J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
P
AUL
M. V
OUTIER
Primitive divisors of Lucas and Lehmer sequences, II
Journal de Théorie des Nombres de Bordeaux, tome
8, n
o2 (1996),
p. 251-274
<http://www.numdam.org/item?id=JTNB_1996__8_2_251_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Primitive divisors of Lucas and Lehmer sequences, II
par PAUL M VOUTIER
RÉSUMÉ. Soit 03B1 et 03B2 deux entiers algébriques complexes conjugués. On
propose un algorithme dont l’objet est de découvrir des éléments des suites de Lucas ou de Lehmer associées à 03B1 et 03B2, n’ayant pas de diviseurs primitifs. On utilise cette algorithme pour démontrer que pour tout 03B1 et 03B2 tel que
h(03B2/03B1)~
4, le n-ième terme des suites de Lucas et de Lehmer admet undiviseur primitif dès que n > 30. Nous donnons en outre une amélioration
d’un résultat de Stewart se rapportant à des suites plus générales.
ABSTRACT Let 03B1 and 03B2 are conjugate complex algebraic integers which
generate Lucas or Lehmer sequences. We present an algorithm to search for elements of such sequences which have no primitive divisors. We use this
algorithm to prove that for all 03B1 and 03B2
with h(03B2/03B1) ~
4, the n-th element ofthese sequences has a primitive divisor for n > 30. In the course of proving
this result, we give an improvement of a result of Stewart concerning more
general sequences.
1. Introduction
Let a and
{3
bealgebraic
numbers such that a+ {3
and arerelatively
prime
non-zero rationalintegers
and is not a root ofunity.
The sequencen
defined
by
ex - (3 for n > 0 is calleda Lucas sequence.
If,
instead ofsupposing
thata +,Q
EZ,
weonly
suppose that(a + (3)2
isa non-zero rational
integer,
stillrelatively
prime
to then we define theLehmer sequence
(u~,)~ o
associated to a and{3
by
We say that a
prime
number p is aprimitive
divisor of a Lucas numberun if p divides but does not divide
(a -
0)2 U2
... U,,- 1 .Similarly,
p is a
primitive
divisor of a Lehmernumber un
if p divides un but not(a2 - ~~)22L3 ...
Un-I.Stewart
[14,
p.80]
showed,
as a consequence of his Theorem1,
thatif n > C then un has a
primitive
divisor,
where C =e 452267
for Lucassequences and C =
e4524fi7
for Lehmersequences. In Theorem
2,
we shallobtain an
improvement
over Theorem 1 of[14]
as well asdecreasing
the size of C.In an earlier article
[16],
we enumerated all Lucas and Lehmer sequenceswhose n-th element has no
primitive
divisor for certain n 12 and all12 n 30. We also
presented
some evidence tosupport
theconjecture
made there that for n >30,
the n-th element of any Lucas or Lehmer sequencealways
has aprimitive
divisor.Here,
wepresent
some further resultsconcerning
thisconjecture.
Our mainresult,
Theorem1,
states that theconjecture
is true if the absolutelogarithmic height
of a is small. In addition toproviding
further evidence for thevalidity
of theconjecture
(or
at least notproviding
acounterexample),
this result will also be useful in a
forthcoming
work where we shall makefurther
improvements
to the size of C.Throughout
this paper, we shall useh(a)
to denote the absoluteloga-rithmic
height
of thealgebraic
number a.THEOREM 1.
Suppose
aand (3
generate
a Lucas or Lehmer sequence withh{~3/a)
4.Thert, for
all n >30,
the n-th elementof
this sequence has aprimitive
divisor.We prove this result
by
using
Stewart’s idea[14,
Section5]
oflooking
atcertain Thue
equations.
For any Lucas or Lehmer sequence there is apair
ofintegers
(p,
q),
dependent only
on the sequence, such that if u~has no
primitive
divisor then(p, q)
is a solution of one of certainfinitely
many Thue
equations
associated to n. We use this to show thatif,
forn >
30,
u~, is without aprimitive
divisor then n must be the denominator of aconvergent
in the continued-fractionexpansion
ofarccos(pl(2q))/(21r).
