The Fibonacci sequence goes F(0)=0, F(1)=1, and for n>1, F(n)=F(n-1)+F(n-2).
Show that for every positive integer m there exists an integer n>0 such that m divides F(n).
(In reply to
re: Proof--Quick fix by Tristan)
ABCDECDECDE... is impossible because C can only be preceded by a unique pair of consecutive fibonacci numbers.
The chart in your previous post provides several counterexamples. For example, modulo 9, 7 is preceded by (4,3) and (8,8).
But I believe you're on the right track...
Edited on June 11, 2005, 9:49 pm
re(2): Proof--Quick fix
