J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
P
IETER
M
OREE
On the prime density of Lucas sequences
Journal de Théorie des Nombres de Bordeaux, tome
8, n
o2 (1996),
p. 449-459
<http://www.numdam.org/item?id=JTNB_1996__8_2_449_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
On
theprime density
of Lucas sequencespar PIETER MOREE
RÉSUMÉ. On donne la densité des nombres
premiers qui
divisentau moins un terme de la suite de Lucas définie par
L0(P)
= 2,L1(P)
= P etLn(P) =
PLn-1(P) + Ln-2 (P)
pourn ~ 2, avec P entier arbitraire.
ABSTRACT. The
density
ofprimes
dividing
at least one term of the Lucas sequence definedby
L0(P)
=2,
L1(P)=
P and
Ln(P)
=PLn-1(P)
+Ln-2(P)
for n ~2,
with P anarbitrary integer,
is determined.1. Introduction
Let P and
Q
be non-zerointegers.
Then the sequence definedby
and for
every n > 2,
Ln (P, Q)
=,
is called a Lucas sequence
(of
the secondkind) .
In this paper we will bemainly
concerned with the caseQ
= -l. For convenience we writeLn(P)
instead of
Ln (P, -1 ) .
If ,S is any set ofprimes,
thenby
S (x )
we denote thenumber of elements in S’ not
exceeding
x. The limitlimz-+oo
S(x)/(x),
ifit
exists,
is called theprime density
of S. It will be denotedby
be a real
quadratic
field with D > 1 and Dsquarefree.
(This
assumption
on D is maintainedthroughout.)
LetD D
denote itsring
ofintegers.
Suppose DD
contains a unit of norm -1.(Thus
ED, thefundament al unit >
1,
has norm-1. )
be aunit0f DD-
In thispaper we are interested in
computing
theprime
density
of the set ofprimes
dividing
at least one term of the sequence + The characteristicpolynomial
associated to the sequencefUn + tl")
is irreducible overQ.
Few
people
seem to have considered thisproblem.
The papers[4, 5]
are theonly
ones known to the author in this direction. In contrast several authors[1,
3, 6,
10]
considered theprime
density
of Lucas sequences of the second kindhaving
reducible characteristicpolynomial (i.e.
sequences of the formOur main result is the
following.
THEOREM 1. Let D be a
squarefree
integer
exceeding
1 such thathas a unit
of
negative
norm. ::1:1 be a unit. Then there 0and e
of
norm -1 such that u =f2Å.
Thesequence Un
has aprime
density.
In case D = 2 it isgiven by
17/24
if
A =0,
5/12
if
A = 1 and2-a/3
otherwise. In case D > 2 theprime
density
equals
21-À/3.
It should be remarked that the
question
whether aquadratic
field has a unit ofnegative
norm is still notcompletely
resolved. If D has aprime
divisor p -
3(mod 4),
then there is no such unit. From this iteasily
followsthat there are at most discriminants D x for which there is a
negative
unit.Stevenhagen
conjectures
that there areasymptotically
such
discriminants,
for someexplict
constant c ~ .58058. Fortopic
see e.g.[9].
Theorem 1 allows one to
compute
thedensity
of the Lucas sequence{Ln(P,1)}
for various P. Forexample
the sequence{Ln(326,1)}
hasdensity
1/3.
Moreinterestingly
Theorem 1 allows one to calculate for everyinteger
P the
prime
density
of the Lucas sequence In this calculation the sequencefLn(2)1"O
={2,2,6,14,34,’" },
the so called Pellsequence,
plays
animportant
role.THEOREM 2. For P a non-zero
integer
let be the Lucasse-quence
defined by Lo (P)
=2,
Ll (P)
= Pand,
for n >
2,
Ln (P) _
PLn_1(P)
+Ln-2(P).
Then theprime
density of
this sequence exists andequals
2/3,
unlessIPI
=Ln(2) for
some odd n >1,
in which case thedensity
is17/24.
On
taking
P = 1 we find that theprime
density
of thesequence of Lucas
numbers
equals
2/3.
This was firstproved by
Lagarias
[4].
