The Period of the Fibonacci Sequence Modulo j

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1 The Period of the Fibonacci Sequence Modulo j Introduction Charles W. Campbell II Math 399 Spring 007 Advisor: Dr. Nic Rogers The main objectives of this project were to examine the Fibonacci sequence modulo j, a positive integer, with the intentions of generalizing the results to general sequences defined by linear recursions to utilize information being acquired during the semester. The main part of the research involved investigating the periodicity of the new sequence obtained after modding out by j. We begin by establishing a motivation for investigating the sequence modulo p a prime p a prime power by first investigating the Fibonacci sequence modulo j, a positive integer. Next we investigate the Fibonacci sequence modulo p a prime then generalize to prime powers. True to the objective the results we obtained were through the use of but not limited to) important ideas from both Number Theory Abstract Algebra including Fermat s Little Theorem, Euler s generalization of Fermat s Little Theorem, Quadratic Reciprocity, Field Theory. Lastly we apply the results established for the Fibonacci sequence to easily prove results for the Lucas sequence modulo p a prime. We begin by defining the sequence itself. Definition The Fibonacci sequence is a linear recursion defined by F n+1 F n 1 + F n for n 1, 1) where F n is the nth Fibonacci number with F 0 0 F 1 F 1. In the study of the Fibonacci sequence, it will be nice to be able to calculate the Fibonacci numbers themselves. There is a closed form equation for doing just that, but before we prove that this equation gives us the correct members of the sequence, we introduce the following identities which help us in the proof. Throughout let φ 1+ φ 1. Identity 1. φ 1 + ) φ. )

2 Identity. φ 1 ) φ. 3) We are now prepared to state prove the first most widely used result, the closed form for generating the Fibonacci numbers. Theorem 1. F n 1 φ n φ n ). Proof. Using induction on n N let Pn) be the statement that F n 1 φ n φ n ). P1) is true as F 1 1 φ φ) P) is true, since by ) 3) we have 1 ) F 1 φ φ ) φ) 1 + φ)) 1 φ φ) F 1 1. Now assume that Pn) is true up to some n N, n > 1, consider Pn + 1). F n+1 F n 1 + F n 1 φ n 1 φ n 1 ) + 1 φ n φ n ) 1 φ n 1 φ n ) φ n 1 φ n )) 1 φ n φ) φ n φ)) by ) 3) 1 φ n 1 φ ) φ n 1 φ )) 1 φ n+1 φ n+1 ). Therefore Pn + 1) is true, so by induction, we conclude that Pn) is true n N.

3 Upon investigating the Fibonacci sequence modulo an integer j, it becomes evident that this modified sequence is periodic in nature. It should be noted that due to the nature of the recurrence relation defining F n, from any point in the sequence, the future sequence is completely determined by two consecutive terms. Thus, considering two arbitrary consecutive members of the sequence modulo j, i.e the ordered pairs F n mod j), F n+1 mod j)), we see that there are j choices for each F n F n+1 so that there are j possibilities for these consecutive numbers. Since there are a finite number of possibilities for these consecutive elements, our sequence must be eventually periodic, i.e. that it comes bac to some member of the sequence repeats thereafter. Again, due to the nature of the recurrence relation, we find that it is reversible, i.e. that we can go bacwards in the sequence. Therefore we must come bac to the beginning of the sequence, thus we find that F n mod j) must be purely periodic. Now that we now it exists, we are ready to define exactly what is meant by the period modulo j. Definition The period of the Fibonacci sequence modulo a positive integer j is the smallest positive integer m such that F m 0 mod j) F m+1 1 mod j). Note that since F n mod j) is purely periodic, if m is the period of F n mod j), then every m-th member of the sequence modulo j must come bac to the starting point. By the definition above, the only members that can possibly come bac to the starting point are multiples of m. This can be summed up in the statement that if m is the period of F n mod j), then for any Z { F 0 mod j) m. 4) F +1 1 mod j) The statement above the corollary of the following theorem will be crucial in proving almost every theorem regarding the period modulo j. Theorem. Let p be a prime let n be a positive integer. If a 1 mod p) then a pn 1 mod p n+1 ). Proof. Suppose that a 1 mod p), let Pn) be the statement that a pn 1 mod p n+1 ). Since a 1 mod p), we now that a 1 + rp for some r Z. Since p rp) i for i p we have that a p 1 + rp) p 1 + ) p rp + 1 p ) p 1 p ) p rp) i 1 + rp + rp) i 1 mod p ) i i i i

