ff n:] o 3 n. L3n Ln(L2n-3(-1)n). If m is odd and m_3, lhen rolLn iff is an oddin[ege. (L n L5n/Ln) =1. I[ k is odd, then (Ln,lkn/Ln) Ik. 2n. If p is an odd pzlme, then plL.
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 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
