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Unformatted text preview: The University of Sydney MATH2068 Number Theory and Cryptography (http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/) Semester 2, 2008 Lecturer: R. Howlett 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657 Tutorial 3 1. The Fibonacci sequence is defined by F = 0, F 1 = 1 and F i = F i- 2 + F i- 1 for all i ≥ 2. (In Question 3 last week we investigated the sequence ( F i ) modulo a prime p .) ( i ) Suppose that a Fibonacci number F n is divisible by some positive integer d , and write F n +1 = k . Show that F n ≡ kF (mod d ) and F n +1 ≡ kF 1 (mod d ), and use induction on i to prove that F n + i ≡ kF i (mod d ) for all integers i ≥ 0. ( ii ) Use Part ( i ) and induction on j to prove that if d | F n then d | F jn for all natural numbers j . Solution. ( i ) Observe that kF = 0. But F n ≡ 0 (mod d ), since d | F n by hypoth- esis. Thus kF ≡ F n (mod d ). And kF 1 = k = F n +1 ; so certainly kF 1 ≡ F n +1 (mod d ). This shows that the statement “ kF i- 1 ≡ F n +( i- 1) and kF i ≡ F n + i (mod d )” is true for i = 1. Assume now that i > 1, and the statement holds with i- 1 in place of i . Thus kF i- 2 ≡ F n +( i- 2) (mod d ) kF i- 1 ≡ F n +( i- 1) (mod d ) and adding these congruences we deduce that kF i = k ( F i- 2 + F i- 1 ) = kF i- 2 + kF i- 1 ≡ F n +( i- 2) + F n +( i- 1) = F n + i . Since also kF i- 1 ≡ F n +( i- 1) (mod d ) it follows that the statement holds for i , and hence for all positive integers, by induction. In particular kF i ≡ F n + i for all natural numbers i , as required. ( ii ) Since we are given that d | F n , Part ( i ) tells us that F n + i ≡ kF i (mod d ) for all i , where k = F n +1 . In particular, if F i ≡ 0 (mod d ) then F n + i ≡ k ≡ (mod d ). That is, for all i , if d | F i then d | F n + i .....
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