4 Thus P1) is true. Now assume that Pn) is true up to some n. So a pn 1 mod p n+1 ) which implies that a pn 1 + sp n+1 for some s Z. Now Consider Pn + 1). ) p p ) p a pn+1 a pn ) p 1+sp n+1 ) p 1+ sp n+1 + sp n+1 ) i 1+sp n+ + 1 i i since p n+ sp n+1 ) i for i p we have that a pn sp n+ + Therefore Pn) is true by induction. p i ) p sp n+1 ) i 1 mod p n+ ). i p i ) p sp n+1 ) i i Corollary 1. Let p be a prime let be a positive integer. If m is the period of F n mod p), then φ mp 1 φ mp 1 1 mod p ). Proof. Since we have F m φm φ m 0 mod p) φ m φ m mod p). F m F m+1 F 1 φm+1 φ m+1 φ φ φm+1 φ m+1 φ + φ φφm 1) φ φ m 1). Substituting φ m for φ m we get F m φφm 1) φ φ m 1) φφm 1) φφ m 1) φm 1)φ φ) Thus φ m 1)F 1 φ m 1) 0 mod p). φ m φ m 1 mod p),

5 so we can apply Theorem to φ m φ m to conclude that φ m ) p 1 φ m ) p 1 1 mod p ). We first investigate the period modulo a positive integer j in order to show that it is dependent on the period modulo the prime powers that divide j. Theorem 3. Let j be a positive integer with j p i i, for p i a prime, let m i denote the period of F n mod p i i ). If m is the period of F n mod j), then Proof. Let j i1 p i i i1 m lcmm 1, m,..., m s ). be a positive integer, let m be the period of F n mod j), let m i be the period of F n mod p i i ). So { F m 0 mod j) F m+1 1 mod j). Applying the Chinese Remainder Theorem to these congruences we have that { F mi 0 mod p i i ) i. F mi +1 1 mod p i i ) By 4) we also now that for r Z { F rmi 0 mod p i i ) F rmi +1 1 mod p i i ) i. Letting r m 1 m i 1 m i+1 m s we have that m rm i F m 0 mod F m +1 1 mod i1 p i i ) i1 m i p i i ) i1 i.

6 But this is not the period because lcmm 1, m,..., m s ) m m, by 4) we have that F m 0 mod p i i ) i1 i. F m +1 1 mod p i i ) Therefore since m is the least positive integer satisfying the above, by definition m m lcmm 1, m,..., m s ). i1 Because the period modulo j depends on the prime powers dividing j, we are motivated to investigate the period modulo a prime power. To facilitate this, we need to loo at the period modulo a prime p. In order to do this we must first determine the conditions under which we can apply Fermat s Little Theorem to φ, φ. The following theorem determines the primes for which φ φ are elements of the field Z/pZ which we denote F p. Theorem 4. Let p be a prime. Then is a quadratic residue modulo p if only if p ±1 mod ), is a quadratic nonresidue modulo p if only if p ± mod ). Proof. Let p be a prime. Utilizing Quadratic Reciprocity we have ) p 1) p ) 1 ) p 1 ) p ) ) 1) p 1 p ) ) 1 1 p 1 mod ) ) p ) 1 1 mod ) p mod ) ) mod ) p 3 mod ) ) ) 4 1 p 4 mod )

7 Therefore is a quadratic residue modulo p if only if p ±1 mod ), is a quadratic nonresidue modulo p if only if p ± mod ). By the above we conclude that for p ±1 mod ), φ φ are elements of F p for p ± mod ), φ φ are not elements of F p. In this second case we need to find an appropriate extension field of F p that contains both φ φ. The following theorems contain results regarding the period modulo p, a prime, Theorem. Let p be a prime m be the period of F n mod p). If p ±1 mod ), then m p 1. Proof. Let p be a prime, let m be the period of F n modulo p, suppose that p ±1 mod. Theorem 4 tells us that is a quadratic residue for this choice of p, so we conclude that φ, φ are elements of F p. Applying Fermat s Little Theorem, we have φ p 1 1 mod p) φp 1 1 mod p). So F p 1 1 φ p 1 φ p 1) 1 1 1) 0 mod p) F p 1 φ p φ p) 1 φ φ) mod p) ) 1 mod p) 1 mod p). Therefore we have that { F p 1 0 mod p) so by 4) we have that m p 1. F p 1 mod p),

8 Theorem 6. Let p be a prime m be the period of F n mod p). If p ± mod ), then m p +, with p+ odd. m Proof. Let p be a prime, let m be the period of F n mod p, suppose that p ± mod. The Legendre symbol satisfies ) a a p 1 mod p). p For our choice of p a, Theorem tells us that p 1 mod p) 1, thus φ φ are not in F p. Therefore we wor in the splitting field for x over F p, F p ) {a + b } a, b F p. Since this field has characteristic p p 1 mod p) 1, we have φ p 1 + ) p φ. ) p ) p 1 + ) p p) 1 1 ) 1 ) p ) 1 + p 1 Similarly φ p 1 ) p 1 ) p ) p 1 ) p 1 φ. ) p 1 1 p) ) 1 + ) 1 p 1 Therefore it is easily seen that F p φp φ p φ φ p 1) mod p),

