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Z[] AND THE FIBONACCI SEQUENCE MODULO n

SAMIN RIASAT

Abstract. It has long been known that the Fibonacci sequence modulo n isperiodic for any integer n > 1. In this paper we present an elementary approachof proving properties of this period by working on Z[] and also deduce somenew results. In the last section a method of proving identities is shown usingexamples.

1. Periodicity Modulo n

We will use the following notations:

n is a positive integer. Fi is the i-th Fibonacci number, with F0 = 0, F1 = 1 and Fi+1 = Fi + Fi1

for all i 1. Li is the i-th Lucas number, with L0 = 2, L1 = 1 and Li+1 = Li + Li1 for

all i 1. = 1+

5

2 , the golden ratio.

And all congruences are taken modulo n unless otherwise stated.

Definition 1.1. For n > 1, k(n) is the smallest positive index such that n | Fk(n).We will often denote k(n) simply by k for brevity.Example: k(2) = 3, k(10) = 15 etc.

Definition 1.2. Forn > 1, (n) is the length of the period of the Fibonacci sequencemodulo n.

Example: (2) = 3, (10) = 60 etc.

The following theorem was stated by Wall in [5].

Theorem 1.3. The Fibonacci sequence mod n is periodic.

Proof. The terms of the Fibonacci sequence mod n can take only n possible values,namely 0, 1, . . . , n 1. Note that if a sub-sequence Fk, Fk+1 repeats at some point,the whole sequence will repeat from that point, since Fk+2 = Fk+1 + Fk and so on.There are at most n2 possible choices for the sub-sequence Fk, Fk+1. So it mustre-appear at some point and hence the sequence will become periodic.

Corollary 1.4. Every positive integer divides infinitely many Fibonacci numbers.

Proof. Since the Fibonacci sequence mod n is periodic for every positive integer n,there exists infinitely many positive integers k such that Fk+1 Fk+2 1. ThenFk 1 1 = 0 for all such k and the conclusion follows immediately.

1

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2 SAMIN RIASAT

Proposition 1.5. (n) = k ordn(Fk+1).Proof. Since Fk

0, we have Fk+1

Fk

1

(0

n

1) and from the

Fibonacci recurrence it follows that Fk+i Fi for all i. Let g = ordn(). Then bythe recursion we obtain

Fgk+1 Fgk+2 g 1.Also, since g is the order, there does not exist g < g such that Fgk+1 Fgk+2 g

1. Hence we conclude that (n) = gk = k ordn(). Corollary 1.6. n | Fm k(n) | m.Proposition 1.7. (n) {k, 2k, 4k} n > 1.Proof. We will work on Z[]. Since Fk 0, we have

k (1 )k

k

+1 (1 )k

k+1 (1 )k+1 (1 )k(2 1) k+1 (1 )k+1

5k

Fk+1 k (mod n).On the other hand, using 1 = 1/ we get

k

1

k 2k (1)k 4k 1. (1.1)

If k 1 then Fk+1 1 and (n) = k. Otherwise, (i) if k is even then F2k+1 2k 1 which implies (n) = 2k. (ii) if k is odd then F4k+1 4k 1 implying(n) = 4k. Hence the conclusion.

Remark: It is not difficult to see that k = Fk + Fk1 holds for all k. ThusFk+1 Fk1 k follows from here as well.

From Fk+1 k (mod n) we can propose a new definition for (n):Definition 1.8. (n) = ordn() n > 1.

Now we present a very short proof of another theorem of Wall in [5].

Theorem 1.9. (n) is even for n > 2.

Proof. Assume that (n) {2k, 4k}. Then (n) = k implies k 1. Hence 2k 1and by (1.1), (1)

k

1. Therefore k is even (since n > 2) and the conclusionfollows. Proposition 1.10. If n > 2 and k(n) is odd, then (n) = 4k(n).

Proof. From the last theorem it follows that (n) = k. Suppose that (n) = 2k. Then(1.1) implies, 1 2k (1)k 1, a contradiction. Therefore (n) = 4k.

