Characteristics of Fibonacci-tye Sequences Yarden Blausa May 018 Abstract This aer resents an exloration of the Fibonacci sequence, as well as multi-nacci sequences and the Lucas sequence. We comare and contrast various characteristics of these sequences, in articular the existence and reetition of rime factors. We show that for the Fibonacci sequence, and for multi-nacci sequences with the same initial conditions, it follows that every rime divides an infinite number of terms of the sequence. By contrast, we show that this is not the case for the Lucas numbers. We rovide conditions for when a rime does divide a Lucas number and give some examles of rimes that do not divide any Lucas number. 1 Introduction The Fibonacci sequence is a famous sequence of integers both in mathematics and in oular culture. It was introduced to the Latin-seaking world in 10, in Fibonacci s Liber Abaci. Fibonacci, or, Leonardo Pisano, was an Italian mathematician born in 117. He grew u traveling with his merchant father and was exosed to the Hindu- Arabic arithmetic system in North Africa. He ublished Liber Abaci in 10, which introduced the Latin-seaking world to the decimal system. In Liber Abaci, he also introduced the Fibonacci sequence: How many airs of rabbits can be bred in one year from one air? A certain erson laces one air of rabbits in a certain lace surrounded on all sides by a wall. We want to know how many airs can be bred from that air in one year, assuming it is their nature that each month they give birth to another air, and in the second month, each new air can also breed (see [3]. This situation can be reresented with the rule, F n = F n 1 + F n, for n with F 0 = 0 and F 1 = 1. The first few terms of the Fibonacci sequence are: 0, 1, 1,, 3,, 8, 13, 1, 34,,... 1
Figure 1: The golden siral The sequence had been noted by Indian mathematicians as early as the sixth century, but Fibonacci was the first to ublish it outside of India. The Fibonacci numbers are also known for aearing in nature. They aear in the form of a siral constructed from quarter circles drawn inside an array of squares with the Fibonacci numbers as dimensions. The Fibonacci siral, sometimes called the golden siral, can be seen in sunflower seeds, hurricanes, and galaxies (see []. Preliminaries All of the following results can be found in [4], along with the roofs. Fermat s Little Theorem is a well-known and useful theorem. It is remarkably short and sweet. We include it here for reference. Theorem.1 (Fermat s Little Theorem. Suose that is a rime and a is an integer. Then (a a 1 1 (mod if a and are relatively rime; (b a a (mod for any a. Another concet we use in our roofs are quadratic residues, which take the idea of erfect squares and extend them to saces Z, where is an odd rime. Definition.1. Suose that is an odd rime and that b and are relatively rime. Then b is a quadratic residue modulo if and only if the equation x b (mod has a solution. If the equation has no solution, then we say that b is a quadratic nonresidue modulo. This definition tells us that if b is a quadratic residue modulo, then it is a erfect square in Z. We want to be able to calculate when a number is a quadratic residue modulo. This is easy with small values of simly by
