SOME PROPERTIES ASSOCIATED WITH SQUARE FIBONACCI NUMBERS
J O H N H. HALTON
Brookhaven National Laboratory f Upton,, Long Island, New York
1.
INTRODUCTION
In 1963, both M o s e r and C a r l i t z [ 1 1 ] and Rollett [ 1 2 ] p o s e d a p r o b l e m .
Conjecture 1,
F 0 = 0,
The only s q u a r e Fibonacci n u m b e r s a r e
F_i = F i = F 2 = 1,
and
F l 2 = 144 .
Wunderlich [ 1 4 ] showed, by an ingenious computational method, that for
3 < m < 1000008, the only s q u a r e F
is F 1 2 ; and the c o n j e c t u r e w a s p r o v e d
analytic ally by Cohn [ 5 , 6, 7 ] , B u r r [ 2 ] , and Wyler [ 1 5 ] ; while a s i m i l a r r e sult for L u c a s n u m b e r s w a s obtained by Cohn [ 6 ] and B r o t h e r Alfred [1]„
Closely a s s o c i a t e d with Conjecture 1 is
Conjecture 2.
When p is p r i m e , the s m a l l e s t F i b o n a c c i n u m b e r d i v i s -
ible by p is not divisible by p2„
It is known (mostly from Wunderlich ? s computation) that Conjecture 2
holds for the f i r s t 3140 p r i m e s
514229*
(p < 28837) and for p = 135721, 141961, and
C l e a r l y , Conjecture 2, t o g e t h e r with C a r m i c h a e P s t h e o r e m (see [ 4 ] ,
T h e o r e m XXIII, and [ 9 ] , T h e o r e m 6), which a s s e r t s that, if m ^ 0, with the
exception of m = 1, 2, 6, and 12,
that F
m
for each F
there is a p r i m e p,
such
i s the s m a l l e s t Fibonacci n u m b e r divisible by p (whence F
i s not
J
^
m
divisible by p 2 and so cannot be a s q u a r e , if Conjecture 2 holds),
Conjecture 1; but not v i c e v e r s a ,
implies
If Conjecture 2 h o l d s , then the divisibility
s e q u e n c e t h e o r e m ( [ 9 ] , T h e o r e m 1) can b e s t r e n g t h e n e d to say that, if p
an odd p r i m e and n ^ 1,
(1)
<*(p,n) = p
or(p) .
In the notation of [ 9 ] , Conjecture 2 for a given p r i m e p s t a t e s
F , v i s not divisible by
*This work performed
Commission.
is
then
p2,
T h i s , by L e m m a 8 and T h e o r e m 1 of [ 9 ] ,
under the a u s p i c e s of the U.
S„ Atomic
that
is
Energy
The p r e s e n t p a p e r i s a r e v i s e d v e r s i o n of a r e p o r t w r i t t e n u n d e r
a u s p i c e s of t h e U. 3* AEC w h i l e t h e a u t h o r . w a s a t Brookhaven
N a t i o n a l L a b o r a t o r y , Upton 9 L . ' I . , N . Y. ( R e p o r t N©eAMD/38^ BNL9300).
347
348
SOME P R O P E R T I E S ASSOCIATED
[Nov.
equivalent to
a(p2)
(2)
= pa(p)
o
Since v.(p) i s the highest power of p dividing F . ,,
(3)
v(p)
By L e m m a 11 of [ 9 ] ,
namely
F>^,
<*(p),
1.
p divides one and only one of F
where
T h u s , if p > 5,
=
this is equivalent to:
\(p) = p - (5/p)
and
(5/p)
s i n c e \(p) i s not d i v i s i b l e , b y p ,
,
F ,
and F
,
is the L e g e n d r e index,,
while it is divisible by
(2) is equivalent to
(4)
F
, v i s not divisible by p 2 ,
and inspection of the c a s e s p = 2, 3, and 5,
shows that the equivalence holds
for t h e s e p r i m e s a l s o .