The
advantage
gained by
this is that theconvergents
of real numbers growquite
quickly
and so theproblem
ofchecking
each n less than 2 -101°
isreduced to
checking
no more thanfifty
such n.In
fact,
we will show thatif un
has noprimitive
divisor,
then theconver-gent
k /n
must be anextremely
good
approximation
toarccos(p/(2q))/(21r),
directly
thatk/n
is notsufficiently
close to the number inquestion
andtherefore,
eliminate n from consideration. In theremaining
cases, a directexamination of Un proves our desired result.
Stewart’s upper bound for n, stated
above,
isquite
large
and would thusgive
rise toextremely long
calculationsjust
to determinetheconvergents.
Fortunately
it is nowpossible
to reduce this upper boundconsiderably.
Because of its benefit to our workhere,
we shall determine such a smallerupper bound. In
fact,
we establish a moregeneral
result which is animprovement
over Theorem 1 in Stewart’s paper(14~,
whoseproof
requires
little more effort than
proving
the morespecific
result whichonly
applies
to Lucas and Lehmer sequences.
THEOREM 2.
(i) Suppose
a arealgebraic
integers
with,Q/a having
degree
di
overQ,
(a,
(3)
=(1)
ccnd not a rootof
unity.
Then there is aprime
ideal P which divides the ideal(an-,C3’~)
but does not divide the ideals(,m -,3m)
for
1 rri nfor
alln > maxf 2(2 di - 1), 4000(d1log(3d1))I2}.
(ii)
If a and,~
generate
a Lucas or Lehmer sequence then the n-th elementof this seqttence has a
primitive
divisor for all n > 2 -1010.
2.Preliminary
Lemmas to Theorem 2We shall first
require
a lower bound for linear forms in twologarithms.
The work of
Laurent, Mignotte
and Nesterenko[8]
will be suitable for ourneeds. We also need a
good
lower bound for theheight
of a non-zeroalgebraic
number which is not a root ofunity.
LEMMA 1.
Suppose that,
is a non-zeroalgebraic
numberof degree
D > 2over Q
which is not a rootof
unity,.
ThenProof.
This isCorollary
1 of(17~.
0Now let us continue.
LEMMA 2.
Let q
be a non-zeroalgebraic
numberof
degree
whichis not a root
of
unity
and letlog -y
denote theprincipal
valueof
itslogarithm.
with
bI
apositive integer, b2
anon-negative integer
andIf
0 thenProof.
First let us suppose that[q[ #
1. We canwrite -y
= where r > 0and -7r 8 ir. Since
r2
= 7 ~7,
we haveso
h(r)
h(¡).
Thus, by
Liouville’sinequality
we haveand the lemma follows
by
Lemma 1 and the fact thath(q) >
log
2 if D =1.We now turn to the case of
1,1
= 1.Since 7
is not a root ofunity,
D > 2.To obtain our lower bound for
IAI
in this case we will use Th6or6me 3of
[8].
However,
this resultrequires
thatbl
andb2
be non-zero, so we mustdeal
specially
with the case ofbl
= 0.By
Liouville’sinequality
and Lemma1,
It is now clear that the lemma holds in this case.
To obtain a
good
constant in our lower bound we show that we mayassume B > 679000. From Liouville’s
inequality,
we obtainWe can use
D/2
here instead of Dsince q £
R(see
Exercise 3.4 of[18)).
So the lemma is true wheneverSince D >
2,
applying
Lemma1,
thisinequality
holds ifUsing Maple,
one can check that this is true for 2 B 679000.We now invoke Th6or6me 3 of
[8].
Let a = max{20, 12.851Iog,1
+D h(7)/2}
and H = max
{17,
Dlog(bl/(2a)
+b2/(25.7~))/2
+ 2.3D +3.25}.
ThenSince a > 20 and
1/(2a) + 1/(25-77r)
0.0374,
H maxf 17,
(D/2) log B
+0.657D +
3.25}.
Moreover,
B > 679000implies
that(D/2)
log B
+0.657D + 3.25 0.66994D
log
B. As this lastquantity
isgreater
than 17 for B >679000,
we have H 0.66994Dlog
B.We also want an upper bound for a in terms of D and
h(y).
