 3-4 3 13 · 32n − 7 · 3n + 1 13 · 3n − 4 26 5 43 · 52n + 13 · 5n + 1 43 · 5n + 6 156 11 43 · 112n + 13 · 11n + 1 43 · 11n + 6 18 3 73 · 32n + 17 · 3n + 1 73 · 3n + 8 112 5 157 · 52n + 25 · 5n + 1 157 · 5n + 12 114 3 211 · 32n + 29 · 3n + 1 211 · 3n + 14 1, 1715 2 241 · 22n + 31 · 2n + 1 241 · 2n + 15 11-15 2 211 · 22n − 29 · 2n + 1 211 · 2n − 15 7-18 11 307 · 112n − 35 · 11n + 1 307 · 11n − 18 11-21 2 421 · 22n − 41 · 2n + 1 421 · 2n − 21 6927 2 757 · 22n + 55 · 2n + 1 757 · 2n + 27 121-28 3 757 · 32n − 55 · 3n + 1 757 · 3n − 28 3, 9
197
In the particular case of u = −2, q = 3, row 2 of Table 7.2, we get
Ln = 32n+1 − 3n+1 + 1 and Nn = 3n+1 − 2.
For n > 3, we need only find some b such that
bLn−1 ≡ 1 (mod Ln) and (bLn−1
3 − 1, Ln) = 1, (7.1)
to establish that Ln is a prime. Note that 3n+1 || Ln − 1. Suppose p is some prime
such that p | Ln. By (7.1) we have
p | bLn−1 − 1 and p - bLn−1
3 − 1.
If ω is the order of b modulo p, then
ω | Ln − 1 and ω -Ln − 1
3.
So 3n+1 | ω and ω | p − 1; thus p ≡ 1 (mod 3n+1). Hence p = k3n+1 + 1 for some
k ∈ N. We then have p ≥ 2 · 3n+1 + 1 and we can conclude that Ln is a prime
since p >√Ln. Having done this we can use Corollary 7.12.1 to establish that Nn
is a prime. This sort of testing of pairs of numbers for primality might have pleased
Lucas.
7.4 The Case of T (N) = N 2 − 1
It is certainly possible to test numbers of the form Aqn±1 for primality by using the
Wn and Cn functions; however, we will confine our attention here to the case where
N = A3n − 1, as this is the analogous form to A2n − 1 mentioned in Chapter 2. We
can produce a theorem similar to Theorem 2.24, except for the necessity condition.
198
Theorem 7.14. Let N = A3n − 1, where 2 | A, A < 3n, n ≥ 2 and (N,R) = 1. If
N | CN+1/CN+13,
then N is prime.
Proof. Let p be any prime divisor of N . Since p | CN+1, we must have some rank of
apparition r(p) in {Cn} such that r(p) | N + 1. Also, since
4CN+1/CN+13
= ∆C2N+1
3
+ 3W 2N+1
3
,
we see that if p | CN+13
, then p | WN+13
. Hence p | EN+13
and by Theorem 6.7,
we know that p ≡ 1 (mod 3n). If p - CN+13
, then r(p) - N+13
. It follows that
3n | r(p). If p | ∆, then r(p) | p or p− 1 and the first case is impossible, as p | N
and r(p) | N + 1 ⇒ r(p) - p. If p is an I prime, then r(p) | p2 + p + 1, but this is
impossible because 9 - p2 + p+ 1. If p is an S prime or a Q prime, then r(p) | p2− 1
and p ≡ ±1 (mod 3n). Thus, any prime divisor of N must be at least as large as
2 · 3n − 1. Since (2 · 3n − 1)2 > N , N must be a prime.
Our next objective will be to produce conditions that are both necessary and
sufficient for N = A3n − 1 to be prime. We first need to produce a result analogous
to Theorem 2.25. We begin with the following theorem.
Theorem 7.15. Let p be an odd prime such that p ≡ −1 (mod 3). Then there exist
P , Q, R such that p is a Q prime if and only if
P ≡ a+ Tr(λ), Q ≡ aTr(λ) + N(λ), R ≡ aN(λ) (mod p),
where a ∈ Z, λ = r1 + r2ρ ∈ Z[ρ], ρ2 + ρ+ 1 = 0 and p - ar2 N(λ).
