697
Internat. J. Math. & Math. Sci.
VOL. 14 NO. 4 (1991) 697-704
LUCAS NUMBERS OF THE FORM PX2, WHERE P IS PRIME
NEVILLE ROBBINS
Mat hemat ics Dept.
San Francisco State University
USA
San F[ancisco, CA 94132
(Received Februaly I, 1989 and in revised form February 8, 1990)
th
Lucas number, where n is a natural number.
Let L denote the n
n
2
px
Using elementary techniques, we find all solutions of the equation: Ln
ABSTRACT.
where p is prime and p (i000.
KEY Wi’RDS AND PHRASES.
Lucas number
IIB39
1985 AMS SUBJECT CLASSIFICATION CODE.
1.
NTRODUCT ION
_
th
Lucas number, that is,
Let L denote the n
n
all Lucas nmers
found
n_
Cohn
J.H.E.
[1],
for
3.
In
Ln_l+Ln_ 2
L1=1, L2=3,
of Cohn [2], it
paper
of
late
a
a
result
a
square. As
which are square or t.wlce
is known that for each integer c 3, there is at. most one Lucas number of the
form cx
Uslnq [3], Definition 2, and (9) below, we see that there are 111
Let. n denote a natural number.
Ln
2.
primes, p, such that (i) 2
< p. 1000,
p[Ln
and (ii) there exists n such that
In this paper, we find all solutions of the equation:
px 2
L
n
(*)
where the prime p satisfies conditions (i) and (ii) above.
We find that only 8
The results aze summarized in Table 3
such values of p yield solutions of (*).
The larger problem of finding all solutions to |*) appears
more difficult; its solution would yield all Lucas numbers which ace prime.
on the last page.
2.
PRELIMINARIES
Let n denote a natural number. Let p denote a p[ime, not necessarily
satisfying conditions (i) and (ii) above.
th
Definition i
-Let Fn denote the n Fibonacci number, that is, F
+
for n_ 3
F
n
k_l and
Definition 2
Let z(n)
F
2
I,
Fn-I Fn-2
Mink:
Definition 3
Let y(n)
1/2z(n) if
nlFk}
21z(n).
For each integer c _3, the equation L
n
If
q is prin,e, then y(q) =n and
Ln=
cx 2 has at most one solution.
y(q2) =qn.
(i)
(2)
No ROBBINS
698
if n-=2 (rood 4)
otherwise
(Ln’L3n/Ln)
Ln
2
x
(3)
ff n:] o 3
n(L2n-3(-1)n)
(5)
L
L3n
rolLn
If m is odd and m_3, lhen
(L
n) =1
n
(6)
is an odd in[ege
iff
(7)
L5n/L
(Ln,lkn/Ln) Ik.
I[ k is odd, then
plLn
If p is an odd pzlme, then
(8)
2n
iff
-6)
(9)
is an odd integez.
L2n-2(-1)n
L2n
LnlLkn iff k
Lmn/Ln
(10)
(-1)
1/2 (m-l) (n+l) +
ate
..-
1/2(m-l)
(-1)
(j-l) (n+l)
1
3
then 2 .y(p)_ 1/2(p+l).
pln
If p is odd and
If p and m
(11)
is odd or n=l.
pm,
odd,
and
ph ILn
then
L(m+1_23)n
(12)
if m is odd.
(13)
ph+k IL k
mnp
(14)
for all k_0.
L n -:3 (rood 4) for all nl.
2
(15)
2 (rood 3) for all n_ 1.
(16)
Lsn
I.
(17)
L(m,n
(18)
If p is a prime such that y(p) exists, then (p,y(p))
If m/(m,n) and n/(m,n) are both odd, then
If y
(p2
Remarks:
py(p), then y(
k-I
pk
p
(Lm,Ln)
(19)
y(p) for all k_-1.
in [i];
(I) follows fzom Theorem 11 in [2] with a=]; (4) is Theorem
(8) follows from Theorem 4 in [5]; (12) follows from (44) in [4]; (14) follows
from Theo[em XI in [6]; (19) follows from (14).
The other identities
ale
elementazy.
THE MAIN RESULTS
3.
If p is a prime such that y(p) exists and L
Y(P)
2
2
(*) has the unique solution: n
u
y(p), x
PROOF:
This follows from hypothesis and (i).