First noticethat ~ 7r - °
Therefore,
12.85~ log7~+Dh(7)/2 Dh(-y)(40.37/(Dh(~y))
+1/2).
Since D >2,
we canapply
Lemma 1. We obtain40.37/(D he,))
+1/2
20.185(log(3D))~+l/2
20.272(log(3D))3.
Therefore,
12.851 log -yi +
Dh(-y)/2
20.272(log(3D))3D h(y)
for all D > 2.Moreover,
thisquantity
is
greater
than20,
so a20.272(log(3D))3D h(7).
Applying
these estimates to(1),
we find that our lemma holds. 0Suppose
that a and(3
arealgebraic
integers
in a number field K ofdegree
d overQ. Letting
Ki
=Q(/3/a),
a number field ofdegree
d1
overQ,
we set(3/a 13, /a,,
where al and,Ql
arealgebraic
integers
inKi
and(al,,31)
=.AI.
We may assume, without loss ofgenerality,
thatWe note
that,
unless we stateotherwise,
log z
shallalways
denote theprincipal
branch of thelogarithmic
function. Now let us prove:LEMMA 3.
(i)
We haveProof.
(i)
We can write =ai(l - ~1/0:1).
By
Liouville’sinequality,
and the result follows.
(ii)
The upper bound followsdirectly
from thetriangle
inequality
and ourassumption
thatFor the lower bound we write the
quantity
inquestion
asApplying
Lemma 2.3 of[10]
with r =1/3
and z = we seethat either
1 1
or
In the first case, the lemma holds so we need
only
consider the second case.Here,
we must haveSince we took the
principal
value of thelogarithm
of(3l/al,
we have -~r 1r and soIkl
n/2
+0.5/(2x)
or12ki
n.As
is,
by
assumption,
not a root ofunity,
A #
0and,
since n >2,
we mayapply
Lemma 2giving
By
Lemma1,
we havelog(1.3)
since
dl
> 2 and n > 2. Our lemma follows. 0LEMMA 4. Let where is the n-th order
cyclotomic polynomial. Suppose
that P is aprime
ideal in K which divides(q,n(a, (3))
for
n >2(2d1 -
1).
Thisimplies
that P divides(an - (3n).
If,
in
addition,
P divides(am -
~3"’~)
for
some rn n, thenProof.
This is Lemma 4 of[13].
0Finally
we need to bound some arithmetic functions which will appearLEMMA 5.
(i)
Letw(n)
denote the numberof
distinctprime
factors of
n.For n >
3,
For n >
3,
where -y
= 0.57721 ... is Euler’s constant.Proof.
(i)
This follows from Th6or6me 11 of[11].
(ii)
This is Theorem 15 of[12~.
D3. Proof of Theorem 2
We may assume that
2,
for otherwise we can write a =OC1 /C2
where cl, c2 E Z with
(cl, c2)
= 1 and so a" -(3n
=(ci - c2 )(,C~/c2)".
Now
Zsigmondy
[20]
and, independently
ofhim,
Birkhoff and Vandiver[2]
have shown that for n > 6 the n-th element of such sequencesalways
has aprimitive
divisor.Therefore,
we may also assume that n >3900(2 log(3 -
2))1211.74 .
since Theorem 2 does not
apply
for smaller n when 2.We note that
Letting
A be the extension of.A1
inK,
we have({3/ (3I)
= since(a, #)
=(1),
and soSince
where denotes the set of all Archimedean absolute values defined
on Ki
up toequivalence.
Applying
Lemma3,
we see that the inner sum in the first term of thisexpression
is at leastCombining
this lower bound withand
we obtain
Notice that n has
2WJ(n)-1
factors m whichsatisfy
= 1 and thesame number of factors m
satisfying
J-L(n/m) = -1.
Now,
by
Lemma 1 and our lower bound for n,By
Lemma4,
if >nd,
then there exists aprime
idealP which divides
(an -
but does not divide(am - ,Q"’ )
for any m n.Using
(3)
and Lemma1~
as well as ourassumptions
that 2 andn > 1.74 .
10~° ,
this condition is satisfied ifFrom Lemma
5,
we find thatfor such n.
Therefore,
(4)
is satisfied forSince
di
>2,
part
(i)
of the theorem holds.(ii)
Let be a Lucas or Lehmer sequencegenerated by
a and(3.
