Divisibility in the Fibonacci Numbers. Stefan Erickson Colorado College January 27, 2006

Divisibility in the Fibonacci Numbers Stefan Erickson Colorado College January 27, 2006

Fibonacci Numbers F n+2 = F n+1 + F n n 1 2 3 4 6 7 8 9 10 11 12 F n 1 1 2 3 8 13 21 34 89 144 n 13 14 1 16 17 18 19 20 F n 233 377 610 987 197 284 4181 676 n 21 22 23 24 2 26 F n 10946 17711 2867 46368 702 121393 n 27 28 29 30 31 32 F n 196418 317811 14229 832040 1346269 2178309

Lucas Numbers L n+2 = L n+1 + L n n 1 2 3 4 6 7 8 9 10 11 12 L n 1 3 4 7 11 18 29 47 76 123 199 322 n 13 14 1 16 17 18 19 20 L n 21 843 1364 2207 371 778 9349 1127 n 21 22 23 24 2 26 L n 24476 39603 64079 103682 167761 271443 n 27 28 29 30 31 32 L n 439204 710647 114981 1860498 3010349 4870847

Fundamental Questions Is there a formula for the n th Fibonacci number? Which Fibonacci numbers divide other Fibonacci numbers? Which primes divide the Fibonacci numbers? What is the entry point for each prime? What about the Lucas numbers?

Guess: F n = r n for some r. Recurrence Relations F n+2 = F n+1 + F n

Recurrence Relations Guess: F n = r n for some r. F n+2 = F n+1 + F n r n+2 = r n+1 + r n r n (r 2 r 1)=0

Recurrence Relations Guess: F n = r n for some r. F n+2 = F n+1 + F n r n+2 = r n+1 + r n r n (r 2 r 1)=0 r 1 =φ= 1+ 2 r 2 =φ = 1 2 = 1.61803398874989484820... = 0.61803398874989484820...

Recurrence Relations Guess: F n = r n for some r. F n+2 = F n+1 + F n r n+2 = r n+1 + r n r n (r 2 r 1)=0 r 1 =φ= 1+ 2 r 2 =φ = 1 2 = 1.61803398874989484820... = 0.61803398874989484820... φ φ = 1 φ n =F n φ+f n 1

Binet s Formula General Solution: F n = c 1 r n 1 + c 2 r n 2

Binet s Formula General Solution: F n = c 1 r n 1 + c 2 r n 2 F 0 =0,F 1 =1 = c 1 = 1,c 2 = 1 F n = 1 ( φ n φ n)

Binet s Formula General Solution: F n = c 1 r n 1 + c 2 r n 2 F 0 =0,F 1 =1 = c 1 = 1,c 2 = 1 F n = 1 ( φ n φ n) L 0 =2,L 1 =1 = c 1 =1,c 2 =1 L n =φ n +φ n

Generating Functions F n x n = n=0 x 1 x x 2

Generating Functions F n x n = n=0 = x 1 x x 2 x (1 φx)(1 φ x)

F n x n = n=0 Generating Functions = x 1 x x 2 x (1 φx)(1 φ x) = 1 1 1 φx 1 1 1 φ x

F n x n = n=0 Generating Functions x 1 x x 2 x = (1 φx)(1 φ x) = 1 1 1 φx 1 1 1 φ x = 1 n=0 (φx) n 1 n=0 (φ x) n

F n x n = n=0 Generating Functions x 1 x x 2 x = (1 φx)(1 φ x) = 1 1 1 φx 1 1 1 φ x = 1 = n=0 n=0 (φx) n 1 1 ( φ n φ n) x n n=0 (φ x) n

Consequences lim n F n+1 F n = lim n L n+1 L n = φ

Consequences lim n F n+1 F n = lim n L n+1 L n = φ lim n L n F n =

Consequences lim n F n+1 F n = lim n L n+1 L n = φ lim n L n F n = F 2n = F n L n

Consequences lim n F n+1 F n = lim n L n+1 L n = φ lim n L n F n = ( Proof. F 2n = F n L n 1 ( φ 2n φ 2n) = 1 ( φ n φ n) (φ n + φ n))

