$gcd(f_k, f_{k+3})$, where $f_k$ is the k'th Fibonacci number I wanted to find the $gcd(f_k, f_{k+3})$, where $f_k$ is the k'th Fibonacci number (i.e. $f_0=0, f_1=1, f_k=f_{k-1} + f_{k-2}$ for $k \geq 2$. So far I've tried to exprss $f_{k+3}$ as $2f_{k+1}+f_k$. Using this it follows that $gcd(f_k, f_{k+3}) = gcd(f_k, 2f_{k+1}$), but I don't see how I can continue from there one. Can someone help me out?
Thanks!

Since $\gcd(f_k, f_{k+1})=1$, we have $\gcd(f_k, 2f_{k+1})=\gcd(f_k, 2)$.
Since $f_3=2$, we have $\gcd(f_k, 2)=2$ iff $n$ is a multiple of $3$. Otherwise, $\gcd(f_k, 2)=1$.