Sincea(3
and(a
+(3)2
arerelatively
prime
non-zero rationalintegers,
there exist two
integers
pand q
such that a and,Q
are the two roots ofX2 -
~/p+2?X+~ Therefore,
a,,Q
=(~/p+2~±
p -
2~)/2
and so eithera/ {3 or (3/a
isequal
to(p+
p2 - 4q2)/(2q).
Therefore we can takedi
= 2and so
part
(i)
of theoremimplies
part
(11).
04.
Preliminary
Lemmas to Theorem 1LEMMA 6. Let a be a
non-negative
real number. E R with -1 ::;Proof.
This result follows fromfinding
the minimum value of the functionon the area in R2 defined
by
0x,
y _
7r, x =F y
which is2/7r2
and thenapplying
thecontrapositive.
DLet us collect here various notations which we shall use
throughout
theremainder of this article.
Notations. Given a
complex-valued
functionf
defined onC,
we useto denote
f (x~ (.
For a
positive integer
n, we let9n(X)
EZ[X]
be the minimalpolynomial
of 2 over
Z;
itsdegree
iscp(n)/2
if n > _ 3. We shallput
Gn(X, Y) =
We let m be the
greatest
oddsquare-free
divisor of n. For such m, weshall write
(xm -
Finally,
for n >1,
we letP(n)
denote thelargest
prime
divisor of n.As we shall see in Section
5,
the crucial result needed in theproof
ofTheorem 1 is a
good
lower bound for for(j, n)
= 1.We will show that we need to obtain an upper bound for the absolute
value of
hm(X)
on the unit circle which we findusing
an idea and a resultof
Bateman,
Pomerance andVaughan
[1].
Let us start
linking
these twopolynomials
now.LEMMA 7. Let
n > 3, 1 j n
with(j, r~)
= 1and (n
=exp(27ri/n) .
Then
Hence,
from which the lemma follows. D
To work with the
cyclotomic
polynomials
we shall need somerelation-ships
whichthey satisfy.
Wegive
these in the next lemma.LEMMA 8.
(i~
Let n be apositive integer
and let m be itsgreatest
oddsquare-free
divisor. Weput
m’ =gcd (2, n) m .
Then(ii)
Let p be aprime
number and n anypositive integer
not divisibleby
p.Then
I / -.,-B r l
(iii)
Let rn, m’ and n be as above. Weput
n’ =n/ gcd(n, 2), hm(X) =
(X"‘ -1)/~",,(X)
andCn
=exp(27ri/n).
Then, for
all j
with(j, n)
=1,
weProof.
(i)
This assertion followseasily
from the two relations:which are
parts
(iv)
and(vi)
ofProposition
5.16 fromChapter
2 ofKarpilovsky’s
book[7].
(ii)
This isagain
fromProposition
5.16 fromChapter
2 of[7}.
(iii)
Applying
part
(i),
we find that= =
Letting
X = which isalways
aprimitive
m-th root ofunity,
we have
(-1)("z-1>("’~~+1>rn~ (;’z 1) -
and the result follows. D
We see now that we have reduced the
problem
ofbounding
frombelow for
primitive
n-th roots ofunity
tobounding
from above. Todeal with this new
problem,
we shall now use ideas from[1].
LEMMA 9. Let m =
pi ... pk where are odd
primes
arranged
inincreasing
order. ThenIn 3 then the
factor of
2 is not needed.Proof.
From Lemma8(ii),
We now use induction on k to prove the lemma.
Since
hi
(X)
=1,
hP,
(X)
= X - 1 and(X) =
(X),
theFork
= 3,
we have *Using
the resultjust
established for h = 2 and a theorem of Carlitz[3]
which shows thatPIP2/2,
we obtainlhplp2p.1,
pip2.
This is the desiredin-equality
for J~ = 3.Suppose
now that the lemma holds for some k > 3. Weapply
thefollowing
estimate ofBateman,
Pomerance andVaughan,
which follows from Theorem 1 of their paper[1]
and holds3,
Thus from
(5),
we haveHence the lemma holds. 0
We need the next lemma to deal with the case
q = 2,
although
we willuse it for all q. A
simple application
of thetriangle inequality
wouldquickly
yield
theinequality
below with31ql/5
replaced by
lql/2.