Finally, (4) i s equivalent to:
F
F ,, i s not divisible by
p2
J
p - l p+l
*
(5)
T h i s p a p e r p r e s e n t s c e r t a i n r e s u l t s obtained in the c o u r s e of investigating
the two C o n j e c t u r e s , the l a t t e r of which i s s t i l l in doubt*
2.
A THEOREM OF M. WARD
We begin with a t h e o r e m posed as a p r o b l e m (published posthumously) by
Ward [ 13].
A different proof from that givenbelow w a s obtained independently
by C a r l i t z [ 3 ] .
T h e o r e m A„
(6)
Let
<fc&) = £
s=i
and
xS/s
1967]
WITH SQUARE FIBONACCI NUMBERS
349
k (x) = (xP" 1 - l ) / p ;
(7)
then, for any p r i m e n u m b e r p — 5,
p 2 divides the s m a l l e s t Fibonacci n u m -
b e r divisible by p if and only if
(8)
c ^ )
Proof,
(|)
= 2 k p ( I ) (mod p) .
We shall show that (8) is t r u e if and only if (5) is false Q
We
shall u s e the congruence (see [ 1 0 ] , page 105) that, when 1 — t — p - 1,
K')
(9)
-If)
= (-Dt
1
(modp)
and F e r m a t T s t h e o r e m (see [ 1 0 ] , p a g e 63), that
(10)
if (a,p) -
1,
ap-1
= 1 (mod p )
T h e identities
(ID
F
2
•.-^IHTff )"!
=
(12)
<14>
F^i
^
= F^
+
F ^
+ (-l)n = F n _ i r n + 1
.
.
and
(15)
v
'
3F2 + 2 ( - l ) n = F2
+ F2
n
n-l
n+l
,
a r e well known (see [ 8 ] , equations (3), (5), (64), (65), (67), and (95) with
m = l)o
F r o m then it follows that (since
(1 ±V5) 2 = 2(3 ± V5) )
350
SOME P R O P E R T I E S ASSOCIATED
[Nov9
,T\"2Q
il6)
=F4n/F2n
= F f f i . i + F 2 n + 1 = 2F2 + F ^
= 5F^ + 2(-l)n = 5 F n _ 1 F n + 1 - 3(-l)n
Now, s i n c e
(7),
p > 5,
p
F ^
.
is odd and -£(p - 1) i s an i n t e g e r .
the factor £ = 6 p [ | ( p - 1)] I
and k (3/2) into i n t e g e r s .
+
By (6)
is p r i m e to p and m a k e s both
T h u s , modulo p ,
fa.
and
v(5/9)
by (6), (9), (16), (7), and '(10),
-i|(t) P < 5 v.v. + 3 >- 2 }
= -5"1P[S(P " Ifl'•V . V>
/P) +
' (^)P"' ' 2kf
• ' • 2kp (l) - S < V V / P ) '
w h e r e f and g
a r e i n t e g e r s p r i m e to p ,
It follows that (8) is t r u e if and only if
(F
and
F
F
+1
F
/p
is an integer.
/p) = 0 (mod p),
and this
c o n t r a d i c t s (5), proving the theorem,,
3.
ANOTHER CONJECTURE
We end the p a p e r with an examination of a conjecture, which i m p l i e s the
f i r s t conjecture (known now to b e t r u e ) , in a r a t h e r different way from Conj e c t u r e 2.
T h e underlying r e s u l t i s
T h e o r e m Bo
i n t e g e r M,
Let p be a p r i m e , and suppose that t h e r e e x i s t s a p o s i t i v e
such that
(i) for no i n t e g e r n, p r i m e to p and g r e a t e r than M, is
o r p t i m e s a s q u a r e ; and
F
a square
1967]
WITH SQUARE FIBONACCI NUMBERS
(ii) if n i s p o s i t i v e and not g r e a t e r than N,
t i m e s a s q u a r e , then F,
Proof.
is a square or
Suppose that (i) and (ii) hold, and that F
J
D divides
Let m = p m i ,
* l
A
and A
is
2
.
