n g 1 n=0 : For odd n the sign of x n depends on the hoi e of the sign of a

LXXXII.4(1997)

Prime divisors of Lu as sequen es

by

Pieter Moree (Bonn)and Peter Stevenhagen (Amsterdam)

1. Introdu tion. Letd>1beasquarefreeintegerandK=Q(

p

d)the

orrespondingrealquadrati eld. Wewrite"=a+b p

dforafundamental

unitintheringofintegersofK,and"forits onjugate. TheLu assequen e

asso iatedwithK istheinteger sequen e

X

K

=fTr

K =Q ("

n

)g 1

n=0

=f"

n

+"

n

g 1

n=0 :

For odd n the sign of x

n

depends on the hoi e of the sign of a. This is

irrelevant forthe divisibilityproperties we will be on erned with, but for

uniquenesssake we takex

1

=2a>0.

The Lu as sequen e X

K

satises these ond orderlinear re urren e

x

n+2

=2ax

n+1 N

K =Q (")x

n

forn0. If we take for K theeld Q( p

5) generated bythe golden ratio,

we obtainthe very lassi alexample of the Lu as sequen e dened by the

\Fibona i re ursion" x

n+2

= x

n+1 +x

n

with initial values x

0

= 2 and

x

1

=1.

In thisnote, we show that thesetof primenumbers pthat dividesome

term of the sequen e X

K

has a natural density Æ

K

and determine it for

ea hK. Morepre isely,we omputethedensityÆ +

K

oftheprimesthatsplit

ompletely in K and divide some term of X

K

and the density Æ

K

of the

primes that are inert in K and dividesome term of X

K

. The arguments

for both kindsof primesare somewhat dierent,and soare the asso iated

densities. It turns out that the determination of Æ

K

is the more diÆ ult

part, unless we are in the \easy ase" in whi h the norm N

K =Q

(") equals

1,when itistriviallydetermined. Clearly,one hasÆ

K

=Æ +

K +Æ

K .

The methodinthisnoteextendsto sequen esfTr

K =Q (

n

)g 1

n=0

=f n

+

 n

g 1

n=0

, where isany algebrai integer ina quadrati eldK. Although

itisabit umbersometo expressthedensityasanexpli itrationalnumber

1991 Mathemati sSubje tClassi ation: Primary11R45;Se ondary11B39.

Keywordsandphrases: Lu assequen e,Chebotarevdensitytheorem.

intermsof,thisyieldsaproofofwhatis alleda\main onje ture"in[5,

p.362℄. Formore details,we referto thetreatment of generalse ond order

\torsion sequen es"in[6℄.

Theorem. Let K =Q( p

d) and a= 1

2 Tr

K =Q

(")>0 be as above. Then

the natural densities Æ +

K and Æ

K

for the sets of prime divisors of the Lu as

sequen e asso iated with K exist. For N

K =Q

(") = 1 the densities are as

follows.

d=2 d>2

Æ +

K

11=24 5=12

Æ

K

1=4 1=4

Æ

K

17=24 2=3

For N

K =Q

(")=1the densitiesdependinthe following wayon whethera+1

and a 1 arerational squares or not.

a 1= a+1= a16=

Æ +

K

5=24 5=24 1=6

Æ

K

1=8 5=24 1=6

Æ

K

1=3 5=12 1=3

ThemainingredientoftheproofistheChebotarevdensitytheorem,and

the basi idea of the method goes ba k to Hasse [2℄. Lagarias [3℄ was the

rst to usethisideaina quadrati setting,forthe lassi alLu as sequen e

mentionedabove, whi h fallsin the ategory N

K =Q

(")= 1. A generaliza-

tiontootherinstan esofunitsofnorm 1isgivenin[4℄. Fortheeasierand

well-studied ase of redu ible se ond order re urren es fr n

+s n

g 1

n=0 with

r;s2Z,orforgeneralizationstohigherorder linearre urren es,thereader

an onsult[1℄.

2. Proof of the Theorem. Let K =Q( p

d) and "= a+b p

d be as

before,andwriteOfortheringofintegersofK. Ifpisaprimethatisunram-

iedinK =Q, thenthekernelofthenormmap

p

=ker[N :(O=pO)



!F



p

℄

isa y li groupof order p d

p

. We set

(2:1) q=q

K

="="=



"

2

ifN

K =Q

(")= 1,

"

2

ifN

K =Q

(")=1.