However,
in thecase
of q
=2,
this would not be sufficient to prove our theorem: withthe lower bound that the
previous
lemmasimply
forthe upper bound we would obtain for the left-hand side of
(11)
would notdecrease with n but
actually
grow with n. To refine this trivialestimate,
it seems we must resort to an
argument
like the one which follows.LEMMA 10. Let n > 30 be a
positive integer
and let p and q be non-zerointegers
withq >
2jp)
2q
andFor 1
j
n/2
=1,
we =Proof.
We first divide the interval(-2, 2)
into four subintervals and di-vide the set ofintegers
less thann/2
which arerelatively
prime
to n into four associated subsets. Let A =(-2,
-1), A’ = {m : n/3
mn/2, (m, n)
=1},B
=(-1,0),8’
=(m : n/4
mn/3, (m, n) _
=
(0,1),C’ = {m :
n/6
mn/4, (m, n)
=1}, D
=(l, 2)
andD’ -
{m :
0 m~/6, (m, n)
=11.
If we let denotethe number of
integers
in the interval(nq/k, n(q
+1)/~C)
which arerel-atively
prime
to n thenIA’l
=cp(6, 2, n)
=cp(n)/2 -
p(3,
0,
n),
18’1
=w (3>
0,
n) -
w (4>
0,
n),
I C’l
=w (4> o>
n) -
w (6>
0,
n)
andI D’I
=V (6,
0,
r~).
Using
Theorems 5-7 of[91,
we have thefollowing
inequalities
for thecardinalities of these sets of
integers:
Let us observe that
p/q
andIf
p/q
E .r4 then p-3,
sinceq > 2,
and either l or k is thelargest
element ofB’.
Thus,
(3}!)
> 3 +4 cos (27rj /n)
foreach j
EC’
U D’ and soSince the
quantity
before~g~(2
on theright-hand
side of thisexpression
is at least aslarge
as the similarquantity
obtained forp/q
EA,
we canignore
this case.If
(3ik)
> 0 then11
-,~~k~/ j~~’~f
1for j
E A’ so
a similaranalysis
tothat above shows that
and
For n
210, n
= 231 and n =462,
we can use these estimates to showby
direct calculation that our lemma holds.To deal with n >
210,
we first show thatmax(ci ,
c2, c3,c4)
1 for suchn.
Using
the aboveexpressions
for thesequantities
we see that this holds ifmin(3~~~~, 31""1)
> n.By
(6),
bothIA’I
and are at least~)/6-2~)/3,
so we need
only
prove that(p(n) -
2 ’ >6 log n
for n > 210.For 210 n 330 = 2 - 3 - 5 -
11,
2w(n)
8n°-389.
Lemma5(ii)
yields
the lower boundp(n)
> n°-719
for n > 210. Sincelog n
forn >
210,
we needonly
show thatno-075 (no .33 -
2)
log
3 > 6 for n in this range. But this iseasily
seen to be true.For 330 n 2310 = 2 - 3 . 5 - 7 -
11,
2- (n)
16no -48
Moreover,
by
Lemma5(i),
for n >2310,
2~’~n~
no .47
Therefore,
for n >
330,
2w~n>
no -48 .
Applying
Lemma5(ii)
again,
we find thatp(n)
> n°- ‘3
for n > 330. Sincelog n
n°-3~
for n >330,
we needonly
show that
no- 17 (no- 25 -
2)
log
3 > 6 for n in this range which is alsoeasily
seen to be true.
Therefore,
max( CI,
C2, C3,C4)
1.So,
from our lower bounds for the absolute values of theproducts
of theis less than
(5/3)cp(n)/2.
Let us first show that
2IA’I(2/3)IC’1 ~
1 and2IV’I(2/3)IB’1
> 1. Thesein-equalities
will show that either the first or the last termsgive
the maximum in thisexpression.
For the first of these two
inequalities
to betrue,
by
(6)
we need toshow that
0.08cp(n) -
0.26 ~2Wt’~> >
0.Similarly,
the secondinequality
requires
that thestronger
inequality
0.08cp(n) -
0.34 -2‘~~"> >
0 holds. Sowe need
only
consider this lastinequality
which we shall rewrite in the form0.08/0.34 ~
2~)/~).