M*
= AB C ,
Then, b y (i),
= BC 2 D,
F
m
1 or p.
t i m e s a s q u a r e , and divisible by F
2
and w r i t e F
>
is a square or p
In contradiction of the t h e o r e m , l e t m > M.
m i s divisible by p«
p
such that k / n i s a power of p;
at all i s a s q u a r e o r p t i m e s a s q u a r e for m
times a square.
where
and F
i s n e i t h e r a s q u a r e n o r p t i m e s a s q u a r e , when k
is the l e a s t i n t e g e r g r e a t e r than M,
then no F
351
mj
This makes F
a square or p
Now, by the well-known identity (see
[8, equation (35)], o r [9]* equation (8)] )
(17)
F
F
m
mj
= y ( P ) F h " 1 F p " h F,
X*. I h i m i m i - l h
w e get that
B(A/D) = BC*D V (A
F h ' 2 F P ~ \ Fh + pF*5"1
A^ I h i
Also,
mi
mi-i
h
mi-l
^ mi-
(F
y
- 1, F
) = 1, so B m u s t divide p;? that i s , B i s 1 o r p;
mi
mi
^
^'
and again D i s 1 o r p . It follows that F
, too, i s a s q u a r e o r p t i m e s
J,
1
a square.
Arguing s i m i l a r l y , we s e e that, if
m = p m ,
then
F
is a
s q u a r e o r p t i m e s a square,,
T h i s will continue until (m ,p) = 1, and then,
S
t
< M. But then, b y (ii), if p m = m . i s the l e a s t such n u m -
by (i) 1 ^ m
S
"
S
b e r g r e a t e r than M,
square,,
s ^ t,
and F
m
S —L
cannot be a s q u a r e o r p t i m e s a
T h i s c o n t r a d i c t i o n shows the c o r r e c t n e s s of t h e t h e o r e m .
Conjecture 3.
T h e r e is no odd i n t e g e r m > 12,
such that F
is a
s q u a r e o r twice a square,,
T h e o r e m C Conjecture 3 i m p l i e s Conjecture 1.
Proof.
M = 12.
Conjecture 3 s t a t e s condition (i) of T h e o r e m B, when p = 2 and
T h e only
F
,
squares are Fj = F2 = ls
c o r r e s p o n d i n g F,
with
1 < m < 12,
F 3 = 2,
a r e F i 6 = 3° 7°47
is a s q u a r e o r twice a s q u a r e .
F 6 = 8,
which a r e s q u a r e s o r twice
and F 1 2 = 144.
5
However,
the
2
and F24 = 2 • 3 • 7 • 2 3 , and n e i t h e r
Thus (ii) holds also, whence the conclusion of
T h e o r e m B , which includes Conjecture 1, i s es tablished.
352
SOME PROBLEMS ASSOCIATED
4.
[Nov.
PYTHAGOREAN RELATIONS
We c l o s e this p a p e r by a r a t h e r c l o s e r examination of Conjecture 3,
using the identities (12) and (13), with the well-known r e s u l t , that the r e l a t i o n
x2 + y2 = z2
(18)
holds between i n t e g e r s if and only if t h e r e a r e i n t e g e r s
p r i m e and of different p a r i t i e s , and an i n t e g e r u,
x = (s 2 - t 2 ) u ,
(19)
y = 2stu,
s and t,
mutually
such that
z = (s 2 + t 2 )u .
and
Conjecture 3 l e a d s us to e x a m i n e the p r o p e r t i e s of F i b o n a c c i n u m b e r s
F
, which a r e s q u a r e s or twice s q u a r e s , for odd i n t e g e r s m.
We obtain
the following r a t h e r r e m a r k a b l e r e s u l t s .
T h e o r e m D,
i n t e g e r s r , s,
even,
If m i s odd,
and t,
(s, t) = 1,
(20)
F
i s a s q u a r e if and only if t h e r e
such that m = 12r ± 1 ,
s i s odd,
t is
and
F 6 r ± 1 = s 2 - t2 .
F 6 r = 2st,
Proof,
s > t ^ 0,
are
Since m is odd, put
m = 4n ± 1,
determining
n
uniquely.