F n = F n 1 + F n 2 Super Cool Formula

Super Cool Formula F n = F n 1 + F n 2 =(F n 2 +F n 3 )+F n 2 =2F n 2 +1F n 3

Super Cool Formula F n = F n 1 + F n 2 =(F n 2 +F n 3 )+F n 2 =2(F n 3 +F n 4 )+F n 3 =2F n 2 +1F n 3 =3F n 3 +2F n 4

Super Cool Formula F n = F n 1 + F n 2 =(F n 2 +F n 3 )+F n 2 =2(F n 3 +F n 4 )+F n 3 =2F n 2 +1F n 3 =3F n 3 +2F n 4 =3(F n 4 +F n )+2F n 4 =F n 4 +3F n

Super Cool Formula F n = F n 1 + F n 2 =(F n 2 +F n 3 )+F n 2 =2(F n 3 +F n 4 )+F n 3 =2F n 2 +1F n 3 =3F n 3 +2F n 4 =3(F n 4 +F n )+2F n 4 =F n 4 +3F n F m+n =F m F n+1 + F m 1 F n L m+n = F m L n+1 + F m 1 L n

Theorem. F n divides F kn for all integers k.

Theorem. F n divides F kn for all integers k. Proof. F n divides F n, so the theorem is true for k =1. Suppose F n divides F kn. Then F (k+1)n = F kn F n+1 + F kn 1 F n is also divisible by F n. By induction, the theorem is true for all k!

Theorem. F n divides F kn for all integers k. Proof. F n divides F n, so the theorem is true for k =1. Suppose F n divides F kn. Then F (k+1)n = F kn F n+1 + F kn 1 F n is also divisible by F n. By induction, the theorem is true for all k! Theorem. L n divides L kn for all odd integers k.

Theorem. F n divides F kn for all integers k. Proof. F n divides F n, so the theorem is true for k =1. Suppose F n divides F kn. Then F (k+1)n = F kn F n+1 + F kn 1 F n is also divisible by F n. By induction, the theorem is true for all k! Theorem. L n divides L kn for all odd integers k. Proof. Same proof, except L (2l+1)n = F 2ln L n+1 + F 2ln 1 L n and L n divides F 2n, which in turn divides F 2ln.

Greatest Common Divisors Theorem. gcd(f m,f n )=F gcd(m,n). Ex: gcd(144, 284) = 8.

Greatest Common Divisors Theorem. gcd(f m,f n )=F gcd(m,n). Ex: gcd(144, 284) = 8. Proof. Let d = gcd(m, n). By the previous theorem, F d divides F m and F n. Thus, F d divides gcd(f m,f n ).

Greatest Common Divisors Theorem. gcd(f m,f n )=F gcd(m,n). Ex: gcd(144, 284) = 8. Proof. Let d = gcd(m, n). By the previous theorem, F d divides F m and F n. Thus, F d divides gcd(f m,f n ). d may be written as am + bn for some integers a and b. Then F d = F am+bn = F am F bn+1 + F am 1 F bn Since F m divides F am and F n divides F bn, F d can be written as a linear combination of F m and F n. Hence, gcd(f m,f n ) divides F d.

Question: Which Fibonacci numbers divide other Fibonacci numbers? Answer: F n divides F m if and only if n divides m.

Question: Which Fibonacci numbers divide other Fibonacci numbers? Answer: F n divides F m if and only if n divides m. For example, the Fibonacci numbers which divide F 30 = 832040 are F 3 =2,F =,F 6 =8,F 10 =, and F 1 = 610.

Question: Which Fibonacci numbers divide other Fibonacci numbers? Answer: F n divides F m if and only if n divides m. For example, the Fibonacci numbers which divide F 30 = 832040 are F 3 =2,F =,F 6 =8,F 10 =, and F 1 = 610. In order for F n to be prime, n must be prime. The converse is not true, since F 19 =4181=37 113.

Prime Divisibility Question: Which primes divide the Fibonacci numbers?

Prime Divisibility Question: Which primes divide the Fibonacci numbers? Answer: All of them! If p, then p divides either F p 1 or F p+1.

Prime Divisibility Question: Which primes divide the Fibonacci numbers? Answer: All of them! If p, then p divides either F p 1 or F p+1. For example, 7 divides F 8 = 21 and 11 divides F 10 =. Note that F = is the only exception.