Taking
P = 2 it is seen that theprime density
of the Pell sequence is17/24.
I would like to thank the referee for
her/his
helpful
comments.2. Outline of the
proofs
The arithmetic of the sequence where
An
= a"+ 1k~,
and a EQ( /15)
is aquadratic integer,
isintimately
connected with that of thesequence where
Wn
:=(a n -
an)/(a -
a).
This sequence can bealternatively
definedby
Wo = 0, W,
=1, Wn
=for n > 2. It is a Lucas sequence
(see
[7,
p.41]
for adefinition)
ofthe first kind. We recall some facts from
[7,
pp.41-60].
Forprimes
pwith
(~, 2N (a) )
=1,
there exists a smallest index 1 such that The indexpa(p)
is called the rank ofapparition
of p in If(p, 2N(a))
=1,
thenplwn
if andonly
if FurthermoreW2n
=WnAn
andUsing
the latter threeproperties
it can beeasily
shown that if
(P,2N(a))
=1,
then p divides{An}
if andonly
ifpa(p)
density
of(An)
is tocompute
thedensity
ofprimes
for whichpa(p)
iseven. The fact
that,
for =1,
pa(p)
divides p-(D/p),
forces us to consider the cases
(D/p)
= 1 and(D/p) =
-1seperately.
Fors = 1, 2,
e ~ 0, j ~ 1 put
x-We show that
N1 (e, j; a)
has aprime
density,
61(e,
j;
a),
and express itin terms of
degrees
of certain Kummerian extensions. Thisapproach
goes back to Hasse[3].
The case s = 2 is rather moredifficult,
except
when a is aunit of
negative
norm, in which case evenelementary arguments
suffice. So assume a = e is aquadratic
unit of norm -1. In this case it is not difficultto show that the
prime
density
of the sequence{An} _ {Ln(Tr(e))}
isgiven by
-The
prime
densities81 (e, j; e)
and62 (e, j; e)
arecomputed
inrespectively
§3
and
§4.
They
are tabulated in Tables I and II. Theentry
e in the last col-umngives
~~°_1 be(e, j; e).
Theentry j
in the last rowgives
££o
The distinction between the case D = 2 and D > 2 is due to the fact that for D >
2,
Q( VÐ)
isonly
a subfield ofQ((2;)
forsome j
if D = 2.Finally
in
§5
proofs
of Theorems 1 and 2 aregiven.
3. The
prime
divisors of Lucas sequencessplitting
in theassoci-ated
quadratic
number fieldLet a E be a
quadratic integer.
In this section theprime
density,
61 (e, j; a),
of the setN1(e,j;a)
will becomputed by
relating
it tothe
degrees
of certain finite extensionsof Q
(Lemma 1).
In Lemma 3 thesedegrees
are thencomputed
in caseN(a) _ -l.
Using
Lemma 1 and Lemma3 one
easily
arrives at Table I.1 and 11.1. The fact that the second column in Table I.1only
contains zero entries is due to the fact that there are noprimes
satisfying
(2/p)
= 1 and p z5(mod 8).
LEMMA 1. Let a E a
quadratic
integer.
Put 0 =a2/N(a).
For 0 r s
put
=~(~/D, Bl~Zr, ~2.).
Let =Q].
Let j >
1exists. In case e =
0,
In case e >
1,
Furthermore
61 (e, j; a)
= 0 in case e >j.
Proof. Some details of the
proof
will besurpressed.
The readerhaving
difficulties
supplying
themissing
details is referred to[5].
If(Dip)
= 1then p
splits
in So(p) =
q3l3
in If(p, 2N(a))
=1,
thenordT(0)
=ordp(0)
=pa(p).
Using
that for alllarge enough
primes
satisfy-ing
(D/P)
=1,
it follows thatNi (e, j; a)
is finite in case e >j.
Then61 (e, j; a)
= 0. Now assume ej.
LetQa(p)
denote the exact powerof 2
dividing pa(P).
