TITLE: How to prove gcd of consecutive Fibonacci numbers is 1?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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 $\rm\,m=n\!+\!1.$