squaring all the elements in Z, but for larger values it is useful to have an arithmetic equation. The notation that we ( use to say whether b is a quadratic residue modulo is the Legendre symbol,, where ( { b 1 b is a quadratic residue mod = 1 b is a quadratic nonresidue mod. ( Euler s Criterion gives us a way to calculate Theorem. (Euler s Criterion. Suose that is an odd rime and that b is an integer. If and b are relatively rime, then ( b = b ( 1/. To reduce the work of checking quadratic residues of larger rimes, we use the following result. Theorem.3 (Quadratic Recirocity Theorem. If and q are distinct odd rimes, then ( ( q ( 1 1 ( 1 1 (q 1 (mod. q The following corollary hels to further break down large numbers. Corollary.3.1. Suose that is an odd rime and that n = are relatively rime, then ( n = k i=1 b ( bi. b. k b i. If n and Finally, we have the secial case of determining whether is a quadratic residue modulo. Theorem.4. If is an odd rime then, ( = { 1, ±1 (mod 8; 1 ±3 (mod 8. To see how the Quadratic Recirocity Theorem hels with large rimes, suose we want to determine whether 607 is a quadratic residue of 03. Using the above results we obtain, ( 607 03 = ( 03 607 = ( 7 607 ( ( 3 = = 607 607 i=1 ( 607 = 3 ( 1 = 1. 3 With these theorems in mind we continue to an exloration of the Fibonacci numbers. 3
3 Fibonacci numbers Recall that the Fibonacci sequence is a recursive sequence where each term is defined by F n = F n 1 + F n for n with the initial condition F 0 = 0 and F 1 = 1. It is useful to be able to find a closed-form formula that gives the nth term of the Fibonacci sequence so that we do not have to recursively generate every term rior to the term that we want to calculate. We do this by solving a second order linear difference equation. Taking our recursive Fibonacci equation and transferring all the terms to the left-hand side gives F n F n 1 F n = 0. We assume that r n solves the equation, so and thus r n r n 1 r n = 0 r r 1 = 0, which is our characteristic equation. By solving the characteristic equation, the two roots are r 1 = 1 + := φ and r = 1 := 1 φ, where φ is the golden ratio. For any c 1 substituting c 1 r1 n for F n in F n F n 1 F n yields zero and for any c substituting c r n for F n in F n F n 1 F n yields zero. This suggests our solution has the form: Using our values for r 1 and r, we now have F n = c 1 r n 1 + c r n, (1 F n = c 1 φ n + c (1 φ n. Incororating the initial condition of F 0 = 0 means that c = c 1 so we have F n = c 1 (φ n (1 φ n. Incororating the initial condition F 1 = 1, we obtain 1 = c 1 (φ (1 φ = c 1 (φ 1 1 c 1 = φ 1 = 1. 4
So our final closed-form formula for the Fibonacci sequence is: F n = φn (1 φ n. ( Now that we have the closed-form formula for the Fibonacci sequence, we move on to an investigation of its rime factors (see [1]. 3.1 Prime factors An interesting question to consider is when do rimes show u as factors of terms in the Fibonacci-tye sequences? Do we eventually get every rime? It turns out every rime divides a Fibonacci number, which is a somewhat surrising feature of the Fibonacci numbers. Looking at the rime factors of some Fibonacci numbers we begin to notice a attern: n F n (factored 8 3 7 10 11 14 13 9 18 3 17 19 4 3 7 3 8 3 13 9 81 30 3 11 31 61 It aears that for rimes of the form = 10k ± 1, it follows that F 1 and for rimes of the form = 10k ± 3, it follows that F +1. To show that this is always true, we first look at some secific examles. First we consider F 1 for = 31. We do this by finding a formula for F n in Z 31. We want to find solutions to the characteristic equation x = x + 1 in Z 31. By multilying both sides by 4, we obtain 4x = 4x + 4 4x 4x + 1 = (x 1 =. We can verify that a solution exists by checking that is a quadratic residue modulo 31. This is obvious since 6 (mod 31. Performing some algebra in Z 31, we get (x 1 = 6 so x 1 ± or x 1 ±. Our roots are therefore α = 13 and β = 1 19. Putting these values back into Equation (1 yields F n c 1 13 n + c 19 n (mod 31.