For 210 n
330,
we saw in aprevious
paragraph
that2Wf">/cp(n)
0.33 C
0.1710.08/0.34.
We also saw thatn-O.25
0.235
0.08/0.34
for n > 330.Therefore,
our desiredinequalities
holds andwe need
only
try
to bound andfrom above.
Notice that
=
cp(4, 0, n)
and that =cp(4,
1,
n).
Lehmer[9,
p.351]
has noted thatE(4, 1, n) =
-E(4, 0, n),
whereE(k, q, n)
denotescp(n) - kcp(k,
q,n),
so we needonly
examineLehmer also
gives precise
information aboutE(4,
0,
n)
in Theorem 6 of[9].
If n > 4 and 4 divides n or n is divisibleby
aprime
congruent
to 1mod 4 then
E(4, 0, n)
= 0 and ourproof
iscomplete.
If neither of these conditions is true thenI E(4,
0,
n)
=
2"(n’)
where n’ is as in the statementof Lemma
8(iii).
Notice that whenE(4, 0, n) 0
0, n
is notcongruent
to 0 mod4,
so n’ is the oddpart
of n.A direct calculation shows that for 210 n
750,
with theexceptions
of n = 231 and 462 which we consideredabove,
0.05.Therefore,
(8/3)~~(3/2)’~o~))/4 ~ (5/3)~(~/2
for 210 n750, n =1=
231,462.
Recalling
that we showedby
calculation that the lemma holds for 30 n210,
for n = 231 and for n =462,
we now know that the lemmaholds for 30 n 750.
Notice that if n 4389 = 3’7’lld9 then either
~E(4, 0, n)~
= 0 or2-(n
8,
since in the latter case n’ is odd and withoutprime
divisorscongruent
to1 mod 4.
Using
Lemma5(ii),
p(n) >
160 and soJE(4, 0,
0.05 for n > 750 and our lemma holds for 30 n 4389.Lemma
5(i),
andpart
(ii)
of this samelemma,
we find thatThe
right-hand
side is amonotone-decreasing
function for n > 10 and so it is less than 0.05 for n > 4389. Therefore(8/3)~~(3/2)~~’/~
(5/3)‘~~"»2
for n >4389,
which shows that the lemma is true. D5. Proof of Theorem 1
Let be a Lucas or Lehmer sequence
generated
by
a Asnoted in the
proof
of Theorem2,
there exist twointegers
p and q such that a and{3
are the two roots ofX2 -
p + 2qX
+ q. Notice that the n-th element of the sequencegenerated
by
ia andI#
isjust
±un.
Therefore,
we can assume that q =a#
ispositive.
Also noticelpl
2q
for otherwise aand /3
are real and Carmichael(4),
Ward[19]
and Durst[5]
have shown thatin this case the n-th element of these sequences has a
primitive
divisor forn > 12.
Let us define the
(3}!) ’s
and,~~~~
as in Lemma 10. Stewart[14,
Section5]
has shown that if the n-th element of this sequence has no
primitive
divisorthen
Since
lpi
2q,
uponapplying
Lemma10,
we obtainfor n > 30.
Therefore,
if we can show thatand so,
by
Theorem 184 of(6~,
k/n
must be aconvergent
in thecontinued-fraction
expansion
ofHence we first want to show that for n
sufficiently
large,
theright-hand
inequality
of(11)
holds. We startby considering
the case of q =2,
as this is the most difficult one.5.1. The case q = 2
Using
the notation of Lemmas 8 and9,
wefind,
from Lemma9,
thatfor m > l.
If m =
1,
but n >1,
we have .Combining
this upper bound with Lemmas 7 and8(iii),
we obtainfor n > 1.
Applying
this lower bound to theright-hand
inequality
of(11)
andsquar-ing
bothsides,
we want to show thatfor n > 30.
To prove that this holds for n
sufficiently large,
we take thelogarithm
of both
sides,
whichyields
From Lemma
5(ii),
we see thatcp(n)
>nO- 8043
for n >3500,
while for the terminvolving
w(n)
we use the fact that2~’ /z
is a monotoneincreasing
function for x >
1/ log 2,
21/1
=22/2
and Lemma5(i).
