Divisibility properties of Fibonacci numbers

Transcription

1 South Asian Journal of Mathematics 2011, Vol. 1 ( 3 ) : ISSN RESEARCH ARTICLE Divisibility properties of Fibonacci numbers K. Raja Rama GANDHI 1 1 Department of Mathematics, Chaitanya Engineering College, Kommadi, Madhrawada, Visakha, India rrmath28@gmail.com Received: ; Accepted: *Corresponding author Abstract The famous mathematical sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,, m, n, m + n,, known as the Fibonacci sequence, has been discovered in many places such as nature, art, and even music. In this paper, we will discuss some of the mathematics behind the sequence and other neat properties of the Fibonacci sequence about modular divisibility s. For the most part, we will discus about 7 F n and other interesting divisibility observations. Key Words MSC 2010 homomorphism, rings, Euclidean algorithm, Bezout identity 05A11 1 Introduction Fibonacci numbers (named after the 13th Century mathematician, Leonardo of Pisa, also called Leonardo Fibonacci) are the elements of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, (1.1) Sometimes this sequence is given as 0, 1, 1, 2, 3, 5, (0 becomes the 0th element of the sequence). Each number is the sum of the two previous numbers. There are other Fibonacci sequences, starting with other numbers: 3, 10, 13, 23, 36, 59, These are sequences of Lucas numbers, which are a kind of generalized version of Fibonacci numbers. It is well known that every third Fibonacci number is even and every fifth Fibonacci number is divisible by 5 (see c.f. [1]) In contrast, some easily derived results may be less well know including no odd Fibonacci number is divisible by primes 17 and 61 and no F n 4( mod 8) or 6( mod 8). We will discuss briefly about 7 F n and others properties (which we given below) by a single proof. The important properties, which we can see in Fibonacci series given below: I. 7 F 8 (1.2) II. For any positive integer k, the following results holds (i) 3 F 4k Citation: K. R. Gandhi, Divisibility properties of Fibonacci numbers, South Asian J Math, 2011, 1(3),

2 South Asian J. Math. Vol. 1 No. 3 (ii) 5 F 5k (1.3) (ii) 8 F 6k (ii) 13 F 7k In this paper, we will discuss some of the mathematics behind the sequence and other neat properties of the Fibonacci sequence about modular divisibility s. For the most part, we will discus about 7 F n and other interesting divisibility observations. 2 A Equivalent Condition of 7 F n Now, we will discuss the property (1.2) above and we will generalize the remaining other properties in a single proof. Look at the below sequence, which F n is reduced to (mod7): 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1 We can notice that, the 8 th and 16 th position of the above reduced series is zero. This proves that 7 F n if only if 8 n. Definition 2.1. A k th order of recurrence relation on some set X is a function a : N X with a 1, a 2,, a k defined for all i > k, a i+k+1 = f(a i, a i+1,, a i+k ). Definition 2.2. Recurrence relation homomorphism, let a n be a k th order recurrence relation on the set X defined by the map f : X k X(with initial values). A map ϕ : X Y is said to be a recurrence relation homomorphism on a, when there exists f : Y k Y satisfying the commutative ϕ f = f ϕ. Theorem 2.3. Suppose we given a recurrence relation homomorphism in the above notation, we claim b n = ϕ(a n ), which is k th order recurrence relation. Proof. For the given k initial values with i > k, we have B i+k+1 = ϕ(f(a i, a i+1,, a i+k )) = f(ϕ(a i ), ϕ(a i+1 ),, ϕ(a i+k ))) = f(b i, b i+1,, b i+k ). We will include an example by discussing the modular arithmetical operations over the image ϕ(z) = Z/3Z and homomorphism of rings. Let Z is a ring and 3Z = 3x Z x Z = {, 6, 3, 0, 3, 6, }. Notice that, all are (mod3) is that 6 = 3 = 0 = 3 = 6 = and = 5 = 2 = 1 = 4 = 6 = and we can proceed further. The first step now is to make an equivalence relation ( ) and this expresses the 0 3, 2 8, 1 5 and this is very easy. Define x y : x + 3z = y + 3z, which we done from ϕ(x) = x + 3z i.e., z = 3 + 3z {, 6, 3, 0, 3, 6, } = {, 3, 0, 3, 6, 9, } true z = 8 + 3z {, 4, 1, 2, 5, 8, } = {, 2, 5, 8, 11, 14, } true. Now, we can define arithmetical operations on ϕ(z) = Z/3Z. (a) ϕ(a) + ϕ(b) = ϕ(a + b) 141