TOREM 1
THEOREM 2
solution with n
2
then
If pe|3,7,11,19,29,47,199,521,2207,9349}, then (*) has a
2
2,4,5,9,7,8,11,13,16,19 respectively; if p=19, then x =4;
x2=l.
in each othe case,
PROOF:
pu
This follows from (2) and Theorem i, since
L2=3, L4=7, L5=]],
L9=19"4, L7;29, L8-47, Ll1=199, L13=521, L16;2207, L19=9349, and y(19)-9
THEOREM 3
PROOF:
Now suppose
x2=9,
L3k
d=l, so (i) it,plies
If k=l, then
x2=4.
2.
px 2 iff either (i) k=p=2,
or (ii) k=3, p=19,
2
Sufficiency is readily shown, since L6=18=2(3) and L9=76=19(2)
2
px
Let d
(Lk,L3k/Lk). If k 2 (rood 4), then (3) implies
L3k
L3/L1
Lk=U2, L3k/Lk
4
pv
2
pv 2 for some u,v.
an impossibility.
If k=3
Now (4) implies k=l or 3.
then
L9/L3
19
pv
2
699
LUCAS NUMBERS
2
If k-=2 (rood 4), then 13) implies d=3, so
2
2
2
3v for some
3pu
or (it) L
3pv
either (i) L
k
k
k
2
wlich
px
18
L
u,v. If (i) holds, then Theorem 2 implies k=2, so
6
2
2
Since 31L
3v
If (it) holds, then (5) implies
implies p=2 and
k
2
we get
(rood 9), so v -1 (rood 3), an impossibility.
2
px is impossible.
TIOREM 4
If p>19 and 31y(p), then L
n
2
so (6) implies ylp)In. Now hypothesis
then
px
PROOF: If L
n
implies 3In, so n=3k for some k. The conclusion now follows from hypothesis
so p=19 and x
L9/19 4.
3u 2, L3k/L
k
L3k/L
L3k
x2=9.
L3k-3
3v2-3
PlLn
and Theorem 3.
(*) has no solution if p/
5
TI|EOREM
{23, 31
79 83 107 167 181 211 227 229
241’271’349’379’383’409’431’439’443’467’499’503’541’571’587’601’631’647’683’691’
739,751,769,811,827,859,863,887,919,947,983,991.
"7"his follows
PR(’f)F:
from hypothesis and Theorem 4, since in each case,
p>19 and according to [31,
2
TI1F/3RFM 6
px
Lsk
31y(p).
ff k
x
2
and p=ll.
11"12.
Now suppose
11
PROOF: Sufficiency is readily shown, since L
5
2
and
Theorem 2 of [11 implies p is odd. Now (7), (1)
hypothesis
px
Lsk
u2
L5k/Lk
Ls/L ii,
imply Lk
then pv
1364/4
2
pv
2
for some u v
so p=ll and x
2
Now (4) implies k;1
L5/11
p=ll
If k=l
3
I k=3, then pv
I.
11"31, an impossibility.
2
If L
px and 51ylp), then n=5
341
TIIEOREM 7
ot
2
L15/L3
x2=l.
llypohess and (6) imply ylp)In. Therefore hypothesis implies
that is, n=5k for some k, so t-he conclusion follows ftom Theorem 6.
PX)F:
51n,
THEORFM 8
(*) has no solution if
pet41,71,101,131,151,191,251,311,331,
401,491,641,911,941,971
This follows from Theorem 7, since in each case, p)11, and
PROOF:
according to [31,
THEORt 9
Let p be an odd prime such that y(p) exists and is odd, and
such that- for every prime divisor, q, of y(p), z(q)
then n
y(p).
PROOF:
n
implies
If L
n
px
2
t-hen (6) implies n
dly(p).
dl
that qld. Therefore qlL
m
such
that is,
"
2 (rood
’4).
my(p) for odd m.
If L
n
px
2,
Now (8)
If d.l, then there exists an odd prime, q,
so (9) implies 2m is an odd integer; since
-)
m is odd, this implies z(q)---2 (rood 4), contrary to hypothesis.
Therefore d=l.
Now !13) implies y(p) 1 so m<n. Therefore hypothesis and (1) imply L
u2
2
pv for some u,v. Now (4 implies m=l or 3. If m=3, then
m
n
so Theorem 3 implies n=9, p=19. But.
