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# tut03sols - 2 The University of Sydney MATH2068 Number...

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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 = 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 0 (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 = 0. But F n 0 (mod d ), since d | F n by hypoth- esis. Thus kF 0 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 0 0 (mod d ). That is, for all i , if d | F i then d | F n + i . Since F 0 = 0 we see that d | F jn is true when j = 0. This starts the induction. Now suppose that j > 0 and that d | F ( j - 1) n . By what we have just shown, it follows that d | F n +( j - 1) n ; that is, d | Fjn , as required.

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