Letp-2dbea primenumber. Lookingat theexpli itform of thenth term

x

n

= "

n

+"

n

of X

K

, we nd that p divides x

n

if and only if we have

q n

= 1 2 (O=pO)



. As q lies in the y li subgroup 

p

 (O=pO)



and 1 is the unique element of order 2 in that group, we nd the basi

hara terization

pdividessome term ofX ,theorder of q2(O=pO)



iseven.

The key idea in determining the densities Æ +

K and Æ

K

is that one an

des ribe the parity of the order of q 2 (O=pO)



in terms of the splitting

behavior of p in some innite algebrai extension of Q. We start with the

easier aseof therationalprimesthatsplit ompletelyinK.

Split ase. Let S +

be the set of odd primes p that split ompletely in

K,and D +

S +

theset ofprimesinS +

that dividesometerm ofX

K .

For k 2 Z

1

, we let S +

k

 S +

be the set of primes p 2 S +

for whi h

p 1 has exa tly k = ord

2

(p 1) fa tors 2. The set S +

k

onsists of the

primes that split ompletely in the eld K(

2

k) obtained by adjoining to

K aprimitive2 k

th root ofunity, butnotintheeld K(

2

k +1) obtainedby

adjoiningtoK aprimitive2 k+1

throotofunity. BytheChebotarevdensity

theorem, the set S +

k

has a natural density inside the set of all primes. It

equalsÆ(S +

k

)=[K(

2 k):Q℄

1

[K(

2

k +1):Q℄ 1

. Clearly,thesumofthese

densities forallk 1 is[K :Q℄ 1

=1=2=Æ(S +

).

Forp2S +

k

,thegroup(O=pO)



isaprodu toftwo y li groupsoforder

p 1, and an element has odd order in (O=pO)



if and onlyif it is a 2 k

th

powerin (O=pO)



. As q 2(O=pO)



is a 2 k

th power ifand onlyif p splits

ompletely inthe eldK(

2 k;

2 k p

q),we on ludethat a prime p2S +

k does

not divideatermofX

K

ifandonlyifitsplits ompletelyinK(

2 k;

2 k p

q),but

notinK(

2 k +1;

2 k p

q). BytheChebotarevdensitytheorem,thesubsetofsu h

primesinS +

k

hasnaturaldensity[K(

2 k;

2 k p

q):Q℄ 1

[K(

2 k +1;

2 k p

q):Q℄ 1

.

The omplement D +

k

= D +

\S +

k

of this set in S +

k

has a density as well,

and we nd that both D +

= S

k1 D

+

k

and its omplement S +

nD +

=

S

k1 (S

+

k nD

+

k )inS

+

are ountabledisjointunionsofsets ofprimeshaving

a natural density. It follows that D +

has lower density P

k1 Æ(D

+

k ), and

that S +

nD +

has lower density P

k1 Æ(S

+

k nD

+

k

). These lower densities

add up to Æ(S +

), sothey areinfa t densities. We on lude thatD +

hasa

naturaldensityÆ +

K

whi hsatises

(2:2)

1

2 Æ

+

K

= X

k1



1

[K(

2 k;

2 k p

q):Q℄

1

[K(

2 k +1;

2 k p

q):Q℄



:

Equation (2.2) redu es the omputation of Æ +

K

to a omputation of eld

degrees in the innite extension K(

2 1

; 2

1 p

q) of Q. To ease notation, we

write

F

k

=K(

2 k +1;

2 k p

q):

Then thekth termof the right handsideof (2.2) equals [F

k :Q℄

1

if

2 k +1

generates a quadrati extensionof K(

2 k;

2 k p

q),and 0 otherwise.

Suppose rst that we have N

K =Q

(") = 1, and onsequently q = "

2

in (2.1). Then q is a square in K(

4

), and also in the eld M = K(

2 1

)

powerinM,sin eM is abelianoverQ and M( 4

q)=M( ")hasa quarti

subeldK(

p

")thatisnotnormaloverQ. ByKummertheory,itfollowsthat

2 k p

q generates a y li extension of degree 2 k 1

of K(

2

k) forevery k  2.