3 L. Yuan: Divisibility properties of Fibonacci numbers (b) ϕ(a) = ϕ( a) (c) ϕ(a).ϕ(b) = ϕ(a.b) To see that, + is actually a function, which is necessary to prove that it respects the equivalence relation in the sense that if ϕ(x) = ϕ(x ) and ϕ(y) = ϕ(y ) then ϕ(x) + ϕ(y) = ϕ(x ) + ϕ(y ) i.e., (x + 3z)(y + 3z) = {, x 6, x 3, x, x + 3, x + 6, } + {, y 6, y 3, y, y + 3, y + 6 }. The same type of calculation proves that negation and multiplication are respectful functions. Since the function respectful of the ring axioms, this proves that Z/3Z is a ring and ϕ is a ring homomorphism. It is very clear that nothing depends on special properties of the number 3 so far and the arguments above are fully general. The standard notation for working in this ring is ϕ(x) ϕ(y), but x y( mod 3), where x is implicitly mapped from Z to Z/3Z if needed. Theorem 2.4. For prime p, every non-zero element of Z/pZ is invertible. Proof. For the number x to be invertible in any ring is that there exists some number x 1 such that x x 1 = 1. In the ring of rationals Q of this number is 1 x (since 1 0) and every non-zero element is invertible. For given (a, b) = 1, there exists x, y such that ax + by = 1 (by Euclidean algorithm), i.e., for (a, b) = 1, x such that ax 1(modb). Note that b is prime every element except 0 is co-prime and this has an inverse. i.e., 1 is not congruent to 0( mod p) this proves Z/pZ is a field. Now, we can compute 2 1 (mod3), it must have the unique number ϕ(x) such that 2x + 3y = 1 x = 2 and y = 1, 2 1 2(mod3). Now we will go back to discuss an example which will tends to our conclusion. Example The reduction(modm) map ϕ : Z Z/mZ as described above is a recurrence homomorphism on any recurrence defined by a polynomial. To understand this, note that, homomorphism on rings such as can be lifted to homomorphism on polynomial rings over the respectful rings. This includes the map from f to f. Example: from the Fibonacci sequence F n reduced to (mod7) F n S n (mod7), which S n is recurrence by the general example above. Theorem 2.5. Suppose an is a recurrence relation of order k, then b n = a n+h for any constant h is also recurrence relation of order k. Proof. If f is the function defining elements of a then f also defines element of b, to see this note that for i > k we have b i = a i+h = f(a i+h, a i+h+1,, a i+h+k ) = f(b i, b i+1,, b i+k ). Theorem 2.6. Suppose the recurrence relation b n is defined as an+h and both being the same order k, if they share the first k initial values then they are equal at everywhere. Proof. Induction on n with the stronger proposition that for all i n, a i = b i.ifn = k, then will include the hypothesis of the theorem. If n k + 1, for every i n, a i = b i, we have a n+1 = f(a n k, a n k+1,, a n ) = f(b n k, b n k+1,, b n ) = b n+1 (since, both are defined by f). Example(s) (a) Define s n = s n+16 they share the first two initial values, so they are equal everywhere. (b) Define s r n = s n + 16r by induction these are all equal and the base case r = 0 is trivial. Also r r + 1 come from an adaptation of the previous example and transitivity. Now, we shall discuss our main theorem. 142

4 South Asian J. Math. Vol. 1 No. 3 Theorem F n 8 n. Proof. As we discussed in our previous examples, we have seen that F n F n+16 r (mod7), Further more a direct computation of the first 16 values of S n (see (1.1)), shows that F n 0(mod7) n = r, r 8 n and x 0(modm) m x. Now, we shall discuss our main theorem, which is given at (1.3) Proof of (2.3) Use the recurrence formula, we have F n = F n 1 + F n 2 = (F n 2 + F n 3 ) + F n 2 (1) F n = 2F n 2 + F n 3 = 2(F n 3 + F n 4 ) + F n 3 (2) F n = 3F n 3 + 2F n 4 = 3(F n 4 + F n 5 ) + 2F n 4 (3) F n = 5F n 4 + 3F n 5 = 5(F n 5 + F n 6 ) + 3F n 5 (4) F n = 8F n 5 + 5F n 6 = 8(F n 6 + F n 7 ) + 5F n 6 (5) F n = 13F n 6 + 8F n 7. From (3), if 3 F k for some k, then 3 F k+4. Clearly, 3 F n 3, so if 3 divides F n 4, then 3 also divides F n. Since 3 F 4, then by induction 3 divides every 4 th Fibonacci number after F 4 ; i.e., 3 F 4k for all k. Similarly from (4), if 5 F k for some k, then 5 F k+5. Since 5 F 5, then by induction 5 F 5k. Now, from (5), if 8 F k for some k, then 8 F k+6 and 8 F 6 and by induction 8 F 6k. Finally, by (6), we say that, if 13 F k for some k, then 13 F k+7 and have 13 F 7 implies 13 F 7k. Finally, we concluded these divisibility properties on Fibonacci numbers. 3 Divisibility Sequence In this second section, we discuss the Fibonacci divisibility s by divisibility sequence. 1. Every third Fibonacci number is even divides every 4th Fibonacci number divides every 5th Fibonacci number divides every 6th Fibonacci number divides every 7th Fibonacci number divides every 8th Fibonacci number divides every 9th Fibonacci number divides every 10th Fibonacci number. 9. 6, 9, 12 and 16 divides every 12th Fibonacci number divides every 14th Fibonacci number and 61 divides every 15th Fibonacci number divides every 20th Fibonacci number. Generally, most of the divisibility properties of Fibonacci numbers follow from the fact that they comprise a divisibility sequence, i.e., m n F m F n. All of the above statements are special cases of this. 143