PutThen the set
Nl (e, j; a)
equals
This,
on itsturn,
can be written as{p :
p ESj,
05# m
1 (mod
~)}
if e = 0and
otherwise. The latter set
equals
with the subset
taken out. The latter set
equals,
with at mostfinitely
manyexceptions,
the set of
primes
thatsplit
completely
in Since forr s,
is normal overQ,
it followsby
theprime
ideal theorem orby
theCheb-otarev
density
theorem that theprime
density
of this setequals
The
density
of the other sets involved arecomputed similarly.
One findsbl (o, j; a) - d, -
and,
in the case e >1,
b1 (e, j; a) -
+ D
In our
computation
of thedegrees
da,b
we will make use of thefollowing
LEMMA 2.
[2,
Satz11
The
field
with a E normal andonly if
N(a)
is aLEMMA 3.
Suppose
that c > 0 is a unitof negative
norm inOD.
(i)
D = 2. We havedo,,
=2,
do,2
= 4 anddo,b
=2b-1
for
b > 3. Furthermored1,1
=4,
dl,b
=do,b
for
b > 2. For b > a >2,
da,b
=2~+b-2.
Finally,
d2 , 2
= 8 anddj,j
=2 2j-2
f or j
> 3.(ii)
D > 2. We havefor
b > a >_1,
da,b
=2a+b-1.
Furthermoredo,b
=2b,
b ~1,
6:1,1 =
4 anddb b
=22b-1
for
b > 2.Proof.
(i).
Sincev’2
EQ((8),
wehave,
forb > 3,
Q(v’2,(2b)
=Q«(2b)
and thusdo,b
=2b-1.
For a =1,
b > 2 we haveQ( v’2, v-a2, (2b) =
Q (,,f2-, i, C 2 b )= ~(~, ~Zb).
Thusdi,b
=do,b
for b > 2. Now assume that b >a > 2. Then
Q(v’2,(-a2)1/2G’(2b) =
Q(~/2,~/~B~)
= I claim thatX2G-l - a
is irreducible over~(~’2b).
Ifit were not,
thenQ( va)
would be a subfield of and hence normal. But since is not
nor-mal
by
Lemma2,
this isimpossible.
Q] =
~(~Zb)~~~(~26~ : Q]
=2a+b-2.
Next consider the field(-(,~2)1/2b’ ~2b)
for b > 3. Note thatBy
taking
composita
sees thatThus
db~b
=rdb_l,b
=22b-2,
Finally
one checks that themissing degrees,
do,,, do,2)
andd2,2,
are as asserted.(ii).
Weonly
deal with the case b > a > 2. The other cases are even more similar to(i)
and left to the reader(see
also[5,
Lemma6]).
We haveQ( v’Ï5,
(-a2)1/24, (26) =
~(~, ~,1~2a~1, ~‘2b)· I
claim that is irreducible Note that the latterfield,
as acompositum
oftwo abelian
fields,
is itself abelian. Hence all its subfields are normal. Nowif
X 2 a ! ~ -
a were reducible would be a subfield of~’2b).
By
Lemma 2 this is seen to beimpossible.
Thedegree
da,b
is thencomputed
as in(i).
p4. The
prime
divisors of Lucas sequences inert in the associatedquadratic
number fieldAs will be seen, in case a is a unit of
negative
norm, theproblem
ofcomputing
thedensity
b2 (e, j; a)
can beeasily
reduced to that ofcomputing
thedensity
of{p : (D/p)
= -1, p - -1 +
2i (mod 2~)}.
For D > 2 thisdensity
iscomputed
in the next lemma.LEMMA 4. Let D > 2 be
squarefree.
For s =1,
2and j >
1put
Then
6(Rs,j),
theprime density
of
Rs,j,
equals
2-1-i.
Proof Consider the set of
primes
Let j >
2. Now(Dip)
= 1 and p -2~)
if andonly
if theprime
p
splits
completely
inQ(VÐ,(2;
+~Z;1).
Similarly
(Dip)
= 1 and p--1(mod
2~)
if andonly ifp splits completely
inQ(B/D,
(2i
+C,~1)
but doesnot
split
completely
inQ(@,
~2~ ).