L
# 3. Theren
fore m=l so n
y(p).
Ln/L
p(2v)2,
then)-)
TItEOREM 10
PROOF:
(’1 has no solution if p
{139,179,239,461,509,599,619,659}.
This follows from hypothesis and Theorem 9, since in each case,
according to [3] and 7], p fulfills the hypothesis of Theorem 9, yet
px
2.
Ly(p)
N. ROBBINS
700
In the ,Drk which follows, we will need the following lemmas:
2(-1)
L2
then (10) ]nplies
L
L231
L21In
L2kn_-" 2_2(-I)
LEMMA 3
1/2(m-l)+
n- 2 (-I)
4
Lpn/pLn
n
_-"
2 (-I)
I
hypothesis implies
n
2J-It
2
E 4 (rood I_, t)
lined
L21n
-2
If k is odd, hen Illl
k=2Jt
If
with j .1 and
odd
Ij.l. n
Now 111) implies
).
so
_
L)
1/2(m-i)_ +
r. (I 2
(rood
L2n
Hypothesis and Lemma 2 imply
9
(mod
1/2(m-l)
Lmn/Ln
Ther efo, e L
mn
(,nl)
(-I)
plLn
/L
n
1/2 (m-1)
and
m(-I
2In,
1/2 (m-I)
2
(rood Ln ).
hen
(rood p).
Lpn/Ln
(mod p2),
Hylx3thesis and Lemma 3 imply
PROOF:
k
2
L).
Lmn/Ln:_ m(-l)
0 (mod 2), then
If p is an odd prime,
_= (-I) 1/2 p-I
2 (mod
11
Bu! L
-2.
j-I
2 (mod
(_I)J-IL (m+l-2j) n
1/2 (m/ l-2j
1/2(m-1)
9=1
:(m-i) /
2
1.2nl.
Lt2n
(rood
Hypothesis and (12) imply
1/2(m-l)
-
If j_ 2,
holds for j=l.
L2)’n
m-i_:
9
(-i)
k
(rood
If
PROOF:
L (I,/|-29)
L2kn 21
iply
1.2kn
so
k
--_ 4-2
then
L2kn --2 21-1)
implies
L2kn L2 J+ir n
then I_emma
5
2
ttypothesis and I101
PROOF:
(-1)
2In,
and
If k:
iemma
j-I
Therefore g
by induct ion hypothesis.
2
2J_1 t
2
if j.2
Lt.)
(I0) implies
-2 l-l)
if j=l
2
(rood
2
2)
(mod L
2
(Induction on j)
PROOF:
implies
t/l
Lpn/Un
p(-l)
1/2(p-I)
p(-l)
1/2(p-l) (rood
L2n).
Now
from which lhe conclusion
immediately fol lows.
5
PROOF:
p is prime, and p---3 (rood 4), then
If 21n, plL
# s
n
n
(rood p). Also, hypoHypothesis and Lemma 4 imply
Lpn/PL
Lpn/PLn-=-]
thesis implies s
LEMMA 6
Let d
(Lm,
2.
2
Let Ln
Ln/Lm).
-I (rood p), so
Lpn/PLn
2
px where p and y(p)
# s2
ate
odd.
Let
.n
mly(p
Then d=l iff m=l.
that is dll, so d=l. Convezsely, if d=],
If m=l, then diL
1
then since hypothesis and (13) imply m (n, hypothesis and (I) imply L
u
m
2
pv
some
for
Now
or
3.
then
u,v.
(4)
If
implies
m=l
m=3,
hypothesis
m
and Theorem 3 imply p=19, n=9=y(19), so mll, an impossibility. Therefore m=l.
PROOF:
2,
Ln/L
701
LUCAS NUMBERS
(Lm,Ln/Lm) , l.
iff
2,
p and y(p) are odd, ml and
ml-),n
then
This follows from hypothesis and Le,ma 6.
PROOF:
]/3VgA
px
If L
n
IF_MMA 7
Let p, q be odd primes such that
8
2hlly(q),
pq[Ln
fox some n.
Then
2hi
where h_0.
Hypothesis and (6) imply n
PROOF:
ky(q) with j, k odd.
jy(p)
he
conclusion now follows.