Using our initial condition F 0 = 0 gives us c 1 = c or F n = c 1 (13 n 19 n. Using the initial condition F 1 = 1 gives c 1 ( 6 = 1 or c 1 =. We now have F n (13 n 19 n (mod 31. Using n = 30 and Fermat s Little Theorem we see that F 30 (13 30 19 30 (1 1 0 (mod 31. Since F 30 0 (mod 31, this shows that 31 divides F 30. The following theorem generalizes this result for all rimes of the form = 10k ± 1. Theorem 3.1. If is a rime of the form = 10k ± 1, then F 1. Proof. Suose is a rime such that = 10k ± 1. We want to find a solution to the characteristic equation x = x + 1 in Z. We can write x = x + 1 as (x 1 =. To show that this has a solution in Z, we need to be a quadratic residue modulo. Note that by the Quadratic Recirocity Theorem, when = 10k + 1, and when = 10k 1, ( ( = = ( ( = = ( 1 = 1 ( 1 = ( 4 = 1. In either case, is a quadratic residue modulo so a solution to (x 1 = exists in Z. There are two solutions to y (mod, namely y and y y. Note that one of these must be odd. Let y be the odd solution. Solving the equation (x 1 = y yields x = y + 1 := α and x = 1 y := β. Note that y = α β. We ut these values back into our general Fibonacci equation with initial condition F 0 = 0 to obtain F n c 1 (α n β n (mod. Using the initial condition F 1 = 1 gives 1 = c 1 (α β or c 1 = (α β 1 = y 1. We now have F n y 1 (α n β n (mod. Using n = 1 and Fermat s Little Theorem we see that F 1 y 1 (α 1 β 1 y 1 (1 1 0 (mod. Since F 1 0 (mod, it follows that divides F 1. We have now covered one of two cases of rimes. We now look at a secific examle of a rime of the form = 10k ± 3 for = 13. 6
Note that is not a quadratic residue modulo 13, so there is no solution to x = x + 1 in Z 13. Since φ is a root of x = x + 1 in R, we must extend the field Z 13 to Z 13 (φ = {a + bφ a, b Z 13 }. We must verify that Z 13 (φ is a field. It is easy to check that Z 13 satisfies the usual roerties for addition and multilication, excet erhas the existence of multilicative inverses. We must show that each nonzero element in Z 13 (φ has an inverse in Z 13 (φ. Each element in Z 13 (φ has the form a + bφ and we need to find an inverse of the form c + dφ such that (a + bφ(c + dφ = 1. Recall that since φ solves the characteristic equation, then φ = φ + 1. Multilying out the left side of the equation we get (a + bφ(c + dφ = ac + bdφ + (ad + bcφ = ac + bd(φ + 1 + (ad + bcφ = ac + bd + (ad + bc + bdφ. For this to equate to 1 + 0φ we need to solve the system of equations: ac + bd = 1 bc + (a + bd = 0 We solve these two equations for c and d using Cramer s Rule: 1 b 0 a + b c = a a = (a + b(a + ab b 1 ; b a + b a 1 b 0 d = a a = b(a + ab b 1. b a + b Therefore for c + dφ to be the inverse of a + bφ we need (a + ab b 1 to always exist. We show that it must exist by way of contradiction. Suose that a + ab b does not have an inverse. This occurs if and only if a + ab b = 0 which means (a + b = b. However, since is not a quadratic residue modulo 13, this is a contradiction. Therefore (a + ab b 1 exists and each nonzero element of Z 13 (φ has an inverse in Z 13 (φ. Lemma 3.. For all n Z +, it follows that φ n = F n 1 + F n φ. Proof. We rove this by induction over n. For our base case note that φ = F 0 +F 1 φ = 0+φ. Suose that φ n = F n 1 +F n φ 7