Then, by (12),
(21)
F
Thus F
= F 4 n ± 1 = F^n + F2n±i .
i s a s q u a r e if and only if F ^ ,
ean t r i p l e t .
2st,
m
Since ( F 2 n , F 2 n ±i) = 1» u = 1,
while F ^ i
2
) = 1,
a
^ d A/F~~ f o r m a P y t h a g o r -
and this p a i r is
2 2
= . ( s ± t ) » T h i s gives that s and t
and of different p a r i t i e s , with s ^ t > 0.
(F , F
F2n±i»
and
a r e mutually p r i m e
By (12), F ^ i
= F2 + F2
.
Since
not both n u m b e r s a r e even, whence F 2 n : ti is e i t h e r odd o r
the s u m of two odd s q u a r e s , which m u s t be of the f o r m
divisible by 4, it follows that
(22)
(s 2 - t 2 )
F 2 n = 2st,
Fm±1
= s 2 - t2 .
8k + 2.
Since 2st
is
1967]
A l s o , by (13),
WITH SQUARE FIBONACCI NUMBERS
353
2st - F ^ F ^ + F n + 1 ) = F ^ F ^ + F n ) .
Since this m u s t b e
divisible'by 4, and (F , 2 F n _ + F ) = ( F , 2 ) ,
= 3r
F
m u s t be even, so that n
(since F 3 = 2); whence m = 12r ± 1, a s s t a t e d in the t h e o r e m , and (22)
b e c o m e s (20).
s 2 - t2 = F 2 + F 2 _
Finally,
s u m of an odd and an even s q u a r e .
i s of the f o r m 4k + 1,
being the
T h u s s m u s t be odd and t even, a s was
asserted.
Since Conjecture 1 i s valid, i t follows f r o m T h e o r e m D that, if r ^ 2,
the equations (20) a r e not satisfied by any i n t e g e r s
T h e o r e m E„
If m i s odd,
a r e i n t e g e r s r , s,
both odd,
and t,
(s, t) = 1,
F
r , s, and 6.
i s twice a s q u a r e if and only if t h e r e
such that m = 12r ± 3 ,
s ^ t
>
0 ,
s and t a r e
and
F 6 r = s 2 - t2, F 6 r ± 3 = 2 s t .
(23)
P r o of.
b y (21),
Fm
We p r o c e e d much a s for T h e o r e m De
Let m = 4n + l a
Then,
and F 2 n ± i m u s t both be odd (since they cannot both be even),
so that F 2 n ± i i s even (since one out of e v e r y consecutive t r i p l e t of Fibonacci
n u m b e r s , one i s even, and i t s index i s a multiple of 3). Thus 2n ± 1 = 6r ± 3 ,
whence m = 12r ± 3,
a s s t a t e d in the t h e o r e m .
s i n c e 2n = 6r ± 2 and 2n ± 1 = 6r ± 1,
F
(24)
6r+2
+
F
6r+l ~
F
6r+3»
F
F 6 r _ 2 + F 6 r „j; = F 6 r ,
It i s e a s i l y v e r i f i e d that,
and
6r+2 "
F
F
6r+i ~
6r
»
F 6 r „ 2 - Fg-p.j = - F 6 r _ 3
.
equation (21) yields that
2
F 1 2 r ^ 3 ~ (F 6r -t2
+
F6r:tl)
+ (F6r±2 " F6r±1)
25
( )
=
Thus F
i s twice a s q u a r e if and only if F 6 r ,
Pythagorean t r i p l e t
by 8,
Clearly,
+ F2
x
6r
6 r±3
F2
F6ri3,
s i n c e F 3 = 2 and
and V 2 F 1 2 r i 3 f o r m a
F 6 = 8,
F 6 r i s divisible
but F 1 2 r ^ 3 and F 6 r ± 3 a r e divisible by 2, but not by 40
2
2
and F 6 r and F 6 r i . 3 a r e of the f o r m s 2(S - T ) and 4ST, w h e r e
(S, T) = 1,
and S and T a r e of opposite p a r i t i e s .