199
Proof. Suppose P , Q, R satisfy the conditions of the theorem. Then clearly
∆ ≡ (a− λ)2(a− λ)2(λ− λ)2
= N(a− λ)2r22(ρ− ρ2)2
= −3m2 (mod p),
where m ∈ Z. Since p ≡ −1 (mod 3) and p - ar2 N(λ), we cannot have p | m. Thus(∆
p
)=
(−3m2
p
)=
(−3
p
)= −1.
Since p - 6R∆, p is a Q prime for f(x) = x3 − Px2 +Qx−R.
Next, suppose that p is a Q prime for f(x) = x3−Px2 +Qx−R. Then Fp2 is the
splitting field of f(x) in Fp[x]. Let α, β, γ be the zeros of f(x) in Fp2 , where αp = α,
βp = γ 6= β = γp. Since F∗p2 = 〈θ〉 for some suitable θ ∈ Fp2 , we put ζ = θp2−1
3 and
note that ζ2 + ζ + 1 = 0. Now since p ≡ −1 (mod 3),(β − γζ − ζ2
)p=γ − βζ2 − ζ
=β − γζ − ζ2
.
Hence β−γζ−ζ2 ∈ Fp. If we put
a ≡ α, b ≡ β − γζ − ζ2
6≡ 0, c ≡ β + γ = P − α ≡ P − a (mod p),
then
β =b+ c
2+ bζ, γ =
b+ c
2+ bζ2.
Putting r1 ≡ (b+ c)/2 (mod p), r2 ≡ b (mod p) we see that
P ≡ a+ Tr(λ), Q ≡ aTr(λ) + N(λ), R ≡ aN(λ) (mod p),
for λ = r1 + r2ρ, p - r2. Since p - R, we must also have p - aN(λ).
200
We can now present our analogue of Theorem 2.25.
Theorem 7.16. Let p be an odd prime such that p ≡ −1 (mod 3). If P , Q, R
satisfy the conditions of Theorem 7.15 and(λp
)36= 1, then
p | Cp+1/C p+13.
Proof. By Theorem 7.15, we know that p is a Q prime and therefore p | Cp+1. Let
α, β, γ be the zeros of f(x) in Fp2 , where αp = α, βp = γ 6= β = γp. Since λp2−1
3 6≡ 1
(mod p), we may assume with no loss of generality that βp2−1
3 6= 1 in Fp2 . Since
βp2−1
3 = (βp−1)p+13 =
(γ
β
) p+13
,
we have βp+13 6= γ
p+13 . Also, since
βpp+13 = γ
p+13 6= β
p+13 ,
we cannot have αp+13 = β
p+13 because β
p+13 6∈ Fp. Similarly α
p+13 6= γ
p+13 . It follows
that C p+136= 0 in Fp2 or p - C p+1
3. Hence p | Cp+1/C p+1
3.
By combining Theorems 7.14 and 7.16 we get the following necessary and suffi-
cient condition for N = A3n − 1 (2 | A, A < 3n) to be prime.
Theorem 7.17. Let N = A3n − 1, where 2 | A and 3 < A < 3n. Futhermore, let
q ≡ 1 (mod 3) be a prime such that q - N and
Nq−13 6≡ 1 (mod q).
Let λ = r1 + r2ρ (ρ2 +ρ+ 1 = 0) be a primary prime divisor of q in Z[ρ] and suppose
that N - r2. Let
P ≡ a+ Tr(λ), Q ≡ aTr(λ) + q, R ≡ aq (mod N),
201
where (a,N) = 1. Then N is a prime if and only if
N | Cp+1/C p+13.
Proof. Since (N,R) = 1, we know by Theorem 7.14 that N is a prime if
N | Cp+1/C p+13.
Next, suppose that N is a prime. We know that
Nq−13 6≡ 1 (mod q),
and since λ is a primary prime divisor of q, we have by the cubic reciprocity law(N
λ
)3
6= 1⇒(λ
N
)3
6= 1.
Thus, N | Cp+1/C p+13
by Theorem 7.16.