Lemma. φ p φ (mod p) if p 1, 4 (mod ) φ (mod p) if p 2, 3 (mod )

Lemma. φ p φ (mod p) if p 1, 4 (mod ) φ (mod p) if p 2, 3 (mod ) Proof. φ p = 1 2 p ( 1+ ) p

Lemma. φ p φ (mod p) if p 1, 4 (mod ) φ (mod p) if p 2, 3 (mod ) Proof. φ p = 1 2 p ( 1+ ) p = 1 ( 1+ ( ) ( p + + p 2 p 1 p 1 )( ) p 1 + ( ) p )

Lemma. φ p φ (mod p) if p 1, 4 (mod ) φ (mod p) if p 2, 3 (mod ) Proof. φ p = 1 2 p ( 1+ ) p = 1 ( 1+ ( ) ( p + + p 2 p 1 2 1 (1+ p 1 2 ) p 1 )( ) p 1 + ( ) p ) (mod p)

Lemma. φ p φ (mod p) if p 1, 4 (mod ) φ (mod p) if p 2, 3 (mod ) Proof. φ p = 1 2 p ( 1+ ) p = 1 ( 1+ ( ) ( p + + p 2 p 1 2 1 (1+ p 1 2 ) 1 2 p 1 ( 1+ ( )( ) p 1 + ( ) p ) p ) ) (mod p)

Lemma. φ p φ (mod p) if p 1, 4 (mod ) φ (mod p) if p 2, 3 (mod ) Proof. φ p = 1 2 p ( 1+ ) p = 1 ( 1+ ( ) ( p + + p 2 p 1 2 1 (1+ p 1 2 ) 1 2 p 1 ( 1+ ( )( ) p 1 + ( ) p ) p ) ) (mod p) ( ) p =1ifp 1,4 (mod ) and ( p ) = 1 ifp 2,3 (mod ).

Theorem. p divides F p 1 if p 1, 4 (mod ). (Ex: 11 F 10 =.) p divides F p+1 if p 2, 3 (mod ). (Ex: 7 F 8 = 21.)

Theorem. p divides F p 1 if p 1, 4 (mod ). (Ex: 11 F 10 =.) p divides F p+1 if p 2, 3 (mod ). (Ex: 7 F 8 = 21.) Proof. If p 1, 4 (mod ), then φ p φ (mod p). Dividing by φ, φ p 1 1 (mod p). F p 1 = 1 ( φ p 1 φ p 1) 1 ( 1 1 ) 0 (mod p).

Theorem. p divides F p 1 if p 1, 4 (mod ). (Ex: 11 F 10 =.) p divides F p+1 if p 2, 3 (mod ). (Ex: 7 F 8 = 21.) Proof. If p 1, 4 (mod ), then φ p φ (mod p). Dividing by φ, φ p 1 1 (mod p). F p 1 = 1 ( φ p 1 φ p 1) 1 ( 1 1 ) 0 (mod p). If p 2, 3 (mod ), then φ p φ (mod p). Multiplying by φ, φ p+1 1 (mod p). F p+1 = 1 ( φ p+1 φ p+1) 1 ( 1+1 ) 0 (mod p).

Theorem. If p k divides F n for some k 1, then p k+1 divides F pn.

Theorem. If p k divides F n for some k 1, then p k+1 divides F pn. Proof. Homework problem.

Theorem. If p k divides F n for some k 1, then p k+1 divides F pn. Proof. Homework problem. Corollary. Every positive integer divides F n for some n. For example, 1000 = 8 12 divides F 6 12 = F 70.

Entry Points The smallest n such that p divides F n is called the entry point of p. We use e F (p) for the entry point of p into the Fibonacci numbers and e L (p) for the entry point of p into the Lucas numbers.

Entry Points The smallest n such that p divides F n is called the entry point of p. We use e F (p) for the entry point of p into the Fibonacci numbers and e L (p) for the entry point of p into the Lucas numbers. From the previous theorem, e F (p) divides either p 1orp+1. For example, we know that 13 divides the 14 th Fibonacci number. However, 13 also divides F 7 = 13, so e F (13) = 7. On the other hand, 13 does not divide any Lucas number. We say that e L (13) is undefined in this case.