for some n. Then, φ n+1 = F n 1 φ + F n φ = F n 1 φ + F n (1 + φ = F n 1 φ + F n + F n φ = F n + (F n 1 + F n φ = F n + F n+1 φ. Therefore by the Princile of Mathematical Induction, φ n = F n 1 + F n φ for all n Z +. So by Lemma 3., φ 13 = F 1 + F 13 φ = 144 + 33φ 1 φ (mod 13. Since φ 13 F 1 + F 13 φ, it follows that F 1 1 and F 13 1. F 14 F 13 + F 1 0 (mod 13. Theorem 3.3. If is a rime of the form = 10k ± 3, then F +1. Therefore, Proof. Suose is a rime of the form = 10k ± 3. We want to find a solution to x = x + 1 in Z. We can write x = x + 1 as (x 1 =. First we will show that there is no solution to (x 1 = in Z. Noting that, ( ( ( when = 10k 3, = = = 1 and when = 10k + 3, ( ( = = ( 3 = 1, we find that is not a quadratic residue modulo, and hence, (x 1 = will not have a solution in Z, therefore we must extend our field to Z (φ = {a + bφ a, b Z }. In our secific examle of = 13, we already showed that Z 13 (φ is a field. The only necessary condition for this argument was that was not a quadratic residue modulo 13. Since is not a quadratic residue modulo, by the same argument it follows that Z (φ is a field with a generic of the form 10k ± 3. We now rewrite ( in Z (φ using inverses instead of division and using only numbers in Z (φ. Since φ = 1+, then = φ 1. We now have the formula: Now note that ( 1 + φf n (φ n (1 φ n (mod. (φ = φ = (φ = (φ + 1, 8
and (φ + 1 = k=0 ( φ k = 1 + k ( φ + 1 ( φ +... + φ. Since ( k for 1 k 1 and we are working in Z (φ, it follows that all of the terms besides 1 and φ disaear, so (φ + 1 φ + 1. It follows that (φ = φ + 1. Therefore, φ is a solution to x = x + 1. However, we already know that the only solutions to x = x + 1 are φ and 1 φ, so φ must be equivalent to one of them. Suose φ φ. Then φ φ 0, so φ is a solution to x x = 0. We know that olynomials of degree have at most roots in a field. Since x x (mod for all x Z by Fermat s Little Theorem, so we already have all roots. Therefore, φ is not a root and φ φ. It follows that φ 1 φ. Now note that (1 φ 1 φ 1 (1 φ φ. Using our substitutions of φ 1 φ and (1 φ φ, we obtain It follows that divides F +1. ( 1 + φf +1 = (φ +1 (1 φ +1 = (φ φ (1 φ (1 φ = ((1 φ φ φ (1 φ = 0. Since all rimes besides and are of the form 10k ± 1 or 10k ± 3, it follows that all rimes divide a Fibonacci number. When we look at the rime factorizations of the Fibonacci terms we see that while the above theorems tell us when a rime divides a Fibonacci number, they do not necessarily tell us the first time that a rime divides a Fibonacci number. For examle, 17 = 0 3, so it is a rime of the form = 10k ± 3. It is true that 17 F 18 but also that 17 F 9. Note that 18 is a multile of 9. The following lemmas bring us to the result that F n if and only if z( n, where z( is the smallest ositive index such that F z(. From here on we will define a sequence S n such that, S n = S n 1 + qs n, where q is some integer and the initial conditions are S 0 = 0 and S 1 = 1. Lemma 3.4. If n and r are ositive integers, it follows that S n+r = S r+1 S n + qs r S n 1. 9
Proof. We will rove this using strong induction over r. We first show that the equality is valid for r = 1. Let n Z +. Note that S n+1 = S S n + qs 1 S n 1 = S n + qs n 1 which is how we define our Fibonacci-tye sequences, so the equation is valid for r = 1. Now suose that the equality is valid for all integers u to r. That is, assume S n+r = S r+1 S n + qs r S n 1. Noting that S n+r+1 = S n+r + qs n+r 1 = S r+1 S n + qs r S n 1 + q(s r S n + qs r 1 S n 1 = S n (S r+1 + qs r + qs n 1 (S r + qs r 1 = S r+ S n + qs r+1 S n 1, we find that the equation is valid for r +1. Therefore, by the Princile of Strong Induction, S n+r = S r+1 S n + qs r S n 1 is valid for all r Z +. Lemma 3.. For each ositive integer n, we find that S n S kn for all n, k Z +. Proof. We rove