Our normalityargument shows that this extensionis linearlydisjoint over

K(

2

k) from K(

2 k +1).

Fork =1, we have a quadrati extension K(

p

q) =K(

4

), whi h oin-

ides with the extension generated by 

2

k +1 = 

4

. This shows that in the

ase ofnorm 1,thetermfork =1 in(2.2) vanishes.

If K is not the real quadrati subeld Q( p

2) of Q( 

2 1

), then 

2 k +1

generates a quadrati extension ofK(

2

k) forall k 2, and F

k

has degree

2
2 k

2 k 1

= 4 k

for these k. We nd1=2 Æ +

K

= P

k2 4

k

= 1=12 and

Æ +

K

=5=12.

For K = Q(

p

2 ) the degree of F

k

is only 2 k

2 k 1

= 2 2k 1

for k  3.

Moreover, the term for k = 2 in (2.2) vanishes sin e K(

4

) = Q(

8 ) now

ontains 

2

k +1 = 

8

. We nd 1=2 Æ +

K

= P

k3 2

1 2k

= 1=24 and Æ +

K

=

11=24.

The diagrams below indi atethe eld degrees in the two situations for

k 2and k 3,respe tively.

F

k

K(

2

k +1) K(

2 k;

2 k p

q)

K(

2 k)

K(

4 )=K(

p

q)

K6=Q(

p

2) 2 G

G

G

G

G

G

G

C

C

C

C

C

C z

z z

z z

z

2 k 1

x x

x x

x x

x

2 k 2

2

F

k

K(

2

k +1) K(

2 k;

2 k p

q)

K(

2 k)

K(

8 )=K(

p

q)

K=Q(

p

2 ) 2 G

G

G

G

G

G

G

C

C

C

C

C

C z

z z

z z

z

2 k 1

x x

x x

x x

x

2 k 3

2

Suppose next that we have N

K =Q

(") = 1, and so in parti ular K 6=

Q( p

2). Theanalysisissimilartotheprevious ase,butwenowhaveq="

2

,

soq is asquare inK. AstheeldK(

4 p

" )isnon-normal ofdegree 8,wesee

thatq ="

2

isa square inM =K(

2 1),

butnotan eighth power. We have

two ases.

Suppose q is not a fourth power in M. Then 2

k p

q generates a y li

extension of degree 2 k 1

of K(

2

k) for every k  1, and this extension is

linearly disjoint from the extension of K(

2

k) generated by 

2

k +1. This is

p

now also have a non-zero term for k = 1 in (2.2). We nd 1=2 Æ +

K

=

P

k1 4

k

=1=3 and Æ +

K

=1=6.

Supposethatq isafourthpowerinM. ThenK(

p

") isasubeldofM.

BesidesK,thequadrati subeldsofK(

p

")arethetwoeldsQ( p

"1=

p

"),

andone of thosetwo is ontainedinQ(

2 1

). FromN

K =Q

(")=1wededu e

(2:3) (

p

"1=

p

" ) 2

=Tr

K =Q

(")2=2(a1);

so this \ex eptional ase" o urs exa tly when one of the elements a1

is a rational square. The eldsK(

4

; 4 p

q) =K(

4

; p

") and K(

8

) oin ide

here, and 2

k p

q generates, forall k 3,a y li extension of degree 2 k 2

of

K(

2

k) that is linearlydisjoint over K(

2

k) from K(

2

k +1). As inthe ase

K =Q( p

2)above,thedegreeofF

k

isonly2 k

2 k 1

=2 2k 1

fork 3. The

term fork =2 vanishes, butfor k =1 we do have a ontribution 1=4. We

nd1=2 Æ +

K

=1=4+ P

k3 2

1 2k

=7=24 andÆ +

K

=5=24 ifeither a+1 or

a 1is asquare. This shows thatthevaluesof Æ +

K

areasasserted.

Inert ase. Letpbea primethatisinertinK =Q. ThenO=pO isaeld

of p 2

elements, and the norm map N : O=pO ! F

p

raises all elements to

thepowerp+1.

