How to prove gcd of consecutive Fibonacci numbers is 1? 
Possible Duplicate:
Prove that two any consecutive terms of Fibonacci sequence are relatively prime 

How to prove it ? Can you help me ?
Let $f_n$ be Fibonacci Sequence.
$$(f_{n},f_{n+1})=1,\quad \forall\,n\in\mathbb{N}.$$
Here $(a,b)$ is the greatest common divisor.
 A: If a $d\in\mathbb{N}$ divides $f_n$ and $f_{n+1}$, then $d$ divides also the difference of these two, $f_{n+1}-f_n=f_{n-1}$. Then using $f_n-f_{n-1}=f_{n-2}$, we get $d\mid f_{n-2}$. Repeating this argument again and again, we'll get $d\mid f_{n-3}$, $d\mid f_{n-4}$, and at the end, $d\mid f_{1}$. But $f_1=1$, therefore $d=1$.
A: Hint: Euclidean algorithm and the recursive definition of the Fibonacci sequence.
A: Hint $\ $ Put $\rm\:a_n = 1 = b_n\:$ in the much more general result below.
Theorem $\ $ If $\rm\:(\color{#c00}{b_n,\,f_n) = 1}\:$ and $\rm\:f_{n+1} = a_n f_n + b_n f_{n-1}\:$ then $\rm\:(f_{n+1},\,f_n) = (f_1,\,f_0).$
Proof $\ $ Clear if $\rm\:n = 0.\:$ Else by Euclid and induction we have
$$\rm (f_{n+1},\,f_n) = (a_n f_n\! +\! \color{}{b_n} f_{n-1},\,f_n) = (\color{#c00}{b_n} f_{n-1},\, \color{#c00}{f_n}) = (f_{n-1},\,f_n) = (f_1,\,f_0)\qquad$$
Remark $ $ Similarly we can prove much more generally that the Fibonacci numbers  $\rm\:f_n\:$ comprise a strong divisibility sequence: $\rm\, (f_m,f_n) = f_{(m,n)},\:$ i.e. $\rm\:gcd(f_m,f_n) = f_{\gcd(m,n)}.$ OP is case $\rm\,m=n\!+\!1.$
A: You could use induction.
First show $(f_2,f_1) = 1$.  Then for $n \geq 2$, assume $(f_n,f_{n-1}) = 1$.  Use this and the recursion $f_{n + 1} = f_n + f_{n - 1}$ to show $(f_{n + 1},f_n) = 1$.