this using induction over k. For our base case note that by Lemma 3.4, we have S n = S n+n = S n+1 S n + qs n S n 1 = S n (S n+1 + qs n 1, so S n S n. Now we assume that S n S kn for some k and note that S (k+1n = S n+kn = S kn+1 S n + qs kn S n 1. Since S n S kn and S n S n, we see that S n S (k+1n. By the Princile of Mathematical Induction, it follows that S n S kn for all n, k Z +. Lemma 3.6. Every air of consecutive Fibonacci numbers F n, F n+1 are relatively rime. Proof. We rove this using induction. For our base case when n = 1 note that gcd(f 1, F = gcd(1, 1 = 1. Now we assume that gcd(f n, F n+1 = 1 for some n Z + and show that gcd(f n+1, F n+ = 1. Since gcd(f n, F n+1 = 1, there exist integers x and y such that F n x + F n+1 y = 1. Also note that since F n+ = F n + F n+1, then F n = F n+ F n+1. Making this substitution for F n we have 1 = (F n+ F n+1 x + F n+1 y = F n+ x F n+1 x + F n+1 y = F n+ x + F n+1 (y x. Since y x is an integer, F n+1 and F n+ are relatively rime. By the Princile of Mathematical Induction, every air of consecutive Fibonacci numbers are relatively rime. Theorem 3.7. Let be a rime. Then F n if and only if z( n, where z( is the smallest ositive index such that F z(. Proof. The existence of z( is guaranteed by the well-ordering roerty, since we know that the set {n : n Z + and F n } is nonemty for each rime. 10
Suose that z( n. Then F n easily follows from Lemma 3.. Now suose that F n.by the Division Algorithm, there exist integers q and r such that n = z(q + r and 0 r < z(. Note that by Lemma 3.4, F n = F z(q+r = F r+1 F z(q + F r F z(q 1. We know that F n and also that F z(q by Lemma 3., so it must be that F r F z(q 1. By Lemma 3.6 we know that does not divide F z(q 1, since the Fibonacci numbers F z(q and F z(q 1 are consecutive. Therefore F r. Since 0 r < z( and z( is the smallest ositive integer such that F z (, it must be that r = 0. It follows that n = z(q, so z( n. We have now shown that not only does every rime divide a Fibonacci number, but that every rime does so an infinite number of times. 3. Periodicity We now know that in our Fibonacci sequence modulo, the zeros reeat at a certain interval. However, this is not necessarily the same interval that the entire sequence reeats. For examle, take the terms of F n (mod 7: 1, 1,, 3,, 1, 6, 0, 6, 6,, 4,, 6, 1, 0, 1, 1,, 3,... We see that a zero aears after just 8 terms, but the entire sequence doesn t reeat until 16 terms. This motivates the following two theorems, which give the eriod of a Fibonacci sequence modulo. Theorem 3.8. If is a rime of the form 10k ± 1, then the Fibonacci sequence modulo reeats after 1 terms. Proof. Once again we use the equation F n y 1 (α n β n (mod in Z. By Fermat s Little Theorem, F n+( 1 y 1 (α n+( 1 β n+( 1 y 1 (α n α 1 β n β 1 y 1 (α n β n F n (mod. Since F n+( 1 F n (mod, the sequence reeats after 1 terms. Theorem 3.9. If is a rime of the form 10k ± 3, where k Z +, then the Fibonacci sequence modulo reeats after ( + 1 terms. Proof. We use the equation F n = (φ n (1 φ n ( 1+φ 1 (mod in Z (φ, along with the equivalences φ 1 φ and (1 φ φ to obtain: φ 1 φ φ +1 φ 1 φ (+1 1, 11