Suppose rst that we are in the ase N

K =Q

(") = 1. Then we have

q = "

2

in (2.1) and "

p+1

= 1 2 (O=pO)



. For p  1mod4 we obtain

q (p+1)=2

= "

p+1

= 1 2 (O=pO)



, whi h shows that the order of q in

(O=pO)



is odd. For p  3mod4 we obtain q (p+1)=2

= "

p+1

= 1 2

(O=pO)



,whi hshowsthattheorder ofqin(O=pO)



iseven. Wendthat

Æ

K

is the density of the primesp 3mod4 that are inertin K =Q, hen e

Æ

K

=1=4.

From nowon we supposeN

K =Q

(")=1. Inparti ular, thisimpliesK 6=

Q( p

2). We have q ="

2

,and onsequently q (p+1)=2

="

p+1

=1 2(O=pO)



forallinertoddprimesp. Thisshowsthatforall inertprimesp1mod4,

the order of q is again odd and p does not dividea term of X

K

. For the

inertprimes p3mod4 we use an approa h thatis similarto thatin the

split ase.

Let S bethesetofoddprimespthatareinertinK =Q, andD S

the set of primes in S that divide some term of X

K

. For k 2 Z

2 , we

let S

k

 S be the set of primes p 2 S for whi h p+1 has exa tly

k =ord

2

(p+1)fa tors2. Thisis asetwitha naturaldensity,and we want

to ompute thedensityof thesubset D

k

=D \S

k

by hara terizing the

primes p 2 D

k

in terms of splitting onditions on p in some niteGalois

extensionF

k

=Q.

AprimepisinS

k

ifandonlyifitsFrobeniussubstitutionintheabelian

groupGal (K( k +1)=Q) istheuniqueelement'thatisnon-trivialonKand

a ts on the 2 k+1

th roots of unity as '(

2

k +1) =  1+2

2 k +1

. As K(

2

k +1) has

degree 2 k+1

over Q, this shows that S

k

has natural density 2 k 1

for all

k. Welet B

k

K(

2

k +1)bethesubeld orrespondingto thesubgrouph'i

of Gal(K(

2 k +1

)=Q). Note that K(

2 k +1

)= B

k

(") is a quadrati extension

of B

k .

IfpisinS

k

,theorderp 2

1=(p 1)(p+1)ofthe y li group(O=pO)



hasexa tlyk+13fa tors 2,andq ="

2

2(O=pO)



hasodd orderifand

onlyif"

2

isa2 k+1

thpowerin(O=pO)



. As 1isa2 k

thpowerin(O=pO)



,

we on lude that p 2 S

k

does not divideany term of X

K

if and only if "

is a2 k

th power in(O=pO)



. Thisleadsto a hara terization of theprimes

p2S

k

outsideD intermsof theirsplittingbehavior intheeld

F

k

=K(

2 k +1;

2 k p

")=B

k (

2 k p

");

theyaretheprimespthatsplit ompletelyinB

k

andhaveextensionsinB

k

that are inert in B

k (")=B

k

and split ompletely in F

k

=B

k

("). This means

thattheFrobeniussymbolofpinthenon-abeliangroupGal(F

k

=Q),whi his

onlydeterminedupto onjuga y,isanelementoforder2inthenormalsub-

groupGal(F

k

=B

k

)thatdoesnotlieinthenormalsubgroupGal(F

k

=B

k (")).

If n

k

denotes the number of su h elements in Gal(F

k

=Q), the Chebotarev

density theoremyieldsan inertanalogueof (2.2):

(2:4)

1

2 Æ

K

= 1

4 +

X

k2 n

k

[F

k :Q℄

= 1

4 +

X

k2 2

k n

k

[F

k :B

k

℄ :

Thistimewehavetodomorethanadegree omputation,sin ewealsoneed

to know thestru tureof thegroupGal(F

k

=B

k ).

Supposerstthatneithera 1nora+1isasquare. Thentheextensions

K(

p

")andK(

2 1)

arelinearlydisjointoverK,andF

k

isa y li extension

of degree 2 k

of B

k

(") = K(

2

k +1) generated by 2

k p

". We an extend the

generator ' of Gal(B

k (")=B

k

) to an element '



2 Gal(F

k

=B

k

) of order 2

bysetting '



( 2

k p

")=1=

2 k p

". As'



a tsbyinversiononbothh"iandh

2 ki,

theGaloisequivarian eof theKummerpairing

Gal(F

k

=B

k (

2

k +1))h"i!h

2 ki

shows that'



ommutes withall elementsinGal(F

k

=B

k

(")). We nd

Gal(F

k

=B

k )



= Gal(F

k

=B

k

("))h'



i



= Z=2

k

ZZ=2Z:

As there are exa tly two elements of order 2 in Gal(F

k

=B

k

) of the form

(;'



), this yields n

k

= 2 in (2.4) for all k  2. We nd 1=2 Æ

K

=

1=4+ P

2 k

2
2 k 1

=1=3 andÆ =1=6 .