7.5 Primality Test
We may now use Theorem 7.17 to produce a primality test, somewhat similar to the
Lucas and Lehmer test for numbers of the form A2n − 1, for numbers of the form
A3n − 1. Of course, this test is not as practical as that of [Wil72b], but it would
have been of some interest to Lucas that Cn could be used to produce such a test.
Let m ∈ Z+ such that (m, 2R) = 1. Compute
S0 ≡Wn
2Rn(mod m) and R0 ≡
∆C2n
4R2n(mod m)
and define
Si ≡W3in
2Rn3i(mod m) and Ri ≡
∆C23in
4R2n3i(mod m).
202
Notice then
Si+1 =1
2R−n3i+1
W3i+1n =1
2R−n3i+1
W3(3in)
=1
2R−n3i+1
[3
∆C23in
4(W3in + 2R3in) +
W 33in
4− 6R3inW3in
4+ 6R3i+1n
]
=1
2R−n3i+1
[3R2(3in)Ri(2R
3inSi + 2R3in) +(2R3inSi)
3
4− 6R3in(2R3inSi)
2
4+ 6R3i+1n
]=
1
2R−n3i+1
[3RiR
3i+1n2(Si + 1) + 2R3i+1nS3i − 6R3i+1nS2
i + 6R3i+1n]
= 3Ri(Si + 1) + S3i − 3S2
i + 3.
Similarly,
Ri+1 =∆C2
3i+1n
4R2(3i+1n).
Using Corollary 3.10.1 we have
C3i+1n =1
4C3in(∆C2
3in + 3W 23in).
Use this and the fact that
W3in ≡ 2R3inSi and ∆C23in ≡ 4R2(3in)Ri (mod m)
to obtain
C3i+1n =1
4C3in(4R2(3in)Ri + 3(2R3inSi)
2).
This yields
C23i+1n =
1
4C2
3in
1
4(4R2(3in)Ri + 3(2R3inSi)
2)2 =C2
3in
4
1
4(4R2(3in)Ri + 12R2(3in)S2
i )2
=C2
3in
4
1
4(4R2(3in)(Ri + 3S2
i ))2 =
C23in
44R4(3in)(Ri + 3S2
i )2.
We manipulate this as follows:
∆C23i+1n
4=
∆C23inR
4(3in)
4(Ri + 3S2
i )2.
203
So then we get
∆C23i+1n
4R2(3i+1)n=
∆C23inR
4(3in)
4R2(3i+1)n(Ri + 3S2
i )2.
Thus
Ri+1 =∆C2
3inR4(3in)
4R6(3i)n(Ri + 3S2
i )2 =
∆C23in
4R2(3i)n(Ri + 3S2
i )2
= Ri(Ri + 3S2i )
2.
We can now employ these observations to produce our primality test. If we satisfy
the conditions of Theorem 7.17 where N = 3nA− 1 and set
S0 ≡WA
2RA(mod N) and R0 ≡
∆C2A
4R2A(mod N),
then
Si ≡W3iA
2R3iA(mod N) and Ri ≡
∆C23iA
4R2(3iA)(mod N).
We can produce the sequence {Si} and {Ri} (mod N) as follows:
Si+1 ≡ 3Ri(Si+1)+S3i −3S2
i +3 (mod N) and Ri+1 ≡ Ri(Ri+3S2i ) (mod N).
Then N is a prime if and only if Rn−1 ≡ −3S2n−1 (mod N).
Chapter 8
Conclusion
8.1 Main Result
The purpose of this thesis was to develop a cubic extension of the Lucas functions
that Lucas himself might have discovered. What has emerged from this work is
a theory of functions that displays a number of pleasing similarities with Lucas’
original work. The main tools in Lucas’ investigation of his functions were the
multiplication formulas (2.14) and (2.15). The multiplication formulas, proved in
Section 3.5, allowed us to obtain arithmetic results that closely resemble those for
the Lucas case. Key results like the laws of repetition and apparition, and Euler’s
criterion, as described in Sections 4.4, 4.5, 5.2, 5.3, 6.2, and 6.3, have analogues
in our extension. Most remarkably, the extension relies on the use of only two
functions1, despite the fact that you would expect three for the cubic case. Further,
when restricted to the quadratic case, our generalization in Section 3.3 satisfyingly
reduces to that of Lucas sequences.