and, Using these substitutions, note that: (1 φ φ (1 φ +1 1 (1 φ (+1 1. F n+(+1 (φ n+(+1 (1 φ n+(+1 ( 1 + φ 1 (φ n φ (+1 (1 φ n (1 φ (+1 ( 1 + φ 1 (φ n (1 φ n ( 1 + φ 1 F n (mod. Since F n+(+1 F n (mod, the sequence reeats after ( + 1 terms. Note that the above theorems give a eriod in which the Fibonacci sequence modulo will reeat, but it is not necessarily the shortest eriod. 4 Multi-nacci numbers The Fibonacci sequence was considered in a math roblem where a mature air of rabbits bears one air of baby rabbits er month. Imagine if instead, a mature air of rabbits bears two airs of baby rabbits er month. Then each month, we would count the number of airs by the number of airs from the last month in addition to two airs for each air from two months earlier that has now matured. We call this sequence the Beta-nacci sequence and define it recursively by B n = B n 1 + B n where n with B 0 = 0 and B 1 = 1. Note that this is a version of our generic multi-nacci sequence, S n = S n 1 + qs n. The first seven terms in some of these multi-nacci sequences is given in Table 1 (see []. As with the Fibonacci sequence, we want to determine if every rime divides a term of a multi-nacci sequence. We take the Gamma-nacci sequence as an examle. The Gamma-nacci sequence is defined by G n = G n 1 + 3G n. To rove that a rime divides the G 1 term, we need a solution to x x 3 = 0 in Z, or, (x 1 = 13. For this to have a solution, we need 13 to be a quadratic residue in Z. By the Division Algorithm we can write = 13q + r 1
n Fibonacci β γ δ ɛ ζ n F n B n G n D n E n Z n 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 4 6 7 4 3 7 9 11 13 11 19 9 41 6 8 1 40 6 96 133 7 13 43 97 181 301 463 Table 1: The first 7 terms of some multi-nacci sequences ( 13 where 0 r < 13. Using Euler s Criterion, we want = 1. By the Quadratic Recirocity Theorem, ( 13 ( = ( 1 1 13 1 13 ( = ( 1 3( 1 ( 13 13 = = ( r. 13 Therefore we just need r to be a quadratic residue modulo 13. The quadratic residues in Z 13 are r = 1, 3, 4, 9, 10, 1. So for rimes of the form 13k ± 1, 13k ± 3, and 13k ± 4, will divide the G 1 term. For other rimes, it follows that r will not be a quadratic residue modulo. This is all we need for the roof that divides the G +1 term. The following theorem generalizes the roof that G +1 to all multi-nacci sequences. Theorem 4.1. For sequences of the form S n = S n 1 +qs n, let b = 4q+1. For all r such that r is a quadratic residue modulo b, rimes of the form = bk + r divides the S 1 and for all other rimes > 4q + 1, it follows that divides the S +1 term. A roof for Theorem 4.1 is identical to the roofs of Theorems 3.1 and 3.3 where b =. Two secial cases of the rime division are for the Beta-nacci sequence defined by B n = B n 1 + B n, and the Zeta-nacci sequence defined by Z n = Z n 1 + 6Z n. 13
The reason that these two sequences are unique is that their characteristic equations are r r = 0 and r r 6 = 0 resectively, which each have two integer roots. This means that for > 3 and > resectively, these equations always have a solution in Z, so for all > 3 and > resectively, it is true that B 1 and Z 1. We can also see this by using the closed form formulas for the Beta-nacci and Zeta-nacci sequences. The closed form formula for the Beta-nacci sequence is: B n = n ( 1 n. 3 When n = 1, Fermat s Little Theorem yields 3B 1 1 ( 1 1 1 1 0, which shows that B 1. The closed form formula for the Zeta-nacci sequence is: Z n = 3n ( n. When n = 1, Fermat s Little Theorem once again yields Z 1 = 3 1 ( 1 = 3 1 1 1 1 0 (mod, which shows that Z 1. Lucas numbers A sequence related to the Fibonacci sequence is the Lucas sequence. It is another sequence defined recursively where each term is given by L n = L n 1 + L n but the initial conditions are L 0 = and L 1 = 1. The first few terms of the Lucas Sequence are, 1, 3, 4, 7, 11, 18, 9, 47, 76,... Note that the only differences between the Lucas and Fibonacci sequences are the initial conditions, in articular the initial condition on the 0 term. Looking at the rime factorizations of the Lucas numbers, it aears that some rimes divide L ( 1/, some divide L (+1/, and some do not divide any Lucas numbers. The following theorems tell us when some rimes divide a Lucas number. Theorem.1. If is a rime of the form 0k +11 or 0k +19, then L ( 1/. 14
