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)

Don't you need to say a bit more to clarify your argument?  Your argument does not work for the Lucas sequence: L(1)=2, L(2)=1, and for n>2, L(n)=L(n-1)+L(n-2).

Posted by McWorter on 2005-06-11 21:46:32