Hi.
How to prove that GCD(F3k, L3k)=2? and If n dosen't divide with 3 GCD(Fn, Ln)=1
Don't even know how to start that:/....Hope someone can help
Ty.
Try to show:
a) $\displaystyle L_n=F_{n+1}+F_{n-1}$
b) $\displaystyle \gcd(F_n,F_{n+1})=1$ for every n.
Using these two properties you can get:
$\displaystyle \gcd(F_n,L_n)=\gcd(F_n,F_{n+1}+F_{n-1})=\gcd(F_n,F_n+2F_{n-1})=\gcd(F_n,2F_{n-1}$. This is either 1 or 2, since Fn and F_n-1 are coprime. So you already know that the only possible values are 1 and 2.
To show that they are equal to what you right, just nottice the pattern 0,0,1,0,0,1,.... when looking to the sequence Fn (or Ln) modulo 2.