Entry Points for the Lucas Numbers If e F (p) is even, then e L (p) = e F(p) 2. If e F (p) is odd, then p does not divide any Lucas number. This follows from F 2n = F n L n. For example, e F (29) = 14 and e L (29) = 7. Question: Which primes divide the Lucas numbers?

One More Formula L 2 n F 2 n =4 ( 1) n

One More Formula L 2 n F 2 n =4 ( 1) n Lemma. If p does not divide L n for any n, then p 1 (mod 4). Proof. Suppose n = e F (p) is odd, that is, p F n for some odd n. L 2 n 4 (mod p) = 1 is a square mod p = p 1 (mod 4).

One More Formula L 2 n F 2 n =4 ( 1) n Lemma. If p does not divide L n for any n, then p 1 (mod 4). Proof. Suppose n = e F (p) is odd, that is, p F n for some odd n. L 2 n 4 (mod p) = 1 is a square mod p = p 1 (mod 4). Corollary. All primes p 3 (mod 4) divide the Lucas numbers.

L 2 n F 2 n =4 ( 1) n Lemma. If p 1 (mod 4) and p 2, 3 (mod ), then p does not divide L n for any n.

L 2 n F 2 n =4 ( 1) n Lemma. If p 1 (mod 4) and p 2, 3 (mod ), then p does not divide L n for any n. Proof. Suppose p L n. Then F 2 n 4 ( 1) n (mod p) = ± is a square mod p. By quadratic reciprocity, ± are not squares when p 1 (mod 4) and p 2, 3 (mod ).

p 1 (mod 4) p 3 (mod 4) p 1, 4 (mod ) p 2, 3 (mod )

p 1 (mod 4) p 3 (mod 4) p 1, 4 (mod ) p L n for some n p 2, 3 (mod ) p L n for some n

p 1 (mod 4) p 3 (mod 4) p 1, 4 (mod ) p L n for some n p 2, 3 (mod ) p L n for any n p L n for some n

p 1 (mod 4) p 3 (mod 4) p 1, 4 (mod )? p L n for some n p 2, 3 (mod ) p L n for any n p L n for some n

Suppose p 1 (mod 4) and p 1, 4 (mod ) such that p L n for some n. We may assume n p 1 2. Then L n = φ n + φ n 0 (mod p) φ n φ n (mod p) φ 2n ( 1) n+1 (mod p)

φ 2n ( 1) n+1 (mod p) If n = p 1 4 is odd, then φ p 1 2 1 (mod p) means φ is a square mod p. Hence, 1 2 of the primes p (mod 8) divide L p 1. 4 If n = p 1 4 is even, then φ p 1 2 1 (mod p) means φ is not a square mod p. Hence, 1 2 of the primes p 1 (mod 8) divide L p 1. 4

φ 2n ( 1) n+1 (mod p) If n = p 1 8 is odd, then φ p 1 4 1 (mod p) means φ is a fourth power mod p. Hence, 1 4 of the primes p 9 (mod 16) divide L p 1. 8 If n = p 1 8 is even, then φ p 1 4 1 (mod p) means φ is a square but not a fourth power mod p. Hence, 1 4 of the primes p 1 (mod 16) divide L p 1. 8

Question: How many primes divide the Lucas numbers? Answer: 1 2 + 1 8 + 1 32 + 1 128 + = 1 2 ( 1 4 n=0 ) n = 2 3 The density of the primes dividing the Lucas numbers is 2 3!

Conclusions F n divides F m if and only if n divides m.

Conclusions F n divides F m if and only if n divides m. L n divides L m if and only if m is an odd multiple of n.

Conclusions F n divides F m if and only if n divides m. L n divides L m if and only if m is an odd multiple of n. Every positive integer divides the Fibonacci numbers.

Conclusions F n divides F m if and only if n divides m. L n divides L m if and only if m is an odd multiple of n. Every positive integer divides the Fibonacci numbers. 2 3 of the primes divide the Lucas numbers.

Conclusions F n divides F m if and only if n divides m. L n divides L m if and only if m is an odd multiple of n. Every positive integer divides the Fibonacci numbers. 2 3 of the primes divide the Lucas numbers. The Fibonacci and Lucas numbers are cool!