9 F p+1 φp+1 p+1 φ φp φ φ p φ φφ φ φ 0 mod p), F p+ F p + F p+1 p 1) mod p). Therefore, for our choice of p, we see that m does not divide p + 1. It is seen that F p+1 φp+1 φ p+1 φp ) φ φp ) φ mod p) φ φ φ φ ) φ φ φ φ mod p) 1) F 1 ) 1) 1) mod p) 1 mod p, F p+ φp+ φ p+ φp ) φ φp ) φ φ φ φ φ 0 mod p), F p+3 F p+1 + F p+ 1 mod p). Therefore, since { F p+ 0 mod p) F p+3 1 mod p), by 4) we have that m p +, since m does not divide p + 1, p+ m must be odd. Now that we have results regarding the period of F n mod p) we are ready to investigate the period modulo a prime power, i.e. the period of F n mod p ).

10 Theorem 7. Let p be a prime, let m denote the period of F n modulo p, let m denote the period of F n modulo p. Then m mp 1. Proof. By Corollary 1 we now that φ mp 1 φ mp 1 1 mod p ). Therefore φ mp 1 φmp 1 F mp mod p ) F mp 1 +1 φmp 1 +1 φ mp 1 +1 φmp 1 )φ φ mp 1 ) φ φ φ 1 mod p ) Therefore by 4) we have that m mp 1. In every nown case of the above theorem m mp 1, but there are thought to be infinitely many primes for which this is not true. Since we now have results regarding the period of F n mod p ) we can apply Theorem 3 to tell us about the period modulo the product of prime powers. Therefore given a positive integer j we can determine the possibilities for the period of F n mod j). That is, given j a positive integer m the period modulo j, we can apply Theorems 3,, 6, 7 to tell us that m divides a certain number t, but we can t be certain if t equals m or not. We now turn our attention to another sequence defined by a recursion relation that is closely related to the Fibonacci sequence, the Lucas sequence. Definition The Lucas sequence is a linear recursion defined by L n+1 L n 1 + L n, for n 1, where L n is the nth Lucas number with L 0 L 1 1. Lie the Fibonacci numbers, there is a closed form equation that maes calculating the Lucas numbers easier. Theorem 8. L n φ n + φ n)

11 Proof. Using induction, let Pn) be the statement L n φ n + φ n). P0) is true since P1) is true as L 0 φ 0 + φ L 1 φ + φ Now assume that Pn) is true for some n N consider Pn + 1). L n+1 φ n+1 + φ n+1 φ n 1 φ ) + φ ) n 1 φ φ n φ) + φ n φ ) φ n 1 + φ n + φ n 1 + φ n φ n 1 + φ n 1) + φ n + φ n) L n 1 + L n Therefore Pn) is true for all n. Lie the Fibonacci sequence, when we investigate the Lucas sequence modulo j an integer, we find that the new sequence is purely periodic for the same reasons as Fibonacci sequence. Definition The period of the Lucas sequence modulo a positive integer j is the smallest positive integer l such that L l mod j) L l+1 1 mod j). Again, for the same reasons as the Fibonacci sequence we have that if l is the period of L n mod j) then { L mod j) l ) L +1 1 mod j) Theorem 9. For p be a prime, let l be the period of L n mod p). If p ±1mod ), then l p 1.

12 Proof. Let p be a prime, let l be the period of L n mod p suppose that p ±1 mod ). Since p ±1 mod ) we now that φ, φ F p thus L p 1 φ p 1 + φ p mod p mod p) L p φ p + φ p φ + φ 1 + Therefore by ) we have that l p mod p). Theorem 10. For p be a prime, let l be the period of L n mod p). If p ± mod ), then l p +. Proof. Let p be a prime, let l be the period of L n mod p) suppose that p ± mod ). Since p ± mod ) we now that φ p φ mod p) φ p φ mod p). Thus we have L p+ φ p+ + φ p+ φ p ) φ + φ p ) φ mod p) φ φ + φ φ φ φ) 1) mod p) mod p) L p+3 φ p+3 + φ p+3 φ p ) φ 3 + φ p ) φ3 mod p) φ φ 3 + φ φ3 φ φ) φ + φ) 1) 1) mod p) 1 modp). Therefore by ) we have that l p +. References [1] Gallian, Joe A. Contemporary Abstract Algebra. Sixth Edition. Houghton Mifflin [] Niven, Zucerman, Montegomery. An Introduction to the Theory of Numbers. Fifth Edition. John Wiley Sons, Inc