On the Fibonacci Sequence. By: Syrous Marivani LSUA. Mathematics Department. Alexandria, LA 71302

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1 On the Fibonacci Sequence By: Syrous Marivani LSUA Mathematics Deartment Alexandria, LA 70 The so-called Fibonacci sequence {(n)} n 0 given by: (n) = (n ) + (n ), () where (0) = 0, and () =. The ollowing table describes the behaviors o {(n)} 0 n 0, mod, where is a rime,. Table n (n) (n)mod (n)mod (n)mod 5 (n)mod 7 (n)mod

2 A moment observation shows that i or examle: n 0(mod ), then (n) 0(mod ), n 0(mod 4), then (n) 0(mod ), n 0(mod 5), then (n) 0(mod 5), n 0(mod 8), then (n) 0(mod 7), n 0(mod 0), then (n) 0(mod ). These observations can be roven, or examle I rove the last assertion. Suose (n) 0(mod ), then by reeated alication o the recurrence relation () it ollows that: (n + 0) = 55 (n + ) + 4 (n). From this it ollows that: (n + 0) 0 (mod ) i and only i (n) 0 (mod ). () Since (0) = 0, and (k) is not divisible by or k 9, then by reeated alication

3 o the latter result it ollows that (n) 0 (mod ), i and only i n 0 (mod 0). These results or rimes, < 00, are summarized in Table. This table gives the values o o such that i n 0 (mod o ), then (n) 0 (mod ). Table and o and o and o This table was obtained by using Male 7. For a rime such as = 6007, o = 6008 and it takes.477 seconds to get this answer. Theorem : I {(n)} n 0 is the Fibonacci sequence, a rime, and o is the least ositive integer such that ( o ) 0 (mod ), then (n) 0 (mod ) i and only i n 0 (mod o ). Proo : One can easily show by induction on k that: (n + k) = (k) (n + ) + (k ) (n). () So i ( o ) 0 (mod ), then: (n + o ) 0 (mod ) i and only i (n) 0 (mod ). (4)

4 Since (n) is not congruent to 0 mod or n < o, then using (4) the conclusion ollows. From now on o denotes the smallest ositive integer such that ( o ) 0 (mod ). Theorem : I is an odd rime such that, or (mod 5), then o divides +, i, or 4 (mod 5) then o divides, and i = 5, then o =. Proo : By [, age 9] we have: Then: Since n n (n) =. (5) 5 ( + ) = ( ) / i= 0 + i 5. i + +, or i ( ) /, i + then ( + ) ( + 5 () / )( + )(mod ). Using Euler s criterion and the Quadratic recirocity law [, Chater 9, Theorems 9. and 9.8] we have: ( + ) 0(mod ) i and only i, or (mod5). (6) Otherwise using Fermat s theorem [, Chater 5, Theorem 5.9] we have: ( Again using (5) we have: + ) (mod ) i and only i, or 4(mod5). (7) () / () 5 (mod ), (8) 4

5 and using [, Chater 5, Theorem5.8] as beore : Since () (mod ) i and only i, or 4(mod 5). (9) then using (7) and (9), it ollows: ( ( ) = ( + ) (), ) 0(mod ) i only i, or 4(mod5). (0) Using (6), (0), and Theorem the assertion o the theorem ollows. The case = 5 ollows rom (8). Corollary : o ( + )/ i and only i, or 7 (mod 0). Proo : Using (8) as beore we have: () (mod ) i, or (mod5). () + Using () with n =, and k =, we have: + () = +. () Suose, or (mod 5) and o ( + )/, then using () and () and Theorem we have: (mod ). This shows that is a quadratic residue mod, and so by [, Chater 9, Theorem 9.4] we should have (mod 4). So o ( +)/ i in addition to, or (mod 5), we should have (mod 4). Then the conclusion ollows rom here. To rove it in the other direction suose, or 7 (mod 0) but o does not divide 5

6 6 ( + )/. Since ( + ) 0 (mod ), using () with k n + = =, then:. ) ( = = + This imlies: (mod ). + () Using () in () we have: (mod ). 5 This imlies 5 is a quadratic residue mod. So (-5 ) =. This imlies (5 ) =, since ) ( =. This is imossible i, or 7 (mod 0). So o divides ( + )/. Corollary : I, or (mod 5), then o does not divide ( + )/ i and only i, or 7 (mod 0). Proo : This ollows as in the revious Corollary excet we should have - (mod 4), and, or (mod 5). Corollary : o ( )/ i and only i or 9 (mod 0). Proo : Suose o ( )/. This imlies, 4 (mod 5) and we have: 9, 7, 5, and in general:. (k) ) ( k k + (4) Letting k = ( )/ in (4), we have:

7 ( ) ( ) / (mod ). (5) Using the results o Theorem, it ollows that ( ) (mod ). Using this result and () with n =, and k = we have: (mod ). (6) Using (5) and (6) we conclude (mod 4). So since also or 4 (mod 5), then or 9 (mod 0). On the other hand suose or 9 (mod 0) and o does not divide ( )/. Since ( ) 0 (mod ) using () with n = k =, we have: this imlies: ( + ) =, + + 5,, and (mod ), and in general: (7) k k ( ) g(k + ) (mod ), (8) or k =0,,,.,( )/, where {g(k)} k is the Lucas sequence. It turns out that: n n n n g(n) = α + β ; (n) = ( α β ) / 5, (9) where α = ( + 5) /, and β = ( 5) /. 7

8 Letting k = ( 5)/ in (8) and using (9) we have: g = ( ) ( ) ( 5) / (mod ). (0) Using the results o Theorem again, ( ) - (mod ). However (0) contradicts this result since, or 9 (mod 0). So o has to divide ( )/. Corollary 4: o does not divide ( )/ i and only i, or 9 (mod 0). Proo : In this case in addition to, or 4 (mod 5) we should have - (mod 4), rom which the conclusion ollows. Using methods similar to these, I think it is ossible to know more about o, or instance, when o ( + ) /, or o ( )/, and so on. Right now I have concentrated my eorts in these directions. Reerences [] N. N. Vorobyov, The Fibonacci Numbers, D.C. Heath and Comany, Boston 96 [] Tom M. Aostol, Introduction to Analytic Number Theory, Sringer-Verlag, New York, Heidelberg, Berlin