2hlIy(P)
for some hl
px 2 where p is an odd prime
Let Ln
2
p=q, x =I, or (ii) n=
q is prime. Then either (i) n=2
2
2
2
x
IqtI for some t_l.
THEOREM II
and L
2
p
h
2hq
h,
L2hq/Iqtl
PROOF:
Hypothesis and (15) imply q---3 (mod 4), so q is odd.
h
that is,
is odd, so (Ii) imilies
L2hlLy(p
.implies y(p)/2
thesis and (6) imply n/y(p) is an odd integer, so (II)
Hypothesis
Hypo-
qlLy(p).
implies Ly(p)ILn,
hence
h, x2=l.
If pq, then hypoIf p=q, then hypothesis and (1) imply n=2
2
2
2 2
for odd m.
my’(q
Now (6) implies n
We
have
so
and
n
(2)
mqy(q).
qy(q),
imply
[.lypothesis
qlLn.
2)
thesis implies
y(q2)
qlLmy(q ). Let qJllLmy(q) Then (14) implies qj+l llLmqy(q)
x/q, we oblain
so qIl(Lmav(q)/Lmv(q)).= Therefore qld, so d:q. Let-ring
2
(Lmv(q)/q) (Lmav(q)/qLmv(q)
)= pt where the factors on he lef side of the
2
pu
equaIion are relatively prime. Therefore either (a) Lmy(q)/q
dlq
{6) implies
or tb} L_.
/q
u,v.
L.
h
so Lemma 5 implies (a) is mpossible.
2
hypothesis and (2) imply y(q)
2
u
Therefore (b) must hold. Now (i), (2) and hypothesis imply m
l,
n
qy(q)
x=qt.
Now
p(qv)2 p(qt)2
__Lv(q
2hq
2hl
If p is an odd prime and
IY(P) where 1- h (-4, hen the
only solutions of (*) are given by Table i below.
THEOREM 12
Table
n
2
4
8
16
28
PROOF:
7
47
2207
14503
1
1
49
This follows from hypothesis and Theorem Ii, since
2
(all primes); also
Note that
14503 7
L8=47 L16=2207
but not odd.)
1553729
2
IIL35312
.
L28
According to [7],
and
L7&/472
L35312/1553729072
pt
t
2.
2.
L2=3
L4=7
2
L6/3 2 (prime
According to the referee,
N. ROBBINS
702
{*) has no solution if
TIIEOREM 13
p{43,67,103,163,223,263,281,283,307,
347,367,449,463,487,523,547,563,569,607,643,727,743,787,823,881,883,907,929,967
This follows f[om hypothesis and Theozem 12, since in each case,
acco[ding to [3], p satisfies the hypothesis of Theo[em ]2 but does not appea[
PROOF:
above.
in Table
2
2
LII k px iff k x
PROOF: Sufficiency is [eadily shcwn,
suppose
Lll k px2. Let d (Lk,Lllk/Lk).
TIHEOREM 14
I and p=199.
since
199"12.
199
L]I
Now
dlll. If d=ll, then
and Theozem 6
then
hypothesis
But
so
since y(ll)=5, (6) implies 51k,
5111k.
2
pv
imply llk=5, an impossibility. If d=l, then (I0) implies
k
fol some u,v. Now (4) implies k=l o[ 3. Theo[em 3 implies k3, so k=l, hence
(8) implies
Lk=U2, L]lk/L
p=199,
x2=1.
x2=l.
2
If L
PX and llly(p), then n=ll, p=199,
n
Hypothesis and (6) imply llln, so the conclusion follows fom
T!{EOREM 15
PROOF:
Theoz em 14.
2
419x is impossible.
L
n
209- 11"19.
According to [31, y1419)
THEORIM 16
P:
The conciusion now
follows fzom Theo[em 15.
THEO 17
L
n
127x
Suppose L
n
thesis and (16) now imply
is impossible.
2
Since y(127)=64
(6) implies
127x
PROOF:
mOREM 18
2
x2_
641n
}lypo-
2 (mod 3), an impossibility.