Proof. The conditions for a rime to divide the L ( 1/ term of the Lucas sequence are that is a quadratic residue modulo and that 1 is a quadratic nonresidue modulo. Note that 1 is a quadratic nonresidue modulo when = 4k + 3 because by Euler s Criterion, ( 1 ( 1 1 1, since 1 is odd for = 4k + 3. To add in the condition that is a quadratic residue modulo, we want 4k + 3 1 (mod or 4k + 3 4 (mod. So, we have: or, 4k + 3 1 (mod 4k k, 4k + 3 4 (mod k 1 k 4. So k = j + or k = m + 4 where k, m Z. We now have or, = 4(j + + 3 = 0j + 11, = 4(j + 4 + 3 = 0j + 19. To get a formula for L n in Z we once again solve the characteristic equation x x 1 = 0 or (x 1 =. Solving the equation (x 1 = y and choosing the odd solution y, once again yields x = 1+y, 1 y. The general solution for our Lucas number formula in Z then looks like ( n ( n 1 + y 1 y L n = c 1 + c. To find our constants, we use the initial conditions L 0 = and L 1 = 1 to obtain the system of equations = c 1 + c 1 = c 1 ( 1 + y ( 1 y + c, which yield c 1 = 1 and c = 1. We now have ( n ( n 1 + y 1 y L n = +. 1
Using n = ( 1/ we have ( ( 1/ ( ( 1/ 1 + y 1 y L ( 1/ = +. Note that by Euler s Criterion, this is equivalent to ( ( 1+y 1 y L ( 1/ = +. Since 1 is a quadratic nonresidue modulo, we see that ( ( ( 1+y 1 y 1 y ( 1 ( 4 4 1 = = = = 1. ( 1+y Therefore, and L ( 1/ 0, as required. ( 1 y must be of oosite signs and it follows that Theorem.. If is a rime of the form 0k + 3 or 0k + 7, then L (+1/. Proof. Suose is a rime such that = 0k + 3 or = 0k + 7. To get a formula for L n in Z, we want a solution to the characteristic x = x + 1 in Z. First we show that there is no solution to (x 1 = in Z. Note that, ( ( ( 3 when = 0k + 3, = = = 1 and when = 0k + 7, ( ( = = ( = 1. In either case, is not a quadratic residue modulo. We therefore have to do a field extension like we did with the Fibonacci sequence. Our formula for L n in Z (φ is L n φ n + (1 φ n (mod. When we substitute n = ( + 1/, which is an even number, and use the identities φ 1 φ and (1 φ φ we obtain: L (+1/ φ (+1/ + (1 φ (+1/ L (+1/ (φ(+1/ + (1 φ (+1/ φ +1 + (φ (+1/ (1 φ (+1/ + (1 φ +1 φ φ + (φ φ (+1/ + (1 φ(1 φ φ (1 φ + + (1 φ φ 1 + 1 0 It follows that divides L (+1/. (mod. 16
Although we have just roved two cases of rime division in the Lucas numbers, not all rimes divide a Lucas number. Since the only thing different about the Lucas sequence is the initial conditions, the sequence has the same eriod modulo as the Fibonacci sequence. That is, for rimes of the form 10k ± 1 the eriod is 1 and for rimes of the form 10k ± 3 the eriod is ( + 1. If divides a Lucas number we will see a 0 in one eriod of the Lucas sequence modulo. To show that not all rimes divide a Lucas number we only need to show an examle of a rime that never divides a Lucas number. We take = and note that the eriod is 4. In one eriod of the Lucas sequence modulo, the terms are, 1, 3, 4,... Since there is no 0 in this eriod, which reeats for the entire sequence, then does not divide a Lucas number. For our next examle we use = 13, which is a rime of the form 10k ± 3. Therefore its eriod is ( + 1 = 8, so we only