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DISTHIBUTION OF FIBONACCI NUMBERS MOD 5 k
HARALDNIEDERRE1TER
Southern Illinois University, Carbondale, Illinois
It was shown by L. Kuipers and Jau-shyong Shiue [2] that the only moduli for which the
Fibonacci sequence \ F }, n = 1, 2, • • • ,
can possibly be uniformly distributed a r e the
powers of 5. In addition, the authors proved the Fibonacci sequence to be uniformly d i s t r i b uted mod 5, and they conjectured that this holds for all other powers of 5 as well.
note, we settle this conjecture in the affirmative.
In this
Thus we show, in particular, that the F i b -
onacci sequence attains values from each residue class mod 5 , and each residue class
occurs with the same frequency. The weaker property of the existence of a complete residue
system mod m in the Fibonacci sequence was investigated e a r l i e r by A. P. Shah [3] and
G. Bruckner [1]„
Theorem.
F o r definitions and terminology we refer to [2].
The Fibonacci sequence
{ F } , n = 1, 2, * • » ,
is uniformly distributed
mod 5 k for all k s> 1.
Before we s t a r t the proof, let us collect some useful preliminaries.
It follows from a
result of D. D. Wall [5, Theorem 5] that { F },
considered mod 5 , has period 4»5 .
n
k
Therefore it will suffice to show that, among the first 4»5 elements of the sequence, we
find exactly four elements, or, equivalently, at most four elements from each residue class
mod 5 . It will also be helpful to know that, for j ^ 1, the l a r g e s t exponent e such that
5 e divides (2j + 1)1 satisfies (see [4]):
00
(1)
e
-i^r]
i=l
*-
u
-
«E a f i = a-fL<i
i=i
We note the formula
i=0
and t, and f n J = 0 for j > n, which can be quickly
t
\ /
r~t-s
r
s
verified by comparing the coefficients of x in (1 + x)
= (1 + x) (1 + x) .
with non-negative integers r , s,
Proof of the Theorem.
We proceed by induction on k.
F o r k = 1, the result was a l -
ready shown in [ 2 ] . Now assume that, for some k ^ 2 and every integer a, the congruence
k-1
k-1
F = a (mod 5
) has exactly four solutions c with 1 < c ^ 4 • 5
. If n is a solution
k
k
k-1
of F s a (mod 5 ), 1 ^ n ^ 4 • 5 , then F = a (mod 5
), hence by periodicity:
n
k-1
n = c (mod 4-5
) for one of the four c f s . We complete the proof by showing that each
k
value of c yields at most one solution n. F o r suppose we also have F
= a (mod 5 ),
373
DISTRIBUTION OF FIBONACCI NUMBERS MOD 5 k
374
k
lr-1
1 < m ^ 4-5 , m = c (mod 4 . 5 v ), and WLOG n > m. Then, in particular,
k
k-1
(mod 5 ) and n = m (mod 4 e 5
). Using the well-known representation
1
F
where ( 1 = 0
n
00
.
Oct. 1972
F
n
= F
m
.
= 2
5=0
V
'
for r > n, we a r r i v e at
k-1
j=0
k-1
V
I
]=o
\
/
4.5^"!
^
Since 2
= 1 (mod 5 ) by the E u l e r - F e r m a t Theorem, we get
k-1
k
^V >( ^l L, \) \ - -( (™$\
, ^ ) - , .0 (mod 5 )
(3)
We claim that, for j — 1, the corresponding t e r m in this sum is divisible by 5 . By (2):
2j+l
^ ((„••!)-(„-•)) - g •• (" 1 %
••?.,) •
We look at 5J ( n ~ m ) . F r o m (1) we see that cancelling out 5Ts from it against 5-* leaves
* 1 '
k-1
at least one power of 5 in the latter number. Since there i s a factor 5
in n - m , we
get the desired divisibility property. Thus, from (3), we a r e left with the term correspondk
k-1
ing to j = 0: n - m = 0 (mod 5 ' ) . Together with n = m (mod 4«5
), this implies n -s
k
m (mod 4-5 ) or n = m.
ACKNOWLEDGEMENT
I wish to thank Prof. L. Kuipers for drawing my attention to this problem.
REFERENCES
1. G. Bruckner, "Fibonacci Sequence modulo a prime p = 3 (mod 4)," Fibonacci Quarterly,
8 (1970), No. 2, pp. 217-220.
2.
L. Kuipers and Jau-shyong Shiue, n A Distribution Property of the Sequence of Fibonacci
N u m b e r s , " Fibonacci Quarterly, Vol. 10, No. 4, pp.
3. A. P . Shah, "Fibonacci Sequence Modulo m , " Fibonacci Quarterly, 6 (1968), No. 2, pp.
139-141.
4. I. M. Vinogradov, Elements of Number Theory, Dover Publications, New York, 1954.
5. D. D. Wall, "Fibonacci Series Modulo m , " Amer. Math. Monthly 67 (1960), pp. 525-532.