If p and y (p) =q aze pz imes, q
3,
)
q2Ly (q) p2Lq
2
qs
and the equation L
m
(considezed as an equation in m) either (A) has no solution oz (B) has the
solution m=y(q) but theze exists a pzime, t, such that tllL /P) and y(q),
q
2
then L
px is impossible.
n
2
px
PROOF: Suppose L
Hypothesis and (6) imply n=mq, m odd, m > 1.
n
Let d
(Lm,Ln/Lm). (8) implies dlq. Lamina 7 implies d)I, so d=q. Theze-
Lq
# ps
2
and eithe (I)
21y(q)
oz (II)
2y(q)
qlLn. If (I) holds, then we get a contradiction via Lamina 8, since pqlLnpqu 2
pqv2 oI (i) L
qu2
If (If) holds, then either (i) L
m
m
m
2
Ln /Lm qv foz some u,v. If (i) holds, then (I) and hypothesis imply m=y(q).
and ty(q). If
Now (B) implies thee exists a pzime, t, such that ill
t=p, hen pll(L/p), so
contrazy to hypothesis. If t#p, %hen t
2
2
so (14)
an impossibility
is, tllpx
so tllx
),
and y(q) Im, so
If (ii) holds instead, then (6) implies y(p)
foIe
Ln/L
(Lq/p)
p21L,
impliestllLav(q ,that
LCM(y(p) ,y(q)
Ira,
qy(q)Is.
Since
so
Im
Im.
But (17) implies I_CM(q,y(q) )=qy(q),
so (14) implies
by hypothesis, we have ql
that is, LCM(q,y(q)
q2Lv(q)__
ILo
ILy(q),
y (q2) =qy (q), hence y(q2)im" Thezefoze hypothesis and (6) mply q21Lm so
Now (19) and (6) imply q2y(q)
that qlu 2, which implies q21u2, hence q31L
m
so m=qk, qy(q)Ik. Let d
Now
implies d lq- Thezefoe
(8)
(Lk,Lm/Lk).
I
ca 2, wheze c
I, p, q, or pq. Since km <n, (I) implies c#p, c#pq.
Lk
If c=l, then (4) implies k=l oz 3, violating qy(q)
and (I) imply k=y(q), again violating qy(q)
Ik.
If c=q, hen hypothesis
703
LUCAS NUMBERS
{*) has no solution if p(,59,359,479,709,719,809,839j.
THEOREM 19
PROOF: In each case, according to [3] and |7], p satisfies the hypothesis of Theorem 18, from which the conclusion follws. Table 2 below
gives the det-ails.
Table 2
P
q
(q)
relevant section of Theo em 18
59
359
479
709
719
809
839
29
179
239
59
359
7
89
119
29
179
50
209
liB, t=19489
i01
419
IIA
IIA
IIA
IIA
I
I IA
Theorem I0 above)
theorem i0 above)
first entry in Table 2)
second entry in Table 2)
(see
(see
(see
(see
(see Theor em 16 above)
Table
3 below, which contains all solutions
We summarize our results in
of (*) with 2<p(1000.
Table 3
p
3
7
ii
19
29
47
199
n
2
4
5
9
7
8
x
2
1
1
4
1
11
1
13
The elated results of M. Goldman [8] follow i,mediately from (I).
521
Remark:
I wish to thank the referee for his numerous helpful
suggestions, as well as for his assistance in obtaining a partial factorization
ACKNOWLF.DGMqT.
of
L35312.
REDES
COHN, J.H.E.
1.
109-113
Square Fibonacci numbers, etc., Fibonacci Quart. v2 n2 91964)
COHN, J.H.E.
(1972) 631-646
Squares in so,e recurrent sequences, Pacific J. Math. v41 n3
2.
3.
ALFRED, Brother U.
1965
Tables of Fibonacci entry points,
Fibonacci Association,
4.
Theorie des fonctions numeriques simplel,ent periodiques,
LUCAS, E.
Amer. J. Math. 1 (1878) 184-240; 289-321
5.
ROBBINS, N. Some identities and divisibility properties of linear second
order recursion sequences, Fibonacci Quart. 20 {1982) 21-24
6.
On the numerical factors of the arithmetic forms
CARMICHAEL, R.D.
Ann. Math. 15 (1913) 30-70
n+ /3n
7.
BRILLHART, J. et al
Tables of Fibonacci and Lucas factorizations,
Math. Comp. 50 (1988) 251-260; $1-S15
2
8.
On Lucas numbers of the form px
GOLDMAN, M.
Math. Reports Canad. Acad. Sci. (June,1988)
where p
3, 7, 47, or 2207