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364
SOME CONGRUENCES FOB FIBONACCI NUMBERS
DEC. 1874
Relying on the known result that the period of divisibility by m^m2 (m],m2 co-prime) is given by D(m^m2) =
LCMdf, z2) (see Wall [6]), we get the results:
LCM (3,5) = 15, and so F^ is the first Fibonacci number to be divisible by 10. Icm (6,25) = 150, and so F-JQQ iS
divisible by 100, LCM( 12,625) = 7,500 and so F7500 is divisible by 104.
This has been an exercise in finding the z numbers. By an extension of the argument we can produce the corresponding k numbers—the period of recurrence of the Fibonacci numbers (mod m2).
REFERENCES
1. S.N. Collings, "Fibonacci Numbers,"Mathematics Teaching, No. 52 (1970), p. 23.
2. D.E. Daykin and L A G . Dresel, "Factorization of Fibonacci Numbers," The Fibonacci Quarterly, Vol. 8, No. 1
(February 1970), pp. 23-30.
3. V.E. Hoggatt, Jr., Fibonacci and Lucas Numbers, Houghton-Mifflin, Boston, 1969, p. 59.
4. A.G. Marshall, "Fibonacci, Modulo n," Mathematics Teaching, No. 46 (1969), p. 29.
5. J. Vinson, "The Relation of the Period Modulo to the Rank of Apparition of m in the Fibonacci Sequence,"
The Fibonacci Quarterly, Vol. 1, No. 2 (April 1963), pp. 37-45.
6. D.D. Wall, "Fibonacci Series Modulo m/''American Math. Monthly, Vol. 67 (1960), pp. 525-532.
7. 0. Wyler, "On Second-Order Recurrences," American Math. Monthly, Vol. 72 (1965), pp. 500-506.
[Continued from page 350.]
[k/2]
(5)
( f,J eI
Fk(x) = £
T=J ( kJ!
~ '
-
) 9k-*(HM
ho
Write
hk(x)
(6)
= (1-akx
ck
= [(r-sh)a]
Following Riordan [ 6 ] , with ao = 2 and hpM=
1 -x,
+ [(sa~r)b]k
- (a0 +
c3 + Ssj5 x(3e + 5s2) = h3(x) - 3e\h1(~x)
2
2
C4-x(2e
+ 20s e + 25s ) = h4(x) - 4ej h2(-x)2
+a3)xg1(-x)\
(a2 +
a4)xg2(-x)}
ho(x) - (a4 - ao)xgo(x) |
+2e \
= h5(x)-5e\h3(-x)-(a3
a2)xg0(-x))
- (a1
4
I
I c5-e1
hf(x)
c2 - x(2e + 5s ) = h2(x) - 2e j h0(-x)
I
<
=
2
1
.
we eventually derive
Cf+s^/5x
(7)
(-1)kx2)gk(x)
+
k
+5e2^h1(x)~(a5-ai)xg1(x)j
+ a5)xg3(-x)^
where
e1 = 2r5~~5r4s
k
k
Substituting values of ak = a +b ,
+ 30r2s2-40r2s3+35rs4~10s5
we have
h7(x)
h3(x)
!
h5(x)
These functions lead back to (2).
[Continued on page 362.]
= V 5 ( r + sx)
= 5(r2-s2x)~
h2(x)
h4(x)
.
3
10exg0(-x)
3
- 5^/5 (r + s x) 4
15exg7(-x)
4
= 25(r - s x) - 40exg2(-x)
= 25^/5 (r5 + s5x)-~ 75exg3(-x)
+
50e2xg0(x)
+ 125e2xg / (x).