In this manner,our
problem
is to show thatFor n >
3500,
the sum of the second and third terms is at mostn°’5g52.
seen to be true for n > 3500. So we have an initial bound of intermediate size.
Notice
though
that we did not make full use of the Lemma 9 in thisargument.
A direct calculation on acomputer
using
the resultgiven
inLemmas
7,8(iii)
and 9 shows that theright-hand inequality
of(11)
holdsfor all n > 1260
when q
= 2.In the case of q =
2,
Lucas and Lehmer sequences can result from p =-3, -l,1
and 3. SinceGn(p,
q)
is aproduct
of terms of the formp - it is
quite
easy to calculateGn(p,
q),
although
care mustbe taken to maintain sufficient accuracy, and so we can check whether un
has
primitive
divisorsby
means of(10).
However,
to check Un for each nup to 1260 in this manner is
quite
time-consuming.
Fortunately,
one canquickly
extract still more information from(11).
Given n, p and q, it is easy to find theinteger
1~ with(k, n)
= 1 which minimizes the farleft-hand side of
(11).
As whenconsidering
1260 n3500,
we can boundfrom above the middle
quantity
in(11).
For q
=2, p
=-3, -1,1, 3
and330 n
1260,
we canverify
in this way that the left-handinequality
in(11)
is violated and so for such n, the n-th element of these sequences has aprimitive
divisor. But we still need to consider 30 n 330. For thesen, we use
(10)
as described earlier in thisparagraph.
For n >
1260,
we have seen that n must be the denominator of acon-vergent
in the continued-fractionexpansion
forThe
question
arises of how to deal with these n. We are fortunate that in these cases the middlequantity
in(11)
isextremely
small. For such n,we
proceed
in the same manner that we checked the left-handinequality
in(11)
holds for 330 n1260,
except
that now we know k too. Theorem 2tells us that we need
only
check thoseconvergents
k /n
with n 2 -l0i° .
For each
convergent
computed
with n 2 -101°,
(p/q -
2I
wasconsiderably
larger
than the bound that the left-handinequality
of(11)
requires
if Un were to be without aprimitive
divisor. In Table1,
forp =
-3,
we list theconvergents
with n > 1260 andgive
thelogarithms
ofthe
required
and actualbounds,
denoteddreq
anddact,
respectively.
Thevalue of
log
ldreql
given
in Table 1 is truncated to itsinteger
part,
whereasthe value of
log
dact
f is
truncated to one decimalplace.
Proceeding
in this same way for p =-l,1
and3,
we are able to concludethat if is the Lucas or Lehmer sequence
generated
by
any of thepairs
(~,,~) _
(1 ± V/--7) / 2,
(V3-.rJ=5)/2, (~~ ~)/2
or(v’7-.rý-1)/2,
Table 1:
(p, q) _ (-3,2)
Verification 5.2. The caseof q
> 2For such
pairs
(p,
q),
weproceed along
the same lines. Theonly
differenceright-hand
inequality
of(11)
is satisfied for n > 1260by
our work in theprevious
section. We can checkdirectly,
as with 1260 n 3500 for q =2,
that theright-hand
inequality
of(11)
holds for> nq
where nq isgiven
inTable 2.
As in the case of 30 n 330
for q
=2,
wedirectly
check those unwith 30 n nq for
primitive
divisors and forlarger n
we compare therequired
and actual differences oflplq -
2I
in(11)
to establishour result. The actual difference is less than the
required
difference for allnq n 2 -
101°
and all 3q
3000(this
corresponds
to allpairs
of aand /3
withh(,Q/a) 4).
By
Theorem2,
Theorem 1 now follows.All the calculations in this article were
performed using
Release 3 ofMaple
V and UBASIC 8.74 on anIBM-compatible
PC with an 486DX2running
at 66 MHz. Intotal,
the calculationsrequired
just
over 100 hourson this machine.
Many
of the calculations wereperformed
using
bothsystems
toprovide
a check on thequantities
obtained and the results werealways
identical up to thespecified
accuracy.References
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Department of Mathematics
City University Northampton Square London, EC1V OHB, UK