With all that in mind, it is difficult to point to a single ‘main’ result. However,
knowing that Lucas’ own goal in generalizing his sequences was to find and implement
new primality tests, Theorem 7.17 and the primality test of Section 7.5 based on it
stand out. The test makes use of {Cn}, a sequence known to Lucas that surely
would have been a part of any generalization he would have done, to test numbers
1It might be argued that we are really considering four functions here because of Dn and En,but these latter functions are simply a convenient way of representing certain divisors of Cn.
204
205
of the form A3n − 1. Certainly, even more important than just the primality test
is Theorem 7.4, a result that is our analogue of Theorem 2.22, which Lucas refered
to as his fundamental theorem. It should be stated, however, that today there exist
many sophisticated methods for primality proving (see, for example, Chapter 4 of
[CP01]). The primality conditions proved here are of mere historical interest and are
perhaps what Lucas had in mind.
8.2 Improvements
In terms of what might be done to refine or improve the results of the thesis, it
would by satisfying to have more elementary proofs of Theorems 4.18 and 4.19. As
it stands, both proofs use facts from algebraic number theory, and these are the only
two instances where such powerful tools are required. However, some key results from
Serret’s Cours d’algebre superieure Vol. II [Ser79] allow for the desired elementary
alternate proof of Theorem 4.18. For the details, the interested reader is directed to
Appendix A. In fact, Lucas was familiar with Serret’s work, so if Lucas had proved
this result, it is imaginable that he would have used the methods seen in Appendix
A.
It would also be interesting to develop further the law of repetition for {Dn} so
that it more closely matches the Lucas case, where if p is a prime and for λ > 0, we
have pλ 6= 2 and pλ || Um, then pλ+µ || Umnpµ when p - n and µ ≥ 0. This is how
the law of repetition is presented for {Un} in Chapter 4 of [Wil98].
206
8.3 Future Work
Beyond the thesis itself, work could be done to try to use {Cn} and {Wn} to im-
plement an RSA-type cryptosystem. Peter Smith’s LUC [Smi93]2, a well-known
example of such a cryptosystem using the Lucas functions, provides the motivation
for this. It might also be possible to use {Cn} and {Wn} to perform a Diffie-Hellman-
like key exchange.
There is also, of course, the possibility of exploring the idea of a quartic extension
as mentioned by Lucas. For such an extension, Cn and Wn would be as described
early in Section 3.3. Despite being an interesting problem, we would expect it to be
difficult to work with the quartic case due to the number of terms involved in the
recurrences.
2See [BBL95] for some useful comments concerning LUC.
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Appendix A
In this appendix we provide an elementary proof of Theorem 4.18. This proof could
easily have been discovered by Lucas through his admitted knowledge of Chapter III
of [Ser79]. We first recapitulate some of the preliminary results in Serret’s work.
Definition A.1. Let p be a prime and let f(x), g(x) ∈ Z[x]. If there exist h(x),
k(x) ∈ Z[x] such that
h(x)g(x) = f(x) + pk(x),
we say that f(x) is divisible by g(x) with respect to modulus p. We write this as
h(x)g(x) ≡ f(x) (mod p).
Definition A.2. If f(x) ∈ Z[x], f(x) is monic and f(x) is not divisible by any
g(x) ∈ Z[x] with respect to modulus p, we say that f(x) is irreducible with respect
to p.
Theorem A.3. If g(x), h(x) ∈ Z[x], g(x), h(x) have no common divisor (of degree
≥ 1) with respect to the prime modulus p, then there exist Y (x), Z(x) ∈ Z[x] such
that
Y (x)h(x)− Z(x)g(x) ≡ 1 (mod p).
The next result does not appear explicitly in [Ser79], but it would have been easy
for Lucas to derive because its proof is entirely analogous to that of the elementary
number theory result which states that if a, b, c ∈ Z, (a, b) = 1, a | c and b | c, then
ab | c. This is proved by Lucas on p. 340 in [Luc91b]. The proof of Theorem A.4
would follow in exactly the same manner by using Theorem A.3.