Now we shall prove the central theorems of this section. From here onwards pwill represent a prime.

Theorem 1.11. If p > 3, n > 1 and n | Fp, then k(n) = p and (n) = 4p.

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Z[] AND THE FIBONACCI SEQUENCE MODULO n 3

Proof. It is well known that gcd(Fi, Fj) = Fgcd(i,j). Hence i 0 (mod p) we havegcd(Fp, Fi) = 1, which implies gcd(n, Fi) = 1. Thus we conclude that p is the

smallest positive integer such that n | Fp, i.e. k(n) = p. Since 3 p, Fp is odd,implying n is odd, and so n > 2. Since k(n) is odd, by Proposition 1.10 weconclude that (n) = 4p.

Theorem 1.12. If n is prime and p > 3, then (n) = 4p n | Fp.Proof. Using Theorem 1.11, we need only prove that (n) = 4p n | Fp. FromProposition 1.7 we have 4p {k(n), 2k(n), 4k(n)}. Hence k(n) {p, 2p, 4p}. Ifk(n) = p we are done. So assume that k(n) = 2p. Since n | F2p = FpLp and n Fpwe must have n | Lp = p + (1/)p. Therefore 2p 1, which implies (n) = 2p,contradiction.

Now suppose that k(n) = 4p. This implies n | F4p = F2pL2p and since n F2p, n |L2p =

2p + (

1/)2p. But then 4p

1, contradiction. Therefore k(n) = p.

Theorem 1.13. If q is a prime and p > 3, then (qn) = 4p qn | Fp.Proof. q = 2; otherwise 3 = (2) | (2n) = 4p, which contradicts p > 3. Thus q isodd. Now Li = Fi+1 + Fi1 implies gcd(Fi, Li) {1, 2} i. The rest of the proof issimilar to that of Theorem 1.12.

Theorem 1.14. Ifp > 3 and(n) = 4p, thenn has a prime factorq with multiplicityr 1 such that qr | Fp.Proof. Let n = pa11 pajj be the prime factorization of n. Then

(n) = lcm((pa11 ), . . . , (pajj )) = 4p. (1.2)

Therefore (paii ) {2, 4, p, 2p, 4p} i. But there is no x such that (x) {2, 4, p}.Hence (paii ) {2p, 4p} i. If (paii ) = 4p for some i, we are done by Theorem1.13. Otherwise, (paii ) = 2p i, which implies from (1.2) (n) = 2p, a contradiction.Hence the result.

As a consequence of these results we arrive at the following conclusion.

Proposition 1.15. If q is an odd prime and r 2, then the following statementsare equivalent, and they imply that q is a Wall-Sun-Sun prime.

(i) qr | Fp.(ii) k(qr) = k(qr1) = = k(q2) = k(q) = p.(iii) (qr) = (qr1) = = (q2) = (q) = 4p.

Proof. The above theorems imply that (i), (ii) and (iii) are equivalent. On the otherhand, it is well known that q | F

q( q5

) for all odd primes q, whereab

is the Legendre

symbol. Since p is the smallest index such that q | Fp, we must have p | q q5

i.e.

Fp | Fq( q5

). Therefore q2 | qr | Fp | Fq( q

5), implying q must be a Wall-Sun-Sun

prime, as desired.

It should, however, be noted that no prime p has yet been found such that q2 | Fp,and the equivalence of statements (i), (ii) and (iii) may suggest a possible approachfor investigating the existence of such primes.

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2. The Range of

In 1913, R. D. Carmichael proved the following theorem:

Theorem 2.1. Every Fibonacci number except F1, F2, F6 and F12 has a prime di-visor which does not divide any smaller Fibonacci number. Such prime divisors arecalled characteristic divisors.

Based on his result, we will attempt to find X, the range of.

Proposition 2.2. (2) = 3 is the only odd element of X.

Proof. This follows directly from Theorem 1.9.