In fact,
Thus u = 2
S > T > 0,
354
SOME P R O P E R T I E S ASSOCIATED
[Nov.
F 6 r = 4ST and F 6 r ± 3 = 2(S 2 - T 2 )
(26)
s i n c e 4ST i s c l e a r l y divisible by 8.
holds, and c l e a r l y s ^ t ^ 0,
,
Put S + T = s and S - T = t;
(s,t) = 1,
and
s
and
then (23)
t a r e both odd,
as
s t a t e d in the t h e o r e m .
We finally note that Conjecture 3 holds if, for
a r e not satisfied by any i n t e g e r s
r,
s,
r ^ 2,
the equations (23)
and t.
REFERENCES
1.
B r o t h e r U. Alfred, "On Square Lucas N u m b e r s , " The Fibonacci Q u a r t e r l y ,
Vol. 2, (1964), pp. 11-12.
2.
S. A. B u r r , "On the O c c u r r e n c e of Squares in L u c a s Sequences, " A. M. S.
Notices (Abstract 63T-302) 10 (1963), p . 367.
3.
L. C a r l i t z , Solution to P r o b l e m H-24, T h e Fibonacci Q u a r t e r l y , Vol. 2,
4.
R. D. C a r m i c h a e l ,
(1964) pp. 205-207.
1
a
5.
"On the N u m e r i c a l F a c t o r s of t h e A r i t h m e t i c F o r m
n
± fi , " Ann, Math. , Vol. 15 (1913-14), pp. 30-70.
J . H. E. Cohn,
"On Square F i b o n a c c i N u m b e r s , " P r o c . ,
1963 N u m b e r
T h e o r y Conf. (Boulder, 1963), pp. 51-53.
6.
J . H. E. Cohn, "Square Fibonacci N u m b e r s , E t c . , " T h e Fibonacci Q u a r t e r l y , Vol. 2, (1964), pp. 109-113.
7.
J . H. E. Cohn, "On Square Fibonacci N u n b e r s , " J o u r . Lond. Math. Soc. ,
Vol. 39, (1964), pp. 537-540.
8.
J . H. Halton, "On a G e n e r a l Fibonacci Identity," T h e Fibonacci Q u a r t e r l y ,
Vol. 3, (1965), pp. 3 1 - 4 3 .
9.
J . H. Halton, "On T h e Divisibility P r o p e r t i e s of F i b o n a c c i N u m b e r s , " The
Fibonacci Q u a r t e r l y , Vol. 4, (1966), O c t . , pp. 217-240.
10.
G. H. Hardy,
E. M. Wright, An Introduction to the T h e o r y of N u m b e r s ,
(Clarendon P r e s s , Oxford, 1962).
11.
L. M o s e r ,
L. C a r l i t z ,
P r o b l e m H - 2 , The F i b o n a c c i Q u a r t e r l y , Vol.
1,
No. 1, (Feb. 1963), p . 46.
12.
A. P . Rollett, P r o b l e m 5080, A m e r .
Math. Monthly, Vol.
70, (1963), p .
216.
13.
M. Ward, P r o b l e m H-24, The F i b o n a c c i Q u a r t e r l y , Vol. 1, No. 4, (Dec.
1963), p. 47.
1967]
.WITH SQUARE FIBONACCI NUMBERS
355
14. M. Wunderlich, "On the Non-Existence of Fibonacci Squares, " Math.
Comp. , Vol. 17 (1963), pp. 455-457.
15. O. Wyler, Solution to Problem 5080, Amer. Math. Monthly, Vol. 71,
(1964), pp. 220-222.
* • * • •
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• • • • •
P I C K I N G
AUAY
AT
1967
CHARLES U. TRIGG
San DiegOp California
1967 can be made from two 2's, three 3 f s, four 4 ! s, five 5 ! s,
six 6 f s ? or seven 7fs with the aid of eighteen toothpicks and
mixed Arabic and Roman number symbolism*
That is,
! 315*7 = \/1 v/ vi vi \/i v/ =
11
jj !!=!!! !',! ill =:viV.v,v