Since both of these number fields arenormal extensions of
Q,
it followsby
the Chebotarevdensity
theorem thatSince
~~(~/D,~z;) ~
~(~,~a~ +(;1)]12
and+ ~))
as ato-tally
real field isstrictly
included in it follows that8(Vj) =
~~(~/D,~2;) :
Q]-’.
Since theonly
realquadratic
subfield ofQ((2i)
is atmost it follows that
[Q (vD, (2j) :
Q] =
(~(~) :
Q] [Q ((2j) : Q]
=2i
and hence6(l§)
=2-j.
Now notice thatRi,;
= Thusb(Ri,j)
=6(l§) -
6(Vj+1)
=2-1-i.
If j
= 1 then note that81
is the set ofprimes
thatsplit
completely
in the normal fieldQ( VÐ,
i).
ThusThe case s = 2 is almost immediate now. D
Remark. From the law of
quadratic reciprocity
one deduces that for oddD,
(D/p)
= 1 if andonly if p -
4D)
for a set of odd¡3
(this
result wasalready conjectured by
Euler).
This set hascp(D)/2
elements.Using
this,
thesupplementary
law ofquadratic reciprocity,
the chinese remainder theorem and theprime
number theorem for arithmeticprogressions,
one cangive
an alternativeproof
of Lemma 4.Let e be a unit of
negative
norm. Now we are in theposition
tocom-pute
62(e,j;,E).
Forprimes
inert in~(~D),
Fp2,
and hencethe Frobenius map acts
by conjugation
on e, that is eP =6 (mod (p)).
Thus,
sinceN(e) _
-1,
we have-1(mod (p)).
Hence if p--1 +
2 (mod 2), ’
>2,
then8 =
(-1)EP+1 -
-1(mod (p)).
Thus-1 +
23 (mod
2~+1)~.
In the case D >2, j >
2,
62 (i, j; C)
=2’-~
by
Lemma 4. If D =2,
then6~ ( j, j; e)
= 0for j >
3 andNZ (2, 2; e)
=p
-3(mod
8)},
that is62(2,2; E)
=1/4.
Incase j
=1,
p -1(mod 4)
and so =(- 1) z!4 eP+l
1(mod (p)).
Since(p +
1)/2
isodd,
N2(0,1;f)
=f p : (D/p) = -1,
p =1(mod 4)}.
If D >2,
then~2(0, l;e)
=1/4
by
Lemma 4. If D = 2 then~2(0, l;c)
={p :
p -
5(mod
8)}
and soagain
62 (0, 1; e)
=1/4.
Thus we arrive at Table 1.2 and Table 11.2. 5. Proofs of Theorems 1 and 2Theorem 1 is
easily
deduced from thefollowing
theorem.THEOREM 3. Let E be a unit
of negative
norm inOD.
Letp,(p)
denote therank
of
apparition
of
p in the sequencele’
+ Considerfor
e > 0 theprime
density,
6(e; f),
of
the setf p :
In case D = 2 itequals
7/24
if e
1,
1/3
if e
= 2 and2-e/3
for e >
3. In case D > 2 itequals
1/3
if e = 0
and21-e/3
if e > 1.
Proof. Let
N (e, e) = {p :
and for s =1,2
letLet
b9 (e; e)
denote theprime
density
ofNs (e; E).
Thus,
with at mostfinitely
many
exceptions,
= NowN1(e; f)
=and
N2(e; f)
=6).
Since the latter is a finitedisjoint
union ofsets of non-zero
density,
we haveb~ (e; E)
=E~1
b2 (e, j; E).
Similarly
wewant to show that
61 (e; e)
=E~1
AsU~° 1 Nl (e, j; E)
is an infiniteunion of sets of non-zero
density,
this needsproof.
Weproceed
as in[4,
p.454].
PutUsing
Lemma 4 thedensity
of this set is seen to be2-j -
b1 (e, j; c)
in caseD = 2
and j ~
3,
and2-1-j -
61 (e, j; c)
in case D > 2. NowThe smallest set in the above inclusion of sets has
density
81 (e, j; e).
Thelargest
set hasprime
density
2-~ + in case D = 2 and m >3,
andprime
density
2-1-m
+81(e,j; f)
in case D > 2.Letting
m --~ oo shows that
81(e;f)
=~~° 1 bl (e, j; E).