B

k (

2 k p

")=F

k

B

k (

p

")

B

k (

2 k p

"+1=

2 k p

")

K(

2 k +1

)=B

k (")

B

k (

p

"

1

p

"

) B

k (

p

"+ 1

p

"

)

B

k u

u u

u u

u u

u

h'



i K

K

K

K

K

K

K

K

~

~

~

~

~

~

2 H

H

H

H

H

H

H 2

k 1

s s

s s

s s

s s

h'i A

A

A

A

A

A

2

u u

u u

u u

u u

Supposenally thatweareintheex eptional asewhereeithera+1 or

a 1 isa square. ThentheextensionF

k

=B

k

inthediagramabove ollapses

to anextension ofdegree 2 k

forall k2. Fork3,we have p

22B

k and

(2.3) showsthatB

k

ontains p

"+1=

p

"ifa+1 isasquareand p

" 1=

p

"

ifa 1is asquare. In therst asewehave an isomorphism

Gal(F

k

=B

k )



= Gal(B

k (

p

")=B

k )h'



i



= Z=2

k 1

ZZ=2Z

and n

k

= 2 as before. In the other ase, F

k

=B

k

is a y li extension of

degree 2 k

with quadrati subextensionB

k

(") =B

k (

p

"). Any extension of

'2Gal(B

k (")=B

k ) to F

k

is then a generator of Gal(F

k

=B

k

), and we have

n

k

=0.

Fork =2,anyoftheelements p

2and p

"1=

p

"generatesthequadrati

extension B

2 (

p

" )=B

2 (

p

2) of B

2

. The extension B

2

B

2 (

4 p

" )=F

2 is of

degree4andhasaquadrati subextensiongeneratedby p

"= p

(a+1)=2+

p

(a 1)=2. Ifa+1isasquare,then p

"hasnorm 1inB

k andF

2

=B

2 isa

y li extension. If a 1isa square,then p

"hasnorm 1inB

k andF

2

=B

2

isa V

4

-extension. The orrespondingvaluesofn

2 aren

2

=0 and n

2

=2.

Fora+1asquarewend1=2 Æ

K

=1=4+0+ P

k3 2

k

2
2 k

=7=24

andÆ

K

=5=24 . Fora 1asquarewendanitesum1=2 Æ

K

=1=4+1=8

and Æ

K

=1=8. Thisnishes theproof ofthe theorem.

Referen es

[1℄ C.Ballot,Densityofprimedivisorsoflinearre urren es,Mem.Amer.Math.So .

551(1995).

[2℄ H.Hasse,



UberdieDi htederPrimzahlenp,f urdieeinevorgegebene ganzrationale

Zahla6=0vongeraderbzw.ungerader Ordnungmodpist,Math.Ann.166(1966),

19{23.

[3℄ J. C. Lagarias, The set of primes dividing the Lu as numbers has density 2=3,

Pa i J.Math.118(1985),449{461;Errata:ibid.162(1994),393{397.

[4℄ P.Moree,On theprime density ofLu assequen es, J.Theor.NombresBordeaux

[5℄ P.Ribenboim,TheNewBookofPrimeNumberRe ords,Springer,NewYork,1995.

[6℄ P.Stevenhagen,Primedensitiesforse ondordertorsion sequen es,preprint.

Max-Plan k-InstitutfurMathematik Fa ulteitWINS

Gottfried-Claren-Str.26 UniversiteitvanAmsterdam

53225Bonn,Germany PlantageMuidergra ht24

E-mail:moreempim.bonn.mpg.de 1018TVAmsterdam,TheNetherlands

E-mail:pshwins.uva.nl

Re eivedon7.4.1997

andinrevisedformon24.7.1997 (3165)