5 L. Yuan: Divisibility properties of Fibonacci numbers In fact F n is strong divisibility sequence [2], i.e., gcd(f m, F n) = F gcd(m,n). This strong property specializes to the above property when m n gcd(m, n) = m). The proof is not difficult. Here is one way. We know that, the Fibonacci addition F n + m = F n+1 F m + F n F m 1 [4]. For n n m this yields F n = F n m+1 F m + F n m F m 1 (modf m ). Now put k = m 1 and apply this theorem to conclude that f n = F n is strong divisibility sequence. Theorem 3.1. Let f n be an integer sequence such that f 0 = 0, f 1 = 1 and such that for all n > m we have f n f k f n m (modf m ) with k < n, (k, m) = 1. Then (f n, f m ) = f(n, m). Proof. By induction on n + m, the theorem is trivially true if n = m or n = 0 or m = 0. So assume w.l.o.g. n > m > 0. Since k + m < n + m, by induction (f k, f m ) = f(k, m) = f 1 = 1. Thus (f n, f m ) = (f k f n m, f m ) = (f n m, f m ) = f (n m,m) = f (n,m) follows by induction (which applies since (n m)+m < n+m), and by the gcd laws (a, b) = (a, b) if a a ( mod b) and (ca, b) = (a, b) if (c, b) = 1. We may find it insightful to simultaneously examine other strong divisibility sequences, example, f n = (x n 1)/(x 1). (3.1) In this case gcd(f m, f n ) = f gcd(m,n) may be interpreted as a q-analog of the integer Bezout identity [3], for example 3 = (15, 21) (x 3 1)/(x 1) = (x 15 + x 9 + 1)(x 15 1)/(x 1) (x 9 + x 3 )(x 21 1)/(x 1). Discussion A nice way to proceed is to exploit the fact that the polynomial sequence (3.1) is a strong divisibility sequence, i.e. (f m, f n ) = f(m, n); where (u, v) denotes the gcd of u and v. So, for example, (f 3, f 5 ) = f (3,5) = f 1 = 1, and f 3 f 15 via (f 3, f 15 ) = f (3,15) = f 3. Combining this with Euclid s lemma quickly yields the sought factor, namely (f 3, f 5 ) = 1, f 3, f 5 f 15 f 3 f 5 f 15 f 3 f15 f 5, i.e., x 2 + x + 1 x 10 + x Note that the above proof shows (a, b) = 1 f a f ab f b for any strong divisibility sequence f n. For example, for the Fibonacci numbers follows f 5 f20 f 4 i.e., = More generally, using these properties and a little number theory and combinatory (inclusion-exclusion) one easily derives the basic factorization properties of cyclostomes polynomials. This is may be not necessary for this paper. But, Every k th number of the sequence is a multiple of F k, which we realized in this paper. Acknowledgements I am heartily thankful to Prof. Quanta, whose encouragement, guidance and online support from the initial to the final level enabled me to develop an understanding of the subject. I also thank the editor Wang for making the paper in latex and encouraging the Math sciences. References 1 Nicolai N. Vorobiev, Fibonacci numbers, Russian Academy of Sciences, 1992 (6th edition) Nauka, Moscow. 2 Susanna S. Epp, Discrete mathematics with applications, Fourth edition, Boston, USA. 3 http : //en.wikipedia.org/wiki/b%c3%a9zout%27s i dentity 4 Bergum G.E., Philippou and Horadam, Application of Fibonacci numbers, Kluwer Academic publishers, Netherlands. 144