need to look at 8 terms of the Lucas sequence modulo 13. Observe that since F n = (φ n (1 φ n ( 1 + φ 1 and L n = φ n + (1 φ n, it follows that F n = F n L n. Therefore if 13 L n, then 13 F n. We know that F 7 = 13, so 7 n, by Theorem 3.7. Since 7 does not divide, 7 divides n. Therefore we only need to look at L n (mod 13 u to 8 where n is a multile of 7. If we check these terms we see that L 7 3, L 14 11, and L 1 10. Since none of these values are 0, it follows that 13 does not divide a Lucas number. Similarly for 17, we only need to check L 9, L 18, and L 7 in Z 17. Since L 9 8, L 18 1, and L 7 9, it follows that 17 does not divide a Lucas number. The fact that not all rimes divide a Lucas number suggests that there is something unique about the Fibonacci sequence and other multi-nacci sequences with the same initial conditions. Particularly, to get the result that all sufficiently large rimes divide every term of a sequence S n, it is necessary that S 0 = 0. The following fact is another interesting characteristic of the Lucas numbers. Theorem.3. For all rimes, L 1 (mod. ( Proof. For the case when 1, we see that in Z, the solutions to the equation (x 1 = are x = 1+y Fermat s Little Theorem,, 1 y L a + b a + b 1 (mod. ( For the case when 1, we see that in Z (φ, 1+y. Let a = and b = 1+y. Then by L φ + (1 φ 1 φ + φ = 1 (mod. 17
For the case when = we see that L = 11 1 (mod. Therefore, for all rimes, it follows that L 1 (mod. The following result combines our knowledge of the Fibonacci and Lucas numbers. Theorem.4. If z( is even, where z( is the smallest integer such that F z(, then L z(/. Proof. Suose z( is even. Then z( = n for some n Z +. Note that F n = F n L n. Since z( is the smallest integer such that F z(, does not divide F n. Therefore L n = L z(/. 6 Conclusion In this aer we exlored characteristics of the Fibonacci numbers and multinacci numbers, contrasted with the Lucas numbers. In articular we focused on rime factors and reetition in these sequences. There are many oortunities for future study on this toic. For examle, we could look at recursive sequences that are non-homogeneous. We also could look at recursive sequences that deend on 3 revious terms rather than two. Another question we left unanswered is whether or not there exists a rime such that z( = z(. We exlored the rime factors of Fibonacci-tye sequences, and when a sequence reeats modulo, but we could also investigate when m divides S n for a generic m that is not necessarily rime. It turns out that when m = q, where and q are rime, then the eriod of the sequence modulo m is the least common multile of and q. This can be used to show when a Fibonacci-tye sequence reeats in the ones digit, since this is equivalent to showing when the sequence reeats modulo 10. Our exloration of Fibonacci-tye sequences has shown that the initial condition S 0 = 0 leads to many interesting results. There is much to be exlored in these numbers and they lead to many questions about other sequences. References [1] Mathematical methods for economic theory. htts://mjo.osborne. economics.utoronto.ca/ index.h/tutorial/index/1/sod/t. [] Elaine J. Hom. What is the fibonacci sequence? htts://www. livescience.com/37470-fibonacci-sequence.html, 018. 18
[3] V. Katz. A history of mathematics, an introduction. Addison-Wesley, second edition, 1998. [4] Patrick Keef and David Guichard. An Introduction to Higher Mathematics. Whitman College. [] Shari Lynn Levine. Suose more rabbits are born. The Fibonacci Quarterly, 6:306 311, 1988. 19