Oncomputing
the densitiesZ~=i{~i(c?.?~)+~2(~J~)}?
onmaking
use of Lemma 1 and Lemma3,
theproof
is thencompleted.
punder
replacing u by ii,
-u or -11and, by assumption, u :A
1,
we may assumew.l.o.g.
that u > 1. Then u =ED
for some N >1,
where16D
is the fundamental unit of Write N =
2arra
with m odd. PutE =
ED .
Then u =f2>’
withN (E)
_ -1. Note that A isunique.
Consider the sequence + as asubsequence
of 1,En + 1*~).
Oneeasily
shows thatp divides + if and
only
if is divisibleby
2À+l.
Hence theprime
density
ofequals
...,.- "
Using
thisexpression
and Theorem3,
Theorem 1 follows. 0Proof of
Theorem 2. Put D =p2
+ 4. Notice that for0,
D is not asquare. We have
Ln(P)
= a"+an with a =(P+
B/D)/2.
If D ==0(mod 4)
then a E if D -1(mod 4)
then a EZ[(1
+B/D)/2].
Thus a E0~.
Furthermore
N (a) =
-1. In order toapply
Theorem 1 we have todeter-mine when
Q( P2 +
4) =
that is we have to find all solutions P tothe Pell
equation
p2 -
2Q2
= -4. The fundamental unit of is 1+ ..,F2.
By
thetheory
of Pellequations
it follows that the solutions(P, Q)
EZ~o
of
2Q2 =
-4 areprecisely given by
1 is odd},
whereXn +
Yn../2
=2 (1
+ý2)n.
Noting
thatit is seen that xn =
Ln(2).
Theorem 2 now follows oninvoking
Theorem1. D
With the
previous
proof
in mind the reader will have fewproblems
inproving
thefollowing
curiosum.THEOREM 4. Let D > 1 be
squarefree.
Suppose
thatX2 - Dy2
= -4 has a solutions. For s =1, 2,
letC,
denote the setof
prime
divisorsof
the setand the set
of prime
divisorsof
the setOne has
b(G4)
= 1.Furthermore,
when D =2,
=7/24,
b(C2)
=5/12
and8(G3)
=7/24.
If
D >2,
thenS(G~)
=1/3
for
1 j
3.Proof.
IfX2 -
Dy2
= -4 has asolution,
then -l. Define sequences ofintegers
and(yn)
by xn
+Yn VD
2el .
Then,
see[8,
p.65],
WI
={x" : n >
1 isodd},
W2
= 1 iseven},
W3
=(yn :
n > 1 is
oddl,
andW4
lYn :
n > 1 iseven}.
Notice that Xn =ED + D
(ED -
Using
this,
oneeasily
sees thatb(Cl) = 8(1; fD),
8(C2)
= 1 -b(0; ED) - 6(1; ED), b(C3)
=6(0;
eD)
andb(C4)
= 1. The result now follows from Theorem 3. 0The case D = 2
Table 1.1
Prime
density,
61
(e, j;,E),
of the set{p :
()
=1,
p - 1
+ 2j(mod 2+1),
whereN(e) _
-1.p
Table 1.2
Prime
density,
82(e,j; f),
of the set-Table 11.1
Prime
density,
61 ( e , j; E),
of the setTable 11.2
Prime
density,
82(e,j; f),
of the setREFERENCES
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Uber die Dichte der Primzahlen p, für die einevorgegebene
ganzra-tionale Zahl
a ~
0 vongerader
bzw.,
ungerader Ordnung
mod. pist,
Math.[4]
J. C.Lagarias,
The set ofprimes
dividing
the Lucas numbers hasdensity
2/3,
Pacific
J. Math. 118(1985),
449-461(Errata,
Pacific
J. Math. 162(1994),
393-397).
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P. Moree,Counting
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1996.
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Aconjecture
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The bookof
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Springer-Verlag,
Berlin etc.,1988.
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P.Ribenboim,
Catalan’sconjecture,
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Boston etc., 1994.[9]
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