Product of Two Consecutive Fibonacci or Lucas Numbers Divisible by their Prime Sum of Indices
aa r X i v : . [ m a t h . N T ] J u l Product of Two Consecutive Fibonacci or LucasNumbers Divisible by their Prime Sum ofIndices
Vladimir PletserOctober 19, 2018
European Space Research and Technology Centre, ESA-ESTEC P.O. Box 299,NL-2200 AG Noordwijk, The Netherlands E-mail: [email protected]
Abstract
We show that the product of two consecutive Fibonacci (respec-tively Lucas) numbers is divisible by the sum of their indices if thissum is a prime number different from and in the form (4 r + 1) (respectively (4 r + 3) ). Keywords : 11B39, 11A41
One of the most interesting divisibility properties of the Fibonacci numbers isthat for each prime p , there is a Fibonacci number F n such that p divides F n (see,e.g. [5]). More specifically, for p = 5 , p divides either F p − if p ≡ ± mod , or F p +1 if p ≡ ± mod . For p = 5 , one has of course p = F p . Although already demonstrated differently in [7, 4], a new demonstration of thefollowing theorem is proposed in this paper.
Theorem 1. If p is prime and r ∈ Z + , p = (4 r + 1) divides the product F r F r +1 , except for p = 5 (1) p = (4 r + 3) divides the product L r +1 L r +2 (2)1 roof. For p prime and r, s, n, m ∈ Z + , for odd primes p = 2 s + 1 , one has L s +1 − L s +1 − L (3)The transformations L n + m − ( − m L n − m = 5 F m F n (4) L n + m + ( − m L n − m = L m L n (5)(relations (17 a, b) in [6] and relations (11) and (23) in [2]) can be used.(i) First, let s be even, s = 2 r . Relation (3) yields respectively from (4) and(5), with m = 2 r and n = 2 r + 1 , L r +1 − F r F r +1 (6) L r +1 + 1 = L r L r +1 (7)If p = 4 r + 1 = 5 is prime, then either p divides F r if p ≡ ± mod
5) =29 , , , . . . , or p divides F r +2 if p ≡ ± mod
5) = 13 , , , . . .On the other hand, one has (relation (13) in [6]) F r = F r L r (8) F r +2 = F r +1 L r +1 (9)Let first p ≡ ± mod , then p divides F r and therefore from (8) also either F r or L r . But p cannot divide L r . Let us assume the contrary. Supposethat p divides L r and also ( L r +1 − , as L p ≡ mod p ) (see e.g. [1], [3]).It would mean from (7) that p should also divide simultaneously ( L r +1 + 1) which makes no sense. Therefore p divides F r and not L r , and also ( L r +1 – .The other case where p ≡ ± mod divides F r +2 is treated similarly.This means that all primes p = 4 r + 1 = 5 divide the product of two consecutiveFibonacci numbers of indices r and r + 1 . More precisely, if p ≡ ± mod
5) =29 , , , . . . , then p divides F r ; if p ≡ ± mod
5) = 13 , , , . . . , then p divides F r +1 .(ii) Second, let s be odd, s = 2 r + 1 . Relation (3) yields respectively from (4)and (5), with m = 2 r + 1 and n = 2 r + 2 , L r +3 + 1 = 5 F r +1 F r +2 (10) L r +3 − L r +1 L r +2 (11)If p = 4 r + 3 is prime, then p divides F r +2 if p ≡ ± mod
5) = 11 , , , . . . ;or p divides F r +4 if p ≡ ± mod
5) = 3 , , , , . . . One has also F r +2 = F r +1 L r +1 (12) F r +4 = F r +2 L r +2 (13)Like above, let first p ≡ ± mod . Then p divides F r +2 and therefore, from(12), also either F r +1 or L r +1 . But p cannot divide F r +1 . Let us assume the2ontrary. Suppose that p divides F r +1 and also ( L r +3 − . It would meanfrom (10) that p should also divide simultaneously ( L r +3 + which makes nosense. Therefore p divides L r +1 and not F r +1 , and also ( L r +3 – . The othercase where p ≡ ± mod divides F r +4 is also treated similarly.This means that all primes p = 4 r + 3 divide the product of two consecutiveLucas numbers of indices r + 1 and r + 2 . More precisely, if p ≡ ± mod
5) =11 , , , . . . , then p divides L r +1 ; if p ≡ ± mod
5) = 3 , , , , . . . , then p divides L r +2 .On the other hand, for p = 5 , one has obviously L = 11 ≡ mod . Dr C. Thiel is acknowledged for helpful discussions.