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: Clueless by McWorter)

I have no reason to doubt that.  From my point of view, n could have easily been bigger than 2003!

I notice, for instance, that the first Fibonacci number divisible by 6 is F(13), 144, if I haven't missed one.  That was calculated, incidentally, without mathematics software.

Posted by Steve Herman on 2005-06-11 19:01:30