# Proof by induction on Fibonacci numbers

I was studying Mathematical Induction when I came across the following problem:

The Fibonacci numbers are the sequence of numbers defined by the linear recurrence equation-

$f_n = f_{n-1} + f_{n-2}$ with $f_1 = f_2 = 1$

Use induction to show that $f_n \; | \; f_{2n}$ ($f_n$ divides $f_{2n}$)

Basis Step is obviously true; but I'm facing difficulty in the Inductive Step. If I assume the inductive hypothesis to be true for some $k$, i.e., $\dfrac{ f_{2k} } { f_{k} } = c$ (For some positive integer $c > 0$), I'm not clear as to how I should proceed further and prove that $P(2k+1)$ is also true.

I'm new here, so if I'm doing anything wrong, please overlook it on the account of my naivety.

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Try looking at some small examples to see what's going on — sometimes you have to strengthen your induction hypothesis and prove a stronger statement. (And also, there are easy ways of proving this statement without induction of course, but in this context I guess what you want/need is a proof by induction.) –  ShreevatsaR Sep 8 '13 at 13:43
@Tarun, please correct the typo where I wrote $\dfrac{ f_{2k} } { f_{2k} } = c$ instead of $\dfrac{ f_{2k} } { f_{k} } = c$. I am not able to edit since the edit requires 6 characters to be changed at least. –  Parth Thakkar Sep 8 '13 at 13:44
–  lab bhattacharjee Sep 8 '13 at 13:45
I don't think that works. The edit checks the stuff that has changed. Retyping won't help (as far as I know). –  Parth Thakkar Sep 8 '13 at 13:46
Hint: Look at the sequence of values of $\frac{f_{2k}}{f_k}$. Do you see a pattern there? That suggests to prove the following fact:
$$\frac{f_{2k+2}} { f_{k+1} } = \frac{f_{2k} } { f_k } + \frac{f_{2k-2} } {f_{k-1} }$$
Check that the first two terms of this series $g_n = \frac{f_{2n}}{f_n}$ are integers, hence conclude by induction that every term is an integer.