Proposition 2.3. 8n + 4 Xn.Proof. For n = 1 we have (8) = 12. Otherwise, let p be a characteristic divisor of

F2n+1. Then k(p) = 2n +1 and from Proposition 1.10, (p) = 4(2n + 1) = 8n + 4,as desired.

Proposition 2.4. 4n + 2 Xn.Proof. For n = 1 we have (4) = 6. Otherwise, let p be a characteristic divisorof F4n+2 = F2n+1L2n+1. Then p | L2n+1 = 2n+1 + (1/)2n+1 which implies4n+2 1 (mod p). Thus (p) = 4n + 2. Proposition 2.5. 8n Xn.Proof. For n = 3 we have (6) = 24. Otherwise, let p be a characteristic divisor ofF4n = F2nL2n. Then p | L2n = 2n + (1/)2n which implies 4n 1 (mod p).Thus 8n

1 (mod p) and we conclude that (p) = 8n.

The above results may be summarized into the following theorem:

Theorem 2.6. The elements of X are precisely 3 and all even numbers > 4.

3. Proving Identities

In this section we will extensively use the following facts, which are very easy toprove, to prove some identities.

If a,b,c,d are integers then

a + b = c + d a = b, c = d. (a + b) + (c + d) = e + f for integers e, f such that e = a + c, f = b + d. (a+b)(c+d) = k+l for integers k, l such that k = ac+bd,l = ad+bc+bd. n = Fn + Fn1.

First Identity.n

i=1

Fn = Fn+2 1. (3.1)

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Z[] AND THE FIBONACCI SEQUENCE MODULO n 5

Proof. Let Sn =n

i=1 Fn. We have

n+1

1

1 1 =n

k=1k

=n

k=1

(Fk + Fk1)

= Sn + Sn1.

On the other hand, n+1 1 = Fn+1 + Fn 1. HenceFn+1 + Fn 1

1 = Sn + Sn1 + 1 Fn+1 + Fn = Sn(2 ) + (Sn1 + 1) Sn1

Fn+1 + Fn = (Sn1 + 1) + Sn Sn1.Therefore we conclude that Sn1 + 1 = Fn+1, as desired.

Second Identity.

Fm+n1 = FmFn + Fm1Fn1, or, Fm+n = FmFn+1 + Fm1Fn. (3.2)

Proof. Since m+n = m n, we getFm+n + Fm+n1

= (Fm + Fm1)(Fn + Fn1)

= (FmFn + Fm1Fn + FmFn1) + FmFn + Fm1Fn1

= (FmFn+1 + Fm1Fn) + FmFn + Fm1Fn1.Hence (3.2) follows.

Third Identity.

Fkn+c =n

i=0

n

i

FikF

nik1 Fc+i. (3.3)

Proof. From kn+c = (k)n c, we can writeFkn+c + Fkn+c1 = (Fk + Fk1)

n c

= n

i=0n

iFi

kiFni

k1c

=n

i=0

n

i

FikF

nik1

c+i

=n

i=0

n

i

FikF

nik1 (Fc+i + Fc+i1)

= n

i=0

n

i

FikF

nik1 Fc+i +

ni=0

n

i

FikF

nik1 Fc+i1

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Thus

Fkn+c =n

i=0n

iFikFnik1 Fc+i.

It is clear that many other identities, if not all, can be proven in similar ways,and new identities may as well be deduced. Finally, the methods discussed here caneasily be generalized to other Fibonacci-like sequences.

References

[1] Wikipedia, Fibonacci number, http://en.wikipedia.org/wiki/Fibonacci number

[2] Wikipedia, Pisano period, http://en.wikipedia.org/wiki/Pisano period[3] Mathworld, Pisano period, http://mathworld.wolfram.com/PisanoPeriod.html[4] Wikipedia, Carmichaels theorem, http://en.wikipedia.org/wiki/Carmichaels theorem[5] D. D. Wall, Fibonacci Series Modulo m, American Mathematical Monthly 67 (1960), 525532.

Samin Riasat, Dhaka, Bangladesh

E-mail address: nayel71@gmail.com