215
216
Theorem A.4. If f(x), g(x), h(x) ∈ Z[x], g(x) and h(x) have no common divisor
(of degree ≥ 1) with respect to the prime modulus p, and g(x), h(x) are both divisors
of f(x) modulo p, then g(x)h(x) is a divisor of f(x) with respect to the modulus p.
The next result (Theorem I in Section 346 of [Ser79]) is most important for our
subsequent work. We give it in a somewhat different form from Serret’s results, but
Serret certainly establishes it in his proof of Theorem I.
Theorem A.5. Let f(x) ∈ Z[x] be irreducible with respect to a prime modulus p and
of degree ν. Then f(x) is a divisor of xpν−1 − 1 with respect to the modulus p.
We now suppose that f(x) = x3 − Px2 + Qx − R and p is a prime such that
p - 6∆R. We know that when p is an S prime we have
f(x) ≡ f1(x)f2(x)f3(x) (mod p),
where f1(x), f2(x), f3(x) ∈ Z[x] are each monic of degree 1 and, because p - ∆, have
no common divisor with respect to p. If p is a Q prime, then
f(x) ≡ f1(x)f2(x) (mod p),
where f1(x), f2(x) ∈ Z[x]; f1(x), f2(x) are monic and irreducible with respect to the
modulus p, deg f1(x) = 1, deg f2(x) = 2. Finally, if p is an I prime, then f(x) is
irreducible with respect to the modulus p.
It is now easy to prove, by making use of Theorems A.4 and A.5, that f(x) is a
divisor of xpµ−1−1 with respect to the modulus p, where µ = 1 when p is an S prime,
µ = 2 when p is a Q prime, and µ = 3 when p is an I prime. Putting m = pµ − 1,
we have
xm − 1 = f(x)g(x) + ph(x)
217
for some g(x), h(x) ∈ Z[x]. Putting x = α, β, γ, where α, β, γ ∈ C are the zeros of
f(x), we get
αm − 1 = ph(α),
βm − 1 = ph(β),
γm − 1 = ph(γ),
from which it follows that
αm − βm = p(h(α)− h(β)),
βm − γm = p(h(β)− h(γ)),
γm − αm = p(h(γ)− h(α)).
Hence
∆C2m = (αm − βm)2(βm − γm)2(γm − αm)2
= p6S(α, β, γ)
where,
S(x, y, z) = (h(x)− h(y))2(h(y)− h(z))2(h(z)− h(x))2.
Since S(x, y, z) is a symmetric polynomial in Z[x, y, z], we must have S(α, β, γ) ∈ Z
by the fundamental theorem of symmetric polynomials, and therefore p3 | Cm. Also
Wm − 6Rm = αm(βm − γm)2 + βm(γm − αm)2 + γm(αm − βm)2
= p2T (α, β, γ),
where
T (x, y, z) = (h(y)− h(z))2 + (h(z)− h(x))2 + (h(x)− h(y))2
+p[h(x)(h(y)− h(z))2 + h(y)(h(z)− h(x))2 + h(z)(h(x)− h(y))2].
218
Since T (x, y, z) is also a symmetric polynomial in Z[x, y, z], we must have T (α, β, γ) ∈
Z and p2 | Wm − 6Rm.
We are now able to present another proof of Theorem 4.18.
Theorem A.6. If p - 6∆R and p | Cn, p | Wn− 6Rn, then p3 | Cn and p2 | Wn−
6Rn.
Proof. Let ω = ω(p). Clearly ω exists and ω divides both n and m by Theorem 5.5.
From the proof of Theorem 5.1, we know that
Cm/Cω ≡ (m/ω)3Rm−ω (mod p).
Thus, since p - m, we get p - Cm/Cω ⇒ p3 | Cω ⇒ p3 | Cn. Also from the proof of
Theorem 5.1, we have
Wm − 6Rm ≡ (m/ω)2Rm(Wω − 6Rω) (mod p2);
thus, we get p2 | Wω − 6Rω. Furthermore,
Wn − 6Rn ≡ (n/ω)2Rn(Wω − 6Rω) (mod p2)
means that p2 | Wn − 6Rn.
Although this proof requires Theorems 5.1 and 5.5, these results did not require
the result of Theorem 4.18 in their respective proofs.